Write a function that computes the Ackermann function A(m, n),
defined as n+1 when m = 0, A(m-1, 1) when n = 0, and
A(m-1, A(m, n-1)) otherwise. Arbitrary precision is preferred
since the function grows extremely quickly.
Task
The '''[[wp:Ackermann function|Ackermann function]]''' is a classic example of a recursive function, notable especially because it is not a [[wp:Primitive_recursive_function|primitive recursive function]]. It grows very quickly in value, as does the size of its call tree.
The Ackermann function is usually defined as follows:
:Its arguments are never negative and it always terminates. Write a function which returns the value of . Arbitrary precision is preferred (since the function grows so quickly), but not required.
;See also:
- [[wp:Conway_chained_arrow_notation#Ackermann_function|Conway chained arrow notation]] for the Ackermann function.
360 Assembly
{{trans|AWK}} The OS/360 linkage is a bit tricky with the S/360 basic instruction set. To simplify, the program is recursive not reentrant.
* Ackermann function 07/09/2015
&LAB XDECO ®,&TARGET
.*-----------------------------------------------------------------*
.* THIS MACRO DISPLAYS THE REGISTER CONTENTS AS A TRUE *
.* DECIMAL VALUE. XDECO IS NOT PART OF STANDARD S360 MACROS! *
*------------------------------------------------------------------*
AIF (T'® EQ 'O').NOREG
AIF (T'&TARGET EQ 'O').NODEST
&LAB B I&SYSNDX BRANCH AROUND WORK AREA
W&SYSNDX DS XL8 CONVERSION WORK AREA
I&SYSNDX CVD ®,W&SYSNDX CONVERT TO DECIMAL
MVC &TARGET,=XL12'402120202020202020202020'
ED &TARGET,W&SYSNDX+2 MAKE FIELD PRINTABLE
BC 2,*+12 BYPASS NEGATIVE
MVI &TARGET+12,C'-' INSERT NEGATIVE SIGN
B *+8 BYPASS POSITIVE
MVI &TARGET+12,C'+' INSERT POSITIVE SIGN
MEXIT
.NOREG MNOTE 8,'INPUT REGISTER OMITTED'
MEXIT
.NODEST MNOTE 8,'TARGET FIELD OMITTED'
MEXIT
MEND
ACKERMAN CSECT
USING ACKERMAN,R12 r12 : base register
LR R12,R15 establish base register
ST R14,SAVER14A save r14
LA R4,0 m=0
LOOPM CH R4,=H'3' do m=0 to 3
BH ELOOPM
LA R5,0 n=0
LOOPN CH R5,=H'8' do n=0 to 8
BH ELOOPN
LR R1,R4 m
LR R2,R5 n
BAL R14,ACKER r1=acker(m,n)
XDECO R1,PG+19
XDECO R4,XD
MVC PG+10(2),XD+10
XDECO R5,XD
MVC PG+13(2),XD+10
XPRNT PG,44 print buffer
LA R5,1(R5) n=n+1
B LOOPN
ELOOPN LA R4,1(R4) m=m+1
B LOOPM
ELOOPM L R14,SAVER14A restore r14
BR R14 return to caller
SAVER14A DS F static save r14
PG DC CL44'Ackermann(xx,xx) = xxxxxxxxxxxx'
XD DS CL12
ACKER CNOP 0,4 function r1=acker(r1,r2)
LR R3,R1 save argument r1 in r3
LR R9,R10 save stackptr (r10) in r9 temp
LA R1,STACKLEN amount of storage required
GETMAIN RU,LV=(R1) allocate storage for stack
USING STACK,R10 make storage addressable
LR R10,R1 establish stack addressability
ST R14,SAVER14B save previous r14
ST R9,SAVER10B save previous r10
LR R1,R3 restore saved argument r1
START ST R1,M stack m
ST R2,N stack n
IF1 C R1,=F'0' if m<>0
BNE IF2 then goto if2
LR R11,R2 n
LA R11,1(R11) return n+1
B EXIT
IF2 C R2,=F'0' else if m<>0
BNE IF3 then goto if3
BCTR R1,0 m=m-1
LA R2,1 n=1
BAL R14,ACKER r1=acker(m)
LR R11,R1 return acker(m-1,1)
B EXIT
IF3 BCTR R2,0 n=n-1
BAL R14,ACKER r1=acker(m,n-1)
LR R2,R1 acker(m,n-1)
L R1,M m
BCTR R1,0 m=m-1
BAL R14,ACKER r1=acker(m-1,acker(m,n-1))
LR R11,R1 return acker(m-1,1)
EXIT L R14,SAVER14B restore r14
L R9,SAVER10B restore r10 temp
LA R0,STACKLEN amount of storage to free
FREEMAIN A=(R10),LV=(R0) free allocated storage
LR R1,R11 value returned
LR R10,R9 restore r10
BR R14 return to caller
LTORG
DROP R12 base no longer needed
STACK DSECT dynamic area
SAVER14B DS F saved r14
SAVER10B DS F saved r10
M DS F m
N DS F n
STACKLEN EQU *-STACK
YREGS
END ACKERMAN
{{out}}
Ackermann( 0, 0) = 1
Ackermann( 0, 1) = 2
Ackermann( 0, 2) = 3
Ackermann( 0, 3) = 4
Ackermann( 0, 4) = 5
Ackermann( 0, 5) = 6
Ackermann( 0, 6) = 7
Ackermann( 0, 7) = 8
Ackermann( 0, 8) = 9
Ackermann( 1, 0) = 2
Ackermann( 1, 1) = 3
Ackermann( 1, 2) = 4
Ackermann( 1, 3) = 5
Ackermann( 1, 4) = 6
Ackermann( 1, 5) = 7
Ackermann( 1, 6) = 8
Ackermann( 1, 7) = 9
Ackermann( 1, 8) = 10
Ackermann( 2, 0) = 3
Ackermann( 2, 1) = 5
Ackermann( 2, 2) = 7
Ackermann( 2, 3) = 9
Ackermann( 2, 4) = 11
Ackermann( 2, 5) = 13
Ackermann( 2, 6) = 15
Ackermann( 2, 7) = 17
Ackermann( 2, 8) = 19
Ackermann( 3, 0) = 5
Ackermann( 3, 1) = 13
Ackermann( 3, 2) = 29
Ackermann( 3, 3) = 61
Ackermann( 3, 4) = 125
Ackermann( 3, 5) = 253
Ackermann( 3, 6) = 509
Ackermann( 3, 7) = 1021
Ackermann( 3, 8) = 2045
68000 Assembly
This implementation is based on the code shown in the computerphile episode in the youtube link at the top of this page (time index 5:00).
;
; Ackermann function for Motorola 68000 under AmigaOs 2+ by Thorham
;
; Set stack space to 60000 for m = 3, n = 5.
;
; The program will print the ackermann values for the range m = 0..3, n = 0..5
;
_LVOOpenLibrary equ -552
_LVOCloseLibrary equ -414
_LVOVPrintf equ -954
m equ 3 ; Nr of iterations for the main loop.
n equ 5 ; Do NOT set them higher, or it will take hours to complete on
; 68k, not to mention that the stack usage will become astronomical.
; Perhaps n can be a little higher... If you do increase the ranges
; then don't forget to increase the stack size.
execBase=4
start
move.l execBase,a6
lea dosName,a1
moveq #36,d0
jsr _LVOOpenLibrary(a6)
move.l d0,dosBase
beq exit
move.l dosBase,a6
lea printfArgs,a2
clr.l d3 ; m
.loopn
clr.l d4 ; n
.loopm
bsr ackermann
move.l d3,0(a2)
move.l d4,4(a2)
move.l d5,8(a2)
move.l #outString,d1
move.l a2,d2
jsr _LVOVPrintf(a6)
addq.l #1,d4
cmp.l #n,d4
ble .loopm
addq.l #1,d3
cmp.l #m,d3
ble .loopn
exit
move.l execBase,a6
move.l dosBase,a1
jsr _LVOCloseLibrary(a6)
rts
;
; ackermann function
;
; in:
;
; d3 = m
; d4 = n
;
; out:
;
; d5 = ans
;
ackermann
move.l d3,-(sp)
move.l d4,-(sp)
tst.l d3
bne .l1
move.l d4,d5
addq.l #1,d5
bra .return
.l1
tst.l d4
bne .l2
subq.l #1,d3
moveq #1,d4
bsr ackermann
bra .return
.l2
subq.l #1,d4
bsr ackermann
move.l d5,d4
subq.l #1,d3
bsr ackermann
.return
move.l (sp)+,d4
move.l (sp)+,d3
rts
;
; variables
;
dosBase
dc.l 0
printfArgs
dcb.l 3
;
; strings
;
dosName
dc.b "dos.library",0
outString
dc.b "ackermann (%ld,%ld) is: %ld",10,0
8th
\ Ackermann function, illustrating use of "memoization".
\ Memoization is a technique whereby intermediate computed values are stored
\ away against later need. It is particularly valuable when calculating those
\ values is time or resource intensive, as with the Ackermann function.
\ make the stack much bigger so this can complete!
100000 stack-size
\ This is where memoized values are stored:
{} var, dict
\ Simple accessor words
: dict! \ "key" val --
dict @ -rot m:! drop ;
: dict@ \ "key" -- val
dict @ swap m:@ nip ;
defer: ack1
\ We just jam the string representation of the two numbers together for a key:
: makeKey \ m n -- m n key
2dup >s swap >s s:+ ;
: ack2 \ m n -- A
makeKey dup
dict@ null?
if \ can't find key in dict
\ m n key null
drop \ m n key
-rot \ key m n
ack1 \ key A
tuck \ A key A
dict! \ A
else \ found value
\ m n key value
>r drop 2drop r>
then ;
: ack \ m n -- A
over not
if
nip n:1+
else
dup not
if
drop n:1- 1 ack2
else
over swap n:1- ack2
swap n:1- swap ack2
then
then ;
' ack is ack1
: ackOf \ m n --
2dup
"Ack(" . swap . ", " . . ") = " . ack . cr ;
0 0 ackOf
0 4 ackOf
1 0 ackOf
1 1 ackOf
2 1 ackOf
2 2 ackOf
3 1 ackOf
3 3 ackOf
4 0 ackOf
\ this last requires a very large data stack. So start 8th with a parameter '-k 100000'
4 1 ackOf
bye
{{out|The output}}
Ack(0, 0) = 1
Ack(0, 4) = 5
Ack(1, 0) = 2
Ack(1, 1) = 3
Ack(2, 1) = 5
Ack(2, 2) = 7
Ack(3, 1) = 13
Ack(3, 3) = 61
Ack(4, 0) = 13
Ack(4, 1) = 65533
ABAP
REPORT zhuberv_ackermann.
CLASS zcl_ackermann DEFINITION.
PUBLIC SECTION.
CLASS-METHODS ackermann IMPORTING m TYPE i
n TYPE i
RETURNING value(v) TYPE i.
ENDCLASS. "zcl_ackermann DEFINITION
CLASS zcl_ackermann IMPLEMENTATION.
METHOD: ackermann.
DATA: lv_new_m TYPE i,
lv_new_n TYPE i.
IF m = 0.
v = n + 1.
ELSEIF m > 0 AND n = 0.
lv_new_m = m - 1.
lv_new_n = 1.
v = ackermann( m = lv_new_m n = lv_new_n ).
ELSEIF m > 0 AND n > 0.
lv_new_m = m - 1.
lv_new_n = n - 1.
lv_new_n = ackermann( m = m n = lv_new_n ).
v = ackermann( m = lv_new_m n = lv_new_n ).
ENDIF.
ENDMETHOD. "ackermann
ENDCLASS. "zcl_ackermann IMPLEMENTATION
PARAMETERS: pa_m TYPE i,
pa_n TYPE i.
DATA: lv_result TYPE i.
START-OF-SELECTION.
lv_result = zcl_ackermann=>ackermann( m = pa_m n = pa_n ).
WRITE: / lv_result.
ActionScript
public function ackermann(m:uint, n:uint):uint
{
if (m == 0)
{
return n + 1;
}
if (n == 0)
{
return ackermann(m - 1, 1);
}
return ackermann(m - 1, ackermann(m, n - 1));
}
Ada
with Ada.Text_IO; use Ada.Text_IO;
procedure Test_Ackermann is
function Ackermann (M, N : Natural) return Natural is
begin
if M = 0 then
return N + 1;
elsif N = 0 then
return Ackermann (M - 1, 1);
else
return Ackermann (M - 1, Ackermann (M, N - 1));
end if;
end Ackermann;
begin
for M in 0..3 loop
for N in 0..6 loop
Put (Natural'Image (Ackermann (M, N)));
end loop;
New_Line;
end loop;
end Test_Ackermann;
The implementation does not care about arbitrary precision numbers because the Ackermann function does not only grow, but also slow quickly, when computed recursively. {{out}} the first 4x7 Ackermann's numbers:
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
Agda
{{works with|Agda|2.5.2}} {{libheader|agda-stdlib v0.13}}
open import Data.Nat
open import Data.Nat.Show
open import IO
module Ackermann where
ack : ℕ -> ℕ -> ℕ
ack zero n = n + 1
ack (suc m) zero = ack m 1
ack (suc m) (suc n) = ack m (ack (suc m) n)
main = run (putStrLn (show (ack 3 9)))
Note the unicode ℕ characters, they can be input in emacs agda mode using "\bN". Running in bash:
agda --compile Ackermann.agda
./Ackermann
{{out}}
4093
ALGOL 68
{{trans|Ada}} {{works with|ALGOL 68|Standard - no extensions to language used}} {{works with|ALGOL 68G|Any - tested with release mk15-0.8b.fc9.i386}} {{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386}}
PROC test ackermann = VOID:
BEGIN
PROC ackermann = (INT m, n)INT:
BEGIN
IF m = 0 THEN
n + 1
ELIF n = 0 THEN
ackermann (m - 1, 1)
ELSE
ackermann (m - 1, ackermann (m, n - 1))
FI
END # ackermann #;
FOR m FROM 0 TO 3 DO
FOR n FROM 0 TO 6 DO
print(ackermann (m, n))
OD;
new line(stand out)
OD
END # test ackermann #;
test ackermann
{{out}}
+1 +2 +3 +4 +5 +6 +7
+2 +3 +4 +5 +6 +7 +8
+3 +5 +7 +9 +11 +13 +15
+5 +13 +29 +61 +125 +253 +509
APL
{{works with|Dyalog APL}}
ackermann←{
0=1⊃⍵:1+2⊃⍵
0=2⊃⍵:∇(¯1+1⊃⍵)1
∇(¯1+1⊃⍵),∇(1⊃⍵),¯1+2⊃⍵
}
AppleScript
on ackermann(m, n)
if m is equal to 0 then return n + 1
if n is equal to 0 then return ackermann(m - 1, 1)
return ackermann(m - 1, ackermann(m, n - 1))
end ackermann
Argile
{{works with|Argile|1.0.0}}
use std
for each (val nat n) from 0 to 6
for each (val nat m) from 0 to 3
print "A("m","n") = "(A m n)
.:A <nat m, nat n>:. -> nat
return (n+1) if m == 0
return (A (m - 1) 1) if n == 0
A (m - 1) (A m (n - 1))
ATS
fun ackermann
{m,n:nat} .<m,n>.
(m: int m, n: int n): Nat =
case+ (m, n) of
| (0, _) => n+1
| (_, 0) =>> ackermann (m-1, 1)
| (_, _) =>> ackermann (m-1, ackermann (m, n-1))
// end of [ackermann]
AutoHotkey
A(m, n) {
If (m > 0) && (n = 0)
Return A(m-1,1)
Else If (m > 0) && (n > 0)
Return A(m-1,A(m, n-1))
Else If (m=0)
Return n+1
}
; Example:
MsgBox, % "A(1,2) = " A(1,2)
AutoIt
Classical version
Func Ackermann($m, $n)
If ($m = 0) Then
Return $n+1
Else
If ($n = 0) Then
Return Ackermann($m-1, 1)
Else
return Ackermann($m-1, Ackermann($m, $n-1))
EndIf
EndIf
EndFunc
Classical + cache implementation
This version works way faster than the classical one: Ackermann(3, 5) runs in 12,7 ms, while the classical version takes 402,2 ms.
Global $ackermann[2047][2047] ; Set the size to whatever you want
Func Ackermann($m, $n)
If ($ackermann[$m][$n] <> 0) Then
Return $ackermann[$m][$n]
Else
If ($m = 0) Then
$return = $n + 1
Else
If ($n = 0) Then
$return = Ackermann($m - 1, 1)
Else
$return = Ackermann($m - 1, Ackermann($m, $n - 1))
EndIf
EndIf
$ackermann[$m][$n] = $return
Return $return
EndIf
EndFunc ;==>Ackermann
AWK
function ackermann(m, n)
{
if ( m == 0 ) {
return n+1
}
if ( n == 0 ) {
return ackermann(m-1, 1)
}
return ackermann(m-1, ackermann(m, n-1))
}
BEGIN {
for(n=0; n < 7; n++) {
for(m=0; m < 4; m++) {
print "A(" m "," n ") = " ackermann(m,n)
}
}
}
Babel
main:
{((0 0) (0 1) (0 2)
(0 3) (0 4) (1 0)
(1 1) (1 2) (1 3)
(1 4) (2 0) (2 1)
(2 2) (2 3) (3 0)
(3 1) (3 2) (4 0))
{ dup
"A(" << { %d " " . << } ... ") = " <<
reverse give
ack
%d cr << } ... }
ack!:
{ dup zero?
{ <-> dup zero?
{ <->
cp
1 -
<- <- 1 - ->
ack ->
ack }
{ <->
1 -
<- 1 ->
ack }
if }
{ zap 1 + }
if }
zero?!: { 0 = }
{{out}}
A(0 0 ) = 1
A(0 1 ) = 2
A(0 2 ) = 3
A(0 3 ) = 4
A(0 4 ) = 5
A(1 0 ) = 2
A(1 1 ) = 3
A(1 2 ) = 4
A(1 3 ) = 5
A(1 4 ) = 6
A(2 0 ) = 3
A(2 1 ) = 5
A(2 2 ) = 7
A(2 3 ) = 9
A(3 0 ) = 5
A(3 1 ) = 13
A(3 2 ) = 29
A(4 0 ) = 13
BASIC
=
Applesoft BASIC
=
100 DIM R%(2900),M(2900),N(2900)
110 FOR M = 0 TO 3
120 FOR N = 0 TO 4
130 GOSUB 200"ACKERMANN
140 PRINT "ACK("M","N") = "ACK
150 NEXT N, M
160 END
200 M(SP) = M
210 N(SP) = N
REM A(M - 1, A(M, N - 1))
220 IF M > 0 AND N > 0 THEN N = N - 1 : R%(SP) = 0 : SP = SP + 1 : GOTO 200
REM A(M - 1, 1)
230 IF M > 0 THEN M = M - 1 : N = 1 : R%(SP) = 1 : SP = SP + 1 : GOTO 200
REM N + 1
240 ACK = N + 1
REM RETURN
250 M = M(SP) : N = N(SP) : IF SP = 0 THEN RETURN
260 FOR SP = SP - 1 TO 0 STEP -1 : IF R%(SP) THEN M = M(SP) : N = N(SP) : NEXT SP : SP = 0 : RETURN
270 M = M - 1 : N = ACK : R%(SP) = 1 : SP = SP + 1 : GOTO 200
=
BASIC256
=
dim stack(5000, 3) # BASIC-256 lacks functions (as of ver. 0.9.6.66)
stack[0,0] = 3 # M
stack[0,1] = 7 # N
lev = 0
gosub ackermann
print "A("+stack[0,0]+","+stack[0,1]+") = "+stack[0,2]
end
ackermann:
if stack[lev,0]=0 then
stack[lev,2] = stack[lev,1]+1
return
end if
if stack[lev,1]=0 then
lev = lev+1
stack[lev,0] = stack[lev-1,0]-1
stack[lev,1] = 1
gosub ackermann
stack[lev-1,2] = stack[lev,2]
lev = lev-1
return
end if
lev = lev+1
stack[lev,0] = stack[lev-1,0]
stack[lev,1] = stack[lev-1,1]-1
gosub ackermann
stack[lev,0] = stack[lev-1,0]-1
stack[lev,1] = stack[lev,2]
gosub ackermann
stack[lev-1,2] = stack[lev,2]
lev = lev-1
return
{{out}}
A(3,7) = 1021
# BASIC256 since 0.9.9.1 supports functions
for m = 0 to 3
for n = 0 to 4
print m + " " + n + " " + ackermann(m,n)
next n
next m
end
function ackermann(m,n)
if m = 0 then
ackermann = n+1
else
if n = 0 then
ackermann = ackermann(m-1,1)
else
ackermann = ackermann(m-1,ackermann(m,n-1))
endif
end if
end function
{{out}}
0 0 1
0 1 2
0 2 3
0 3 4
0 4 5
1 0 2
1 1 3
1 2 4
1 3 5
1 4 6
2 0 3
2 1 5
2 2 7
2 3 9
2 4 11
3 0 5
3 1 13
3 2 29
3 3 61
3 4 125
=
BBC BASIC
=
PRINT FNackermann(3, 7)
END
DEF FNackermann(M%, N%)
IF M% = 0 THEN = N% + 1
IF N% = 0 THEN = FNackermann(M% - 1, 1)
= FNackermann(M% - 1, FNackermann(M%, N%-1))
=
QuickBasic
= {{works with|QuickBasic|4.5}} BASIC runs out of stack space very quickly. The call ack(3, 4) gives a stack error.
DECLARE FUNCTION ack! (m!, n!)
FUNCTION ack (m!, n!)
IF m = 0 THEN ack = n + 1
IF m > 0 AND n = 0 THEN
ack = ack(m - 1, 1)
END IF
IF m > 0 AND n > 0 THEN
ack = ack(m - 1, ack(m, n - 1))
END IF
END FUNCTION
Batch File
Had trouble with this, so called in the gurus at StackOverflow. Thanks to Patrick Cuff for pointing out where I was going wrong.
::Ackermann.cmd
@echo off
set depth=0
:ack
if %1==0 goto m0
if %2==0 goto n0
:else
set /a n=%2-1
set /a depth+=1
call :ack %1 %n%
set t=%errorlevel%
set /a depth-=1
set /a m=%1-1
set /a depth+=1
call :ack %m% %t%
set t=%errorlevel%
set /a depth-=1
if %depth%==0 ( exit %t% ) else ( exit /b %t% )
:m0
set/a n=%2+1
if %depth%==0 ( exit %n% ) else ( exit /b %n% )
:n0
set /a m=%1-1
set /a depth+=1
call :ack %m% 1
set t=%errorlevel%
set /a depth-=1
if %depth%==0 ( exit %t% ) else ( exit /b %t% )
Because of the exit statements, running this bare closes one's shell, so this test routine handles the calling of Ackermann.cmd
::Ack.cmd
@echo off
cmd/c ackermann.cmd %1 %2
echo Ackermann(%1, %2)=%errorlevel%
A few test runs:
D:\Documents and Settings\Bruce>ack 0 4
Ackermann(0, 4)=5
D:\Documents and Settings\Bruce>ack 1 4
Ackermann(1, 4)=6
D:\Documents and Settings\Bruce>ack 2 4
Ackermann(2, 4)=11
D:\Documents and Settings\Bruce>ack 3 4
Ackermann(3, 4)=125
bc
Requires a bc that supports long names and the print statement. {{works with|OpenBSD bc}} {{Works with|GNU bc}}
define ack(m, n) {
if ( m == 0 ) return (n+1);
if ( n == 0 ) return (ack(m-1, 1));
return (ack(m-1, ack(m, n-1)));
}
for (n=0; n<7; n++) {
for (m=0; m<4; m++) {
print "A(", m, ",", n, ") = ", ack(m,n), "\n";
}
}
quit
BCPL
GET "libhdr"
LET ack(m, n) = m=0 -> n+1,
n=0 -> ack(m-1, 1),
ack(m-1, ack(m, n-1))
LET start() = VALOF
{ FOR i = 0 TO 6 FOR m = 0 TO 3 DO
writef("ack(%n, %n) = %n*n", m, n, ack(m,n))
RESULTIS 0
}
beeswax
Iterative slow version:
>M?f@h@gMf@h3yzp if m>0 and n>0 => replace m,n with m-1,m,n-1
>@h@g'b?1f@h@gM?f@hzp if m>0 and n=0 => replace m,n with m-1,1
_ii>Ag~1?~Lpz1~2h@g'd?g?Pfzp if m=0 => replace m,n with n+1
>I;
b < <
A functional and recursive realization of the version above. Functions are realized by direct calls of functions via jumps (instruction J) to the entry points of two distinct functions:
1st function _ii (input function) with entry point at (row,col) = (4,1)
2nd function Ag~1.... (Ackermann function) with entry point at (row,col) = (1,1)
Each block of 1FJ or 1fFJ in the code is a call of the Ackermann function itself.
Ag~1?Lp1~2@g'p?g?Pf1FJ Ackermann function. if m=0 => run Ackermann function (m, n+1)
xI; x@g'p??@Mf1fFJ if m>0 and n=0 => run Ackermann (m-1,1)
xM@~gM??f~f@f1FJ if m>0 and n>0 => run Ackermann(m,Ackermann(m-1,n-1))
_ii1FJ input function. Input m,n, then execute Ackermann(m,n)
Highly optimized and fast version, returns A(4,1)/A(5,0) almost instantaneously:
>Mf@Ph?@g@2h@Mf@Php if m>4 and n>0 => replace m,n with m-1,m,n-1
>~4~L#1~2hg'd?1f@hgM?f2h p if m>4 and n=0 => replace m,n with m-1,1
# q < /n+3 times \
#X~4K#?2Fg?PPP>@B@M"pb if m=4 => replace m,n with 2^(2^(....)))-3
# >~3K#?g?PPP~2BMMp>@MMMp if m=3 => replace m,n with 2^(n+3)-3
_ii>Ag~1?~Lpz1~2h@gX'#?g?P p M if m=0 => replace m,n with n+1
z I ~>~1K#?g?PP p if m=1 => replace m,n with n+2
f ; >2K#?g?~2.PPPp if m=2 => replace m,n with 2n+3
z b < < <
d <
Higher values than A(4,1)/(5,0) lead to UInt64 wraparound, support for numbers bigger than 2^64-1 is not implemented in these solutions.
Befunge
===Befunge-93=== {{trans|ERRE}} Since Befunge-93 doesn't have recursive capabilities we need to use an iterative algorithm.
&>&>vvg0>#0\#-:#1_1v
@v:\<vp0 0:-1<\+<
^>00p>:#^_$1+\:#^_$.
===Befunge-98=== {{works with|CCBI|2.1}}
r[1&&{0
>v
j
u>.@
1> \:v
^ v:\_$1+
\^v_$1\1-
u^>1-0fp:1-\0fg101-
The program reads two integers (first m, then n) from command line, idles around funge space, then outputs the result of the Ackerman function. Since the latter is calculated truly recursively, the execution time becomes unwieldy for most m>3.
Bracmat
Three solutions are presented here. The first one is a purely recursive version, only using the formulas at the top of the page. The value of A(4,1) cannot be computed due to stack overflow. It can compute A(3,9) (4093), but probably not A(3,10)
( Ack
= m n
. !arg:(?m,?n)
& ( !m:0&!n+1
| !n:0&Ack$(!m+-1,1)
| Ack$(!m+-1,Ack$(!m,!n+-1))
)
);
The second version is a purely non-recursive solution that easily can compute A(4,1). The program uses a stack for Ackermann function calls that are to be evaluated, but that cannot be computed given the currently known function values - the "known unknowns". The currently known values are stored in a hash table. The Hash table also contains incomplete Ackermann function calls, namely those for which the second argument is not known yet - "the unknown unknowns". These function calls are associated with "known unknowns" that are going to provide the value of the second argument. As soon as such an associated known unknown becomes known, the unknown unknown becomes a known unknown and is pushed onto the stack.
Although all known values are stored in the hash table, the converse is not true: an element in the hash table is either a "known known" or an "unknown unknown" associated with an "known unknown".
( A
= m n value key eq chain
, find insert future stack va val
. ( chain
= key future skey
. !arg:(?key.?future)
& str$!key:?skey
& (cache..insert)$(!skey..!future)
&
)
& (find=.(cache..find)$(str$!arg))
& ( insert
= key value future v futureeq futurem skey
. !arg:(?key.?value)
& str$!key:?skey
& ( (cache..find)$!skey:(?key.?v.?future)
& (cache..remove)$!skey
& (cache..insert)$(!skey.!value.)
& ( !future:(?futurem.?futureeq)
& (!futurem,!value.!futureeq)
|
)
| (cache..insert)$(!skey.!value.)&
)
)
& !arg:(?m,?n)
& !n+1:?value
& :?eq:?stack
& whl
' ( (!m,!n):?key
& ( find$!key:(?.#%?value.?future)
& insert$(!eq.!value) !future
| !m:0
& !n+1:?value
& ( !eq:&insert$(!key.!value)
| insert$(!key.!value) !stack:?stack
& insert$(!eq.!value)
)
| !n:0
& (!m+-1,1.!key)
(!eq:|(!key.!eq))
| find$(!m,!n+-1):(?.?val.?)
& ( !val:#%
& ( find$(!m+-1,!val):(?.?va.?)
& !va:#%
& insert$(!key.!va)
| (!m+-1,!val.!eq)
(!m,!n.!eq)
)
|
)
| chain$(!m,!n+-1.!m+-1.!key)
& (!m,!n+-1.)
(!eq:|(!key.!eq))
)
!stack
: (?m,?n.?eq) ?stack
)
& !value
)
& new$hash:?cache
{{out|Some results}}
A$(0,0):1
A$(3,13):65533
A$(3,14):131069
A$(4,1):65533
The last solution is a recursive solution that employs some extra formulas, inspired by the Common Lisp solution further down.
( AckFormula
= m n
. !arg:(?m,?n)
& ( !m:0&!n+1
| !m:1&!n+2
| !m:2&2*!n+3
| !m:3&2^(!n+3)+-3
| !n:0&AckFormula$(!m+-1,1)
| AckFormula$(!m+-1,AckFormula$(!m,!n+-1))
)
)
{{Out|Some results}}
AckFormula$(4,1):65533
AckFormula$(4,2):2003529930406846464979072351560255750447825475569751419265016973.....22087777506072339445587895905719156733
The last computation costs about 0,03 seconds.
Brat
ackermann = { m, n |
when { m == 0 } { n + 1 }
{ m > 0 && n == 0 } { ackermann(m - 1, 1) }
{ m > 0 && n > 0 } { ackermann(m - 1, ackermann(m, n - 1)) }
}
p ackermann 3, 4 #Prints 125
C
Straightforward implementation per Ackermann definition:
#include <stdio.h>
int ackermann(int m, int n)
{
if (!m) return n + 1;
if (!n) return ackermann(m - 1, 1);
return ackermann(m - 1, ackermann(m, n - 1));
}
int main()
{
int m, n;
for (m = 0; m <= 4; m++)
for (n = 0; n < 6 - m; n++)
printf("A(%d, %d) = %d\n", m, n, ackermann(m, n));
return 0;
}
{{out}}
A(0, 0) = 1
A(0, 1) = 2
A(0, 2) = 3
A(0, 3) = 4
A(0, 4) = 5
A(0, 5) = 6
A(1, 0) = 2
A(1, 1) = 3
A(1, 2) = 4
A(1, 3) = 5
A(1, 4) = 6
A(2, 0) = 3
A(2, 1) = 5
A(2, 2) = 7
A(2, 3) = 9
A(3, 0) = 5
A(3, 1) = 13
A(3, 2) = 29
A(4, 0) = 13
A(4, 1) = 65533
Ackermann function makes a lot of recursive calls, so the above program is a bit naive. We need to be slightly less naive, by doing some simple caching:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int m_bits, n_bits;
int *cache;
int ackermann(int m, int n)
{
int idx, res;
if (!m) return n + 1;
if (n >= 1<<n_bits) {
printf("%d, %d\n", m, n);
idx = 0;
} else {
idx = (m << n_bits) + n;
if (cache[idx]) return cache[idx];
}
if (!n) res = ackermann(m - 1, 1);
else res = ackermann(m - 1, ackermann(m, n - 1));
if (idx) cache[idx] = res;
return res;
}
int main()
{
int m, n;
m_bits = 3;
n_bits = 20; /* can save n values up to 2**20 - 1, that's 1 meg */
cache = malloc(sizeof(int) * (1 << (m_bits + n_bits)));
memset(cache, 0, sizeof(int) * (1 << (m_bits + n_bits)));
for (m = 0; m <= 4; m++)
for (n = 0; n < 6 - m; n++)
printf("A(%d, %d) = %d\n", m, n, ackermann(m, n));
return 0;
}
{{out}}
A(0, 0) = 1
A(0, 1) = 2
A(0, 2) = 3
A(0, 3) = 4
A(0, 4) = 5
A(0, 5) = 6
A(1, 0) = 2
A(1, 1) = 3
A(1, 2) = 4
A(1, 3) = 5
A(1, 4) = 6
A(2, 0) = 3
A(2, 1) = 5
A(2, 2) = 7
A(2, 3) = 9
A(3, 0) = 5
A(3, 1) = 13
A(3, 2) = 29
A(4, 0) = 13
A(4, 1) = 65533
Whee. Well, with some extra work, we calculated one more n value, big deal, right?
But see, A(4, 2) = A(3, A(4, 1)) = A(3, 65533) = A(2, A(3, 65532)) = ... you can see how fast it blows up. In fact, no amount of caching will help you calculate large m values; on the machine I use A(4, 2) segfaults because the recursions run out of stack space--not a whole lot I can do about it. At least it runs out of stack space quickly, unlike the first solution...
C#
Basic Version
using System;
class Program
{
public static long Ackermann(long m, long n)
{
if(m > 0)
{
if (n > 0)
return Ackermann(m - 1, Ackermann(m, n - 1));
else if (n == 0)
return Ackermann(m - 1, 1);
}
else if(m == 0)
{
if(n >= 0)
return n + 1;
}
throw new System.ArgumentOutOfRangeException();
}
static void Main()
{
for (long m = 0; m <= 3; ++m)
{
for (long n = 0; n <= 4; ++n)
{
Console.WriteLine("Ackermann({0}, {1}) = {2}", m, n, Ackermann(m, n));
}
}
}
}
{{out}}
Ackermann(0, 0) = 1
Ackermann(0, 1) = 2
Ackermann(0, 2) = 3
Ackermann(0, 3) = 4
Ackermann(0, 4) = 5
Ackermann(1, 0) = 2
Ackermann(1, 1) = 3
Ackermann(1, 2) = 4
Ackermann(1, 3) = 5
Ackermann(1, 4) = 6
Ackermann(2, 0) = 3
Ackermann(2, 1) = 5
Ackermann(2, 2) = 7
Ackermann(2, 3) = 9
Ackermann(2, 4) = 11
Ackermann(3, 0) = 5
Ackermann(3, 1) = 13
Ackermann(3, 2) = 29
Ackermann(3, 3) = 61
Ackermann(3, 4) = 125
Efficient Version
using System;
using System.Numerics;
using System.IO;
using System.Diagnostics;
namespace Ackermann_Function
{
class Program
{
static void Main(string[] args)
{
int _m = 0;
int _n = 0;
Console.Write("m = ");
try
{
_m = Convert.ToInt32(Console.ReadLine());
}
catch (Exception)
{
Console.WriteLine("Please enter a number.");
}
Console.Write("n = ");
try
{
_n = Convert.ToInt32(Console.ReadLine());
}
catch (Exception)
{
Console.WriteLine("Please enter a number.");
}
//for (long m = 0; m <= 10; ++m)
//{
// for (long n = 0; n <= 10; ++n)
// {
// DateTime now = DateTime.Now;
// Console.WriteLine("Ackermann({0}, {1}) = {2}", m, n, Ackermann(m, n));
// Console.WriteLine("Time taken:{0}", DateTime.Now - now);
// }
//}
DateTime now = DateTime.Now;
Console.WriteLine("Ackermann({0}, {1}) = {2}", _m, _n, Ackermann(_m, _n));
Console.WriteLine("Time taken:{0}", DateTime.Now - now);
File.WriteAllText("number.txt", Ackermann(_m, _n).ToString());
Process.Start("number.txt");
Console.ReadKey();
}
public class OverflowlessStack<T>
{
internal sealed class SinglyLinkedNode
{
private const int ArraySize = 2048;
T[] _array;
int _size;
public SinglyLinkedNode Next;
public SinglyLinkedNode()
{
_array = new T[ArraySize];
}
public bool IsEmpty { get { return _size == 0; } }
public SinglyLinkedNode Push(T item)
{
if (_size == ArraySize - 1)
{
SinglyLinkedNode n = new SinglyLinkedNode();
n.Next = this;
n.Push(item);
return n;
}
_array[_size++] = item;
return this;
}
public T Pop()
{
return _array[--_size];
}
}
private SinglyLinkedNode _head = new SinglyLinkedNode();
public T Pop()
{
T ret = _head.Pop();
if (_head.IsEmpty && _head.Next != null)
_head = _head.Next;
return ret;
}
public void Push(T item)
{
_head = _head.Push(item);
}
public bool IsEmpty
{
get { return _head.Next == null && _head.IsEmpty; }
}
}
public static BigInteger Ackermann(BigInteger m, BigInteger n)
{
var stack = new OverflowlessStack<BigInteger>();
stack.Push(m);
while (!stack.IsEmpty)
{
m = stack.Pop();
skipStack:
if (m == 0)
n = n + 1;
else if (m == 1)
n = n + 2;
else if (m == 2)
n = n * 2 + 3;
else if (n == 0)
{
--m;
n = 1;
goto skipStack;
}
else
{
stack.Push(m - 1);
--n;
goto skipStack;
}
}
return n;
}
}
}
Possibly the most efficient implementation of Ackermann in C#. It successfully runs Ack(4,2) when executed in Visual Studio. Don't forget to add a reference to System.Numerics.
C++
Basic version
#include <iostream>
unsigned int ackermann(unsigned int m, unsigned int n) {
if (m == 0) {
return n + 1;
}
if (n == 0) {
return ackermann(m - 1, 1);
}
return ackermann(m - 1, ackermann(m, n - 1));
}
int main() {
for (unsigned int m = 0; m < 4; ++m) {
for (unsigned int n = 0; n < 10; ++n) {
std::cout << "A(" << m << ", " << n << ") = " << ackermann(m, n) << "\n";
}
}
}
Efficient version
{{trans|D}} C++11 with boost's big integer type. Compile with: g++ -std=c++11 -I /path/to/boost ackermann.cpp.
#include <iostream>
#include <sstream>
#include <string>
#include <boost/multiprecision/cpp_int.hpp>
using big_int = boost::multiprecision::cpp_int;
big_int ipow(big_int base, big_int exp) {
big_int result(1);
while (exp) {
if (exp & 1) {
result *= base;
}
exp >>= 1;
base *= base;
}
return result;
}
big_int ackermann(unsigned m, unsigned n) {
static big_int (*ack)(unsigned, big_int) =
[](unsigned m, big_int n)->big_int {
switch (m) {
case 0:
return n + 1;
case 1:
return n + 2;
case 2:
return 3 + 2 * n;
case 3:
return 5 + 8 * (ipow(big_int(2), n) - 1);
default:
return n == 0 ? ack(m - 1, big_int(1)) : ack(m - 1, ack(m, n - 1));
}
};
return ack(m, big_int(n));
}
int main() {
for (unsigned m = 0; m < 4; ++m) {
for (unsigned n = 0; n < 10; ++n) {
std::cout << "A(" << m << ", " << n << ") = " << ackermann(m, n) << "\n";
}
}
std::cout << "A(4, 1) = " << ackermann(4, 1) << "\n";
std::stringstream ss;
ss << ackermann(4, 2);
auto text = ss.str();
std::cout << "A(4, 2) = (" << text.length() << " digits)\n"
<< text.substr(0, 80) << "\n...\n"
<< text.substr(text.length() - 80) << "\n";
}
```txt
A(0, 0) = 1
A(0, 1) = 2
A(0, 2) = 3
A(0, 3) = 4
A(0, 4) = 5
A(0, 5) = 6
A(0, 6) = 7
A(0, 7) = 8
A(0, 8) = 9
A(0, 9) = 10
A(1, 0) = 2
A(1, 1) = 3
A(1, 2) = 4
A(1, 3) = 5
A(1, 4) = 6
A(1, 5) = 7
A(1, 6) = 8
A(1, 7) = 9
A(1, 8) = 10
A(1, 9) = 11
A(2, 0) = 3
A(2, 1) = 5
A(2, 2) = 7
A(2, 3) = 9
A(2, 4) = 11
A(2, 5) = 13
A(2, 6) = 15
A(2, 7) = 17
A(2, 8) = 19
A(2, 9) = 21
A(3, 0) = 5
A(3, 1) = 13
A(3, 2) = 29
A(3, 3) = 61
A(3, 4) = 125
A(3, 5) = 253
A(3, 6) = 509
A(3, 7) = 1021
A(3, 8) = 2045
A(3, 9) = 4093
A(4, 1) = 65533
A(4, 2) = (19729 digits)
2003529930406846464979072351560255750447825475569751419265016973710894059556311
...
4717124577965048175856395072895337539755822087777506072339445587895905719156733
Chapel
proc A(m:int, n:int):int {
if m == 0 then
return n + 1;
else if n == 0 then
return A(m - 1, 1);
else
return A(m - 1, A(m, n - 1));
}
Clay
ackermann(m, n) {
if(m == 0)
return n + 1;
if(n == 0)
return ackermann(m - 1, 1);
return ackermann(m - 1, ackermann(m, n - 1));
}
CLIPS
'''Functional solution'''
(deffunction ackerman
(?m ?n)
(if (= 0 ?m)
then (+ ?n 1)
else (if (= 0 ?n)
then (ackerman (- ?m 1) 1)
else (ackerman (- ?m 1) (ackerman ?m (- ?n 1)))
)
)
)
{{out|Example usage}}
CLIPS> (ackerman 0 4)
5
CLIPS> (ackerman 1 4)
6
CLIPS> (ackerman 2 4)
11
CLIPS> (ackerman 3 4)
125
'''Fact-based solution'''
(deffacts solve-items
(solve 0 4)
(solve 1 4)
(solve 2 4)
(solve 3 4)
)
(defrule acker-m-0
?compute <- (compute 0 ?n)
=>
(retract ?compute)
(assert (ackerman 0 ?n (+ ?n 1)))
)
(defrule acker-n-0-pre
(compute ?m&:(> ?m 0) 0)
(not (ackerman =(- ?m 1) 1 ?))
=>
(assert (compute (- ?m 1) 1))
)
(defrule acker-n-0
?compute <- (compute ?m&:(> ?m 0) 0)
(ackerman =(- ?m 1) 1 ?val)
=>
(retract ?compute)
(assert (ackerman ?m 0 ?val))
)
(defrule acker-m-n-pre-1
(compute ?m&:(> ?m 0) ?n&:(> ?n 0))
(not (ackerman ?m =(- ?n 1) ?))
=>
(assert (compute ?m (- ?n 1)))
)
(defrule acker-m-n-pre-2
(compute ?m&:(> ?m 0) ?n&:(> ?n 0))
(ackerman ?m =(- ?n 1) ?newn)
(not (ackerman =(- ?m 1) ?newn ?))
=>
(assert (compute (- ?m 1) ?newn))
)
(defrule acker-m-n
?compute <- (compute ?m&:(> ?m 0) ?n&:(> ?n 0))
(ackerman ?m =(- ?n 1) ?newn)
(ackerman =(- ?m 1) ?newn ?val)
=>
(retract ?compute)
(assert (ackerman ?m ?n ?val))
)
(defrule acker-solve
(solve ?m ?n)
(not (compute ?m ?n))
(not (ackerman ?m ?n ?))
=>
(assert (compute ?m ?n))
)
(defrule acker-solved
?solve <- (solve ?m ?n)
(ackerman ?m ?n ?result)
=>
(retract ?solve)
(printout t "A(" ?m "," ?n ") = " ?result crlf)
)
When invoked, each required A(m,n) needed to solve the requested (solve ?m ?n) facts gets generated as its own fact. Below shows the invocation of the above, as well as an excerpt of the final facts list. Regardless of how many input (solve ?m ?n) requests are made, each possible A(m,n) is only solved once.
CLIPS> (reset)
CLIPS> (facts)
f-0 (initial-fact)
f-1 (solve 0 4)
f-2 (solve 1 4)
f-3 (solve 2 4)
f-4 (solve 3 4)
For a total of 5 facts.
CLIPS> (run)
A(3,4) = 125
A(2,4) = 11
A(1,4) = 6
A(0,4) = 5
CLIPS> (facts)
f-0 (initial-fact)
f-15 (ackerman 0 1 2)
f-16 (ackerman 1 0 2)
f-18 (ackerman 0 2 3)
...
f-632 (ackerman 1 123 125)
f-633 (ackerman 2 61 125)
f-634 (ackerman 3 4 125)
For a total of 316 facts.
CLIPS>
Clojure
(defn ackermann [m n]
(cond (zero? m) (inc n)
(zero? n) (ackermann (dec m) 1)
:else (ackermann (dec m) (ackermann m (dec n)))))
COBOL
IDENTIFICATION DIVISION.
PROGRAM-ID. Ackermann.
DATA DIVISION.
LINKAGE SECTION.
01 M USAGE UNSIGNED-LONG.
01 N USAGE UNSIGNED-LONG.
01 Return-Val USAGE UNSIGNED-LONG.
PROCEDURE DIVISION USING M N Return-Val.
EVALUATE M ALSO N
WHEN 0 ALSO ANY
ADD 1 TO N GIVING Return-Val
WHEN NOT 0 ALSO 0
SUBTRACT 1 FROM M
CALL "Ackermann" USING BY CONTENT M BY CONTENT 1
BY REFERENCE Return-Val
WHEN NOT 0 ALSO NOT 0
SUBTRACT 1 FROM N
CALL "Ackermann" USING BY CONTENT M BY CONTENT N
BY REFERENCE Return-Val
SUBTRACT 1 FROM M
CALL "Ackermann" USING BY CONTENT M
BY CONTENT Return-Val BY REFERENCE Return-Val
END-EVALUATE
GOBACK
.
CoffeeScript
ackermann = (m, n) ->
if m is 0 then n + 1
else if m > 0 and n is 0 then ackermann m - 1, 1
else ackermann m - 1, ackermann m, n - 1
Common Lisp
(defun ackermann (m n)
(cond ((zerop m) (1+ n))
((zerop n) (ackermann (1- m) 1))
(t (ackermann (1- m) (ackermann m (1- n))))))
More elaborately:
(defun ackermann (m n)
(case m ((0) (1+ n))
((1) (+ 2 n))
((2) (+ n n 3))
((3) (- (expt 2 (+ 3 n)) 3))
(otherwise (ackermann (1- m) (if (zerop n) 1 (ackermann m (1- n)))))))
(loop for m from 0 to 4 do
(loop for n from (- 5 m) to (- 6 m) do
(format t "A(~d, ~d) = ~d~%" m n (ackermann m n))))
{{out}}
A(0, 5) = 6
A(0, 6) = 7
A(1, 4) = 6
A(1, 5) = 7
A(2, 3) = 9
A(2, 4) = 11
A(3, 2) = 29
A(3, 3) = 61
A(4, 1) = 65533
A(4, 2) = 2003529930 <... skipping a few digits ...> 56733
Coq
Require Import Arith.
Fixpoint A m := fix A_m n :=
match m with
| 0 => n + 1
| S pm =>
match n with
| 0 => A pm 1
| S pn => A pm (A_m pn)
end
end.
Require Import Utf8.
Section FOLD.
Context {A: Type} (f: A → A) (a: A).
Fixpoint fold (n: nat) : A :=
match n with
| O => a
| S n' => f (fold n')
end.
End FOLD.
Definition ackermann : nat → nat → nat :=
fold (λ g, fold g (g (S O))) S.
Component Pascal
BlackBox Component Builder
MODULE NpctAckerman;
IMPORT StdLog;
VAR
m,n: INTEGER;
PROCEDURE Ackerman (x,y: INTEGER):INTEGER;
BEGIN
IF x = 0 THEN RETURN y + 1
ELSIF y = 0 THEN RETURN Ackerman (x - 1 , 1)
ELSE
RETURN Ackerman (x - 1 , Ackerman (x , y - 1))
END
END Ackerman;
PROCEDURE Do*;
BEGIN
FOR m := 0 TO 3 DO
FOR n := 0 TO 6 DO
StdLog.Int (Ackerman (m, n));StdLog.Char (' ')
END;
StdLog.Ln
END;
StdLog.Ln
END Do;
END NpctAckerman.
Execute: ^Q NpctAckerman.Do
```txt
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
Crystal
{{trans|Ruby}}
def ack(m, n)
if m == 0
n + 1
elsif n == 0
ack(m-1, 1)
else
ack(m-1, ack(m, n-1))
end
end
#Example:
(0..3).each do |m|
puts (0..6).map { |n| ack(m, n) }.join(' ')
end
{{out}}
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
D
Basic version
ulong ackermann(in ulong m, in ulong n) pure nothrow @nogc {
if (m == 0)
return n + 1;
if (n == 0)
return ackermann(m - 1, 1);
return ackermann(m - 1, ackermann(m, n - 1));
}
void main() {
assert(ackermann(2, 4) == 11);
}
More Efficient Version
{{trans|Mathematica}}
import std.stdio, std.bigint, std.conv;
BigInt ipow(BigInt base, BigInt exp) pure nothrow {
auto result = 1.BigInt;
while (exp) {
if (exp & 1)
result *= base;
exp >>= 1;
base *= base;
}
return result;
}
BigInt ackermann(in uint m, in uint n) pure nothrow
out(result) {
assert(result >= 0);
} body {
static BigInt ack(in uint m, in BigInt n) pure nothrow {
switch (m) {
case 0: return n + 1;
case 1: return n + 2;
case 2: return 3 + 2 * n;
//case 3: return 5 + 8 * (2 ^^ n - 1);
case 3: return 5 + 8 * (ipow(2.BigInt, n) - 1);
default: return (n == 0) ?
ack(m - 1, 1.BigInt) :
ack(m - 1, ack(m, n - 1));
}
}
return ack(m, n.BigInt);
}
void main() {
foreach (immutable m; 1 .. 4)
foreach (immutable n; 1 .. 9)
writefln("ackermann(%d, %d): %s", m, n, ackermann(m, n));
writefln("ackermann(4, 1): %s", ackermann(4, 1));
immutable a = ackermann(4, 2).text;
writefln("ackermann(4, 2)) (%d digits):\n%s...\n%s",
a.length, a[0 .. 94], a[$ - 96 .. $]);
}
{{out}}
ackermann(1, 1): 3
ackermann(1, 2): 4
ackermann(1, 3): 5
ackermann(1, 4): 6
ackermann(1, 5): 7
ackermann(1, 6): 8
ackermann(1, 7): 9
ackermann(1, 8): 10
ackermann(2, 1): 5
ackermann(2, 2): 7
ackermann(2, 3): 9
ackermann(2, 4): 11
ackermann(2, 5): 13
ackermann(2, 6): 15
ackermann(2, 7): 17
ackermann(2, 8): 19
ackermann(3, 1): 13
ackermann(3, 2): 29
ackermann(3, 3): 61
ackermann(3, 4): 125
ackermann(3, 5): 253
ackermann(3, 6): 509
ackermann(3, 7): 1021
ackermann(3, 8): 2045
ackermann(4, 1): 65533
ackermann(4, 2)) (19729 digits):
2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880...
699146577530041384717124577965048175856395072895337539755822087777506072339445587895905719156733
Dart
no caching, the implementation takes ages even for A(4,1)
int A(int m, int n) => m==0 ? n+1 : n==0 ? A(m-1,1) : A(m-1,A(m,n-1));
main() {
print(A(0,0));
print(A(1,0));
print(A(0,1));
print(A(2,2));
print(A(2,3));
print(A(3,3));
print(A(3,4));
print(A(3,5));
print(A(4,0));
}
Dc
This needs a modern Dc with r (swap) and # (comment). It easily can be adapted to an older Dc, but it will impact readability a lot.
[ # todo: n 0 -- n+1 and break 2 levels
+ 1 + # n+1
q
] s1
[ # todo: m 0 -- A(m-1,1) and break 2 levels
+ 1 - # m-1
1 # m-1 1
lA x # A(m-1,1)
q
] s2
[ # todo: m n -- A(m,n)
r d 0=1 # n m(!=0)
r d 0=2 # m(!=0) n(!=0)
Sn # m(!=0)
d 1 - r # m-1 m
Ln 1 - # m-1 m n-1
lA x # m-1 A(m,n-1)
lA x # A(m-1,A(m,n-1))
] sA
3 9 lA x f
{{out}}
4093
Delphi
function Ackermann(m,n:Int64):Int64;
begin
if m = 0 then
Result := n + 1
else if n = 0 then
Result := Ackermann(m-1, 1)
else
Result := Ackermann(m-1, Ackermann(m, n - 1));
end;
DWScript
function Ackermann(m, n : Integer) : Integer;
begin
if m = 0 then
Result := n+1
else if n = 0 then
Result := Ackermann(m-1, 1)
else Result := Ackermann(m-1, Ackermann(m, n-1));
end;
Dylan
define method ack(m == 0, n :: <integer>)
n + 1
end;
define method ack(m :: <integer>, n :: <integer>)
ack(m - 1, if (n == 0) 1 else ack(m, n - 1) end)
end;
E
def A(m, n) {
return if (m <=> 0) { n+1 } \
else if (m > 0 && n <=> 0) { A(m-1, 1) } \
else { A(m-1, A(m,n-1)) }
}
EasyLang
## Egel
```Egel
def ackermann =
[ 0 N -> N + 1
| M 0 -> ackermann (M - 1) 1
| M N -> ackermann (M - 1) (ackermann M (N - 1)) ]
Eiffel
note
description: "Example of Ackerman function"
synopsis: "[
The EIS link below (in Eiffel Studio) will launch in either an in-IDE browser or
and external browser (your choice). The protocol informs Eiffel Studio about what
program to use to open the `src' reference, which can be URI, PDF, or DOC. See
second EIS for more information.
]"
EIS: "name=Ackermann_function", "protocol=URI", "tag=rosetta_code",
"src=https://rosettacode.org/wiki/Ackermann_function"
EIS: "name=eis_protocols", "protocol=URI", "tag=eiffel_docs",
"src=https://docs.eiffel.com/book/eiffelstudio/protocols"
class
APPLICATION
create
make
feature {NONE} -- Initialization
make
do
print ("%N A(0,0):" + ackerman (0, 0).out)
print ("%N A(1,0):" + ackerman (1, 0).out)
print ("%N A(0,1):" + ackerman (0, 1).out)
print ("%N A(1,1):" + ackerman (1, 1).out)
print ("%N A(2,0):" + ackerman (2, 0).out)
print ("%N A(2,1):" + ackerman (2, 1).out)
print ("%N A(2,2):" + ackerman (2, 2).out)
print ("%N A(0,2):" + ackerman (0, 2).out)
print ("%N A(1,2):" + ackerman (1, 2).out)
print ("%N A(3,3):" + ackerman (3, 3).out)
print ("%N A(3,4):" + ackerman (3, 4).out)
end
feature -- Access
ackerman (m, n: NATURAL): NATURAL
do
if m = 0 then
Result := n + 1
elseif n = 0 then
Result := ackerman (m - 1, 1)
else
Result := ackerman (m - 1, ackerman (m, n - 1))
end
end
end
Ela
ack 0 n = n+1
ack m 0 = ack (m - 1) 1
ack m n = ack (m - 1) <| ack m <| n - 1
Elena
ELENA 4.x :
import extensions;
ackermann(m,n)
{
if(n < 0 || m < 0)
{
InvalidArgumentException.raise()
};
m =>
0 { ^n + 1 }
: {
n =>
0 { ^ackermann(m - 1,1) }
: { ^ackermann(m - 1,ackermann(m,n-1)) }
}
}
public program()
{
for(int i:=0, i <= 3, i += 1)
{
for(int j := 0, j <= 5, j += 1)
{
console.printLine("A(",i,",",j,")=",ackermann(i,j))
}
};
console.readChar()
}
{{out}}
A(0,0)=1
A(0,1)=2
A(0,2)=3
A(0,3)=4
A(0,4)=5
A(0,5)=6
A(1,0)=2
A(1,1)=3
A(1,2)=4
A(1,3)=5
A(1,4)=6
A(1,5)=7
A(2,0)=3
A(2,1)=5
A(2,2)=7
A(2,3)=9
A(2,4)=11
A(2,5)=13
A(3,0)=5
A(3,1)=13
A(3,2)=29
A(3,3)=61
A(3,4)=125
A(3,5)=253
Elixir
defmodule Ackermann do
def ack(0, n), do: n + 1
def ack(m, 0), do: ack(m - 1, 1)
def ack(m, n), do: ack(m - 1, ack(m, n - 1))
end
Enum.each(0..3, fn m ->
IO.puts Enum.map_join(0..6, " ", fn n -> Ackermann.ack(m, n) end)
end)
{{out}}
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
Emacs Lisp
(defun ackermann (m n)
(cond ((zerop m) (1+ n))
((zerop n) (ackermann (1- m) 1))
(t (ackermann (1- m)
(ackermann m (1- n))))))
Erlang
-module(ackermann).
-export([ackermann/2]).
ackermann(0, N) ->
N+1;
ackermann(M, 0) ->
ackermann(M-1, 1);
ackermann(M, N) when M > 0 andalso N > 0 ->
ackermann(M-1, ackermann(M, N-1)).
ERRE
Iterative version, using a stack. First version used various GOTOs statement, now removed and substituted with the new ERRE control statements.
PROGRAM ACKERMAN
!
! computes Ackermann function
! (second version for rosettacode.org)
!
!$INTEGER
DIM STACK[10000]
!$INCLUDE="PC.LIB"
PROCEDURE ACK(M,N->N)
LOOP
CURSOR_SAVE(->CURX%,CURY%)
LOCATE(8,1)
PRINT("Livello Stack:";S;" ")
LOCATE(CURY%,CURX%)
IF M<>0 THEN
IF N<>0 THEN
STACK[S]=M
S+=1
N-=1
ELSE
M-=1
N+=1
END IF
CONTINUE LOOP
ELSE
N+=1
S-=1
END IF
IF S<>0 THEN
M=STACK[S]
M-=1
CONTINUE LOOP
ELSE
EXIT PROCEDURE
END IF
END LOOP
END PROCEDURE
BEGIN
PRINT(CHR$(12);)
FOR X=0 TO 3 DO
FOR Y=0 TO 9 DO
S=1
ACK(X,Y->ANS)
PRINT(ANS;)
END FOR
PRINT
END FOR
END PROGRAM
Prints a list of Ackermann function values: from A(0,0) to A(3,9). Uses a stack to avoid overflow. Formating options to make this pretty are available, but for this example only basic output is used.
1 2 3 4 5 6 7 8 9 10
2 3 4 5 6 7 8 9 10 11
3 5 7 9 11 13 15 17 19 21
5 13 29 61 125 253 509 1021 2045 4093
Stack Level: 1
Euphoria
This is based on the [[VBScript]] example.
function ack(atom m, atom n)
if m = 0 then
return n + 1
elsif m > 0 and n = 0 then
return ack(m - 1, 1)
else
return ack(m - 1, ack(m, n - 1))
end if
end function
for i = 0 to 3 do
for j = 0 to 6 do
printf( 1, "%5d", ack( i, j ) )
end for
puts( 1, "\n" )
end for
Euler Math Toolbox
>M=zeros(1000,1000);
>function map A(m,n) ...
$ global M;
$ if m==0 then return n+1; endif;
$ if n==0 then return A(m-1,1); endif;
$ if m<=cols(M) and n<=cols(M) then
$ M[m,n]=A(m-1,A(m,n-1));
$ return M[m,n];
$ else return A(m-1,A(m,n-1));
$ endif;
$endfunction
>shortestformat; A((0:3)',0:5)
1 2 3 4 5 6
2 3 4 5 6 7
3 5 7 9 11 13
5 13 29 61 125 253
Ezhil
நிரல்பாகம் அகெர்மன்(முதலெண், இரண்டாமெண்)
@((முதலெண் < 0) || (இரண்டாமெண் < 0)) ஆனால்
பின்கொடு -1
முடி
@(முதலெண் == 0) ஆனால்
பின்கொடு இரண்டாமெண்+1
முடி
@((முதலெண் > 0) && (இரண்டாமெண் == 00)) ஆனால்
பின்கொடு அகெர்மன்(முதலெண் - 1, 1)
முடி
பின்கொடு அகெர்மன்(முதலெண் - 1, அகெர்மன்(முதலெண், இரண்டாமெண் - 1))
முடி
அ = int(உள்ளீடு("ஓர் எண்ணைத் தாருங்கள், அது பூஜ்ஜியமாகவோ, அதைவிடப் பெரியதாக இருக்கலாம்: "))
ஆ = int(உள்ளீடு("அதேபோல் இன்னோர் எண்ணைத் தாருங்கள், இதுவும் பூஜ்ஜியமாகவோ, அதைவிடப் பெரியதாகவோ இருக்கலாம்: "))
விடை = அகெர்மன்(அ, ஆ)
@(விடை < 0) ஆனால்
பதிப்பி "தவறான எண்களைத் தந்துள்ளீர்கள்!"
இல்லை
பதிப்பி "நீங்கள் தந்த எண்களுக்கான அகர்மென் மதிப்பு: ", விடை
முடி
=={{header|F_Sharp|F#}}== The following program implements the Ackermann function in F# but is not tail-recursive and so runs out of stack space quite fast.
let rec ackermann m n =
match m, n with
| 0, n -> n + 1
| m, 0 -> ackermann (m - 1) 1
| m, n -> ackermann (m - 1) ackermann m (n - 1)
do
printfn "%A" (ackermann (int fsi.CommandLineArgs.[1]) (int fsi.CommandLineArgs.[2]))
Transforming this into continuation passing style avoids limited stack space by permitting tail-recursion.
let ackermann M N =
let rec acker (m, n, k) =
match m,n with
| 0, n -> k(n + 1)
| m, 0 -> acker ((m - 1), 1, k)
| m, n -> acker (m, (n - 1), (fun x -> acker ((m - 1), x, k)))
acker (M, N, (fun x -> x))
Factor
USING: kernel math locals combinators ;
IN: ackermann
:: ackermann ( m n -- u )
{
{ [ m 0 = ] [ n 1 + ] }
{ [ n 0 = ] [ m 1 - 1 ackermann ] }
[ m 1 - m n 1 - ackermann ackermann ]
} cond ;
Falcon
function ackermann( m, n )
if m == 0: return( n + 1 )
if n == 0: return( ackermann( m - 1, 1 ) )
return( ackermann( m - 1, ackermann( m, n - 1 ) ) )
end
for M in [ 0:4 ]
for N in [ 0:7 ]
>> ackermann( M, N ), " "
end
>
end
The above will output the below. Formating options to make this pretty are available, but for this example only basic output is used.
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
FALSE
[$$[%
\$$[%
1-\$@@a;! { i j -> A(i-1, A(i, j-1)) }
1]?0=[
%1 { i 0 -> A(i-1, 1) }
]?
\1-a;!
1]?0=[
%1+ { 0 j -> j+1 }
]?]a: { j i }
3 3 a;! . { 61 }
Fantom
class Main
{
// assuming m,n are positive
static Int ackermann (Int m, Int n)
{
if (m == 0)
return n + 1
else if (n == 0)
return ackermann (m - 1, 1)
else
return ackermann (m - 1, ackermann (m, n - 1))
}
public static Void main ()
{
(0..3).each |m|
{
(0..6).each |n|
{
echo ("Ackerman($m, $n) = ${ackermann(m, n)}")
}
}
}
}
{{out}}
Ackerman(0, 0) = 1
Ackerman(0, 1) = 2
Ackerman(0, 2) = 3
Ackerman(0, 3) = 4
Ackerman(0, 4) = 5
Ackerman(0, 5) = 6
Ackerman(0, 6) = 7
Ackerman(1, 0) = 2
Ackerman(1, 1) = 3
Ackerman(1, 2) = 4
Ackerman(1, 3) = 5
Ackerman(1, 4) = 6
Ackerman(1, 5) = 7
Ackerman(1, 6) = 8
Ackerman(2, 0) = 3
Ackerman(2, 1) = 5
Ackerman(2, 2) = 7
Ackerman(2, 3) = 9
Ackerman(2, 4) = 11
Ackerman(2, 5) = 13
Ackerman(2, 6) = 15
Ackerman(3, 0) = 5
Ackerman(3, 1) = 13
Ackerman(3, 2) = 29
Ackerman(3, 3) = 61
Ackerman(3, 4) = 125
Ackerman(3, 5) = 253
Ackerman(3, 6) = 509
FBSL
Mixed-language solution using pure FBSL, Dynamic Assembler, and Dynamic C layers of FBSL v3.5 concurrently. '''The following is a single script'''; the breaks are caused by switching between RC's different syntax highlighting schemes:
#APPTYPE CONSOLE
TestAckermann()
PAUSE
SUB TestAckermann()
FOR DIM m = 0 TO 3
FOR DIM n = 0 TO 10
PRINT AckermannF(m, n), " ";
NEXT
PRINT
NEXT
END SUB
FUNCTION AckermannF(m AS INTEGER, n AS INTEGER) AS INTEGER
IF NOT m THEN RETURN n + 1
IF NOT n THEN RETURN AckermannA(m - 1, 1)
RETURN AckermannC(m - 1, AckermannF(m, n - 1))
END FUNCTION
DYNC AckermannC(m AS INTEGER, n AS INTEGER) AS INTEGER
int Ackermann(int m, int n)
{
if (!m) return n + 1;
if (!n) return Ackermann(m - 1, 1);
return Ackermann(m - 1, Ackermann(m, n - 1));
}
int main(int m, int n)
{
return Ackermann(m, n);
}
END DYNC
DYNASM AckermannA(m AS INTEGER, n AS INTEGER) AS INTEGER
ENTER 0, 0
INVOKE Ackermann, m, n
LEAVE
RET
@Ackermann
ENTER 0, 0
.IF DWORD PTR [m] .THEN
JMP @F
.ENDIF
MOV EAX, n
INC EAX
JMP xit
@@
.IF DWORD PTR [n] .THEN
JMP @F
.ENDIF
MOV EAX, m
DEC EAX
INVOKE Ackermann, EAX, 1
JMP xit
@@
MOV EAX, n
DEC EAX
INVOKE Ackermann, m, EAX
MOV ECX, m
DEC ECX
INVOKE Ackermann, ECX, EAX
@xit
LEAVE
RET 8
END DYNASM
{{out}}
1 2 3 4 5 6 7 8 9 10 11
2 3 4 5 6 7 8 9 10 11 12
3 5 7 9 11 13 15 17 19 21 23
5 13 29 61 125 253 509 1021 2045 4093 8189
Press any key to continue...
Forth
: acker ( m n -- u )
over 0= IF nip 1+ EXIT THEN
swap 1- swap ( m-1 n -- )
dup 0= IF 1+ recurse EXIT THEN
1- over 1+ swap recurse recurse ;
{{out|Example of use}}
FORTH> 0 0 acker . 1 ok
FORTH> 3 4 acker . 125 ok
An optimized version:
: ackermann ( m n -- u )
over ( case statement)
0 over = if drop nip 1+ else
1 over = if drop nip 2 + else
2 over = if drop nip 2* 3 + else
3 over = if drop swap 5 + swap lshift 3 - else
drop swap 1- swap dup
if
1- over 1+ swap recurse recurse exit
else
1+ recurse exit \ allow tail recursion
then
then then then then
;
Fortran
{{works with|Fortran|90 and later}}
PROGRAM EXAMPLE
IMPLICIT NONE
INTEGER :: i, j
DO i = 0, 3
DO j = 0, 6
WRITE(*, "(I10)", ADVANCE="NO") Ackermann(i, j)
END DO
WRITE(*,*)
END DO
CONTAINS
RECURSIVE FUNCTION Ackermann(m, n) RESULT(ack)
INTEGER :: ack, m, n
IF (m == 0) THEN
ack = n + 1
ELSE IF (n == 0) THEN
ack = Ackermann(m - 1, 1)
ELSE
ack = Ackermann(m - 1, Ackermann(m, n - 1))
END IF
END FUNCTION Ackermann
END PROGRAM EXAMPLE
Free Pascal
See [[#Delphi]] or [[#Pascal]].
FreeBASIC
' version 28-10-2016
' compile with: fbc -s console
' to do A(4, 2) the stack size needs to be increased
' compile with: fbc -s console -t 2000
Function ackerman (m As Long, n As Long) As Long
If m = 0 Then ackerman = n +1
If m > 0 Then
If n = 0 Then
ackerman = ackerman(m -1, 1)
Else
If n > 0 Then
ackerman = ackerman(m -1, ackerman(m, n -1))
End If
End If
End If
End Function
' ------=< MAIN >=------
Dim As Long m, n
Print
For m = 0 To 4
Print Using "###"; m;
For n = 0 To 10
' A(4, 1) or higher will run out of stack memory (default 1M)
' change n = 1 to n = 2 to calculate A(4, 2), increase stack!
If m = 4 And n = 1 Then Exit For
Print Using "######"; ackerman(m, n);
Next
Print
Next
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
{{out}}
0 1 2 3 4 5 6 7 8 9 10 11
1 2 3 4 5 6 7 8 9 10 11 12
2 3 5 7 9 11 13 15 17 19 21 23
3 5 13 29 61 125 253 509 1021 2045 4093 8189
4 13
FunL
def
ackermann( 0, n ) = n + 1
ackermann( m, 0 ) = ackermann( m - 1, 1 )
ackermann( m, n ) = ackermann( m - 1, ackermann(m, n - 1) )
for m <- 0..3, n <- 0..4
printf( 'Ackermann( %d, %d ) = %d\n', m, n, ackermann(m, n) )
{{out}}
Ackermann( 0, 0 ) = 1
Ackermann( 0, 1 ) = 2
Ackermann( 0, 2 ) = 3
Ackermann( 0, 3 ) = 4
Ackermann( 0, 4 ) = 5
Ackermann( 1, 0 ) = 2
Ackermann( 1, 1 ) = 3
Ackermann( 1, 2 ) = 4
Ackermann( 1, 3 ) = 5
Ackermann( 1, 4 ) = 6
Ackermann( 2, 0 ) = 3
Ackermann( 2, 1 ) = 5
Ackermann( 2, 2 ) = 7
Ackermann( 2, 3 ) = 9
Ackermann( 2, 4 ) = 11
Ackermann( 3, 0 ) = 5
Ackermann( 3, 1 ) = 13
Ackermann( 3, 2 ) = 29
Ackermann( 3, 3 ) = 61
Ackermann( 3, 4 ) = 125
Futhark
fun ackermann(m: int, n: int): int =
if m == 0 then n + 1
else if n == 0 then ackermann(m-1, 1)
else ackermann(m - 1, ackermann(m, n-1))
FutureBasic
include "ConsoleWindow"
def tab 1
begin globals
dim as container gC
end globals
def fn Ackerman( m as long, n as long ) as long
local fn Ackerman( m as long, n as long ) as long
dim as long result
if m == 0 then result = n + 1 : exit fn
if ( n == 0 )
result = fn Ackerman( m - 1, 1 )
exit fn
end if
result = fn Ackerman( m - 1, fn Ackerman(m, n - 1) )
end fn = result
dim as long n, m
/*
Cache response in global string container to speed
processing rather printing each iteration.
*/
for m = 0 to 3
for n = 0 to 10
gC += "fn ackerman(" + str$(m) + "," + str$(n) + " ) =" + Str$( fn Ackerman( m, n ) ) + chr$(13)
next
next
print gC
Output:
fn ackerman( 0, 0 ) = 1
fn ackerman( 0, 1 ) = 2
fn ackerman( 0, 2 ) = 3
fn ackerman( 0, 3 ) = 4
fn ackerman( 0, 4 ) = 5
fn ackerman( 0, 5 ) = 6
fn ackerman( 0, 6 ) = 7
fn ackerman( 0, 7 ) = 8
fn ackerman( 0, 8 ) = 9
fn ackerman( 0, 9 ) = 10
fn ackerman( 0, 10 ) = 11
fn ackerman( 1, 0 ) = 2
fn ackerman( 1, 1 ) = 3
fn ackerman( 1, 2 ) = 4
fn ackerman( 1, 3 ) = 5
fn ackerman( 1, 4 ) = 6
fn ackerman( 1, 5 ) = 7
fn ackerman( 1, 6 ) = 8
fn ackerman( 1, 7 ) = 9
fn ackerman( 1, 8 ) = 10
fn ackerman( 1, 9 ) = 11
fn ackerman( 1, 10 ) = 12
fn ackerman( 2, 0 ) = 3
fn ackerman( 2, 1 ) = 5
fn ackerman( 2, 2 ) = 7
fn ackerman( 2, 3 ) = 9
fn ackerman( 2, 4 ) = 11
fn ackerman( 2, 5 ) = 13
fn ackerman( 2, 6 ) = 15
fn ackerman( 2, 7 ) = 17
fn ackerman( 2, 8 ) = 19
fn ackerman( 2, 9 ) = 21
fn ackerman( 2, 10 ) = 23
fn ackerman( 3, 0 ) = 5
fn ackerman( 3, 1 ) = 13
fn ackerman( 3, 2 ) = 29
fn ackerman( 3, 3 ) = 61
fn ackerman( 3, 4 ) = 125
fn ackerman( 3, 5 ) = 253
fn ackerman( 3, 6 ) = 509
fn ackerman( 3, 7 ) = 1021
fn ackerman( 3, 8 ) = 2045
fn ackerman( 3, 9 ) = 4093
fn ackerman( 3, 10 ) = 8189
Gambas
Public Function Ackermann(m As Float, n As Float) As Float
If m = 0 Then
Return n + 1
End If
If n = 0 Then
Return Ackermann(m - 1, 1)
End If
Return Ackermann(m - 1, Ackermann(m, n - 1))
End
Public Sub Main()
Dim m, n As Float
For m = 0 To 3
For n = 0 To 4
Print "Ackermann("; m; ", "; n; ") = "; Ackermann(m, n)
Next
Next
End
GAP
ack := function(m, n)
if m = 0 then
return n + 1;
elif (m > 0) and (n = 0) then
return ack(m - 1, 1);
elif (m > 0) and (n > 0) then
return ack(m - 1, ack(m, n - 1));
else
return fail;
fi;
end;
Genyris
def A (m n)
cond
(equal? m 0)
+ n 1
(equal? n 0)
A (- m 1) 1
else
A (- m 1)
A m (- n 1)
GML
Define a script resource named ackermann and paste this code inside:
///ackermann(m,n)
var m, n;
m = argument0;
n = argument1;
if(m=0)
{
return (n+1)
}
else if(n == 0)
{
return (ackermann(m-1,1,1))
}
else
{
return (ackermann(m-1,ackermann(m,n-1,2),1))
}
gnuplot
A (m, n) = m == 0 ? n + 1 : n == 0 ? A (m - 1, 1) : A (m - 1, A (m, n - 1))
print A (0, 4)
print A (1, 4)
print A (2, 4)
print A (3, 4)
{{out}} 5 6 11 stack overflow
Go
Classic version
func Ackermann(m, n uint) uint {
switch 0 {
case m:
return n + 1
case n:
return Ackermann(m - 1, 1)
}
return Ackermann(m - 1, Ackermann(m, n - 1))
}
Expanded version
func Ackermann2(m, n uint) uint {
switch {
case m == 0:
return n + 1
case m == 1:
return n + 2
case m == 2:
return 2*n + 3
case m == 3:
return 8 << n - 3
case n == 0:
return Ackermann2(m - 1, 1)
}
return Ackermann2(m - 1, Ackermann2(m, n - 1))
}
Expanded version with arbitrary precision
package main
import (
"fmt"
"math/big"
"math/bits" // Added in Go 1.9
)
var one = big.NewInt(1)
var two = big.NewInt(2)
var three = big.NewInt(3)
var eight = big.NewInt(8)
func Ackermann2(m, n *big.Int) *big.Int {
if m.Cmp(three) <= 0 {
switch m.Int64() {
case 0:
return new(big.Int).Add(n, one)
case 1:
return new(big.Int).Add(n, two)
case 2:
r := new(big.Int).Lsh(n, 1)
return r.Add(r, three)
case 3:
if nb := n.BitLen(); nb > bits.UintSize {
// n is too large to represent as a
// uint for use in the Lsh method.
panic(TooBigError(nb))
// If we tried to continue anyway, doing
// 8*2^n-3 as bellow, we'd use hundreds
// of megabytes and lots of CPU time
// without the Exp call even returning.
r := new(big.Int).Exp(two, n, nil)
r.Mul(eight, r)
return r.Sub(r, three)
}
r := new(big.Int).Lsh(eight, uint(n.Int64()))
return r.Sub(r, three)
}
}
if n.BitLen() == 0 {
return Ackermann2(new(big.Int).Sub(m, one), one)
}
return Ackermann2(new(big.Int).Sub(m, one),
Ackermann2(m, new(big.Int).Sub(n, one)))
}
type TooBigError int
func (e TooBigError) Error() string {
return fmt.Sprintf("A(m,n) had n of %d bits; too large", int(e))
}
func main() {
show(0, 0)
show(1, 2)
show(2, 4)
show(3, 100)
show(3, 1e6)
show(4, 1)
show(4, 2)
show(4, 3)
}
func show(m, n int64) {
defer func() {
// Ackermann2 could/should have returned an error
// instead of a panic. But here's how to recover
// from the panic, and report "expected" errors.
if e := recover(); e != nil {
if err, ok := e.(TooBigError); ok {
fmt.Println("Error:", err)
} else {
panic(e)
}
}
}()
fmt.Printf("A(%d, %d) = ", m, n)
a := Ackermann2(big.NewInt(m), big.NewInt(n))
if a.BitLen() <= 256 {
fmt.Println(a)
} else {
s := a.String()
fmt.Printf("%d digits starting/ending with: %s...%s\n",
len(s), s[:20], s[len(s)-20:],
)
}
}
{{out}}
A(0, 0) = 1
A(1, 2) = 4
A(2, 4) = 11
A(3, 100) = 10141204801825835211973625643005
A(3, 1000000) = 301031 digits starting/ending with: 79205249834367186005...39107225301976875005
A(4, 1) = 65533
A(4, 2) = 19729 digits starting/ending with: 20035299304068464649...45587895905719156733
A(4, 3) = Error: A(m,n) had n of 65536 bits; too large
Golfscript
{
:_n; :_m;
_m 0= {_n 1+}
{_n 0= {_m 1- 1 ack}
{_m 1- _m _n 1- ack ack}
if}
if
}:ack;
Groovy
def ack ( m, n ) {
assert m >= 0 && n >= 0 : 'both arguments must be non-negative'
m == 0 ? n + 1 : n == 0 ? ack(m-1, 1) : ack(m-1, ack(m, n-1))
}
Test program:
def ackMatrix = (0..3).collect { m -> (0..8).collect { n -> ack(m, n) } }
ackMatrix.each { it.each { elt -> printf "%7d", elt }; println() }
{{out}}
1 2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9 10
3 5 7 9 11 13 15 17 19
5 13 29 61 125 253 509 1021 2045
Note: In the default groovyConsole configuration for WinXP, "ack(4,1)" caused a stack overflow error!
Haskell
ack :: Int -> Int -> Int
ack 0 n = succ n
ack m 0 = ack (pred m) 1
ack m n = ack (pred m) (ack m (pred n))
main :: IO ()
main = mapM_ print $ uncurry ack <$> [(0, 0), (3, 4)]
{{out}}
1
125
Generating a list instead:
import Data.List (mapAccumL)
-- everything here are [Int] or [[Int]], which would overflow
-- * had it not overrun the stack first *
ackermann = iterate ack [1..] where
ack a = s where
s = snd $ mapAccumL f (tail a) (1 : zipWith (-) s (1:s))
f a b = (aa, head aa) where aa = drop b a
main = mapM_ print $ map (\n -> take (6 - n) $ ackermann !! n) [0..5]
Haxe
class RosettaDemo
{
static public function main()
{
Sys.print(ackermann(3, 4));
}
static function ackermann(m : Int, n : Int)
{
if (m == 0)
{
return n + 1;
}
else if (n == 0)
{
return ackermann(m-1, 1);
}
return ackermann(m-1, ackermann(m, n-1));
}
}
=={{header|Icon}} and {{header|Unicon}}== {{libheader|Icon Programming Library}} Taken from the public domain Icon Programming Library's acker in memrfncs, written by Ralph E. Griswold.
procedure acker(i, j)
static memory
initial {
memory := table()
every memory[0 to 100] := table()
}
if i = 0 then return j + 1
if j = 0 then /memory[i][j] := acker(i - 1, 1)
else /memory[i][j] := acker(i - 1, acker(i, j - 1))
return memory[i][j]
end
procedure main()
every m := 0 to 3 do {
every n := 0 to 8 do {
writes(acker(m, n) || " ")
}
write()
}
end
{{out}}
1 2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9 10
3 5 7 9 11 13 15 17 19
5 13 29 61 125 253 509 1021 2045
Idris
A : Nat -> Nat -> Nat
A Z n = S n
A (S m) Z = A m (S Z)
A (S m) (S n) = A m (A (S m) n)
Ioke
{{trans|Clojure}}
ackermann = method(m,n,
cond(
m zero?, n succ,
n zero?, ackermann(m pred, 1),
ackermann(m pred, ackermann(m, n pred)))
)
J
As posted at the [[j:Essays/Ackermann%27s%20Function|J wiki]]
ack=: c1`c1`c2`c3 @. (#.@,&*) M.
c1=: >:@] NB. if 0=x, 1+y
c2=: <:@[ ack 1: NB. if 0=y, (x-1) ack 1
c3=: <:@[ ack [ ack <:@] NB. else, (x-1) ack x ack y-1
{{out|Example use}}
0 ack 3
4
1 ack 3
5
2 ack 3
9
3 ack 3
61
J's stack was too small for me to compute 4 ack 1.
Alternative Primitive Recursive Version
This version works by first generating verbs (functions) and then applying them to compute the rows of the related Buck function; then the Ackermann function is obtained in terms of the Buck function. It uses extended precision to be able to compute 4 Ack 2.
The Ackermann function derived in this fashion is primitive recursive. This is possible because in J (as in some other languages) functions, or representations of them, are first-class values.
Ack=. 3 -~ [ ({&(2 4$'>: 2x&+') ::(,&'&1'&'2x&*'@:(-&2))"0@:[ 128!:2 ]) 3 + ]
{{out|Example use}}
0 1 2 3 Ack 0 1 2 3 4 5 6 7
1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
3 5 7 9 11 13 15 17
5 13 29 61 125 253 509 1021
3 4 Ack 0 1 2
5 13 ...
13 65533 2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880933348101038234342907263181822949382118812668869506364761547029165041871916351587966347219442930927982084309104855990570159318959639524863372367203002916...
4 # @: ": @: Ack 2 NB. Number of digits of 4 Ack 2
19729
5 Ack 0
65533
A structured derivation of Ack follows:
o=. @: NB. Composition of verbs (functions)
x=. o[ NB. Composing the left noun (argument)
(rows2up=. ,&'&1'&'2x&*') o i. 4
2x&*
2x&*&1
2x&*&1&1
2x&*&1&1&1
NB. 2's multiplication, exponentiation, tetration, pentation, etc.
0 1 2 (BuckTruncated=. (rows2up x apply ]) f.) 0 1 2 3 4 5
0 2 4 6 8 ...
1 2 4 8 16 ...
1 2 4 16 65536 2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880933348101038234342907263181822949382118812668869506364761547029165041871916351587966347219442930927982084309104855990570159318959639524863372367203...
NB. Buck truncated function (missing the first two rows)
BuckTruncated NB. Buck truncated function-level code
,&'&1'&'2x&*'@:[ 128!:2 ]
(rows01=. {&('>:',:'2x&+')) 0 1 NB. The missing first two rows
>:
2x&+
Buck=. (rows01 :: (rows2up o (-&2)))"0 x apply ]
(Ack=. (3 -~ [ Buck 3 + ])f.) NB. Ackermann function-level code
3 -~ [ ({&(2 4$'>: 2x&+') ::(,&'&1'&'2x&*'@:(-&2))"0@:[ 128!:2 ]) 3 + ]
Java
[[Category:Arbitrary precision]]
import java.math.BigInteger;
public static BigInteger ack(BigInteger m, BigInteger n) {
return m.equals(BigInteger.ZERO)
? n.add(BigInteger.ONE)
: ack(m.subtract(BigInteger.ONE),
n.equals(BigInteger.ZERO) ? BigInteger.ONE : ack(m, n.subtract(BigInteger.ONE)));
}
{{works with|Java|8+}}
@FunctionalInterface
public interface FunctionalField<FIELD extends Enum<?>> {
public Object untypedField(FIELD field);
@SuppressWarnings("unchecked")
public default <VALUE> VALUE field(FIELD field) {
return (VALUE) untypedField(field);
}
}
import java.util.function.BiFunction;
import java.util.function.Function;
import java.util.function.Predicate;
import java.util.function.UnaryOperator;
import java.util.stream.Stream;
public interface TailRecursive {
public static <INPUT, INTERMEDIARY, OUTPUT> Function<INPUT, OUTPUT> new_(Function<INPUT, INTERMEDIARY> toIntermediary, UnaryOperator<INTERMEDIARY> unaryOperator, Predicate<INTERMEDIARY> predicate, Function<INTERMEDIARY, OUTPUT> toOutput) {
return input ->
$.new_(
Stream.iterate(
toIntermediary.apply(input),
unaryOperator
),
predicate,
toOutput
)
;
}
public static <INPUT1, INPUT2, INTERMEDIARY, OUTPUT> BiFunction<INPUT1, INPUT2, OUTPUT> new_(BiFunction<INPUT1, INPUT2, INTERMEDIARY> toIntermediary, UnaryOperator<INTERMEDIARY> unaryOperator, Predicate<INTERMEDIARY> predicate, Function<INTERMEDIARY, OUTPUT> toOutput) {
return (input1, input2) ->
$.new_(
Stream.iterate(
toIntermediary.apply(input1, input2),
unaryOperator
),
predicate,
toOutput
)
;
}
public enum $ {
$$;
private static <INTERMEDIARY, OUTPUT> OUTPUT new_(Stream<INTERMEDIARY> stream, Predicate<INTERMEDIARY> predicate, Function<INTERMEDIARY, OUTPUT> function) {
return stream
.filter(predicate)
.map(function)
.findAny()
.orElseThrow(RuntimeException::new)
;
}
}
}
import java.math.BigInteger;
import java.util.Stack;
import java.util.function.BinaryOperator;
import java.util.stream.Collectors;
import java.util.stream.Stream;
public interface Ackermann {
public static Ackermann new_(BigInteger number1, BigInteger number2, Stack<BigInteger> stack, boolean flag) {
return $.new_(number1, number2, stack, flag);
}
public static void main(String... arguments) {
$.main(arguments);
}
public BigInteger number1();
public BigInteger number2();
public Stack<BigInteger> stack();
public boolean flag();
public enum $ {
$$;
private static final BigInteger ZERO = BigInteger.ZERO;
private static final BigInteger ONE = BigInteger.ONE;
private static final BigInteger TWO = BigInteger.valueOf(2);
private static final BigInteger THREE = BigInteger.valueOf(3);
private static final BigInteger FOUR = BigInteger.valueOf(4);
private static Ackermann new_(BigInteger number1, BigInteger number2, Stack<BigInteger> stack, boolean flag) {
return (FunctionalAckermann) field -> {
switch (field) {
case number1: return number1;
case number2: return number2;
case stack: return stack;
case flag: return flag;
default: throw new UnsupportedOperationException(
field instanceof Field
? "Field checker has not been updated properly."
: "Field is not of the correct type."
);
}
};
}
private static final BinaryOperator<BigInteger> ACKERMANN =
TailRecursive.new_(
(BigInteger number1, BigInteger number2) ->
new_(
number1,
number2,
Stream.of(number1).collect(
Collectors.toCollection(Stack::new)
),
false
)
,
ackermann -> {
BigInteger number1 = ackermann.number1();
BigInteger number2 = ackermann.number2();
Stack<BigInteger> stack = ackermann.stack();
if (!stack.empty() && !ackermann.flag()) {
number1 = stack.pop();
}
switch (number1.intValue()) {
case 0:
return new_(
number1,
number2.add(ONE),
stack,
false
);
case 1:
return new_(
number1,
number2.add(TWO),
stack,
false
);
case 2:
return new_(
number1,
number2.multiply(TWO).add(THREE),
stack,
false
);
default:
if (ZERO.equals(number2)) {
return new_(
number1.subtract(ONE),
ONE,
stack,
true
);
} else {
stack.push(number1.subtract(ONE));
return new_(
number1,
number2.subtract(ONE),
stack,
true
);
}
}
},
ackermann -> ackermann.stack().empty(),
Ackermann::number2
)::apply
;
private static void main(String... arguments) {
System.out.println(ACKERMANN.apply(FOUR, TWO));
}
private enum Field {
number1,
number2,
stack,
flag
}
@FunctionalInterface
private interface FunctionalAckermann extends FunctionalField<Field>, Ackermann {
@Override
public default BigInteger number1() {
return field(Field.number1);
}
@Override
public default BigInteger number2() {
return field(Field.number2);
}
@Override
public default Stack<BigInteger> stack() {
return field(Field.stack);
}
@Override
public default boolean flag() {
return field(Field.flag);
}
}
}
}
{{Iterative version}}
/*
* Source https://stackoverflow.com/a/51092690/5520417
*/
package matematicas;
import java.math.BigInteger;
import java.util.HashMap;
import java.util.Stack;
/**
* @author rodri
*
*/
public class IterativeAckermannMemoryOptimization extends Thread {
/**
* Max percentage of free memory that the program will use. Default is 10% since
* the majority of the used devices are mobile and therefore it is more likely
* that the user will have more opened applications at the same time than in a
* desktop device
*/
private static Double SYSTEM_MEMORY_LIMIT_PERCENTAGE = 0.1;
/**
* Attribute of the type IterativeAckermann
*/
private IterativeAckermann iterativeAckermann;
/**
* @param iterativeAckermann
*/
public IterativeAckermannMemoryOptimization(IterativeAckermann iterativeAckermann) {
super();
this.iterativeAckermann = iterativeAckermann;
}
/**
* @return
*/
public IterativeAckermann getIterativeAckermann() {
return iterativeAckermann;
}
/**
* @param iterativeAckermann
*/
public void setIterativeAckermann(IterativeAckermann iterativeAckermann) {
this.iterativeAckermann = iterativeAckermann;
}
public static Double getSystemMemoryLimitPercentage() {
return SYSTEM_MEMORY_LIMIT_PERCENTAGE;
}
/**
* Principal method of the thread. Checks that the memory used doesn't exceed or
* equal the limit, and informs the user when that happens.
*/
@Override
public void run() {
String operating_system = System.getProperty("os.name").toLowerCase();
if ( operating_system.equals("windows") || operating_system.equals("linux") || operating_system.equals("macintosh") ) {
SYSTEM_MEMORY_LIMIT_PERCENTAGE = 0.25;
}
while ( iterativeAckermann.getConsumed_heap() >= SYSTEM_MEMORY_LIMIT_PERCENTAGE * Runtime.getRuntime().freeMemory() ) {
try {
wait();
}
catch ( InterruptedException e ) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
if ( ! iterativeAckermann.isAlive() )
iterativeAckermann.start();
else
notifyAll();
}
}
public class IterativeAckermann extends Thread {
/*
* Adjust parameters conveniently
*/
/**
*
*/
private static final int HASH_SIZE_LIMIT = 636;
/**
*
*/
private BigInteger m;
/**
*
*/
private BigInteger n;
/**
*
*/
private Integer hash_size;
/**
*
*/
private Long consumed_heap;
/**
* @param m
* @param n
* @param invalid
* @param invalid2
*/
public IterativeAckermann(BigInteger m, BigInteger n, Integer invalid, Long invalid2) {
super();
this.m = m;
this.n = n;
this.hash_size = invalid;
this.consumed_heap = invalid2;
}
/**
*
*/
public IterativeAckermann() {
// TODO Auto-generated constructor stub
super();
m = null;
n = null;
hash_size = 0;
consumed_heap = 0l;
}
/**
* @return
*/
public static BigInteger getLimit() {
return LIMIT;
}
/**
* @author rodri
*
* @param <T1>
* @param <T2>
*/
/**
* @author rodri
*
* @param <T1>
* @param <T2>
*/
static class Pair<T1, T2> {
/**
*
*/
/**
*
*/
T1 x;
/**
*
*/
/**
*
*/
T2 y;
/**
* @param x_
* @param y_
*/
/**
* @param x_
* @param y_
*/
Pair(T1 x_, T2 y_) {
x = x_;
y = y_;
}
/**
*
*/
/**
*
*/
@Override
public int hashCode() {
return x.hashCode() ^ y.hashCode();
}
/**
*
*/
/**
*
*/
@Override
public boolean equals(Object o_) {
if ( o_ == null ) {
return false;
}
if ( o_.getClass() != this.getClass() ) {
return false;
}
Pair<?, ?> o = (Pair<?, ?>) o_;
return x.equals(o.x) && y.equals(o.y);
}
}
/**
*
*/
private static final BigInteger LIMIT = new BigInteger("6");
/**
* @param m
* @param n
* @return
*/
/**
*
*/
@Override
public void run() {
while ( hash_size >= HASH_SIZE_LIMIT ) {
try {
this.wait();
}
catch ( InterruptedException e ) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
for ( BigInteger i = BigInteger.ZERO; i.compareTo(LIMIT) == - 1; i = i.add(BigInteger.ONE) ) {
for ( BigInteger j = BigInteger.ZERO; j.compareTo(LIMIT) == - 1; j = j.add(BigInteger.ONE) ) {
IterativeAckermann iterativeAckermann = new IterativeAckermann(i, j, null, null);
System.out.printf("Ackmermann(%d, %d) = %d\n", i, j, iterativeAckermann.iterative_ackermann(i, j));
}
}
}
/**
* @return
*/
public BigInteger getM() {
return m;
}
/**
* @param m
*/
public void setM(BigInteger m) {
this.m = m;
}
/**
* @return
*/
public BigInteger getN() {
return n;
}
/**
* @param n
*/
public void setN(BigInteger n) {
this.n = n;
}
/**
* @return
*/
public Integer getHash_size() {
return hash_size;
}
/**
* @param hash_size
*/
public void setHash_size(Integer hash_size) {
this.hash_size = hash_size;
}
/**
* @return
*/
public Long getConsumed_heap() {
return consumed_heap;
}
/**
* @param consumed_heap
*/
public void setConsumed_heap(Long consumed_heap) {
this.consumed_heap = consumed_heap;
}
/**
* @param m
* @param n
* @return
*/
public BigInteger iterative_ackermann(BigInteger m, BigInteger n) {
if ( m.compareTo(BigInteger.ZERO) != - 1 && m.compareTo(BigInteger.ZERO) != - 1 )
try {
HashMap<Pair<BigInteger, BigInteger>, BigInteger> solved_set = new HashMap<Pair<BigInteger, BigInteger>, BigInteger>(900000);
Stack<Pair<BigInteger, BigInteger>> to_solve = new Stack<Pair<BigInteger, BigInteger>>();
to_solve.push(new Pair<BigInteger, BigInteger>(m, n));
while ( ! to_solve.isEmpty() ) {
Pair<BigInteger, BigInteger> head = to_solve.peek();
if ( head.x.equals(BigInteger.ZERO) ) {
solved_set.put(head, head.y.add(BigInteger.ONE));
to_solve.pop();
}
else if ( head.y.equals(BigInteger.ZERO) ) {
Pair<BigInteger, BigInteger> next = new Pair<BigInteger, BigInteger>(head.x.subtract(BigInteger.ONE), BigInteger.ONE);
BigInteger result = solved_set.get(next);
if ( result == null ) {
to_solve.push(next);
}
else {
solved_set.put(head, result);
to_solve.pop();
}
}
else {
Pair<BigInteger, BigInteger> next0 = new Pair<BigInteger, BigInteger>(head.x, head.y.subtract(BigInteger.ONE));
BigInteger result0 = solved_set.get(next0);
if ( result0 == null ) {
to_solve.push(next0);
}
else {
Pair<BigInteger, BigInteger> next = new Pair<BigInteger, BigInteger>(head.x.subtract(BigInteger.ONE), result0);
BigInteger result = solved_set.get(next);
if ( result == null ) {
to_solve.push(next);
}
else {
solved_set.put(head, result);
to_solve.pop();
}
}
}
}
this.hash_size = solved_set.size();
System.out.println("Hash Size: " + hash_size);
consumed_heap = (Runtime.getRuntime().totalMemory() / (1024 * 1024));
System.out.println("Consumed Heap: " + consumed_heap + "m");
setHash_size(hash_size);
setConsumed_heap(consumed_heap);
return solved_set.get(new Pair<BigInteger, BigInteger>(m, n));
}
catch ( OutOfMemoryError e ) {
// TODO: handle exception
e.printStackTrace();
}
throw new IllegalArgumentException("The arguments must be non-negative integers.");
}
/**
* @param args
*/
/**
* @param args
*/
public static void main(String[] args) {
IterativeAckermannMemoryOptimization iterative_ackermann_memory_optimization = new IterativeAckermannMemoryOptimization(
new IterativeAckermann());
iterative_ackermann_memory_optimization.start();
}
}
JavaScript
ES5
function ack(m, n) {
return m === 0 ? n + 1 : ack(m - 1, n === 0 ? 1 : ack(m, n - 1));
}
ES6
(() => {
'use strict';
// ackermann :: Int -> Int -> Int
const ackermann = m => n => {
const go = (m, n) =>
0 === m ? (
succ(n)
) : go(pred(m), 0 === n ? (
1
) : go(m, pred(n)));
return go(m, n);
};
// TEST -----------------------------------------------
const main = () => console.log(JSON.stringify(
[0, 1, 2, 3].map(
flip(ackermann)(3)
)
));
// GENERAL FUNCTIONS ----------------------------------
// flip :: (a -> b -> c) -> b -> a -> c
const flip = f =>
x => y => f(y)(x);
// pred :: Enum a => a -> a
const pred = x => x - 1;
// succ :: Enum a => a -> a
const succ = x => 1 + x;
// MAIN ---
return main();
})();
{{Out}}
[4,5,9,61]
Joy
From here
DEFINE ack == [ [ [pop null] popd succ ]
[ [null] pop pred 1 ack ]
[ [dup pred swap] dip pred ack ack ] ]
cond.
another using a combinator
DEFINE ack == [ [ [pop null] [popd succ] ]
[ [null] [pop pred 1] [] ]
[ [[dup pred swap] dip pred] [] [] ] ]
condnestrec.
Whenever there are two definitions with the same name, the last one is the one that is used, when invoked.
jq
{{ works with|jq|1.4}}
Without Memoization
# input: [m,n]
def ack:
.[0] as $m | .[1] as $n
| if $m == 0 then $n + 1
elif $n == 0 then [$m-1, 1] | ack
else [$m-1, ([$m, $n-1 ] | ack)] | ack
end ;
'''Example:'''
range(0;5) as $i
| range(0; if $i > 3 then 1 else 6 end) as $j
| "A(\($i),\($j)) = \( [$i,$j] | ack )"
{{out}}
# jq -n -r -f ackermann.jq
A(0,0) = 1
A(0,1) = 2
A(0,2) = 3
A(0,3) = 4
A(0,4) = 5
A(0,5) = 6
A(1,0) = 2
A(1,1) = 3
A(1,2) = 4
A(1,3) = 5
A(1,4) = 6
A(1,5) = 7
A(2,0) = 3
A(2,1) = 5
A(2,2) = 7
A(2,3) = 9
A(2,4) = 11
A(2,5) = 13
A(3,0) = 5
A(3,1) = 13
A(3,2) = 29
A(3,3) = 61
A(3,4) = 125
A(3,5) = 253
A(4,0) = 13
With Memoization and Optimization
# input: [m,n, cache]
# output [value, updatedCache]
def ack:
# input: [value,cache]; output: [value, updatedCache]
def cache(key): .[1] += { (key): .[0] };
def pow2: reduce range(0; .) as $i (1; .*2);
.[0] as $m | .[1] as $n | .[2] as $cache
| if $m == 0 then [$n + 1, $cache]
elif $m == 1 then [$n + 2, $cache]
elif $m == 2 then [2 * $n + 3, $cache]
elif $m == 3 then [8 * ($n|pow2) - 3, $cache]
else
(.[0:2]|tostring) as $key
| $cache[$key] as $value
| if $value then [$value, $cache]
elif $n == 0 then
([$m-1, 1, $cache] | ack)
| cache($key)
else
([$m, $n-1, $cache ] | ack)
| [$m-1, .[0], .[1]] | ack
| cache($key)
end
end;
def A(m;n): [m,n,{}] | ack | .[0];
'''Example:'''
A(4,1)
{{out}}
65533
Jsish
From javascript entry.
/* Ackermann function, in Jsish */
function ack(m, n) {
return m === 0 ? n + 1 : ack(m - 1, n === 0 ? 1 : ack(m, n - 1));
}
if (Interp.conf('unitTest')) {
Interp.conf({maxDepth:4096});
; ack(1,3);
; ack(2,3);
; ack(3,3);
; ack(1,5);
; ack(2,5);
; ack(3,5);
}
/*
=!EXPECTSTART!=
ack(1,3) ==> 5
ack(2,3) ==> 9
ack(3,3) ==> 61
ack(1,5) ==> 7
ack(2,5) ==> 13
ack(3,5) ==> 253
=!EXPECTEND!=
*/
{{out}}
prompt$ jsish --U Ackermann.jsi
ack(1,3) ==> 5
ack(2,3) ==> 9
ack(3,3) ==> 61
ack(1,5) ==> 7
ack(2,5) ==> 13
ack(3,5) ==> 253
Julia
function ack(m,n)
if m == 0
return n + 1
elseif n == 0
return ack(m-1,1)
else
return ack(m-1,ack(m,n-1))
end
end
'''One-liner:'''
ack2(m::Integer, n::Integer) = m == 0 ? n + 1 : n == 0 ? ack2(m - 1, 1) : ack2(m - 1, ack2(m, n - 1))
'''Using memoization''', source:
using Memoize
@memoize ack3(m::Integer, n::Integer) = m == 0 ? n + 1 : n == 0 ? ack3(m - 1, 1) : ack3(m - 1, ack3(m, n - 1))
'''Benchmarking''':
julia> @time ack2(4,1)
elapsed time: 71.98668457 seconds (96 bytes allocated)
65533
julia> @time ack3(4,1)
elapsed time: 0.49337724 seconds (30405308 bytes allocated)
65533
K
See the K wiki
ack:{:[0=x;y+1;0=y;_f[x-1;1];_f[x-1;_f[x;y-1]]]}
ack[2;2]
Kdf9 Usercode
V6; W0;
YS26000;
RESTART; J999; J999;
PROGRAM; (main program);
V1 = B1212121212121212; (radix 10 for FRB);
V2 = B2020202020202020; (high bits for decimal digits);
V3 = B0741062107230637; ("A[3," in Flexowriter code);
V4 = B0727062200250007; ("7] = " in Flexowriter code);
V5 = B7777777777777777;
ZERO; NOT; =M1; (Q1 := 0/0/-1);
SETAYS0; =M2; I2=2; (Q2 := 0/2/AYS0: M2 is the stack pointer);
SET 3; =RC7; (Q7 := 3/1/0: C7 = m);
SET 7; =RC8; (Q8 := 7/1/0: C8 = n);
JSP1; (call Ackermann function);
V1; REV; FRB; (convert result to base 10);
V2; OR; (convert decimal digits to characters);
V5; REV;
SHLD+24; =V5; ERASE; (eliminate leading zeros);
SETAV5; =RM9;
SETAV3; =I9;
POAQ9; (write result to Flexowriter);
999; ZERO; OUT; (terminate run);
P1; (To compute A[m, n]);
99;
J1C7NZ; (to 1 if m ± 0);
I8; =+C8; (n := n + 1);
C8; (result to NEST);
EXIT 1; (return);
*1;
J2C8NZ; (to 2 if n ± 0);
I8; =C8; (n := 1);
DC7; (m := m - 1);
J99; (tail recursion for A[m-1, 1]);
*2;
LINK; =M0M2; (push return address);
C7; =M0M2QN; (push m);
DC8; (n := n - 1);
JSP1; (full recursion for A[m, n-1]);
=C8; (n := A[m, n-1]);
M1M2; =C7; (m := top of stack);
DC7; (m := m - 1);
M-I2; (pop stack);
M0M2; =LINK; (return address := top of stack);
J99; (tail recursion for A[m-1, A[m, n-1]]);
FINISH;
Klong
ack::{:[0=x;y+1:|0=y;.f(x-1;1);.f(x-1;.f(x;y-1))]}
ack(2;2)
Kotlin
fun A(m: Long, n: Long): Long = when {
m == 0L -> n + 1
m > 0L -> when {
n == 0L -> A(m - 1, 1)
n > 0L -> A(m - 1, A(m, n - 1))
else -> throw IllegalArgumentException("illegal n")
}
else -> throw IllegalArgumentException("illegal m")
}
fun main(args: Array<String>) {
val M: Long = 4
val N: Long = 20
val r = 0..N
for (m in 0..M) {
print("\nA($m, $r) =")
var able = true
r.forEach {
try {
if (able) {
val a = A(m, it)
print(" %6d".format(a))
} else
print(" ?")
} catch(e: Throwable) {
print(" ?")
able = false
}
}
}
}
{{out}}
A(0, 0..20) = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
A(1, 0..20) = 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
A(2, 0..20) = 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43
A(3, 0..20) = 5 13 29 61 125 253 509 1021 2045 4093 8189 16381 32765 ? ? ? ? ? ? ? ?
A(4, 0..20) = 13 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
Lasso
#!/usr/bin/lasso9
define ackermann(m::integer, n::integer) => {
if(#m == 0) => {
return ++#n
else(#n == 0)
return ackermann(--#m, 1)
else
return ackermann(#m-1, ackermann(#m, --#n))
}
}
with x in generateSeries(1,3),
y in generateSeries(0,8,2)
do stdoutnl(#x+', '#y+': ' + ackermann(#x, #y))
{{out}}
1, 0: 2
1, 2: 4
1, 4: 6
1, 6: 8
1, 8: 10
2, 0: 3
2, 2: 7
2, 4: 11
2, 6: 15
2, 8: 19
3, 0: 5
3, 2: 29
3, 4: 125
3, 6: 509
3, 8: 2045
LFE
(defun ackermann
((0 n) (+ n 1))
((m 0) (ackermann (- m 1) 1))
((m n) (ackermann (- m 1) (ackermann m (- n 1)))))
Liberty BASIC
Print Ackermann(1, 2)
Function Ackermann(m, n)
Select Case
Case (m < 0) Or (n < 0)
Exit Function
Case (m = 0)
Ackermann = (n + 1)
Case (m > 0) And (n = 0)
Ackermann = Ackermann((m - 1), 1)
Case (m > 0) And (n > 0)
Ackermann = Ackermann((m - 1), Ackermann(m, (n - 1)))
End Select
End Function
LiveCode
function ackermann m,n
switch
Case m = 0
return n + 1
Case (m > 0 And n = 0)
return ackermann((m - 1), 1)
Case (m > 0 And n > 0)
return ackermann((m - 1), ackermann(m, (n - 1)))
end switch
end ackermann
Logo
to ack :i :j
if :i = 0 [output :j+1]
if :j = 0 [output ack :i-1 1]
output ack :i-1 ack :i :j-1
end
Logtalk
ack(0, N, V) :-
!,
V is N + 1.
ack(M, 0, V) :-
!,
M2 is M - 1,
ack(M2, 1, V).
ack(M, N, V) :-
M2 is M - 1,
N2 is N - 1,
ack(M, N2, V2),
ack(M2, V2, V).
LOLCODE
{{trans|C}}
HAI 1.3
HOW IZ I ackermann YR m AN YR n
NOT m, O RLY?
YA RLY, FOUND YR SUM OF n AN 1
OIC
NOT n, O RLY?
YA RLY, FOUND YR I IZ ackermann YR DIFF OF m AN 1 AN YR 1 MKAY
OIC
FOUND YR I IZ ackermann YR DIFF OF m AN 1 AN YR...
I IZ ackermann YR m AN YR DIFF OF n AN 1 MKAY MKAY
IF U SAY SO
IM IN YR outer UPPIN YR m TIL BOTH SAEM m AN 5
IM IN YR inner UPPIN YR n TIL BOTH SAEM n AN DIFF OF 6 AN m
VISIBLE "A(" m ", " n ") = " I IZ ackermann YR m AN YR n MKAY
IM OUTTA YR inner
IM OUTTA YR outer
KTHXBYE
Lua
function ack(M,N)
if M == 0 then return N + 1 end
if N == 0 then return ack(M-1,1) end
return ack(M-1,ack(M, N-1))
end
Lucid
ack(m,n)
where
ack(m,n) = if m eq 0 then n+1
else if n eq 0 then ack(m-1,1)
else ack(m-1, ack(m, n-1)) fi
fi;
end
Luck
function ackermann(m: int, n: int): int = (
if m==0 then n+1
else if n==0 then ackermann(m-1,1)
else ackermann(m-1,ackermann(m,n-1))
)
M2000 Interpreter
Module Checkit {
Def ackermann(m,n) =If(m=0-> n+1, If(n=0-> ackermann(m-1,1), ackermann(m-1,ackermann(m,n-1))))
For m = 0 to 3 {For n = 0 to 4 {Print m;" ";n;" ";ackermann(m,n)}}
}
Checkit
Module Checkit {
Module Inner (ack) {
For m = 0 to 3 {For n = 0 to 4 {Print m;" ";n;" ";ack(m,n)}}
}
Inner lambda (m,n) ->If(m=0-> n+1, If(n=0-> lambda(m-1,1),lambda(m-1,lambda(m,n-1))))
}
Checkit
M4
define(`ack',`ifelse($1,0,`incr($2)',`ifelse($2,0,`ack(decr($1),1)',`ack(decr($1),ack($1,decr($2)))')')')dnl
ack(3,3)
{{out}}
61
Maple
Strictly by the definition given above, we can code this as follows.
Ackermann := proc( m :: nonnegint, n :: nonnegint )
option remember; # optional automatic memoization
if m = 0 then
n + 1
elif n = 0 then
thisproc( m - 1, 1 )
else
thisproc( m - 1, thisproc( m, n - 1 ) )
end if
end proc:
In Maple, the keyword
thisproc
refers to the currently executing procedure (closure) and is used when writing recursive procedures. (You could also use the name of the procedure, Ackermann in this case, but then a concurrently executing task or thread could re-assign that name while the recursive procedure is executing, resulting in an incorrect result.)
To make this faster, you can use known expansions for small values of . (See [[wp:Ackermann_function|Wikipedia:Ackermann function]])
Ackermann := proc( m :: nonnegint, n :: nonnegint )
option remember; # optional automatic memoization
if m = 0 then
n + 1
elif m = 1 then
n + 2
elif m = 2 then
2 * n + 3
elif m = 3 then
8 * 2^n - 3
elif n = 0 then
thisproc( m - 1, 1 )
else
thisproc( m - 1, thisproc( m, n - 1 ) )
end if
end proc:
This makes it possible to compute Ackermann( 4, 1 ) and Ackermann( 4, 2 ) essentially instantly, though Ackermann( 4, 3 ) is still out of reach.
To compute Ackermann( 1, i ) for i from 1 to 10 use
> map2( Ackermann, 1, [seq]( 1 .. 10 ) );
[3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
To get the first 10 values for m = 2 use
> map2( Ackermann, 2, [seq]( 1 .. 10 ) );
[5, 7, 9, 11, 13, 15, 17, 19, 21, 23]
For Ackermann( 4, 2 ) we get a very long number with
> length( Ackermann( 4, 2 ) );
19729
digits.
=={{header|Mathematica}} / {{header|Wolfram Language}}== Two possible implementations would be:
$RecursionLimit=Infinity
Ackermann1[m_,n_]:=
If[m==0,n+1,
If[ n==0,Ackermann1[m-1,1],
Ackermann1[m-1,Ackermann1[m,n-1]]
]
]
Ackermann2[0,n_]:=n+1;
Ackermann2[m_,0]:=Ackermann1[m-1,1];
Ackermann2[m_,n_]:=Ackermann1[m-1,Ackermann1[m,n-1]]
Note that the second implementation is quite a bit faster, as doing 'if' comparisons is slower than the built-in pattern matching algorithms. Examples:
Flatten[#,1]&@Table[{"Ackermann2["<>ToString[i]<>","<>ToString[j]<>"] =",Ackermann2[i,j]},{i,3},{j,8}]//Grid
gives back:
Ackermann2[1,1] = 3
Ackermann2[1,2] = 4
Ackermann2[1,3] = 5
Ackermann2[1,4] = 6
Ackermann2[1,5] = 7
Ackermann2[1,6] = 8
Ackermann2[1,7] = 9
Ackermann2[1,8] = 10
Ackermann2[2,1] = 5
Ackermann2[2,2] = 7
Ackermann2[2,3] = 9
Ackermann2[2,4] = 11
Ackermann2[2,5] = 13
Ackermann2[2,6] = 15
Ackermann2[2,7] = 17
Ackermann2[2,8] = 19
Ackermann2[3,1] = 13
Ackermann2[3,2] = 29
Ackermann2[3,3] = 61
Ackermann2[3,4] = 125
Ackermann2[3,5] = 253
Ackermann2[3,6] = 509
Ackermann2[3,7] = 1021
Ackermann2[3,8] = 2045
If we would like to calculate Ackermann[4,1] or Ackermann[4,2] we have to optimize a little bit:
Clear[Ackermann3]
$RecursionLimit=Infinity;
Ackermann3[0,n_]:=n+1;
Ackermann3[1,n_]:=n+2;
Ackermann3[2,n_]:=3+2n;
Ackermann3[3,n_]:=5+8 (2^n-1);
Ackermann3[m_,0]:=Ackermann3[m-1,1];
Ackermann3[m_,n_]:=Ackermann3[m-1,Ackermann3[m,n-1]]
Now computing Ackermann[4,1] and Ackermann[4,2] can be done quickly (<0.01 sec): Examples 2:
Ackermann3[4, 1]
Ackermann3[4, 2]
gives back:
MATLAB
function A = ackermannFunction(m,n)
if m == 0
A = n+1;
elseif (m > 0) && (n == 0)
A = ackermannFunction(m-1,1);
else
A = ackermannFunction( m-1,ackermannFunction(m,n-1) );
end
end
Maxima
ackermann(m, n) := if integerp(m) and integerp(n) then ackermann[m, n] else 'ackermann(m, n)$
ackermann[m, n] := if m = 0 then n + 1
elseif m = 1 then 2 + (n + 3) - 3
elseif m = 2 then 2 * (n + 3) - 3
elseif m = 3 then 2^(n + 3) - 3
elseif n = 0 then ackermann[m - 1, 1]
else ackermann[m - 1, ackermann[m, n - 1]]$
tetration(a, n) := if integerp(n) then block([b: a], for i from 2 thru n do b: a^b, b) else 'tetration(a, n)$
/* this should evaluate to zero */
ackermann(4, n) - (tetration(2, n + 3) - 3);
subst(n = 2, %);
ev(%, nouns);
MAXScript
Use with caution. Will cause a stack overflow for m > 3.
fn ackermann m n =
(
if m == 0 then
(
return n + 1
)
else if n == 0 then
(
ackermann (m-1) 1
)
else
(
ackermann (m-1) (ackermann m (n-1))
)
)
min
{{works with|min|0.19.3}}
(
:n :m
(
((m 0 ==) (n 1 +))
((n 0 ==) (m 1 - 1 ackermann))
((true) (m 1 - m n 1 - ackermann ackermann))
) case
) :ackermann
MiniScript
ackermann = function(m, n)
if m == 0 then return n+1
if n == 0 then return ackermann(m - 1, 1)
return ackermann(m - 1, ackermann(m, n - 1))
end function
for m in range(0, 3)
for n in range(0, 4)
print "(" + m + ", " + n + "): " + ackermann(m, n)
end for
end for
=={{header|MK-61/52}}==
## mLite
```haskell
fun ackermann( 0, n ) = n + 1
| ( m, 0 ) = ackermann( m - 1, 1 )
| ( m, n ) = ackermann( m - 1, ackermann(m, n - 1) )
Test code providing tuples from (0,0) to (3,8)
fun jota x = map (fn x = x-1) ` iota x
fun test_tuples (x, y) = append_map (fn a = map (fn b = (b, a)) ` jota x) ` jota y
map ackermann (test_tuples(4,9))
Result
[1, 2, 3, 5, 2, 3, 5, 13, 3, 4, 7, 29, 4, 5, 9, 61, 5, 6, 11, 125, 6, 7, 13, 253, 7, 8, 15, 509, 8, 9, 17, 1021, 9, 10, 19, 2045]
ML/I
ML/I loves recursion, but runs out of its default amount of storage with larger numbers than those tested here!
Program
MCSKIP "WITH" NL
"" Ackermann function
"" Will overflow when it reaches implementation-defined signed integer limit
MCSKIP MT,<>
MCINS %.
MCDEF ACK WITHS ( , )
AS <MCSET T1=%A1.
MCSET T2=%A2.
MCGO L1 UNLESS T1 EN 0
%%T2.+1.MCGO L0
%L1.MCGO L2 UNLESS T2 EN 0
ACK(%%T1.-1.,1)MCGO L0
%L2.ACK(%%T1.-1.,ACK(%T1.,%%T2.-1.))>
"" Macro ACK now defined, so try it out
a(0,0) => ACK(0,0)
a(0,1) => ACK(0,1)
a(0,2) => ACK(0,2)
a(0,3) => ACK(0,3)
a(0,4) => ACK(0,4)
a(0,5) => ACK(0,5)
a(1,0) => ACK(1,0)
a(1,1) => ACK(1,1)
a(1,2) => ACK(1,2)
a(1,3) => ACK(1,3)
a(1,4) => ACK(1,4)
a(2,0) => ACK(2,0)
a(2,1) => ACK(2,1)
a(2,2) => ACK(2,2)
a(2,3) => ACK(2,3)
a(3,0) => ACK(3,0)
a(3,1) => ACK(3,1)
a(3,2) => ACK(3,2)
a(4,0) => ACK(4,0)
{{out}}
a(0,0) => 1
a(0,1) => 2
a(0,2) => 3
a(0,3) => 4
a(0,4) => 5
a(0,5) => 6
a(1,0) => 2
a(1,1) => 3
a(1,2) => 4
a(1,3) => 5
a(1,4) => 6
a(2,0) => 3
a(2,1) => 5
a(2,2) => 7
a(2,3) => 9
a(3,0) => 5
a(3,1) => 13
a(3,2) => 29
a(4,0) => 13
Mercury
This is the Ackermann function with some (obvious) elements elided. The ack/3 predicate is implemented in terms of the ack/2 function. The ack/2 function is implemented in terms of the ack/3 predicate. This makes the code both more concise and easier to follow than would otherwise be the case. The integer type is used instead of int because the problem statement stipulates the use of bignum integers if possible.
:- func ack(integer, integer) = integer.
ack(M, N) = R :- ack(M, N, R).
:- pred ack(integer::in, integer::in, integer::out) is det.
ack(M, N, R) :-
( ( M < integer(0)
; N < integer(0) ) -> throw(bounds_error)
; M = integer(0) -> R = N + integer(1)
; N = integer(0) -> ack(M - integer(1), integer(1), R)
; ack(M - integer(1), ack(M, N - integer(1)), R) ).
=={{header|Modula-2}}==
MODULE ackerman;
IMPORT ASCII, NumConv, InOut;
VAR m, n : LONGCARD;
string : ARRAY [0..19] OF CHAR;
OK : BOOLEAN;
PROCEDURE Ackerman (x, y : LONGCARD) : LONGCARD;
BEGIN
IF x = 0 THEN RETURN y + 1
ELSIF y = 0 THEN RETURN Ackerman (x - 1 , 1)
ELSE
RETURN Ackerman (x - 1 , Ackerman (x , y - 1))
END
END Ackerman;
BEGIN
FOR m := 0 TO 3 DO
FOR n := 0 TO 6 DO
NumConv.Num2Str (Ackerman (m, n), 10, string, OK);
IF OK THEN
InOut.WriteString (string)
ELSE
InOut.WriteString ("* Error in number * ")
END;
InOut.Write (ASCII.HT)
END;
InOut.WriteLn
END;
InOut.WriteLn
END ackerman.
{{out}}
jan@Beryllium:~/modula/rosetta$ ackerman
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
=={{header|Modula-3}}== The type CARDINAL is defined in Modula-3 as [0..LAST(INTEGER)], in other words, it can hold all positive integers.
MODULE Ack EXPORTS Main;
FROM IO IMPORT Put;
FROM Fmt IMPORT Int;
PROCEDURE Ackermann(m, n: CARDINAL): CARDINAL =
BEGIN
IF m = 0 THEN
RETURN n + 1;
ELSIF n = 0 THEN
RETURN Ackermann(m - 1, 1);
ELSE
RETURN Ackermann(m - 1, Ackermann(m, n - 1));
END;
END Ackermann;
BEGIN
FOR m := 0 TO 3 DO
FOR n := 0 TO 6 DO
Put(Int(Ackermann(m, n)) & " ");
END;
Put("\n");
END;
END Ack.
{{out}}
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
MUMPS
Ackermann(m,n) ;
If m=0 Quit n+1
If m>0,n=0 Quit $$Ackermann(m-1,1)
If m>0,n>0 Quit $$Ackermann(m-1,$$Ackermann(m,n-1))
Set $Ecode=",U13-Invalid parameter for Ackermann: m="_m_", n="_n_","
Write $$Ackermann(1,8) ; 10
Write $$Ackermann(2,8) ; 19
Write $$Ackermann(3,5) ; 253
Nemerle
In Nemerle, we can state the Ackermann function as a lambda. By using pattern-matching, our definition strongly resembles the mathematical notation.
using System;
using Nemerle.IO;
def ackermann(m, n) {
def A = ackermann;
match(m, n) {
| (0, n) => n + 1
| (m, 0) when m > 0 => A(m - 1, 1)
| (m, n) when m > 0 && n > 0 => A(m - 1, A(m, n - 1))
| _ => throw Exception("invalid inputs");
}
}
for(mutable m = 0; m < 4; m++) {
for(mutable n = 0; n < 5; n++) {
print("ackermann($m, $n) = $(ackermann(m, n))\n");
}
}
A terser version using implicit match (which doesn't use the alias A internally):
def ackermann(m, n) {
| (0, n) => n + 1
| (m, 0) when m > 0 => ackermann(m - 1, 1)
| (m, n) when m > 0 && n > 0 => ackermann(m - 1, ackermann(m, n - 1))
| _ => throw Exception("invalid inputs");
}
Or, if we were set on using the A notation, we could do this:
def ackermann = {
def A(m, n) {
| (0, n) => n + 1
| (m, 0) when m > 0 => A(m - 1, 1)
| (m, n) when m > 0 && n > 0 => A(m - 1, A(m, n - 1))
| _ => throw Exception("invalid inputs");
}
A
}
NetRexx
/* NetRexx */
options replace format comments java crossref symbols binary
numeric digits 66
parse arg j_ k_ .
if j_ = '' | j_ = '.' | \j_.datatype('w') then j_ = 3
if k_ = '' | k_ = '.' | \k_.datatype('w') then k_ = 5
loop m_ = 0 to j_
say
loop n_ = 0 to k_
say 'ackermann('m_','n_') =' ackermann(m_, n_).right(5)
end n_
end m_
return
method ackermann(m, n) public static
select
when m = 0 then rval = n + 1
when n = 0 then rval = ackermann(m - 1, 1)
otherwise rval = ackermann(m - 1, ackermann(m, n - 1))
end
return rval
NewLISP
#! /usr/local/bin/newlisp
(define (ackermann m n)
(cond ((zero? m) (inc n))
((zero? n) (ackermann (dec m) 1))
(true (ackermann (- m 1) (ackermann m (dec n))))))
In case of stack overflow error, you have to start your program with a proper "-s <value>" flag
as "newlisp -s 100000 ./ackermann.lsp".
See http://www.newlisp.org/newlisp_manual.html#stack_size
Nim
from strutils import parseInt
proc ackermann(m, n: int64): int64 =
if m == 0:
result = n + 1
elif n == 0:
result = ackermann(m - 1, 1)
else:
result = ackermann(m - 1, ackermann(m, n - 1))
proc getNumber(): int =
try:
result = stdin.readLine.parseInt
except ValueError:
echo "An integer, please!"
result = getNumber()
if result < 0:
echo "Please Enter a non-negative Integer: "
result = getNumber()
echo "First non-negative Integer please: "
let first = getNumber()
echo "Second non-negative Integer please: "
let second = getNumber()
echo "Result: ", $ackermann(first, second)
Nit
Source: the official Nit’s repository.
# Task: Ackermann function
#
# A simple straightforward recursive implementation.
module ackermann_function
fun ack(m, n: Int): Int
do
if m == 0 then return n + 1
if n == 0 then return ack(m-1,1)
return ack(m-1, ack(m, n-1))
end
for m in [0..3] do
for n in [0..6] do
print ack(m,n)
end
print ""
end
Output:
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
5
7
9
11
13
15
5
13
29
61
125
253
509
Objeck
{{trans|C#|C sharp}}
class Ackermann {
function : Main(args : String[]) ~ Nil {
for(m := 0; m <= 3; ++m;) {
for(n := 0; n <= 4; ++n;) {
a := Ackermann(m, n);
if(a > 0) {
"Ackermann({$m}, {$n}) = {$a}"->PrintLine();
};
};
};
}
function : Ackermann(m : Int, n : Int) ~ Int {
if(m > 0) {
if (n > 0) {
return Ackermann(m - 1, Ackermann(m, n - 1));
}
else if (n = 0) {
return Ackermann(m - 1, 1);
};
}
else if(m = 0) {
if(n >= 0) {
return n + 1;
};
};
return -1;
}
}
Ackermann(0, 0) = 1
Ackermann(0, 1) = 2
Ackermann(0, 2) = 3
Ackermann(0, 3) = 4
Ackermann(0, 4) = 5
Ackermann(1, 0) = 2
Ackermann(1, 1) = 3
Ackermann(1, 2) = 4
Ackermann(1, 3) = 5
Ackermann(1, 4) = 6
Ackermann(2, 0) = 3
Ackermann(2, 1) = 5
Ackermann(2, 2) = 7
Ackermann(2, 3) = 9
Ackermann(2, 4) = 11
Ackermann(3, 0) = 5
Ackermann(3, 1) = 13
Ackermann(3, 2) = 29
Ackermann(3, 3) = 61
Ackermann(3, 4) = 125
OCaml
let rec a m n =
if m=0 then (n+1) else
if n=0 then (a (m-1) 1) else
(a (m-1) (a m (n-1)))
or:
let rec a = function
| 0, n -> (n+1)
| m, 0 -> a(m-1, 1)
| m, n -> a(m-1, a(m, n-1))
with memoization using an hash-table:
let h = Hashtbl.create 4001
let a m n =
try Hashtbl.find h (m, n)
with Not_found ->
let res = a (m, n) in
Hashtbl.add h (m, n) res;
(res)
taking advantage of the memoization we start calling small values of '''m''' and '''n''' in order to reduce the recursion call stack:
let a m n =
for _m = 0 to m do
for _n = 0 to n do
ignore(a _m _n);
done;
done;
(a m n)
Arbitrary precision
With arbitrary-precision integers (Big_int module):
open Big_int
let one = unit_big_int
let zero = zero_big_int
let succ = succ_big_int
let pred = pred_big_int
let eq = eq_big_int
let rec a m n =
if eq m zero then (succ n) else
if eq n zero then (a (pred m) one) else
(a (pred m) (a m (pred n)))
compile with: ocamlopt -o acker nums.cmxa acker.ml === Tail-Recursive === Here is a [[:Category:Recursion|tail-recursive]] version:
let rec find_option h v =
try Some(Hashtbl.find h v)
with Not_found -> None
let rec a bounds caller todo m n =
match m, n with
| 0, n ->
let r = (n+1) in
( match todo with
| [] -> r
| (m,n)::todo ->
List.iter (fun k ->
if not(Hashtbl.mem bounds k)
then Hashtbl.add bounds k r) caller;
a bounds [] todo m n )
| m, 0 ->
a bounds caller todo (m-1) 1
| m, n ->
match find_option bounds (m, n-1) with
| Some a_rec ->
let caller = (m,n)::caller in
a bounds caller todo (m-1) a_rec
| None ->
let todo = (m,n)::todo
and caller = [(m, n-1)] in
a bounds caller todo m (n-1)
let a = a (Hashtbl.create 42 (* arbitrary *) ) [] [] ;;
This one uses the arbitrary precision, the tail-recursion, and the optimisation explain on the Wikipedia page about (m,n) = (3,_).
open Big_int
let one = unit_big_int
let zero = zero_big_int
let succ = succ_big_int
let pred = pred_big_int
let add = add_big_int
let sub = sub_big_int
let eq = eq_big_int
let three = succ(succ one)
let power = power_int_positive_big_int
let eq2 (a1,a2) (b1,b2) =
(eq a1 b1) && (eq a2 b2)
module H = Hashtbl.Make
(struct
type t = Big_int.big_int * Big_int.big_int
let equal = eq2
let hash (x,y) = Hashtbl.hash
(Big_int.string_of_big_int x ^ "," ^
Big_int.string_of_big_int y)
(* probably not a very good hash function *)
end)
let rec find_option h v =
try Some (H.find h v)
with Not_found -> None
let rec a bounds caller todo m n =
let may_tail r =
let k = (m,n) in
match todo with
| [] -> r
| (m,n)::todo ->
List.iter (fun k ->
if not (H.mem bounds k)
then H.add bounds k r) (k::caller);
a bounds [] todo m n
in
match m, n with
| m, n when eq m zero ->
let r = (succ n) in
may_tail r
| m, n when eq n zero ->
let caller = (m,n)::caller in
a bounds caller todo (pred m) one
| m, n when eq m three ->
let r = sub (power 2 (add n three)) three in
may_tail r
| m, n ->
match find_option bounds (m, pred n) with
| Some a_rec ->
let caller = (m,n)::caller in
a bounds caller todo (pred m) a_rec
| None ->
let todo = (m,n)::todo in
let caller = [(m, pred n)] in
a bounds caller todo m (pred n)
let a = a (H.create 42 (* arbitrary *)) [] [] ;;
let () =
let m, n =
try
big_int_of_string Sys.argv.(1),
big_int_of_string Sys.argv.(2)
with _ ->
Printf.eprintf "usage: %s <int> <int>\n" Sys.argv.(0);
exit 1
in
let r = a m n in
Printf.printf "(a %s %s) = %s\n"
(string_of_big_int m)
(string_of_big_int n)
(string_of_big_int r);
;;
=={{header|Oberon-2}}==
MODULE ackerman;
IMPORT Out;
VAR m, n : INTEGER;
PROCEDURE Ackerman (x, y : INTEGER) : INTEGER;
BEGIN
IF x = 0 THEN RETURN y + 1
ELSIF y = 0 THEN RETURN Ackerman (x - 1 , 1)
ELSE
RETURN Ackerman (x - 1 , Ackerman (x , y - 1))
END
END Ackerman;
BEGIN
FOR m := 0 TO 3 DO
FOR n := 0 TO 6 DO
Out.Int (Ackerman (m, n), 10);
Out.Char (9X)
END;
Out.Ln
END;
Out.Ln
END ackerman.
Octave
function r = ackerman(m, n)
if ( m == 0 )
r = n + 1;
elseif ( n == 0 )
r = ackerman(m-1, 1);
else
r = ackerman(m-1, ackerman(m, n-1));
endif
endfunction
for i = 0:3
disp(ackerman(i, 4));
endfor
Oforth
: A( m n -- p )
m ifZero: [ n 1+ return ]
m 1- n ifZero: [ 1 ] else: [ A( m, n 1- ) ] A
;
OOC
ack: func (m: Int, n: Int) -> Int {
if (m == 0) {
n + 1
} else if (n == 0) {
ack(m - 1, 1)
} else {
ack(m - 1, ack(m, n - 1))
}
}
main: func {
for (m in 0..4) {
for (n in 0..10) {
"ack(#{m}, #{n}) = #{ack(m, n)}" println()
}
}
}
ooRexx
loop m = 0 to 3
loop n = 0 to 6
say "Ackermann("m", "n") =" ackermann(m, n)
end
end
::routine ackermann
use strict arg m, n
-- give us some precision room
numeric digits 10000
if m = 0 then return n + 1
else if n = 0 then return ackermann(m - 1, 1)
else return ackermann(m - 1, ackermann(m, n - 1))
{{out}}
Ackermann(0, 0) = 1
Ackermann(0, 1) = 2
Ackermann(0, 2) = 3
Ackermann(0, 3) = 4
Ackermann(0, 4) = 5
Ackermann(0, 5) = 6
Ackermann(0, 6) = 7
Ackermann(1, 0) = 2
Ackermann(1, 1) = 3
Ackermann(1, 2) = 4
Ackermann(1, 3) = 5
Ackermann(1, 4) = 6
Ackermann(1, 5) = 7
Ackermann(1, 6) = 8
Ackermann(2, 0) = 3
Ackermann(2, 1) = 5
Ackermann(2, 2) = 7
Ackermann(2, 3) = 9
Ackermann(2, 4) = 11
Ackermann(2, 5) = 13
Ackermann(2, 6) = 15
Ackermann(3, 0) = 5
Ackermann(3, 1) = 13
Ackermann(3, 2) = 29
Ackermann(3, 3) = 61
Ackermann(3, 4) = 125
Ackermann(3, 5) = 253
Ackermann(3, 6) = 509
Order
#include <order/interpreter.h>
#define ORDER_PP_DEF_8ack ORDER_PP_FN( \
8fn(8X, 8Y, \
8cond((8is_0(8X), 8inc(8Y)) \
(8is_0(8Y), 8ack(8dec(8X), 1)) \
(8else, 8ack(8dec(8X), 8ack(8X, 8dec(8Y)))))))
ORDER_PP(8to_lit(8ack(3, 4))) // 125
Oz
Oz has arbitrary precision integers.
declare
fun {Ack M N}
if M == 0 then N+1
elseif N == 0 then {Ack M-1 1}
else {Ack M-1 {Ack M N-1}}
end
end
in
{Show {Ack 3 7}}
PARI/GP
Naive implementation.
A(m,n)={
if(m,
if(n,
A(m-1, A(m,n-1))
,
A(m-1,1)
)
,
n+1
)
};
Pascal
Program Ackerman;
function ackermann(m, n: Integer) : Integer;
begin
if m = 0 then
ackermann := n+1
else if n = 0 then
ackermann := ackermann(m-1, 1)
else
ackermann := ackermann(m-1, ackermann(m, n-1));
end;
var
m, n : Integer;
begin
for n := 0 to 6 do
for m := 0 to 3 do
WriteLn('A(', m, ',', n, ') = ', ackermann(m,n));
end.
Perl
We memoize calls to ''A'' to make ''A''(2, ''n'') and ''A''(3, ''n'') feasible for larger values of ''n''.
{
my @memo;
sub A {
my( $m, $n ) = @_;
$memo[ $m ][ $n ] and return $memo[ $m ][ $n ];
$m or return $n + 1;
return $memo[ $m ][ $n ] = (
$n
? A( $m - 1, A( $m, $n - 1 ) )
: A( $m - 1, 1 )
);
}
}
An implementation using the conditional statements 'if', 'elsif' and 'else':
sub A {
my ($m, $n) = @_;
if ($m == 0) { $n + 1 }
elsif ($n == 0) { A($m - 1, 1) }
else { A($m - 1, A($m, $n - 1)) }
}
An implementation using ternary chaining:
sub A {
my ($m, $n) = @_;
$m == 0 ? $n + 1 :
$n == 0 ? A($m - 1, 1) :
A($m - 1, A($m, $n - 1))
}
Adding memoization and extra terms:
use Memoize; memoize('ack2');
use bigint try=>"GMP";
sub ack2 {
my ($m, $n) = @_;
$m == 0 ? $n + 1 :
$m == 1 ? $n + 2 :
$m == 2 ? 2*$n + 3 :
$m == 3 ? 8 * (2**$n - 1) + 5 :
$n == 0 ? ack2($m-1, 1)
: ack2($m-1, ack2($m, $n-1));
}
print "ack2(3,4) is ", ack2(3,4), "\n";
print "ack2(4,1) is ", ack2(4,1), "\n";
print "ack2(4,2) has ", length(ack2(4,2)), " digits\n";
{{output}}
ack2(3,4) is 125
ack2(4,1) is 65533
ack2(4,2) has 19729 digits
An optimized version, which uses @_ as a stack,
instead of recursion. Very fast.
use strict;
use warnings;
use Math::BigInt;
use constant two => Math::BigInt->new(2);
sub ack {
my $n = pop;
while( @_ ) {
my $m = pop;
if( $m > 3 ) {
push @_, (--$m) x $n;
push @_, reverse 3 .. --$m;
$n = 13;
} elsif( $m == 3 ) {
if( $n < 29 ) {
$n = ( 1 << ( $n + 3 ) ) - 3;
} else {
$n = two ** ( $n + 3 ) - 3;
}
} elsif( $m == 2 ) {
$n = 2 * $n + 3;
} elsif( $m >= 0 ) {
$n = $n + $m + 1;
} else {
die "negative m!";
}
}
$n;
}
print "ack(3,4) is ", ack(3,4), "\n";
print "ack(4,1) is ", ack(4,1), "\n";
print "ack(4,2) has ", length(ack(4,2)), " digits\n";
Perl 6
{{works with|Rakudo|2018.03}}
sub A(Int $m, Int $n) {
if $m == 0 { $n + 1 }
elsif $n == 0 { A($m - 1, 1) }
else { A($m - 1, A($m, $n - 1)) }
}
An implementation using multiple dispatch:
multi sub A(0, Int $n) { $n + 1 }
multi sub A(Int $m, 0 ) { A($m - 1, 1) }
multi sub A(Int $m, Int $n) { A($m - 1, A($m, $n - 1)) }
Note that in either case, Int is defined to be arbitrary precision in Perl 6.
Here's a caching version of that, written in the sigilless style, with liberal use of Unicode, and the extra optimizing terms to make A(4,2) possible:
proto A(Int \𝑚, Int \𝑛) { (state @)[𝑚][𝑛] //= {*} }
multi A(0, Int \𝑛) { 𝑛 + 1 }
multi A(1, Int \𝑛) { 𝑛 + 2 }
multi A(2, Int \𝑛) { 3 + 2 * 𝑛 }
multi A(3, Int \𝑛) { 5 + 8 * (2 ** 𝑛 - 1) }
multi A(Int \𝑚, 0 ) { A(𝑚 - 1, 1) }
multi A(Int \𝑚, Int \𝑛) { A(𝑚 - 1, A(𝑚, 𝑛 - 1)) }
# Testing:
say A(4,1);
say .chars, " digits starting with ", .substr(0,50), "..." given A(4,2);
{{out}}
65533
19729 digits starting with 20035299304068464649790723515602557504478254755697...
Phix
native version
function ack(integer m, integer n)
if m=0 then
return n+1
elsif m=1 then
return n+2
elsif m=2 then
return 2*n+3
elsif m=3 then
return power(2,n+3)-3
elsif m>0 and n=0 then
return ack(m-1,1)
else
return ack(m-1,ack(m,n-1))
end if
end function
constant limit = 23,
fmtlens = {1,2,2,2,3,3,3,4,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8}
atom t0 = time()
printf(1," 0")
for j=1 to limit do
string fmt = sprintf(" %%%dd",fmtlens[j+1])
printf(1,fmt,j)
end for
printf(1,"\n")
for i=0 to 5 do
printf(1,"%d:",i)
for j=0 to iff(i>=4?5-i:limit) do
string fmt = sprintf(" %%%dd",fmtlens[j+1])
printf(1,fmt,{ack(i,j)})
end for
printf(1,"\n")
end for
{{out}}
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
0: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
1: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
2: 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
3: 5 13 29 61 125 253 509 1021 2045 4093 8189 16381 32765 65533 131069 262141 524285 1048573 2097149 4194301 8388605 16777213 33554429 67108861
4: 13 65533
5: 65533
ack(4,2) and above fail with power function overflow. ack(3,100) will get you an answer, but only accurate to 16 or so digits.
gmp version
{{trans|Go}} {{libheader|mpfr}}
-- demo\rosetta\Ackermann.exw
include mpfr.e
procedure ack(integer m, mpz n)
if m=0 then
mpz_add_ui(n, n, 1) -- return n+1
elsif m=1 then
mpz_add_ui(n, n, 2) -- return n+2
elsif m=2 then
mpz_mul_si(n, n, 2)
mpz_add_ui(n, n, 3) -- return 2*n+3
elsif m=3 then
if not mpz_fits_integer(n) then
-- As per Go: 2^MAXINT would most certainly run out of memory.
-- (think about it: a million digits is fine but pretty daft;
-- however a billion digits requires > addressable memory.)
integer bn = mpz_sizeinbase(n, 2)
throw(sprintf("A(m,n) had n of %d bits; too large",bn))
end if
integer ni = mpz_get_integer(n)
mpz_set_si(n, 8)
mpz_mul_2exp(n, n, ni) -- (n:=8*2^ni)
mpz_sub_ui(n, n, 3) -- return power(2,n+3)-3
elsif mpz_cmp_si(n,0)=0 then
mpz_set_si(n, 1)
ack(m-1,n) -- return ack(m-1,1)
else
mpz_sub_ui(n, n, 1)
ack(m,n)
ack(m-1,n) -- return ack(m-1,ack(m,n-1))
end if
end procedure
constant limit = 23,
fmtlens = {1,2,2,2,3,3,3,4,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8},
extras = {{3,100},{3,1e6},{4,2},{4,3}}
procedure ackermann_tests()
atom t0 = time()
atom n = mpz_init()
printf(1," 0")
for j=1 to limit do
string fmt = sprintf(" %%%dd",fmtlens[j+1])
printf(1,fmt,j)
end for
printf(1,"\n")
for i=0 to 5 do
printf(1,"%d:",i)
for j=0 to iff(i>=4?5-i:limit) do
mpz_set_si(n, j)
ack(i,n)
string fmt = sprintf(" %%%ds",fmtlens[j+1])
printf(1,fmt,{mpz_get_str(n)})
end for
printf(1,"\n")
end for
printf(1,"\n")
for i=1 to length(extras) do
integer {em, en} = extras[i]
mpz_set_si(n, en)
string res
try
ack(em,n)
res = mpz_get_str(n)
integer lr = length(res)
if lr>50 then
res[21..-21] = "..."
res &= sprintf(" (%d digits)",lr)
end if
catch e
-- ack(4,3), ack(5,1) and ack(6,0) all fail,
-- just as they should do
res = "***ERROR***: "&e[E_USER]
end try
printf(1,"ack(%d,%d) %s\n",{em,en,res})
end for
n = mpz_free(n)
printf(1,"\n")
printf(1,"ackermann_tests completed (%s)\n\n",{elapsed(time()-t0)})
end procedure
ackermann_tests()
{{out}}
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
0: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
1: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
2: 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
3: 5 13 29 61 125 253 509 1021 2045 4093 8189 16381 32765 65533 131069 262141 524285 1048573 2097149 4194301 8388605 16777213 33554429 67108861
4: 13 65533
5: 65533
ack(3,100) 10141204801825835211973625643005
ack(3,1000000) 79205249834367186005...39107225301976875005 (301031 digits)
ack(4,2) 20035299304068464649...45587895905719156733 (19729 digits)
ack(4,3) ***ERROR***: A(m,n) had n of 65536 bits; too large
ackermann_tests completed (0.2s)
PHP
function ackermann( $m , $n )
{
if ( $m==0 )
{
return $n + 1;
}
elseif ( $n==0 )
{
return ackermann( $m-1 , 1 );
}
return ackermann( $m-1, ackermann( $m , $n-1 ) );
}
echo ackermann( 3, 4 );
// prints 125
PicoLisp
(de ack (X Y)
(cond
((=0 X) (inc Y))
((=0 Y) (ack (dec X) 1))
(T (ack (dec X) (ack X (dec Y)))) ) )
Piet
Rendered as wikitable:
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| style="background-color:#000000; color:#000000;" | ww | style="background-color:#ff0000; color:#ff0000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c00000; color:#c00000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#0000c0; color:#0000c0;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c00000; color:#c00000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ff0000; color:#ff0000;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ff0000; color:#ff0000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffc0c0; color:#ffc0c0;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#0000ff; color:#0000ff;" | ww | style="background-color:#c0c0ff; color:#c0c0ff;" | ww | style="background-color:#00c0c0; color:#00c0c0;" | ww | style="background-color:#00ffff; color:#00ffff;" | ww | style="background-color:#00c0c0; color:#00c0c0;" | ww | style="background-color:#00c0c0; color:#00c0c0;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ff0000; color:#ff0000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c0ffff; color:#c0ffff;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ff0000; color:#ff0000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#0000c0; color:#0000c0;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#00ffff; color:#00ffff;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffc0ff; color:#ffc0ff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ff0000; color:#ff0000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#00c000; color:#00c000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c0c000; color:#c0c000;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c00000; color:#c00000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ff0000; color:#ff0000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c0ffc0; color:#c0ffc0;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c000c0; color:#c000c0;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c000c0; color:#c000c0;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffc0c0; color:#ffc0c0;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c000c0; color:#c000c0;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ff00ff; color:#ff00ff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffff00; color:#ffff00;" | ww | style="background-color:#ffff00; color:#ffff00;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c000c0; color:#c000c0;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#c0c0ff; color:#c0c0ff;" | ww | style="background-color:#c0c0ff; color:#c0c0ff;" | ww | style="background-color:#c0ffff; color:#c0ffff;" | ww | style="background-color:#00c0c0; color:#00c0c0;" | ww | style="background-color:#ff00ff; color:#ff00ff;" | ww | style="background-color:#ffc0ff; color:#ffc0ff;" | ww | style="background-color:#c000c0; color:#c000c0;" | ww | style="background-color:#c000c0; color:#c000c0;" | ww | style="background-color:#00ff00; color:#00ff00;" | ww | style="background-color:#ff00ff; color:#ff00ff;" | ww | style="background-color:#ffffc0; color:#ffffc0;" | ww | style="background-color:#c0c000; color:#c0c000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffc0ff; color:#ffc0ff;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ff00ff; color:#ff00ff;" | ww |-
| style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#000000; color:#000000;" | ww | style="background-color:#ff0000; color:#ff0000;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#ffffff; color:#ffffff;" | ww | style="background-color:#c0ffff; color:#c0ffff;" | ww | style="background-color:#00c000; color:#00c000;" | ww | style="background-color:#00ff00; color:#00ff00;" | ww | style="background-color:#c0c0ff; color:#c0c0ff;" | ww | style="background-color:#0000c0; color:#0000c0;" | ww | style="background-color:#0000ff; color:#0000ff;" | ww | style="background-color:#0000ff; color:#0000ff;" | ww | style="background-color:#0000ff; color:#0000ff;" | ww | style="background-color:#c0ffff; color:#c0ffff;" | ww | style="background-color:#00c0c0; color:#00c0c0;" | ww |-
|}
This is a naive implementation that does not use any optimization. Find the explanation at https://rosettacode.org/wiki/User:Albedo. Computing the Ackermann function for (4,1) is possible, but takes quite a while because the stack grows very fast to large dimensions.
Example output:
? 3
? 5
253
Pike
int main(){
write(ackermann(3,4) + "\n");
}
int ackermann(int m, int n){
if(m == 0){
return n + 1;
} else if(n == 0){
return ackermann(m-1, 1);
} else {
return ackermann(m-1, ackermann(m, n-1));
}
}
PL/I
Ackerman: procedure (m, n) returns (fixed (30)) recursive;
declare (m, n) fixed (30);
if m = 0 then return (n+1);
else if m > 0 & n = 0 then return (Ackerman(m-1, 1));
else if m > 0 & n > 0 then return (Ackerman(m-1, Ackerman(m, n-1)));
return (0);
end Ackerman;
PL/SQL
DECLARE
FUNCTION ackermann(pi_m IN NUMBER,
pi_n IN NUMBER) RETURN NUMBER IS
BEGIN
IF pi_m = 0 THEN
RETURN pi_n + 1;
ELSIF pi_n = 0 THEN
RETURN ackermann(pi_m - 1, 1);
ELSE
RETURN ackermann(pi_m - 1, ackermann(pi_m, pi_n - 1));
END IF;
END ackermann;
BEGIN
FOR n IN 0 .. 6 LOOP
FOR m IN 0 .. 3 LOOP
dbms_output.put_line('A(' || m || ',' || n || ') = ' || ackermann(m, n));
END LOOP;
END LOOP;
END;
{{out}}
A(0,0) = 1
A(1,0) = 2
A(2,0) = 3
A(3,0) = 5
A(0,1) = 2
A(1,1) = 3
A(2,1) = 5
A(3,1) = 13
A(0,2) = 3
A(1,2) = 4
A(2,2) = 7
A(3,2) = 29
A(0,3) = 4
A(1,3) = 5
A(2,3) = 9
A(3,3) = 61
A(0,4) = 5
A(1,4) = 6
A(2,4) = 11
A(3,4) = 125
A(0,5) = 6
A(1,5) = 7
A(2,5) = 13
A(3,5) = 253
A(0,6) = 7
A(1,6) = 8
A(2,6) = 15
A(3,6) = 509
PostScript
/ackermann{
/n exch def
/m exch def %PostScript takes arguments in the reverse order as specified in the function definition
m 0 eq{
n 1 add
}if
m 0 gt n 0 eq and
{
m 1 sub 1 ackermann
}if
m 0 gt n 0 gt and{
m 1 sub m n 1 sub ackermann ackermann
}if
}def
{{libheader|initlib}}
/A {
[/.m /.n] let
{
{.m 0 eq} {.n succ} is?
{.m 0 gt .n 0 eq and} {.m pred 1 A} is?
{.m 0 gt .n 0 gt and} {.m pred .m .n pred A A} is?
} cond
end}.
Potion
ack = (m, n):
if (m == 0): n + 1
. elsif (n == 0): ack(m - 1, 1)
. else: ack(m - 1, ack(m, n - 1)).
.
4 times(m):
7 times(n):
ack(m, n) print
" " print.
"\n" print.
PowerBASIC
FUNCTION PBMAIN () AS LONG
DIM m AS QUAD, n AS QUAD
m = ABS(VAL(INPUTBOX$("Enter a whole number.")))
n = ABS(VAL(INPUTBOX$("Enter another whole number.")))
MSGBOX STR$(Ackermann(m, n))
END FUNCTION
FUNCTION Ackermann (m AS QUAD, n AS QUAD) AS QUAD
IF 0 = m THEN
FUNCTION = n + 1
ELSEIF 0 = n THEN
FUNCTION = Ackermann(m - 1, 1)
ELSE ' m > 0; n > 0
FUNCTION = Ackermann(m - 1, Ackermann(m, n - 1))
END IF
END FUNCTION
PowerShell
{{trans|PHP}}
function ackermann ([long] $m, [long] $n) {
if ($m -eq 0) {
return $n + 1
}
if ($n -eq 0) {
return (ackermann ($m - 1) 1)
}
return (ackermann ($m - 1) (ackermann $m ($n - 1)))
}
Building an example table (takes a while to compute, though, especially for the last three numbers; also it fails with the last line in Powershell v1 since the maximum recursion depth is only 100 there):
foreach ($m in 0..3) {
foreach ($n in 0..6) {
Write-Host -NoNewline ("{0,5}" -f (ackermann $m $n))
}
Write-Host
}
{{out}}
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
===A More "PowerShelly" Way===
function Get-Ackermann ([int64]$m, [int64]$n)
{
if ($m -eq 0)
{
return $n + 1
}
if ($n -eq 0)
{
return Get-Ackermann ($m - 1) 1
}
return (Get-Ackermann ($m - 1) (Get-Ackermann $m ($n - 1)))
}
Save the result to an array (for possible future use?), then display it using the Format-Wide cmdlet:
$ackermann = 0..3 | ForEach-Object {$m = $_; 0..6 | ForEach-Object {Get-Ackermann $m $_}}
$ackermann | Format-Wide {"{0,3}" -f $_} -Column 7 -Force
{{Out}}
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
Processing
int ackermann(int m, n)
{
if (m == 0)
return n + 1;
else if (m > 0 && n == 0)
return ackermann(m - 1, 1);
else
return ackermann( m - 1, ackermann(m, n - 1) );
}
Prolog
{{works with|SWI Prolog}}
:- table ack/3. % memoization reduces the execution time of ack(4,1,X) from several
% minutes to about one second on a typical desktop computer.
ack(0, N, Ans) :- Ans is N+1.
ack(M, 0, Ans) :- M>0, X is M-1, ack(X, 1, Ans).
ack(M, N, Ans) :- M>0, N>0, X is M-1, Y is N-1, ack(M, Y, Ans2), ack(X, Ans2, Ans).
Pure
A 0 n = n+1;
A m 0 = A (m-1) 1 if m > 0;
A m n = A (m-1) (A m (n-1)) if m > 0 && n > 0;
PureBasic
Procedure.q Ackermann(m, n)
If m = 0
ProcedureReturn n + 1
ElseIf n = 0
ProcedureReturn Ackermann(m - 1, 1)
Else
ProcedureReturn Ackermann(m - 1, Ackermann(m, n - 1))
EndIf
EndProcedure
Debug Ackermann(3,4)
Pure Data
#N canvas 741 265 450 436 10;
#X obj 83 111 t b l;
#X obj 115 163 route 0;
#X obj 115 185 + 1;
#X obj 83 380 f;
#X obj 161 186 swap;
#X obj 161 228 route 0;
#X obj 161 250 - 1;
#X obj 161 208 pack;
#X obj 115 314 t f f;
#X msg 161 272 \$1 1;
#X obj 115 142 t l;
#X obj 207 250 swap;
#X obj 273 271 - 1;
#X obj 207 272 t f f;
#X obj 207 298 - 1;
#X obj 207 360 pack;
#X obj 239 299 pack;
#X obj 83 77 inlet;
#X obj 83 402 outlet;
#X connect 0 0 3 0;
#X connect 0 1 10 0;
#X connect 1 0 2 0;
#X connect 1 1 4 0;
#X connect 2 0 8 0;
#X connect 3 0 18 0;
#X connect 4 0 7 0;
#X connect 4 1 7 1;
#X connect 5 0 6 0;
#X connect 5 1 11 0;
#X connect 6 0 9 0;
#X connect 7 0 5 0;
#X connect 8 0 3 1;
#X connect 8 1 15 1;
#X connect 9 0 10 0;
#X connect 10 0 1 0;
#X connect 11 0 13 0;
#X connect 11 1 12 0;
#X connect 12 0 16 1;
#X connect 13 0 14 0;
#X connect 13 1 16 0;
#X connect 14 0 15 0;
#X connect 15 0 10 0;
#X connect 16 0 10 0;
#X connect 17 0 0 0;
Purity
data Iter = f =
FoldNat <const $f One, $f>
data Ackermann = FoldNat <const Succ, Iter>
Python
{{works with|Python|2.5}}
def ack1(M, N):
return (N + 1) if M == 0 else (
ack1(M-1, 1) if N == 0 else ack1(M-1, ack1(M, N-1)))
Another version:
from functools import lru_cache
@lru_cache(None)
def ack2(M, N):
if M == 0:
return N + 1
elif N == 0:
return ack2(M - 1, 1)
else:
return ack2(M - 1, ack2(M, N - 1))
{{out|Example of use}}
>>
import sys
>>> sys.setrecursionlimit(3000)
>>> ack1(0,0)
1
>>> ack1(3,4)
125
>>> ack2(0,0)
1
>>> ack2(3,4)
125
From the Mathematica ack3 example:
def ack2(M, N):
return (N + 1) if M == 0 else (
(N + 2) if M == 1 else (
(2*N + 3) if M == 2 else (
(8*(2**N - 1) + 5) if M == 3 else (
ack2(M-1, 1) if N == 0 else ack2(M-1, ack2(M, N-1))))))
Results confirm those of Mathematica for ack(4,1) and ack(4,2)
R
ackermann <- function(m, n) {
if ( m == 0 ) {
n+1
} else if ( n == 0 ) {
ackermann(m-1, 1)
} else {
ackermann(m-1, ackermann(m, n-1))
}
}
for ( i in 0:3 ) {
print(ackermann(i, 4))
}
Racket
#lang racket
(define (ackermann m n)
(cond [(zero? m) (add1 n)]
[(zero? n) (ackermann (sub1 m) 1)]
[else (ackermann (sub1 m) (ackermann m (sub1 n)))]))
REBOL
ackermann: func [m n] [
case [
m = 0 [n + 1]
n = 0 [ackermann m - 1 1]
true [ackermann m - 1 ackermann m n - 1]
]
]
REXX
no optimization
/*REXX program calculates and displays some values for the Ackermann function. */
/*╔════════════════════════════════════════════════════════════════════════╗
║ Note: the Ackermann function (as implemented here) utilizes deep ║
║ recursive and is limited by the largest number that can have ║
║ "1" (unity) added to a number (successfully and accurately). ║
╚════════════════════════════════════════════════════════════════════════╝*/
high=24
do j=0 to 3; say
do k=0 to high % (max(1, j))
call tell_Ack j, k
end /*k*/
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell_Ack: parse arg mm,nn; calls=0 /*display an echo message to terminal. */
#=right(nn,length(high))
say 'Ackermann('mm", "#')='right(ackermann(mm, nn), high),
left('', 12) 'calls='right(calls, high)
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
ackermann: procedure expose calls /*compute value of Ackermann function. */
parse arg m,n; calls=calls+1
if m==0 then return n+1
if n==0 then return ackermann(m-1, 1)
return ackermann(m-1, ackermann(m, n-1) )
'''output'''
Ackermann(0, 0)= 1 calls= 1
Ackermann(0, 1)= 2 calls= 1
Ackermann(0, 2)= 3 calls= 1
Ackermann(0, 3)= 4 calls= 1
Ackermann(0, 4)= 5 calls= 1
Ackermann(0, 5)= 6 calls= 1
Ackermann(0, 6)= 7 calls= 1
Ackermann(0, 7)= 8 calls= 1
Ackermann(0, 8)= 9 calls= 1
Ackermann(0, 9)= 10 calls= 1
Ackermann(0,10)= 11 calls= 1
Ackermann(0,11)= 12 calls= 1
Ackermann(0,12)= 13 calls= 1
Ackermann(0,13)= 14 calls= 1
Ackermann(0,14)= 15 calls= 1
Ackermann(0,15)= 16 calls= 1
Ackermann(0,16)= 17 calls= 1
Ackermann(0,17)= 18 calls= 1
Ackermann(0,18)= 19 calls= 1
Ackermann(0,19)= 20 calls= 1
Ackermann(0,20)= 21 calls= 1
Ackermann(0,21)= 22 calls= 1
Ackermann(0,22)= 23 calls= 1
Ackermann(0,23)= 24 calls= 1
Ackermann(0,24)= 25 calls= 1
Ackermann(1, 0)= 2 calls= 2
Ackermann(1, 1)= 3 calls= 4
Ackermann(1, 2)= 4 calls= 6
Ackermann(1, 3)= 5 calls= 8
Ackermann(1, 4)= 6 calls= 10
Ackermann(1, 5)= 7 calls= 12
Ackermann(1, 6)= 8 calls= 14
Ackermann(1, 7)= 9 calls= 16
Ackermann(1, 8)= 10 calls= 18
Ackermann(1, 9)= 11 calls= 20
Ackermann(1,10)= 12 calls= 22
Ackermann(1,11)= 13 calls= 24
Ackermann(1,12)= 14 calls= 26
Ackermann(1,13)= 15 calls= 28
Ackermann(1,14)= 16 calls= 30
Ackermann(1,15)= 17 calls= 32
Ackermann(1,16)= 18 calls= 34
Ackermann(1,17)= 19 calls= 36
Ackermann(1,18)= 20 calls= 38
Ackermann(1,19)= 21 calls= 40
Ackermann(1,20)= 22 calls= 42
Ackermann(1,21)= 23 calls= 44
Ackermann(1,22)= 24 calls= 46
Ackermann(1,23)= 25 calls= 48
Ackermann(1,24)= 26 calls= 50
Ackermann(2, 0)= 3 calls= 5
Ackermann(2, 1)= 5 calls= 14
Ackermann(2, 2)= 7 calls= 27
Ackermann(2, 3)= 9 calls= 44
Ackermann(2, 4)= 11 calls= 65
Ackermann(2, 5)= 13 calls= 90
Ackermann(2, 6)= 15 calls= 119
Ackermann(2, 7)= 17 calls= 152
Ackermann(2, 8)= 19 calls= 189
Ackermann(2, 9)= 21 calls= 230
Ackermann(2,10)= 23 calls= 275
Ackermann(2,11)= 25 calls= 324
Ackermann(2,12)= 27 calls= 377
Ackermann(3, 0)= 5 calls= 15
Ackermann(3, 1)= 13 calls= 106
Ackermann(3, 2)= 29 calls= 541
Ackermann(3, 3)= 61 calls= 2432
Ackermann(3, 4)= 125 calls= 10307
Ackermann(3, 5)= 253 calls= 42438
Ackermann(3, 6)= 509 calls= 172233
Ackermann(3, 7)= 1021 calls= 693964
Ackermann(3, 8)= 2045 calls= 2785999
===optimized for m ≤ 2===
/*REXX program calculates and displays some values for the Ackermann function. */
high=24
do j=0 to 3; say
do k=0 to high % (max(1, j))
call tell_Ack j, k
end /*k*/
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell_Ack: parse arg mm,nn; calls=0 /*display an echo message to terminal. */
#=right(nn,length(high))
say 'Ackermann('mm", "#')='right(ackermann(mm, nn), high),
left('', 12) 'calls='right(calls, high)
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
ackermann: procedure expose calls /*compute value of Ackermann function. */
parse arg m,n; calls=calls+1
if m==0 then return n + 1
if n==0 then return ackermann(m-1, 1)
if m==2 then return n + 3 + n
return ackermann(m-1, ackermann(m, n-1) )
'''output'''
Ackermann(0, 0)= 1 calls= 1
Ackermann(0, 1)= 2 calls= 1
Ackermann(0, 2)= 3 calls= 1
Ackermann(0, 3)= 4 calls= 1
Ackermann(0, 4)= 5 calls= 1
Ackermann(0, 5)= 6 calls= 1
Ackermann(0, 6)= 7 calls= 1
Ackermann(0, 7)= 8 calls= 1
Ackermann(0, 8)= 9 calls= 1
Ackermann(0, 9)= 10 calls= 1
Ackermann(0,10)= 11 calls= 1
Ackermann(0,11)= 12 calls= 1
Ackermann(0,12)= 13 calls= 1
Ackermann(0,13)= 14 calls= 1
Ackermann(0,14)= 15 calls= 1
Ackermann(0,15)= 16 calls= 1
Ackermann(0,16)= 17 calls= 1
Ackermann(0,17)= 18 calls= 1
Ackermann(0,18)= 19 calls= 1
Ackermann(0,19)= 20 calls= 1
Ackermann(0,20)= 21 calls= 1
Ackermann(0,21)= 22 calls= 1
Ackermann(0,22)= 23 calls= 1
Ackermann(0,23)= 24 calls= 1
Ackermann(0,24)= 25 calls= 1
Ackermann(1, 0)= 2 calls= 2
Ackermann(1, 1)= 3 calls= 4
Ackermann(1, 2)= 4 calls= 6
Ackermann(1, 3)= 5 calls= 8
Ackermann(1, 4)= 6 calls= 10
Ackermann(1, 5)= 7 calls= 12
Ackermann(1, 6)= 8 calls= 14
Ackermann(1, 7)= 9 calls= 16
Ackermann(1, 8)= 10 calls= 18
Ackermann(1, 9)= 11 calls= 20
Ackermann(1,10)= 12 calls= 22
Ackermann(1,11)= 13 calls= 24
Ackermann(1,12)= 14 calls= 26
Ackermann(1,13)= 15 calls= 28
Ackermann(1,14)= 16 calls= 30
Ackermann(1,15)= 17 calls= 32
Ackermann(1,16)= 18 calls= 34
Ackermann(1,17)= 19 calls= 36
Ackermann(1,18)= 20 calls= 38
Ackermann(1,19)= 21 calls= 40
Ackermann(1,20)= 22 calls= 42
Ackermann(1,21)= 23 calls= 44
Ackermann(1,22)= 24 calls= 46
Ackermann(1,23)= 25 calls= 48
Ackermann(1,24)= 26 calls= 50
Ackermann(2, 0)= 3 calls= 5
Ackermann(2, 1)= 5 calls= 1
Ackermann(2, 2)= 7 calls= 1
Ackermann(2, 3)= 9 calls= 1
Ackermann(2, 4)= 11 calls= 1
Ackermann(2, 5)= 13 calls= 1
Ackermann(2, 6)= 15 calls= 1
Ackermann(2, 7)= 17 calls= 1
Ackermann(2, 8)= 19 calls= 1
Ackermann(2, 9)= 21 calls= 1
Ackermann(2,10)= 23 calls= 1
Ackermann(2,11)= 25 calls= 1
Ackermann(2,12)= 27 calls= 1
Ackermann(3, 0)= 5 calls= 2
Ackermann(3, 1)= 13 calls= 4
Ackermann(3, 2)= 29 calls= 6
Ackermann(3, 3)= 61 calls= 8
Ackermann(3, 4)= 125 calls= 10
Ackermann(3, 5)= 253 calls= 12
Ackermann(3, 6)= 509 calls= 14
Ackermann(3, 7)= 1021 calls= 16
Ackermann(3, 8)= 2045 calls= 18
===optimized for m ≤ 4=== This REXX version takes advantage that some of the lower numbers for the Ackermann function have direct formulas.
If the '''numeric digits 100''' were to be increased to '''20000''', then the value of '''Ackermann(4,2)'''
(the last line of output) would be presented with the full '''19,729''' decimal digits.
/*REXX program calculates and displays some values for the Ackermann function. */
numeric digits 100 /*use up to 100 decimal digit integers.*/
/*╔═════════════════════════════════════════════════════════════╗
║ When REXX raises a number to an integer power (via the ** ║
║ operator, the power can be positive, zero, or negative). ║
║ Ackermann(5,1) is a bit impractical to calculate. ║
╚═════════════════════════════════════════════════════════════╝*/
high=24
do j=0 to 4; say
do k=0 to high % (max(1, j))
call tell_Ack j, k
if j==4 & k==2 then leave /*there's no sense in going overboard. */
end /*k*/
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell_Ack: parse arg mm,nn; calls=0 /*display an echo message to terminal. */
#=right(nn,length(high))
say 'Ackermann('mm", "#')='right(ackermann(mm, nn), high),
left('', 12) 'calls='right(calls, high)
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
ackermann: procedure expose calls /*compute value of Ackermann function. */
parse arg m,n; calls=calls+1
if m==0 then return n + 1
if m==1 then return n + 2
if m==2 then return n + 3 + n
if m==3 then return 2**(n+3) - 3
if m==4 then do; #=2 /* [↓] Ugh! ··· and still more ughs.*/
do (n+3)-1 /*This is where the heavy lifting is. */
#=2**#
end
return #-3
end
if n==0 then return ackermann(m-1, 1)
return ackermann(m-1, ackermann(m, n-1) )
Output note: none of the numbers shown below use recursion to compute.
'''output'''
Ackermann(0, 0)= 1 calls= 1
Ackermann(0, 1)= 2 calls= 1
Ackermann(0, 2)= 3 calls= 1
Ackermann(0, 3)= 4 calls= 1
Ackermann(0, 4)= 5 calls= 1
Ackermann(0, 5)= 6 calls= 1
Ackermann(0, 6)= 7 calls= 1
Ackermann(0, 7)= 8 calls= 1
Ackermann(0, 8)= 9 calls= 1
Ackermann(0, 9)= 10 calls= 1
Ackermann(0,10)= 11 calls= 1
Ackermann(0,11)= 12 calls= 1
Ackermann(0,12)= 13 calls= 1
Ackermann(0,13)= 14 calls= 1
Ackermann(0,14)= 15 calls= 1
Ackermann(0,15)= 16 calls= 1
Ackermann(0,16)= 17 calls= 1
Ackermann(0,17)= 18 calls= 1
Ackermann(0,18)= 19 calls= 1
Ackermann(0,19)= 20 calls= 1
Ackermann(0,20)= 21 calls= 1
Ackermann(0,21)= 22 calls= 1
Ackermann(0,22)= 23 calls= 1
Ackermann(0,23)= 24 calls= 1
Ackermann(0,24)= 25 calls= 1
Ackermann(1, 0)= 2 calls= 1
Ackermann(1, 1)= 3 calls= 1
Ackermann(1, 2)= 4 calls= 1
Ackermann(1, 3)= 5 calls= 1
Ackermann(1, 4)= 6 calls= 1
Ackermann(1, 5)= 7 calls= 1
Ackermann(1, 6)= 8 calls= 1
Ackermann(1, 7)= 9 calls= 1
Ackermann(1, 8)= 10 calls= 1
Ackermann(1, 9)= 11 calls= 1
Ackermann(1,10)= 12 calls= 1
Ackermann(1,11)= 13 calls= 1
Ackermann(1,12)= 14 calls= 1
Ackermann(1,13)= 15 calls= 1
Ackermann(1,14)= 16 calls= 1
Ackermann(1,15)= 17 calls= 1
Ackermann(1,16)= 18 calls= 1
Ackermann(1,17)= 19 calls= 1
Ackermann(1,18)= 20 calls= 1
Ackermann(1,19)= 21 calls= 1
Ackermann(1,20)= 22 calls= 1
Ackermann(1,21)= 23 calls= 1
Ackermann(1,22)= 24 calls= 1
Ackermann(1,23)= 25 calls= 1
Ackermann(1,24)= 26 calls= 1
Ackermann(2, 0)= 3 calls= 1
Ackermann(2, 1)= 5 calls= 1
Ackermann(2, 2)= 7 calls= 1
Ackermann(2, 3)= 9 calls= 1
Ackermann(2, 4)= 11 calls= 1
Ackermann(2, 5)= 13 calls= 1
Ackermann(2, 6)= 15 calls= 1
Ackermann(2, 7)= 17 calls= 1
Ackermann(2, 8)= 19 calls= 1
Ackermann(2, 9)= 21 calls= 1
Ackermann(2,10)= 23 calls= 1
Ackermann(2,11)= 25 calls= 1
Ackermann(2,12)= 27 calls= 1
Ackermann(3, 0)= 5 calls= 1
Ackermann(3, 1)= 13 calls= 1
Ackermann(3, 2)= 29 calls= 1
Ackermann(3, 3)= 61 calls= 1
Ackermann(3, 4)= 125 calls= 1
Ackermann(3, 5)= 253 calls= 1
Ackermann(3, 6)= 509 calls= 1
Ackermann(3, 7)= 1021 calls= 1
Ackermann(3, 8)= 2045 calls= 1
Ackermann(4, 0)= 13 calls= 1
Ackermann(4, 1)= 65533 calls= 1
Ackermann(4, 2)=89506130880933368E+19728 calls= 1
Ring
{{trans|C#}}
for m = 0 to 3
for n = 0 to 4
see "Ackermann(" + m + ", " + n + ") = " + Ackermann(m, n) + nl
next
next
func Ackermann m, n
if m > 0
if n > 0
return Ackermann(m - 1, Ackermann(m, n - 1))
but n = 0
return Ackermann(m - 1, 1)
ok
but m = 0
if n >= 0
return n + 1
ok
ok
Raise("Incorrect Numerical input !!!")
{{out}}
Ackermann(0, 0) = 1
Ackermann(0, 1) = 2
Ackermann(0, 2) = 3
Ackermann(0, 3) = 4
Ackermann(0, 4) = 5
Ackermann(1, 0) = 2
Ackermann(1, 1) = 3
Ackermann(1, 2) = 4
Ackermann(1, 3) = 5
Ackermann(1, 4) = 6
Ackermann(2, 0) = 3
Ackermann(2, 1) = 5
Ackermann(2, 2) = 7
Ackermann(2, 3) = 9
Ackermann(2, 4) = 11
Ackermann(3, 0) = 5
Ackermann(3, 1) = 13
Ackermann(3, 2) = 29
Ackermann(3, 3) = 61
Ackermann(3, 4) = 125
=={{header|Risc-V}}==
the basic recursive function, because memorization and other improvements would blow the clarity.
## Ruby
{{trans|Ada}}
```ruby
def ack(m, n)
if m == 0
n + 1
elsif n == 0
ack(m-1, 1)
else
ack(m-1, ack(m, n-1))
end
end
Example:
(0..3).each do |m|
puts (0..6).map { |n| ack(m, n) }.join(' ')
end
{{out}}
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
Run BASIC
print ackermann(1, 2)
function ackermann(m, n)
if (m = 0) then ackermann = (n + 1)
if (m > 0) and (n = 0) then ackermann = ackermann((m - 1), 1)
if (m > 0) and (n > 0) then ackermann = ackermann((m - 1), ackermann(m, (n - 1)))
end function
Rust
fn ack(m: isize, n: isize) -> isize {
if m == 0 {
n + 1
} else if n == 0 {
ack(m - 1, 1)
} else {
ack(m - 1, ack(m, n - 1))
}
}
fn main() {
let a = ack(3, 4);
println!("{}", a); // 125
}
Or:
fn ack(m: u64, n: u64) -> u64 {
match (m, n) {
(0, n) => n + 1,
(m, 0) => ack(m - 1, 1),
(m, n) => ack(m - 1, ack(m, n - 1)),
}
}
Sather
class MAIN is
ackermann(m, n:INT):INT
pre m >= 0 and n >= 0
is
if m = 0 then return n + 1; end;
if n = 0 then return ackermann(m-1, 1); end;
return ackermann(m-1, ackermann(m, n-1));
end;
main is
n, m :INT;
loop n := 0.upto!(6);
loop m := 0.upto!(3);
#OUT + "A(" + m + ", " + n + ") = " + ackermann(m, n) + "\n";
end;
end;
end;
end;
Instead of INT, the class INTI could be used, even though we need to use a workaround since in the GNU Sather v1.2.3 compiler the INTI literals are not implemented yet.
class MAIN is
ackermann(m, n:INTI):INTI is
zero ::= 0.inti; -- to avoid type conversion each time
one ::= 1.inti;
if m = zero then return n + one; end;
if n = zero then return ackermann(m-one, one); end;
return ackermann(m-one, ackermann(m, n-one));
end;
main is
n, m :INT;
loop n := 0.upto!(6);
loop m := 0.upto!(3);
#OUT + "A(" + m + ", " + n + ") = " + ackermann(m.inti, n.inti) + "\n";
end;
end;
end;
end;
Scala
def ack(m: BigInt, n: BigInt): BigInt = {
if (m==0) n+1
else if (n==0) ack(m-1, 1)
else ack(m-1, ack(m, n-1))
}
{{out|Example}}
scala> for ( m <- 0 to 3; n <- 0 to 6 ) yield ack(m,n)
res0: Seq.Projection[BigInt] = RangeG(1, 2, 3, 4, 5, 6, 7, 2, 3, 4, 5, 6, 7, 8, 3, 5, 7, 9, 11, 13, 15, 5, 13, 29, 61, 125, 253, 509)
Memoized version using a mutable hash map:
val ackMap = new mutable.HashMap[(BigInt,BigInt),BigInt]
def ackMemo(m: BigInt, n: BigInt): BigInt = {
ackMap.getOrElseUpdate((m,n), ack(m,n))
}
Scheme
(define (A m n)
(cond
((= m 0) (+ n 1))
((= n 0) (A (- m 1) 1))
(else (A (- m 1) (A m (- n 1))))))
An improved solution that uses a lazy data structure, streams, and defines [[Knuth up-arrow]]s to calculate iterative exponentiation:
(define (A m n)
(letrec ((A-stream
(cons-stream
(ints-from 1) ;; m = 0
(cons-stream
(ints-from 2) ;; m = 1
(cons-stream
;; m = 2
(stream-map (lambda (n)
(1+ (* 2 (1+ n))))
(ints-from 0))
(cons-stream
;; m = 3
(stream-map (lambda (n)
(- (knuth-up-arrow 2 (- m 2) (+ n 3)) 3))
(ints-from 0))
;; m = 4...
(stream-tail A-stream 3)))))))
(stream-ref (stream-ref A-stream m) n)))
(define (ints-from n)
(letrec ((ints-rec (cons-stream n (stream-map 1+ ints-rec))))
ints-rec))
(define (knuth-up-arrow a n b)
(let loop ((n n) (b b))
(cond ((= b 0) 1)
((= n 1) (expt a b))
(else (loop (-1+ n) (loop n (-1+ b)))))))
Scilab
{{out}}
```txt
ackermann(0,0)=1 calls=1
ackermann(0,1)=2 calls=1
ackermann(0,2)=3 calls=1
ackermann(0,3)=4 calls=1
ackermann(0,4)=5 calls=1
ackermann(0,5)=6 calls=1
ackermann(0,6)=7 calls=1
ackermann(1,0)=2 calls=2
ackermann(1,1)=3 calls=4
ackermann(1,2)=4 calls=6
ackermann(1,3)=5 calls=8
ackermann(1,4)=6 calls=10
ackermann(1,5)=7 calls=12
ackermann(1,6)=8 calls=14
ackermann(2,0)=3 calls=5
ackermann(2,1)=5 calls=14
ackermann(2,2)=7 calls=27
ackermann(2,3)=9 calls=44
ackermann(2,4)=11 calls=65
ackermann(2,5)=13 calls=90
ackermann(2,6)=15 calls=119
ackermann(3,0)=5 calls=15
ackermann(3,1)=13 calls=106
ackermann(3,2)=29 calls=541
ackermann(3,3)=61 calls=2432
ackermann(3,4)=125 calls=10307
ackermann(3,5)=253 calls=42438
ackermann(3,6)=509 calls=172233
Seed7
const func integer: ackermann (in integer: m, in integer: n) is func
result
var integer: result is 0;
begin
if m = 0 then
result := succ(n);
elsif n = 0 then
result := ackermann(pred(m), 1);
else
result := ackermann(pred(m), ackermann(m, pred(n)));
end if;
end func;
Original source: https://seed7.sourceforge.net/algorith/math.htm#ackermann
SETL
program ackermann;
(for m in [0..3])
print(+/ [rpad('' + ack(m, n), 4): n in [0..6]]);
end;
proc ack(m, n);
return {[0,n+1]}(m) ? ack(m-1, {[0,1]}(n) ? ack(m, n-1));
end proc;
end program;
Shen
(define ack
0 N -> (+ N 1)
M 0 -> (ack (- M 1) 1)
M N -> (ack (- M 1)
(ack M (- N 1))))
Sidef
func A(m, n) {
m == 0 ? (n + 1)
: (n == 0 ? (A(m - 1, 1))
: (A(m - 1, A(m, n - 1))));
}
Alternatively, using multiple dispatch:
func A((0), n) { n + 1 }
func A(m, (0)) { A(m - 1, 1) }
func A(m, n) { A(m-1, A(m, n-1)) }
Calling the function:
say A(3, 2); # prints: 29
Simula
as modified by R. Péter and R. Robinson:
BEGIN
INTEGER procedure
Ackermann(g, p); SHORT INTEGER g, p;
Ackermann:= IF g = 0 THEN p+1
ELSE Ackermann(g-1, IF p = 0 THEN 1
ELSE Ackermann(g, p-1));
INTEGER g, p;
FOR p := 0 STEP 3 UNTIL 13 DO BEGIN
g := 4 - p/3;
outtext("Ackermann("); outint(g, 0);
outchar(','); outint(p, 2); outtext(") = ");
outint(Ackermann(g, p), 0); outimage
END
END
{{Output}}
Ackermann(4, 0) = 13
Ackermann(3, 3) = 61
Ackermann(2, 6) = 15
Ackermann(1, 9) = 11
Ackermann(0,12) = 13
Slate
m@(Integer traits) ackermann: n@(Integer traits)
[
m isZero
ifTrue: [n + 1]
ifFalse:
[n isZero
ifTrue: [m - 1 ackermann: n]
ifFalse: [m - 1 ackermann: (m ackermann: n - 1)]]
].
Smalltalk
|ackermann|
ackermann := [ :n :m |
(n = 0) ifTrue: [ (m + 1) ]
ifFalse: [
(m = 0) ifTrue: [ ackermann value: (n-1) value: 1 ]
ifFalse: [
ackermann value: (n-1)
value: ( ackermann value: n
value: (m-1) )
]
]
].
(ackermann value: 0 value: 0) displayNl.
(ackermann value: 3 value: 4) displayNl.
SmileBASIC
DEF ACK(M,N)
IF M==0 THEN
RETURN N+1
ELSEIF M>0 AND N==0 THEN
RETURN ACK(M-1,1)
ELSE
RETURN ACK(M-1,ACK(M,N-1))
ENDIF
END
SNOBOL4
{{works with|Macro Spitbol}} Both Snobol4+ and CSnobol stack overflow, at ack(3,3) and ack(3,4), respectively.
define('ack(m,n)') :(ack_end)
ack ack = eq(m,0) n + 1 :s(return)
ack = eq(n,0) ack(m - 1,1) :s(return)
ack = ack(m - 1,ack(m,n - 1)) :(return)
ack_end
* # Test and display ack(0,0) .. ack(3,6)
L1 str = str ack(m,n) ' '
n = lt(n,6) n + 1 :s(L1)
output = str; str = ''
n = 0; m = lt(m,3) m + 1 :s(L1)
end
{{out}}
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
SNUSP
/==!/==atoi=@@@-@-----#
| | Ackermann function
| | /
### ===
\!==\!====\ recursion:
$,@/>,@/==ack=!\?\<+# | | | A(0,j) -> j+1
j i \<?\+>-@/# | | A(i,0) -> A(i-1,1)
\@\>@\->@/@\<-@/# A(i,j) -> A(i-1,A(i,j-1))
| | |
# # | | | /+<<<-\
/-<<+>>\!=/ \=====|==!/
### ==
?\>>>=?/<<#
? ? | \<<<+>+>>-/
\>>+<<-/!
### ====
/
# #
One could employ [[:Category:Recursion|tail recursion]] elimination by replacing "@/#" with "/" in two places above.
SPAD
{{works with|FriCAS, OpenAxiom, Axiom}}
NNI ==> NonNegativeInteger
A:(NNI,NNI) -> NNI
A(m,n) ==
m=0 => n+1
m>0 and n=0 => A(m-1,1)
m>0 and n>0 => A(m-1,A(m,n-1))
-- Example
matrix [[A(i,j) for i in 0..3] for j in 0..3]
{{out}}
+1 2 3 5 +
| |
|2 3 5 13|
(1) | |
|3 4 7 29|
| |
+4 5 9 61+
Type: Matrix(NonNegativeInteger)
SQL PL
{{works with|Db2 LUW}} version 9.7 or higher. With SQL PL:
--#SET TERMINATOR @
SET SERVEROUTPUT ON@
CREATE OR REPLACE FUNCTION ACKERMANN(
IN M SMALLINT,
IN N BIGINT
) RETURNS BIGINT
BEGIN
DECLARE RET BIGINT;
DECLARE STMT STATEMENT;
IF (M = 0) THEN
SET RET = N + 1;
ELSEIF (N = 0) THEN
PREPARE STMT FROM 'SET ? = ACKERMANN(? - 1, 1)';
EXECUTE STMT INTO RET USING M;
ELSE
PREPARE STMT FROM 'SET ? = ACKERMANN(? - 1, ACKERMANN(?, ? - 1))';
EXECUTE STMT INTO RET USING M, M, N;
END IF;
RETURN RET;
END @
BEGIN
DECLARE M SMALLINT DEFAULT 0;
DECLARE N SMALLINT DEFAULT 0;
DECLARE MAX_LEVELS CONDITION FOR SQLSTATE '54038';
DECLARE CONTINUE HANDLER FOR MAX_LEVELS BEGIN END;
WHILE (N <= 6) DO
WHILE (M <= 3) DO
CALL DBMS_OUTPUT.PUT_LINE('ACKERMANN(' || M || ', ' || N || ') = ' || ACKERMANN(M, N));
SET M = M + 1;
END WHILE;
SET M = 0;
SET N = N + 1;
END WHILE;
END @
Output:
db2 -td@
db2 => CREATE OR REPLACE FUNCTION ACKERMANN(
...
db2 (cont.) => END @
DB20000I The SQL command completed successfully.
db2 => BEGIN
db2 (cont.) => END
...
DB20000I The SQL command completed successfully.
ACKERMANN(0, 0) = 1
ACKERMANN(1, 0) = 2
ACKERMANN(2, 0) = 3
ACKERMANN(3, 0) = 5
ACKERMANN(0, 1) = 2
ACKERMANN(1, 1) = 3
ACKERMANN(2, 1) = 5
ACKERMANN(3, 1) = 13
ACKERMANN(0, 2) = 3
ACKERMANN(1, 2) = 4
ACKERMANN(2, 2) = 7
ACKERMANN(3, 2) = 29
ACKERMANN(0, 3) = 4
ACKERMANN(1, 3) = 5
ACKERMANN(2, 3) = 9
ACKERMANN(3, 3) = 61
ACKERMANN(0, 4) = 5
ACKERMANN(1, 4) = 6
ACKERMANN(2, 4) = 11
ACKERMANN(0, 5) = 6
ACKERMANN(1, 5) = 7
ACKERMANN(2, 5) = 13
ACKERMANN(0, 6) = 7
ACKERMANN(1, 6) = 8
ACKERMANN(2, 6) = 15
The maximum levels of cascade calls in Db2 are 16, and in some cases when executing the Ackermann function, it arrives to this limit (SQL0724N). Thus, the code catches the exception and continues with the next try.
Standard ML
fun a (0, n) = n+1
| a (m, 0) = a (m-1, 1)
| a (m, n) = a (m-1, a (m, n-1))
Stata
mata
function ackermann(m,n) {
if (m==0) {
return(n+1)
} else if (n==0) {
return(ackermann(m-1,1))
} else {
return(ackermann(m-1,ackermann(m,n-1)))
}
}
for (i=0; i<=3; i++) printf("%f\n",ackermann(i,4))
5
6
11
125
end
Swift
func ackerman(m:Int, n:Int) -> Int {
if m == 0 {
return n+1
} else if n == 0 {
return ackerman(m-1, 1)
} else {
return ackerman(m-1, ackerman(m, n-1))
}
}
Tcl
Simple
{{trans|Ruby}}
proc ack {m n} {
if {$m == 0} {
expr {$n + 1}
} elseif {$n == 0} {
ack [expr {$m - 1}] 1
} else {
ack [expr {$m - 1}] [ack $m [expr {$n - 1}]]
}
}
With Tail Recursion
With Tcl 8.6, this version is preferred (though the language supports tailcall optimization, it does not apply it automatically in order to preserve stack frame semantics):
proc ack {m n} {
if {$m == 0} {
expr {$n + 1}
} elseif {$n == 0} {
tailcall ack [expr {$m - 1}] 1
} else {
tailcall ack [expr {$m - 1}] [ack $m [expr {$n - 1}]]
}
}
===To Infinity… and Beyond!=== If we want to explore the higher reaches of the world of Ackermann's function, we need techniques to really cut the amount of computation being done. {{works with|Tcl|8.6}}
package require Tcl 8.6
# A memoization engine, from https://wiki.tcl-lang.org/18152
oo::class create cache {
filter Memoize
variable ValueCache
method Memoize args {
# Do not filter the core method implementations
if {[lindex [self target] 0] eq "::oo::object"} {
return [next {*}$args]
}
# Check if the value is already in the cache
set key [self target],$args
if {[info exist ValueCache($key)]} {
return $ValueCache($key)
}
# Compute value, insert into cache, and return it
return [set ValueCache($key) [next {*}$args]]
}
method flushCache {} {
unset ValueCache
# Skip the cacheing
return -level 2 ""
}
}
# Make an object, attach the cache engine to it, and define ack as a method
oo::object create cached
oo::objdefine cached {
mixin cache
method ack {m n} {
if {$m==0} {
expr {$n+1}
} elseif {$m==1} {
# From the Mathematica version
expr {$m+2}
} elseif {$m==2} {
# From the Mathematica version
expr {2*$n+3}
} elseif {$m==3} {
# From the Mathematica version
expr {8*(2**$n-1)+5}
} elseif {$n==0} {
tailcall my ack [expr {$m-1}] 1
} else {
tailcall my ack [expr {$m-1}] [my ack $m [expr {$n-1}]]
}
}
}
# Some small tweaks...
interp recursionlimit {} 100000
interp alias {} ack {} cacheable ack
But even with all this, you still run into problems calculating as that's kind-of large…
TSE SAL
// library: math: get: ackermann: recursive <description></description> <version>1.0.0.0.5</version> <version control></version control>
(filenamemacro=getmaare.s) [kn, ri, tu, 27-12-2011 14:46:59]
INTEGER PROC FNMathGetAckermannRecursiveI( INTEGER mI, INTEGER nI )
IF ( mI == 0 )
RETURN( nI + 1 )
ENDIF
IF ( nI == 0 )
RETURN( FNMathGetAckermannRecursiveI( mI - 1, 1 ) )
ENDIF
RETURN( FNMathGetAckermannRecursiveI( mI - 1, FNMathGetAckermannRecursiveI( mI, nI - 1 ) ) )
END
PROC Main()
STRING s1[255] = "2"
STRING s2[255] = "3"
IF ( NOT ( Ask( "math: get: ackermann: recursive: m = ", s1, _EDIT_HISTORY_ ) ) AND ( Length( s1 ) > 0 ) ) RETURN() ENDIF
IF ( NOT ( Ask( "math: get: ackermann: recursive: n = ", s2, _EDIT_HISTORY_ ) ) AND ( Length( s2 ) > 0 ) ) RETURN() ENDIF
Message( FNMathGetAckermannRecursiveI( Val( s1 ), Val( s2 ) ) ) // gives e.g. 9
END
TI-83 BASIC
This program assumes the variables N and M are the arguments of the function, and that the list L1 is empty. It stores the result in the system variable ANS. (Program names can be no longer than 8 characters, so I had to truncate the function's name.)
PROGRAM:ACKERMAN
:If not(M
:Then
:N+1→N
:Return
:Else
:If not(N
:Then
:1→N
:M-1→M
:prgmACKERMAN
:Else
:N-1→N
:M→L1(1+dim(L1
:prgmACKERMAN
:Ans→N
:L1(dim(L1))-1→M
:dim(L1)-1→dim(L1
:prgmACKERMAN
:End
:End
Here is a handler function that makes the previous function easier to use. (You can name it whatever you want.)
PROGRAM:AHANDLER
:0→dim(L1
:Prompt M
:Prompt N
:prgmACKERMAN
:Disp Ans
=={{header|TI-89 BASIC}}==
Define A(m,n) = when(m=0, n+1, when(n=0, A(m-1,1), A(m-1, A(m, n-1))))
TorqueScript
function ackermann(%m,%n)
{
if(%m==0)
return %n+1;
if(%m>0&&%n==0)
return ackermann(%m-1,1);
if(%m>0&&%n>0)
return ackermann(%m-1,ackermann(%m,%n-1));
}
TXR
{{trans|Scheme}} with memoization.
(defmacro defmemofun (name (. args) . body)
(let ((hash (gensym "hash-"))
(argl (gensym "args-"))
(hent (gensym "hent-"))
(uniq (copy-str "uniq")))
^(let ((,hash (hash :equal-based)))
(defun ,name (,*args)
(let* ((,argl (list ,*args))
(,hent (inhash ,hash ,argl ,uniq)))
(if (eq (cdr ,hent) ,uniq)
(set (cdr ,hent) (block ,name (progn ,*body)))
(cdr ,hent)))))))
(defmemofun ack (m n)
(cond
((= m 0) (+ n 1))
((= n 0) (ack (- m 1) 1))
(t (ack (- m 1) (ack m (- n 1))))))
(each ((i (range 0 3)))
(each ((j (range 0 4)))
(format t "ack(~a, ~a) = ~a\n" i j (ack i j))))
{{out}}
ack(0, 0) = 1
ack(0, 1) = 2
ack(0, 2) = 3
ack(0, 3) = 4
ack(0, 4) = 5
ack(1, 0) = 2
ack(1, 1) = 3
ack(1, 2) = 4
ack(1, 3) = 5
ack(1, 4) = 6
ack(2, 0) = 3
ack(2, 1) = 5
ack(2, 2) = 7
ack(2, 3) = 9
ack(2, 4) = 11
ack(3, 0) = 5
ack(3, 1) = 13
ack(3, 2) = 29
ack(3, 3) = 61
ack(3, 4) = 125
UNIX Shell
{{works with|Bash}}
ack() {
local m=$1
local n=$2
if [ $m -eq 0 ]; then
echo -n $((n+1))
elif [ $n -eq 0 ]; then
ack $((m-1)) 1
else
ack $((m-1)) $(ack $m $((n-1)))
fi
}
Example:
for ((m=0;m<=3;m++)); do
for ((n=0;n<=6;n++)); do
ack $m $n
echo -n " "
done
echo
done
{{out}}
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
Ursala
Anonymous recursion is the usual way of doing things like this.
#import std
#import nat
ackermann =
~&al^?\successor@ar ~&ar?(
^R/~&f ^/predecessor@al ^|R/~& ^|/~& predecessor,
^|R/~& ~&\1+ predecessor@l)
test program for the first 4 by 7 numbers:
#cast %nLL
test = block7 ackermann*K0 iota~~/4 7
{{out}}
<
<1,2,3,4,5,6,7>,
<2,3,4,5,6,7,8>,
<3,5,7,9,11,13,15>,
<5,13,29,61,125,253,509>>
V
{{trans|Joy}}
[ack
[ [pop zero?] [popd succ]
[zero?] [pop pred 1 ack]
[true] [[dup pred swap] dip pred ack ack ]
] when].
using destructuring view
[ack
[ [pop zero?] [ [m n : [n succ]] view i]
[zero?] [ [m n : [m pred 1 ack]] view i]
[true] [ [m n : [m pred m n pred ack ack]] view i]
] when].
VBA
Private Function Ackermann_function(m As Variant, n As Variant) As Variant
Dim result As Variant
Debug.Assert m >= 0
Debug.Assert n >= 0
If m = 0 Then
result = CDec(n + 1)
Else
If n = 0 Then
result = Ackermann_function(m - 1, 1)
Else
result = Ackermann_function(m - 1, Ackermann_function(m, n - 1))
End If
End If
Ackermann_function = CDec(result)
End Function
Public Sub main()
Debug.Print " n=",
For j = 0 To 7
Debug.Print j,
Next j
Debug.Print
For i = 0 To 3
Debug.Print "m=" & i,
For j = 0 To 7
Debug.Print Ackermann_function(i, j),
Next j
Debug.Print
Next i
End Sub
{{out}}
n= 0 1 2 3 4 5 6 7
m=0 1 2 3 4 5 6 7 8
m=1 2 3 4 5 6 7 8 9
m=2 3 5 7 9 11 13 15 17
m=3 5 13 29 61 125 253 509 1021
VBScript
Based on BASIC version. Uncomment all the lines referring to depth and see just how deep the recursion goes.
;Implementation
option explicit
'~ dim depth
function ack(m, n)
'~ wscript.stdout.write depth & " "
if m = 0 then
'~ depth = depth + 1
ack = n + 1
'~ depth = depth - 1
elseif m > 0 and n = 0 then
'~ depth = depth + 1
ack = ack(m - 1, 1)
'~ depth = depth - 1
'~ elseif m > 0 and n > 0 then
else
'~ depth = depth + 1
ack = ack(m - 1, ack(m, n - 1))
'~ depth = depth - 1
end if
end function
;Invocation
wscript.echo ack( 1, 10 )
'~ depth = 0
wscript.echo ack( 2, 1 )
'~ depth = 0
wscript.echo ack( 4, 4 )
{{out}}
12
5
C:\foo\ackermann.vbs(16, 3) Microsoft VBScript runtime error: Out of stack space: 'ack'
Visual Basic
{{trans|Rexx}} {{works with|Visual Basic|VB6 Standard}}
Option Explicit
Dim calls As Long
Sub main()
Const maxi = 4
Const maxj = 9
Dim i As Long, j As Long
For i = 0 To maxi
For j = 0 To maxj
Call print_acker(i, j)
Next j
Next i
End Sub 'main
Sub print_acker(m As Long, n As Long)
calls = 0
Debug.Print "ackermann("; m; ","; n; ")=";
Debug.Print ackermann(m, n), "calls="; calls
End Sub 'print_acker
Function ackermann(m As Long, n As Long) As Long
calls = calls + 1
If m = 0 Then
ackermann = n + 1
Else
If n = 0 Then
ackermann = ackermann(m - 1, 1)
Else
ackermann = ackermann(m - 1, ackermann(m, n - 1))
End If
End If
End Function 'ackermann
{{Out}}
ackermann( 0 , 0 )= 1 calls= 1
ackermann( 0 , 1 )= 2 calls= 1
ackermann( 0 , 2 )= 3 calls= 1
ackermann( 0 , 3 )= 4 calls= 1
ackermann( 0 , 4 )= 5 calls= 1
ackermann( 0 , 5 )= 6 calls= 1
ackermann( 0 , 6 )= 7 calls= 1
ackermann( 0 , 7 )= 8 calls= 1
ackermann( 0 , 8 )= 9 calls= 1
ackermann( 0 , 9 )= 10 calls= 1
ackermann( 1 , 0 )= 2 calls= 2
ackermann( 1 , 1 )= 3 calls= 4
ackermann( 1 , 2 )= 4 calls= 6
ackermann( 1 , 3 )= 5 calls= 8
ackermann( 1 , 4 )= 6 calls= 10
ackermann( 1 , 5 )= 7 calls= 12
ackermann( 1 , 6 )= 8 calls= 14
ackermann( 1 , 7 )= 9 calls= 16
ackermann( 1 , 8 )= 10 calls= 18
ackermann( 1 , 9 )= 11 calls= 20
ackermann( 2 , 0 )= 3 calls= 5
ackermann( 2 , 1 )= 5 calls= 14
ackermann( 2 , 2 )= 7 calls= 27
ackermann( 2 , 3 )= 9 calls= 44
ackermann( 2 , 4 )= 11 calls= 65
ackermann( 2 , 5 )= 13 calls= 90
ackermann( 2 , 6 )= 15 calls= 119
ackermann( 2 , 7 )= 17 calls= 152
ackermann( 2 , 8 )= 19 calls= 189
ackermann( 2 , 9 )= 21 calls= 230
ackermann( 3 , 0 )= 5 calls= 15
ackermann( 3 , 1 )= 13 calls= 106
ackermann( 3 , 2 )= 29 calls= 541
ackermann( 3 , 3 )= 61 calls= 2432
ackermann( 3 , 4 )= 125 calls= 10307
ackermann( 3 , 5 )= 253 calls= 42438
ackermann( 3 , 6 )= 509 calls= 172233
ackermann( 3 , 7 )= 1021 calls= 693964
ackermann( 3 , 8 )= 2045 calls= 2785999
ackermann( 3 , 9 )= 4093 calls= 11164370
ackermann( 4 , 0 )= 13 calls= 107
ackermann( 4 , 1 )= out of stack space
Vlang
fn ackermann(m int, n int) int {
switch 0 {
case m:
return n + 1
case n:
return ackermann(m - 1, 1)
}
return ackermann(m - 1, ackermann(m, n - 1))
}
Wart
def (ackermann m n)
(if m=0
n+1
n=0
(ackermann m-1 1)
:else
(ackermann m-1 (ackermann m n-1)))
WDTE
let memo a m n =
true {
== m 0 => + n 1;
== n 0 => a (- m 1) 1;
true => a (- m 1) (a m (- n 1));
};
Wren
// To use recursion definition and declaration must be on separate lines
var Ackermann
Ackermann = Fn.new {|m, n|
if (m == 0) return n + 1
if (n == 0) return Ackermann.call(m - 1, 1)
return Ackermann.call(m - 1, Ackermann.call(m, n - 1))
}
XLISP
(defun ackermann (m n)
(cond
((= m 0) (+ n 1))
((= n 0) (ackermann (- m 1) 1))
(t (ackermann (- m 1) (ackermann m (- n 1))))))
Test it:
(print (ackermann 3 9))
Output (after a very perceptible pause):
4093
That worked well. Test it again:
(print (ackermann 4 1))
Output (after another pause):
Abort: control stack overflow
happened in: #<Code ACKERMANN>
XPL0
include c:\cxpl\codes;
func Ackermann(M, N);
int M, N;
[if M=0 then return N+1;
if N=0 then return Ackermann(M-1, 1);
return Ackermann(M-1, Ackermann(M, N-1));
]; \Ackermann
int M, N;
[for M:= 0 to 3 do
[for N:= 0 to 7 do
[IntOut(0, Ackermann(M, N)); ChOut(0,9\tab\)];
CrLf(0);
];
]
Recursion overflows the stack if either M or N is extended by a single count. {{out}}
1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
3 5 7 9 11 13 15 17
5 13 29 61 125 253 509 1021
XSLT
The following named template calculates the Ackermann function:
<xsl:template name="ackermann">
<xsl:param name="m"/>
<xsl:param name="n"/>
<xsl:choose>
<xsl:when test="$m = 0">
<xsl:value-of select="$n+1"/>
</xsl:when>
<xsl:when test="$n = 0">
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m - 1"/>
<xsl:with-param name="n" select="'1'"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:variable name="p">
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m"/>
<xsl:with-param name="n" select="$n - 1"/>
</xsl:call-template>
</xsl:variable>
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m - 1"/>
<xsl:with-param name="n" select="$p"/>
</xsl:call-template>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
Here it is as part of a template
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="https://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:template match="arguments">
<xsl:for-each select="args">
<div>
<xsl:value-of select="m"/>, <xsl:value-of select="n"/>:
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="m"/>
<xsl:with-param name="n" select="n"/>
</xsl:call-template>
</div>
</xsl:for-each>
</xsl:template>
<xsl:template name="ackermann">
<xsl:param name="m"/>
<xsl:param name="n"/>
<xsl:choose>
<xsl:when test="$m = 0">
<xsl:value-of select="$n+1"/>
</xsl:when>
<xsl:when test="$n = 0">
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m - 1"/>
<xsl:with-param name="n" select="'1'"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:variable name="p">
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m"/>
<xsl:with-param name="n" select="$n - 1"/>
</xsl:call-template>
</xsl:variable>
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m - 1"/>
<xsl:with-param name="n" select="$p"/>
</xsl:call-template>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
Which will transform this input
<?xml version="1.0" ?>
<?xml-stylesheet type="text/xsl" href="ackermann.xslt"?>
<arguments>
<args>
<m>0</m>
<n>0</n>
</args>
<args>
<m>0</m>
<n>1</n>
</args>
<args>
<m>0</m>
<n>2</n>
</args>
<args>
<m>0</m>
<n>3</n>
</args>
<args>
<m>0</m>
<n>4</n>
</args>
<args>
<m>0</m>
<n>5</n>
</args>
<args>
<m>0</m>
<n>6</n>
</args>
<args>
<m>0</m>
<n>7</n>
</args>
<args>
<m>0</m>
<n>8</n>
</args>
<args>
<m>1</m>
<n>0</n>
</args>
<args>
<m>1</m>
<n>1</n>
</args>
<args>
<m>1</m>
<n>2</n>
</args>
<args>
<m>1</m>
<n>3</n>
</args>
<args>
<m>1</m>
<n>4</n>
</args>
<args>
<m>1</m>
<n>5</n>
</args>
<args>
<m>1</m>
<n>6</n>
</args>
<args>
<m>1</m>
<n>7</n>
</args>
<args>
<m>1</m>
<n>8</n>
</args>
<args>
<m>2</m>
<n>0</n>
</args>
<args>
<m>2</m>
<n>1</n>
</args>
<args>
<m>2</m>
<n>2</n>
</args>
<args>
<m>2</m>
<n>3</n>
</args>
<args>
<m>2</m>
<n>4</n>
</args>
<args>
<m>2</m>
<n>5</n>
</args>
<args>
<m>2</m>
<n>6</n>
</args>
<args>
<m>2</m>
<n>7</n>
</args>
<args>
<m>2</m>
<n>8</n>
</args>
<args>
<m>3</m>
<n>0</n>
</args>
<args>
<m>3</m>
<n>1</n>
</args>
<args>
<m>3</m>
<n>2</n>
</args>
<args>
<m>3</m>
<n>3</n>
</args>
<args>
<m>3</m>
<n>4</n>
</args>
<args>
<m>3</m>
<n>5</n>
</args>
<args>
<m>3</m>
<n>6</n>
</args>
<args>
<m>3</m>
<n>7</n>
</args>
<args>
<m>3</m>
<n>8</n>
</args>
</arguments>
into this output
0, 0: 1
0, 1: 2
0, 2: 3
0, 3: 4
0, 4: 5
0, 5: 6
0, 6: 7
0, 7: 8
0, 8: 9
1, 0: 2
1, 1: 3
1, 2: 4
1, 3: 5
1, 4: 6
1, 5: 7
1, 6: 8
1, 7: 9
1, 8: 10
2, 0: 3
2, 1: 5
2, 2: 7
2, 3: 9
2, 4: 11
2, 5: 13
2, 6: 15
2, 7: 17
2, 8: 19
3, 0: 5
3, 1: 13
3, 2: 29
3, 3: 61
3, 4: 125
3, 5: 253
3, 6: 509
3, 7: 1021
3, 8: 2045
Yabasic
sub ack(M,N)
if M = 0 return N + 1
if N = 0 return ack(M-1,1)
return ack(M-1,ack(M, N-1))
end sub
print ack(3, 4)
What smart code can get. Fast as lightning! {{trans|Phix}}
sub ack(m, n)
if m=0 then
return n+1
elsif m=1 then
return n+2
elsif m=2 then
return 2*n+3
elsif m=3 then
return 2^(n+3)-3
elsif m>0 and n=0 then
return ack(m-1,1)
else
return ack(m-1,ack(m,n-1))
end if
end sub
sub Ackermann()
local i, j
for i=0 to 3
for j=0 to 10
print ack(i,j) using "#####";
next
print
next
print "ack(4,1) ";: print ack(4,1) using "#####"
end sub
Ackermann()
Yorick
func ack(m, n) {
if(m == 0)
return n + 1;
else if(n == 0)
return ack(m - 1, 1);
else
return ack(m - 1, ack(m, n - 1));
}
Example invocation:
for(m = 0; m <= 3; m++) {
for(n = 0; n <= 6; n++)
write, format="%d ", ack(m, n);
write, "";
}
{{out}}
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
{{omit from|LaTeX}} {{omit from|Make}} {{omit from|PlainTeX}}
ZED
Source -> https://ideone.com/53FzPA Compiled -> https://ideone.com/OlS7zL
(A) m n
comment:
(=) m 0
(add1) n
(A) m n
comment:
(=) n 0
(A) (sub1) m 1
(A) m n
comment:
#true
(A) (sub1) m (A) m (sub1) n
(add1) n
comment:
#true
(003) "+" n 1
(sub1) n
comment:
#true
(003) "-" n 1
(=) n1 n2
comment:
#true
(003) "=" n1 n2
ZX Spectrum Basic
{{trans|BASIC256}}
10 DIM s(2000,3)
20 LET s(1,1)=3: REM M
30 LET s(1,2)=7: REM N
40 LET lev=1
50 GO SUB 100
60 PRINT "A(";s(1,1);",";s(1,2);") = ";s(1,3)
70 STOP
100 IF s(lev,1)=0 THEN LET s(lev,3)=s(lev,2)+1: RETURN
110 IF s(lev,2)=0 THEN LET lev=lev+1: LET s(lev,1)=s(lev-1,1)-1: LET s(lev,2)=1: GO SUB 100: LET s(lev-1,3)=s(lev,3): LET lev=lev-1: RETURN
120 LET lev=lev+1
130 LET s(lev,1)=s(lev-1,1)
140 LET s(lev,2)=s(lev-1,2)-1
150 GO SUB 100
160 LET s(lev,1)=s(lev-1,1)-1
170 LET s(lev,2)=s(lev,3)
180 GO SUB 100
190 LET s(lev-1,3)=s(lev,3)
200 LET lev=lev-1
210 RETURN
{{out}}
A(3,7) = 1021