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This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
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{{task|Puzzles}}
You are given a box with eight holes labelled A-to-H, connected by fifteen straight lines in the pattern as shown below:
'''A''' '''B'''
/│\ /│\
/ │ X │ \
/ │/ \│ \
'''C''' ─ '''D''' ─ '''E''' ─ '''F'''
\ │\ /│ /
\ │ X │ /
\│/ \│/
'''G''' '''H'''
You are also given eight pegs numbered 1-to-8.
;Objective: Place the eight pegs in the holes so that the (absolute) difference between any two numbers connected by any line is greater than one.
;Example: In this attempt:
'''4''' '''7'''
/│\ /│\
/ │ X │ \
/ │/ \│ \
'''8''' ─ '''1''' ─ '''6''' ─ '''2'''
\ │\ /│ /
\ │ X │ /
\│/ \│/
'''3''' '''5'''
Note that '''7''' and '''6''' are connected and have a difference of '''1''', so it is ''not'' a solution.
;Task Produce and show here ''one'' solution to the puzzle.
;Related tasks: :* [[A* search algorithm]] :* [[Solve a Holy Knight's tour]] :* [[Knight's tour]] :* [[N-queens problem]] :* [[Solve a Hidato puzzle]] :* [[Solve a Holy Knight's tour]] :* [[Solve a Hopido puzzle]] :* [[Solve a Numbrix puzzle]] :* [[4-rings or 4-squares puzzle]]
;See also [https://www.youtube.com/watch?v=AECElyEyZBQ No Connection Puzzle] (youtube).
Ada
This solution is a bit longer than it actually needs to be; however, it uses tasks to find the solution and the used types and solution-generating functions are well-separated, making it more amenable to other solutions or altering it to display all solutions.
With
Ada.Text_IO,
Connection_Types,
Connection_Combinations;
procedure main is
Result : Connection_Types.Partial_Board renames Connection_Combinations;
begin
Ada.Text_IO.Put_Line( Connection_Types.Image(Result) );
end;
Pragma Ada_2012;
Package Connection_Types with Pure is
-- Name of the nodes.
Type Node is (A, B, C, D, E, F, G, H);
-- Type for indicating if a node is connected.
Type Connection_List is array(Node) of Boolean
with Size => 8, Object_Size => 8, Pack;
Function "&"( Left : Connection_List; Right : Node ) return Connection_List;
-- The actual map of the network connections.
Network : Constant Array (Node) of Connection_List:=
(
A => (C|D|E => True, others => False),
B => (D|E|F => True, others => False),
C => (A|D|G => True, others => False),
D => (C|A|B|E|H|G => True, others => False),
E => (D|A|B|F|H|G => True, others => False),
F => (B|E|H => True, others => False),
G => (C|D|E => True, others => False),
H => (D|E|F => True, others => False)
);
-- Values of the nodes.
Type Peg is range 1..8;
-- Indicator for which values have been assigned.
Type Used_Peg is array(Peg) of Boolean
with Size => 8, Object_Size => 8, Pack;
Function "&"( Left : Used_Peg; Right : Peg ) return Used_Peg;
-- Type describing the layout of the network.
Type Partial_Board is array(Node range <>) of Peg;
Subtype Board is Partial_Board(Node);
-- Determines if the given board is a solution or partial-solution.
Function Is_Solution ( Input : Partial_Board ) return Boolean;
-- Displays the board as text.
Function Image ( Input : Partial_Board ) Return String;
End Connection_Types;
Pragma Ada_2012;
with Connection_Types;
use Connection_Types;
Function Connection_Combinations return Partial_Board;
Pragma Ada_2012;
Package Body Connection_Types is
New_Line : Constant String := ASCII.CR & ASCII.LF;
---------------------
-- Solution Test --
---------------------
Function Is_Solution( Input : Partial_Board ) return Boolean is
(for all Index in Input'Range =>
(for all Connection in Node'Range =>
(if Network(Index)(Connection) and Connection in Input'Range
then abs (Input(Index) - Input(Connection)) > 1
)
)
);
------------------------
-- Concat Operators --
------------------------
Function "&"( Left : Used_Peg; Right : Peg ) return Used_Peg is
begin
return Result : Used_Peg := Left do
Result(Right):= True;
end return;
end "&";
Function "&"(Left : Connection_List; Right : Node) return Connection_List is
begin
Return Result : Connection_List := Left do
Result(Right):= True;
end return;
end "&";
-----------------------
-- IMAGE FUNCTIONS --
-----------------------
Function Image(Input : Peg) Return Character is
( Peg'Image(Input)(2) );
Function Image(Input : Peg) Return String is
( 1 => Image(Input) );
Function Image(Input : Partial_Board; Item : Node) Return String is
( 1 => (if Item not in Input'Range then '*' else Image(Input(Item)) ));
Function Image( Input : Partial_Board ) Return String is
A : String renames Image(Input, Connection_Types.A);
B : String renames Image(Input, Connection_Types.B);
C : String renames Image(Input, Connection_Types.C);
D : String renames Image(Input, Connection_Types.D);
E : String renames Image(Input, Connection_Types.E);
F : String renames Image(Input, Connection_Types.F);
G : String renames Image(Input, Connection_Types.G);
H : String renames Image(Input, Connection_Types.H);
begin
return
" "&A&" "&B & New_Line &
" /|\ /|\" & New_Line &
" / | X | \" & New_Line &
" / |/ \| \" & New_Line &
" "&C&" - "&D&" - "&E&" - "&F & New_Line &
" \ |\ /| /" & New_Line &
" \ | X | /" & New_Line &
" \|/ \|/" & New_Line &
" "&G&" "&H & New_Line;
end Image;
End Connection_Types;
Function Connection_Combinations return Partial_Board is
begin
Return Result : Board do
declare
-- The Generate task takes two parameters
-- (1) a list of pegs already in use, and
-- (2) a partial-board
-- and, if the state given is a viable yet incomplete solution, it
-- takes a peg and adds it to the state creating a new task with
-- that peg in its used list.
--
-- When a complete solution is found it is copied into result.
task type Generate(
Pegs : not null access Used_Peg:= new Used_Peg'(others => False);
State : not null access Partial_Board:= new Partial_Board'(Node'Last..Node'First => <>)
) is
end Generate;
-- An access to Generate and array thereof, for creating the
-- children tasks.
type Generator is access all Generate;
type Generators is array(Peg range <>) of Generator;
-- Gen handles the actual creation of a new task and state.
Function Gen(P : Peg; G : not null access Generate) return Generator is
begin
return (if G.Pegs(P) then null
else new Generate(
Pegs => new Used_Peg'(G.Pegs.all & P),
State => New Partial_Board'(G.All.State.All & P)
)
);
end;
task body Generate is
begin
if Is_Solution(State.All) then
-- If the state is a partial board, we make children to
-- complete the calculations.
if State'Length <= Node'Pos(Node'Last) then
declare
Subtasks : Constant Generators:=
(
Gen(1, Generate'Access),
Gen(2, Generate'Access),
Gen(3, Generate'Access),
Gen(4, Generate'Access),
Gen(5, Generate'Access),
Gen(6, Generate'Access),
Gen(7, Generate'Access),
Gen(8, Generate'Access)
);
begin
null;
end;
else
Result:= State.All;
end if;
else
-- The current state is not a solution, so we do not continue it.
Null;
end if;
end Generate;
Master : Generate;
begin
null;
end;
End return;
End Connection_Combinations;
{{out}}
4 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 6
APL
{{works with|Dyalog APL 17.0 Unicode}}
perms←{
⍝∇ 20100513/20140818 ra⌈ --()--
1=⍴⍴⍵:⍵[∇ ''⍴⍴⍵]
↑{0∊⍴⍵:⍵ ⋄ (⍺[1]⌷⍵),(1↓⍺)∇ ⍵~⍺[1]⌷⍵}∘(⍳⍵)¨↓⍉1+(⌽⍳⍵)⊤¯1+⍳!⍵
}
solution←{
links← (3 4 5) (4 5 6) (1 4 7) (1 2 3 5 7 8) (1 2 4 6 7 8) (2 5 8) (3 4 5) (4 5 6) ⍝ node i connects with nodes i⊃links
tries←8 perms 8
fails←{1∊{1∊⍵∊¯1 0 1}¨|⍺-¨⍺∘{⍺[⍵]}¨⍵}
⍝ ⍴⍸~tries fails ⍤1⊢links
⍝ 16
solns←⍸~tries fails ⍤1⊢links
tries[''⍴solns;]
}
ARM Assembly
{{works with|as|Raspberry Pi}}
/* ARM assembly Raspberry PI */
/* program noconnpuzzle.s */
/************************************/
/* Constantes */
/************************************/
.equ STDOUT, 1 @ Linux output console
.equ EXIT, 1 @ Linux syscall
.equ WRITE, 4 @ Linux syscall
.equ NBBOX, 8
.equ POSA, 5
/*********************************/
/* Initialized data */
/*********************************/
.data
sMessDeb: .ascii "a="
sMessValeur_a: .fill 11, 1, ' ' @ size => 11
.ascii "b="
sMessValeur_b: .fill 11, 1, ' ' @ size => 11
.ascii "c="
sMessValeur_c: .fill 11, 1, ' ' @ size => 11
.ascii "d="
sMessValeur_d: .fill 11, 1, ' ' @ size => 11
.ascii "\n"
.ascii "e="
sMessValeur_e: .fill 11, 1, ' ' @ size => 11
.ascii "f="
sMessValeur_f: .fill 11, 1, ' ' @ size => 11
.ascii "g="
sMessValeur_g: .fill 11, 1, ' ' @ size => 11
.ascii "h="
sMessValeur_h: .fill 11, 1, ' ' @ size => 11
szCarriageReturn: .asciz "\n************************\n"
szMessLine1: .asciz " \n"
szMessLine2: .asciz " /|\\ /|\\ \n"
szMessLine3: .asciz " / | X | \\ \n"
szMessLine4: .asciz " / |/ \\| \\ \n"
szMessLine5: .asciz " - - | - \n"
szMessLine6: .asciz " \\ |\\ /| / \n"
szMessLine7: .asciz " \\ | X | / \n"
szMessLine8: .asciz " \\|/ \\|/ \n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
.align 4
iValues_a: .skip 4 * NBBOX
iValues_b: .skip 4 * NBBOX - 1
iValues_c: .skip 4 * NBBOX - 2
iValues_d: .skip 4 * NBBOX - 3
iValues_e: .skip 4 * NBBOX - 4
iValues_f: .skip 4 * NBBOX - 5
iValues_g: .skip 4 * NBBOX - 6
iValues_h: .skip 4 * NBBOX - 7
sConvValue: .skip 12
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: @ entry of program
mov r0,#1
mov r1,#8
bl searchPb
100: @ standard end of the program
mov r0, #0 @ return code
mov r7, #EXIT @ request to exit program
svc #0 @ perform the system call
iAdrszCarriageReturn: .int szCarriageReturn
/******************************************************************/
/* search problem unique solution */
/******************************************************************/
/* r0 contains start digit */
/* r1 contains end digit */
searchPb:
push {r0-r12,lr} @ save registers
@ init
ldr r3,iAdriValues_a @ area value a
mov r4,#0
1: @ loop init value a
str r0,[r3,r4,lsl #2]
add r4,#1
add r0,#1
cmp r0,r1
ble 1b
mov r12,#-1
2:
add r12,#1 @ increment indice a
cmp r12,#NBBOX-1
bgt 90f
ldr r0,iAdriValues_a @ area value a
ldr r1,iAdriValues_b @ area value b
mov r2,r12 @ indice a
mov r3,#NBBOX @ number of origin values
bl prepValues
mov r11,#-1
3:
add r11,#1 @ increment indice b
cmp r11,#NBBOX - 2
bgt 2b
ldr r0,iAdriValues_b @ area value b
ldr r1,iAdriValues_c @ area value c
mov r2,r11 @ indice b
mov r3,#NBBOX -1 @ number of origin values
bl prepValues
mov r10,#-1
4:
add r10,#1
cmp r10,#NBBOX - 3
bgt 3b
ldr r0,iAdriValues_a
ldr r0,[r0,r12,lsl #2]
ldr r1,iAdriValues_c
ldr r1,[r1,r10,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 4b
ldr r0,iAdriValues_c
ldr r1,iAdriValues_d
mov r2,r10
mov r3,#NBBOX - 2
bl prepValues
mov r9,#-1
5:
add r9,#1
cmp r9,#NBBOX - 4
bgt 4b
@ control d / a b c
ldr r0,iAdriValues_d
ldr r0,[r0,r9,lsl #2]
ldr r1,iAdriValues_a
ldr r1,[r1,r12,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 5b
ldr r1,iAdriValues_b
ldr r1,[r1,r11,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 5b
ldr r1,iAdriValues_c
ldr r1,[r1,r10,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 5b
ldr r0,iAdriValues_d
ldr r1,iAdriValues_e
mov r2,r9
mov r3,#NBBOX - 3
bl prepValues
mov r8,#-1
6:
add r8,#1
cmp r8,#NBBOX - 5
bgt 5b
@ control e / a b d
ldr r0,iAdriValues_e
ldr r0,[r0,r8,lsl #2]
ldr r1,iAdriValues_a
ldr r1,[r1,r12,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 6b
ldr r1,iAdriValues_b
ldr r1,[r1,r11,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 6b
ldr r1,iAdriValues_d
ldr r1,[r1,r9,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 6b
ldr r0,iAdriValues_e
ldr r1,iAdriValues_f
mov r2,r8
mov r3,#NBBOX - 4
bl prepValues
mov r7,#-1
7:
add r7,#1
cmp r7,#NBBOX - 6
bgt 6b
@ control f / b e
ldr r0,iAdriValues_f
ldr r0,[r0,r7,lsl #2]
ldr r1,iAdriValues_b
ldr r1,[r1,r11,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 7b
ldr r1,iAdriValues_e
ldr r1,[r1,r8,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 7b
ldr r0,iAdriValues_f
ldr r1,iAdriValues_g
mov r2,r7
mov r3,#NBBOX - 5
bl prepValues
mov r6,#-1
8:
add r6,#1
cmp r6,#NBBOX - 7
bgt 7b
@ control g / c d e
ldr r0,iAdriValues_g
ldr r0,[r0,r6,lsl #2]
ldr r1,iAdriValues_c
ldr r1,[r1,r10,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 8b
ldr r1,iAdriValues_d
ldr r1,[r1,r9,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 8b
ldr r1,iAdriValues_e
ldr r1,[r1,r8,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 8b
ldr r0,iAdriValues_g
ldr r1,iAdriValues_h
mov r2,r6
mov r3,#NBBOX - 6
bl prepValues
mov r5,#-1
9:
add r5,#1
cmp r5,#NBBOX - 8
bgt 8b
@ control h / d e f
ldr r0,iAdriValues_h
ldr r0,[r0,r5,lsl #2]
ldr r1,iAdriValues_d
ldr r1,[r1,r9,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 9b
ldr r1,iAdriValues_e
ldr r1,[r1,r8,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 9b
ldr r1,iAdriValues_f
ldr r1,[r1,r7,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 9b
@ solution ok display text
ldr r0,iAdriValues_a
ldr r0,[r0,r12,lsl #2]
ldr r1,iAdrsMessValeur_a
bl conversion10
ldr r0,iAdriValues_b
ldr r0,[r0,r11,lsl #2]
ldr r1,iAdrsMessValeur_b
bl conversion10
ldr r0,iAdriValues_c
ldr r0,[r0,r10,lsl #2]
ldr r1,iAdrsMessValeur_c
bl conversion10
ldr r0,iAdriValues_d
ldr r0,[r0,r9,lsl #2]
ldr r1,iAdrsMessValeur_d
bl conversion10
ldr r0,iAdriValues_e
ldr r0,[r0,r8,lsl #2]
ldr r1,iAdrsMessValeur_e
bl conversion10
ldr r0,iAdriValues_f
ldr r0,[r0,r7,lsl #2]
ldr r1,iAdrsMessValeur_f
bl conversion10
ldr r0,iAdriValues_g
ldr r0,[r0,r6,lsl #2]
ldr r1,iAdrsMessValeur_g
bl conversion10
ldr r0,iAdriValues_h
ldr r0,[r0,r5,lsl #2]
ldr r1,iAdrsMessValeur_h
bl conversion10
ldr r0,iAdrsMessDeb
bl affichageMess
@ display design
ldr r0,iAdriValues_a
ldr r0,[r0,r12,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine1
strb r2,[r0,#POSA]
ldr r0,iAdriValues_b
ldr r0,[r0,r11,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine1
strb r2,[r0,#POSA+4]
bl affichageMess
ldr r0,iAdrszMessLine2
bl affichageMess
ldr r0,iAdrszMessLine3
bl affichageMess
ldr r0,iAdrszMessLine4
bl affichageMess
ldr r0,iAdriValues_c
ldr r0,[r0,r10,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine5
strb r2,[r0,#POSA-4]
ldr r0,iAdriValues_d
ldr r0,[r0,r9,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine5
strb r2,[r0,#POSA]
ldr r0,iAdriValues_e
ldr r0,[r0,r8,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine5
strb r2,[r0,#POSA+4]
ldr r0,iAdriValues_f
ldr r0,[r0,r7,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine5
strb r2,[r0,#POSA+8]
bl affichageMess
ldr r0,iAdrszMessLine6
bl affichageMess
ldr r0,iAdrszMessLine7
bl affichageMess
ldr r0,iAdrszMessLine8
bl affichageMess
ldr r0,iAdriValues_g
ldr r0,[r0,r6,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine1
strb r2,[r0,#POSA]
ldr r0,iAdriValues_h
ldr r0,[r0,r5,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine1
strb r2,[r0,#POSA+4]
bl affichageMess
//b 9b @ loop for other solution
90:
100:
pop {r0-r12,lr} @ restaur registers
bx lr @return
iAdriValues_a: .int iValues_a
iAdriValues_b: .int iValues_b
iAdriValues_c: .int iValues_c
iAdriValues_d: .int iValues_d
iAdriValues_e: .int iValues_e
iAdriValues_f: .int iValues_f
iAdriValues_g: .int iValues_g
iAdriValues_h: .int iValues_h
iAdrsMessValeur_a: .int sMessValeur_a
iAdrsMessValeur_b: .int sMessValeur_b
iAdrsMessValeur_c: .int sMessValeur_c
iAdrsMessValeur_d: .int sMessValeur_d
iAdrsMessValeur_e: .int sMessValeur_e
iAdrsMessValeur_f: .int sMessValeur_f
iAdrsMessValeur_g: .int sMessValeur_g
iAdrsMessValeur_h: .int sMessValeur_h
iAdrsMessDeb: .int sMessDeb
iAdrsConvValue: .int sConvValue
iAdrszMessLine1: .int szMessLine1
iAdrszMessLine2: .int szMessLine2
iAdrszMessLine3: .int szMessLine3
iAdrszMessLine4: .int szMessLine4
iAdrszMessLine5: .int szMessLine5
iAdrszMessLine6: .int szMessLine6
iAdrszMessLine7: .int szMessLine7
iAdrszMessLine8: .int szMessLine8
/******************************************************************/
/* copy value area and substract value of indice */
/******************************************************************/
/* r0 contains the address of values origin */
/* r1 contains the address of values destination */
/* r2 contains value indice to substract */
/* r3 contains origin values number */
prepValues:
push {r1-r6,lr} @ save registres
mov r4,#0 @ indice origin value
mov r5,#0 @ indice destination value
1:
cmp r4,r2 @ substract indice ?
beq 2f @ yes -> jump
ldr r6,[r0,r4,lsl #2] @ no -> copy value
str r6,[r1,r5,lsl #2]
add r5,#1 @ increment destination indice
2:
add r4,#1 @ increment origin indice
cmp r4,r3 @ end ?
blt 1b
100:
pop {r1-r6,lr} @ restaur registres
bx lr @return
/******************************************************************/
/* display text with size calculation */
/******************************************************************/
/* r0 contains the address of the message */
affichageMess:
push {r0,r1,r2,r7,lr} @ save registres
mov r2,#0 @ counter length
1: @ loop length calculation
ldrb r1,[r0,r2] @ read octet start position + index
cmp r1,#0 @ if 0 its over
addne r2,r2,#1 @ else add 1 in the length
bne 1b @ and loop
@ so here r2 contains the length of the message
mov r1,r0 @ address message in r1
mov r0,#STDOUT @ code to write to the standard output Linux
mov r7, #WRITE @ code call system "write"
svc #0 @ call systeme
pop {r0,r1,r2,r7,lr} @ restaur des 2 registres */
bx lr @ return
/******************************************************************/
/* Converting a register to a decimal unsigned */
/******************************************************************/
/* r0 contains value and r1 address area */
/* r0 return size of result (no zero final in area) */
/* area size => 11 bytes */
.equ LGZONECAL, 10
conversion10:
push {r1-r4,lr} @ save registers
mov r3,r1
mov r2,#LGZONECAL
1: @ start loop
bl divisionpar10U @ unsigned r0 <- dividende. quotient ->r0 reste -> r1
add r1,#48 @ digit
strb r1,[r3,r2] @ store digit on area
cmp r0,#0 @ stop if quotient = 0
subne r2,#1 @ else previous position
bne 1b @ and loop
@ and move digit from left of area
mov r4,#0
2:
ldrb r1,[r3,r2]
strb r1,[r3,r4]
add r2,#1
add r4,#1
cmp r2,#LGZONECAL
ble 2b
@ and move spaces in end on area
mov r0,r4 @ result length
mov r1,#' ' @ space
3:
strb r1,[r3,r4] @ store space in area
add r4,#1 @ next position
cmp r4,#LGZONECAL
ble 3b @ loop if r4 <= area size
100:
pop {r1-r4,lr} @ restaur registres
bx lr @return
/***************************************************/
/* division par 10 unsigned */
/***************************************************/
/* r0 dividende */
/* r0 quotient */
/* r1 remainder */
divisionpar10U:
push {r2,r3,r4, lr}
mov r4,r0 @ save value
ldr r3,iMagicNumber @ r3 <- magic_number raspberry 1 2
umull r1, r2, r3, r0 @ r1<- Lower32Bits(r1*r0) r2<- Upper32Bits(r1*r0)
mov r0, r2, LSR #3 @ r2 <- r2 >> shift 3
add r2,r0,r0, lsl #2 @ r2 <- r0 * 5
sub r1,r4,r2, lsl #1 @ r1 <- r4 - (r2 * 2) = r4 - (r0 * 10)
pop {r2,r3,r4,lr}
bx lr @ leave function
iMagicNumber: .int 0xCCCCCCCD
a=3 b=4 c=7 d=1
e=8 f=2 g=5 h=6
************************
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
AutoHotkey
oGrid := [[ "", "X", "X"] ; setup oGrid
,[ "X", "X", "X", "X"]
,[ "", "X", "X"]]
oNeighbor := [], oCell := [], oRoute := [] , oVisited := [] ; initialize objects
for row, oRow in oGrid
for col, val in oRow
if val ; for each valid cell in oGrid
oNeighbor[row, col] := Neighbors(row, col, oGrid) ; list valid no-connection neighbors
Solve:
for row, oRow in oGrid
for col , val in oRow
if val ; for each valid cell in oGrid
if (oSolution := SolveNoConnect(row, col, 1)).8 ; solve for this cell
break, Solve ; if solution found stop
; show solution
for i , val in oSolution
oCell[StrSplit(val, ":").1 , StrSplit(val, ":").2] := i
A := oCell[1, 2] , B := oCell[1, 3]
C := oCell[2, 1], D := oCell[2, 2] , E := oCell[2, 3], F := oCell[2, 4]
G := oCell[3, 2] , H := oCell[3, 3]
sol =
(
%A% %B%
/|\ /|\
/ | X | \
/ |/ \| \
%C% - %D% - %E% - %F%
\ |\ /| /
\ | X | /
\|/ \|/
%G% %H%
)
MsgBox % sol
return
;-----------------------------------------------------------------------
SolveNoConnect(row, col, val){
global
oRoute.push(row ":" col) ; save route
oVisited[row, col] := true ; mark this cell visited
if oRoute[8] ; if solution found
return true ; end recursion
for each, nn in StrSplit(oNeighbor[row, col], ",") ; for each no-connection neighbor of cell
{
rowX := StrSplit(nn, ":").1 , colX := StrSplit(nn, ":").2 ; get coords of this neighbor
if !oVisited[rowX, colX] ; if not previously visited
{
oVisited[rowX, colX] := true ; mark this cell visited
val++ ; increment
if (SolveNoConnect(rowX, colX, val)) ; recurse
return oRoute ; if solution found return route
}
}
oRoute.pop() ; Solution not found, backtrack oRoute
oVisited[row, col] := false ; Solution not found, remove mark
}
;-----------------------------------------------------------------------
Neighbors(row, col, oGrid){ ; return distant neighbors of oGrid[row,col]
for r , oRow in oGrid
for c, v in oRow
if (v="X") && (abs(row-r) > 1 || abs(col-c) > 1)
list .= r ":"c ","
if (row<>2) && oGrid[row, col]
list .= oGrid[row, col+1] ? row ":" col+1 "," : oGrid[row, col-1] ? row ":" col-1 "," : ""
return Trim(list, ",")
}
Outputs:
3 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 6
Chapel
type hole = int;
param A : hole = 1;
param B : hole = A+1;
param C : hole = B+1;
param D : hole = C+1;
param E : hole = D+1;
param F : hole = E+1;
param G : hole = F+1;
param H : hole = G+1;
param starting : int = 0;
const holes : domain(hole) = { A,B,C,D,E,F,G,H };
const graph : [holes] domain(hole) = [ A => { C,D,E },
B => { D,E,F },
C => { A,D,G },
D => { A,B,C,E,G,H },
E => { A,B,D,F,G,H },
F => { B,E,H },
G => { C,D,E },
H => { D,E,F }
];
proc check( configuration : [] int, idx : hole ) : bool {
var good = true;
for adj in graph[idx] {
if adj >= idx then continue;
if abs( configuration[idx] - configuration[adj] ) <= 1 {
good = false;
break;
}
}
return good;
}
proc solve( configuration : [] int, pegs : domain(int), idx : hole = A ) : bool {
for value in pegs {
configuration[idx] = value;
if check( configuration, idx ) {
if idx < holes.size {
var prePegs = pegs;
if solve( configuration, prePegs - value, idx + 1 ){
return true;
}
} else {
return true;
}
}
}
configuration[idx] = starting;
return false;
}
proc printBoard( configuration : [] int ){
return
"\n " + configuration[A] + " " + configuration[B]+ "\n" +
" /|\\ /|\\ \n"+
" / | X | \\ \n"+
" / |/ \\| \\ \n"+
" " + configuration[C] +" - " + configuration[D] + " - " + configuration[E] + " - " + configuration[F] + " \n"+
" \\ |\\ /| / \n"+
" \\ | X | / \n"+
" \\|/ \\|/ \n"+
" " + configuration[G] + " " + configuration[H]+ "\n";
}
proc main(){
var configuration : [holes] int;
for idx in holes do configuration[idx] = starting;
var pegs : domain(int) = {1,2,3,4,5,6,7,8};
solve( configuration, pegs );
writeln( printBoard( configuration ) );
}
4 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 6
D
void main() @safe {
import std.stdio, std.math, std.algorithm, std.traits, std.string;
enum Peg { A, B, C, D, E, F, G, H }
immutable Peg[2][15] connections =
[[Peg.A, Peg.C], [Peg.A, Peg.D], [Peg.A, Peg.E],
[Peg.B, Peg.D], [Peg.B, Peg.E], [Peg.B, Peg.F],
[Peg.C, Peg.D], [Peg.D, Peg.E], [Peg.E, Peg.F],
[Peg.G, Peg.C], [Peg.G, Peg.D], [Peg.G, Peg.E],
[Peg.H, Peg.D], [Peg.H, Peg.E], [Peg.H, Peg.F]];
immutable board = r"
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H";
Peg[EnumMembers!Peg.length] perm = [EnumMembers!Peg];
do if (connections[].all!(con => abs(perm[con[0]] - perm[con[1]]) > 1))
return board.tr("ABCDEFGH", "%(%d%)".format(perm)).writeln;
while (perm[].nextPermutation);
}
{{out}}
2 3
/|\ /|\
/ | X | \
/ |/ \| \
6 - 0 - 7 - 1
\ |\ /| /
\ | X | /
\|/ \|/
4 5
Alternative version
Using a simple backtracking. {{trans|Go}}
import std.stdio, std.algorithm, std.conv, std.string, std.typecons;
// Holes A=0, B=1, ..., H=7
// With connections:
const board = r"
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H";
struct Connection { uint a, b; }
immutable Connection[] connections = [
{0, 2}, {0, 3}, {0, 4}, // A to C,D,E
{1, 3}, {1, 4}, {1, 5}, // B to D,E,F
{6, 2}, {6, 3}, {6, 4}, // G to C,D,E
{7, 3}, {7, 4}, {7, 5}, // H to D,E,F
{2, 3}, {3, 4}, {4, 5}, // C-D, D-E, E-F
];
alias Pegs = uint[8];
int absDiff(in uint a, in uint b) pure nothrow @safe @nogc {
return (a > b) ? (a - b) : (b - a);
}
/** Solution is a simple recursive brute force solver,
it stops at the first found solution.
It returns the solution, the number of positions tested,
and the number of pegs swapped. */
Tuple!(Pegs,"p", uint,"tests", uint,"swaps") solve() pure nothrow @safe @nogc {
uint tests = 0, swaps = 0;
Pegs p = [1, 2, 3, 4, 5, 6, 7, 8];
bool recurse(in uint i) nothrow @safe @nogc {
if (i >= p.length.signed - 1) {
tests++;
return connections.all!(c => absDiff(p[c.a], p[c.b]) > 1);
}
// Try each remain peg from.
foreach (immutable j; i .. p.length) {
swaps++;
swap(p[i], p[j]);
if (recurse(i + 1))
return true;
swap(p[i], p[j]);
}
return false;
}
recurse(0);
return typeof(return)(p, tests, swaps);
}
void main() {
immutable sol = solve();
board.tr("ABCDEFGH", "%(%d%)".format(sol.p)).writeln;
writeln("Tested ", sol.tests, " positions and did ", sol.swaps, " swaps.");
}
{{out}}
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.
Elixir
{{trans|Ruby}} This solution uses HLPsolver from [[Solve_a_Hidato_puzzle#Elixir | here]]
# It solved if connected A and B, connected G and H (according to the video).
# require HLPsolver
adjacent = for i <- -2..2, j <- -2..2, not(i in -1..1 and j in -1..1), do: {i,j}
layout = ~S"""
A - B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G - H
"""
board = """
. 0 0 .
0 1 0 0
. 0 0 .
"""
HLPsolver.solve(board, adjacent, false)
|> Enum.sort |> Enum.map(fn {_,cell} -> cell.value end)
|> Enum.zip(~w[A B C D E F G H])
|> Enum.reduce(layout, fn {n,c},acc -> String.replace(acc, c, to_string(n)) end)
|> IO.puts
{{out}}
4 - 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 - 5
Factor
USING: assocs interpolate io kernel math math.combinatorics
math.ranges math.parser multiline pair-rocket sequences
sequences.generalizations ;
STRING: diagram
${} ${}
/|\ /|\
/ | X | \
/ |/ \| \
${} - ${} - ${} - ${}
\ |\ /| /
\ | X | /
\|/ \|/
${} ${}
;
CONSTANT: adjacency
H{
0 => { 2 3 4 }
1 => { 3 4 5 }
2 => { 0 3 6 }
3 => { 0 1 2 4 6 7 }
4 => { 0 1 3 5 6 7 }
5 => { 1 4 7 }
6 => { 2 3 4 }
7 => { 3 4 5 }
}
: any-consecutive? ( seq n -- ? ) [ - abs 1 = ] curry any? ;
: neighbors ( elt seq i -- seq elt )
adjacency at swap nths swap ;
: solution? ( permutation-seq -- ? )
dup [ neighbors any-consecutive? ] with find-index nip not ;
: find-solution ( -- seq )
8 [1,b] [ solution? ] find-permutation ;
: display-solution ( seq -- )
[ number>string ] map 8 firstn diagram interpolate>string
print ;
: main ( -- ) find-solution display-solution ;
MAIN: main
{{out}}
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Go
A simple recursive brute force solution.
package main
import (
"fmt"
"strings"
)
func main() {
p, tests, swaps := Solution()
fmt.Println(p)
fmt.Println("Tested", tests, "positions and did", swaps, "swaps.")
}
// Holes A=0, B=1, …, H=7
// With connections:
const conn = `
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H`
var connections = []struct{ a, b int }{
{0, 2}, {0, 3}, {0, 4}, // A to C,D,E
{1, 3}, {1, 4}, {1, 5}, // B to D,E,F
{6, 2}, {6, 3}, {6, 4}, // G to C,D,E
{7, 3}, {7, 4}, {7, 5}, // H to D,E,F
{2, 3}, {3, 4}, {4, 5}, // C-D, D-E, E-F
}
type pegs [8]int
// Valid checks if the pegs are a valid solution.
// If the absolute difference between any pair of connected pegs is
// greater than one it is a valid solution.
func (p *pegs) Valid() bool {
for _, c := range connections {
if absdiff(p[c.a], p[c.b]) <= 1 {
return false
}
}
return true
}
// Solution is a simple recursive brute force solver,
// it stops at the first found solution.
// It returns the solution, the number of positions tested,
// and the number of pegs swapped.
func Solution() (p *pegs, tests, swaps int) {
var recurse func(int) bool
recurse = func(i int) bool {
if i >= len(p)-1 {
tests++
return p.Valid()
}
// Try each remain peg from p[i:] in p[i]
for j := i; j < len(p); j++ {
swaps++
p[i], p[j] = p[j], p[i]
if recurse(i + 1) {
return true
}
p[i], p[j] = p[j], p[i]
}
return false
}
p = &pegs{1, 2, 3, 4, 5, 6, 7, 8}
recurse(0)
return
}
func (p *pegs) String() string {
return strings.Map(func(r rune) rune {
if 'A' <= r && r <= 'H' {
return rune(p[r-'A'] + '0')
}
return r
}, conn)
}
func absdiff(a, b int) int {
if a > b {
return a - b
}
return b - a
}
{{out}}
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.
Haskell
import Data.List (intercalate, permutations)
solution :: [Int]
solution@(a:b:c:d:e:f:g:h:_) = head $ filter isSolution (permutations [1 .. 8])
where
isSolution :: [Int] -> Bool
isSolution (a:b:c:d:e:f:g:h:_) =
all ((> 1) . abs) $
zipWith
(-)
[a, c, g, e, a, c, g, e, b, d, h, f, b, d, h, f]
[d, d, d, d, c, g, e, a, e, e, e, e, d, h, f, b]
main :: IO ()
main =
(putStrLn . unlines) $
let rightShift s
| length s > 3 = s
| otherwise = " " ++ s
in intercalate
"\n"
(zipWith (\x y -> x : (" = " ++ show y)) ['A' .. 'H'] solution) :
((rightShift . unwords . fmap show) <$> [[], [a, b], [c, d, e, f], [g, h]])
{{Out}}
A = 3
B = 4
C = 7
D = 1
E = 8
F = 2
G = 5
H = 6
3 4
7 1 8 2
5 6
```
## J
Supporting code:
```J
holes=:;:'A B C D E F G H'
connections=:".;._2]0 :0
holes e.;:'C D E' NB. A
holes e.;:'D E F' NB. B
holes e.;:'A D G' NB. C
holes e.;:'A B C E G H' NB. D
holes e.;:'A B D F G H' NB. E
holes e.;:'B E H' NB. F
holes e.;:'C D E' NB. G
holes e.;:'D E F' NB. H
)
assert (-:|:) connections NB. catch typos
pegs=: 1+(A.&i.~ !)8
attempt=: [: <./@(-.&0)@,@:| connections * -/~
box=:0 :0
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
)
disp=:verb define
rplc&(,holes;&":&>y) box
)
```
Intermezzo:
```J
(#~ 1 vals = range(1, 9).mapToObj(i -> i).collect(toList());
do {
Collections.shuffle(vals);
for (int i = 0; i < pegs.length; i++)
pegs[i] = vals.get(i);
} while (!solved());
printResult();
}
static boolean solved() {
for (int i = 0; i < links.length; i++)
for (int peg : links[i])
if (abs(pegs[i] - peg) == 1)
return false;
return true;
}
static void printResult() {
System.out.printf(" %s %s%n", pegs[0], pegs[1]);
System.out.printf("%s %s %s %s%n", pegs[2], pegs[3], pegs[4], pegs[5]);
System.out.printf(" %s %s%n", pegs[6], pegs[7]);
}
}
```
(takes about 500 shuffles on average)
```txt
4 5
2 8 1 7
6 3
```
## JavaScript
### ES6
{{Trans|Haskell}}
```JavaScript
(() => {
'use strict';
// GENERIC FUNCTIONS ------------------------------------------------------
// abs :: Num a => a -> a
const abs = Math.abs;
// all :: (a -> Bool) -> [a] -> Bool
const all = (f, xs) => xs.every(f);
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) => [].concat.apply([], xs.map(f));
// delete_ :: Eq a => a -> [a] -> [a]
const delete_ = (x, xs) =>
deleteBy((a, b) => a === b, x, xs);
// deleteBy :: (a -> a -> Bool) -> a -> [a] -> [a]
const deleteBy = (f, x, xs) =>
xs.length > 0 ? (
f(x, xs[0]) ? (
xs.slice(1)
) : [xs[0]].concat(deleteBy(f, x, xs.slice(1)))
) : [];
// enumFromTo :: Enum a => a -> a -> [a]
const enumFromTo = (m, n) => {
const [tm, tn] = [typeof m, typeof n];
return tm !== tn ? undefined : (() => {
const
blnS = (tm === 'string'),
[base, end] = [m, n].map(blnS ? (s => s.codePointAt(0)) : id);
return Array.from({
length: Math.floor(end - base) + 1
}, (_, i) => blnS ? String.fromCodePoint(base + i) : m + i);
})();
};
// id :: a -> a
const id = x => x;
// justifyRight :: Int -> Char -> Text -> Text
const justifyRight = (n, cFiller, strText) =>
n > strText.length ? (
(cFiller.repeat(n) + strText)
.slice(-n)
) : strText;
// permutations :: [a] -> [[a]]
const permutations = xs =>
xs.length ? concatMap(x => concatMap(ys => [
[x].concat(ys)
],
permutations(delete_(x, xs))), xs) : [
[]
];
// show :: a -> String
const show = x => JSON.stringify(x);
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
};
// unwords :: [String] -> String
const unwords = xs => xs.join(' ');
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = (f, xs, ys) => {
const ny = ys.length;
return (xs.length <= ny ? xs : xs.slice(0, ny))
.map((x, i) => f(x, ys[i]));
};
// CONNECTION PUZZLE ------------------------------------------------------
// universe :: [[Int]]
const universe = permutations(enumFromTo(1, 8));
// isSolution :: [Int] -> Bool
const isSolution = ([a, b, c, d, e, f, g, h]) =>
all(x => abs(x) > 1, [a - d, c - d, g - d, e - d, a - c, c - g, g - e,
e - a, b - e, d - e, h - e, f - e, b - d, d - h, h - f, f - b
]);
// firstSolution :: [Int]
const firstSolution = universe[until(
i => isSolution(universe[i]),
i => i + 1,
0
)];
// TEST -------------------------------------------------------------------
// [Int]
const [a, b, c, d, e, f, g, h] = firstSolution;
return unlines(
zipWith(
(a, n) => a + ' = ' + n.toString(),
enumFromTo('A', 'H'),
firstSolution
)
.concat(
[
[],
[a, b],
[c, d, e, f],
[g, h]
].map(xs => justifyRight(5, ' ', unwords(xs.map(show))))
)
);
})();
```
{{Out}}
```txt
A = 3
B = 4
C = 7
D = 1
E = 8
F = 2
G = 5
H = 6
3 4
7 1 8 2
5 6
```
## jq
{{works with|jq|1.4}}
We present a generate-and-test solver for a slightly more general version of the problem, in which there are N pegs and holes, and in which the connectedness of holes is defined by an array such that holes i and j are connected if and only if [i,j] is a member of the
array.
The jq index origin is 0, and so in the following, the pegs and
holes are internally numbered from 0 to (N-1) inclusive. That is, we interpret a permutation, p, of 0 .. (N-1) as meaning that the i-th peg is numbered p[i], for i in 0 .. (N-1).
However the pretty-print function shows solutions using the 1-to-8 numbering scheme for pegs, and the A-to-H lettering scheme for holes.
'''Part 1: Generic functions'''
```jq
# Short-circuit determination of whether (a|condition)
# is true for all a in array:
def forall(array; condition):
def check:
. as $ix
| if $ix == (array|length) then true
elif (array[$ix] | condition) then ($ix + 1) | check
else false
end;
0 | check;
# permutations of 0 .. (n-1)
def permutations(n):
# Given a single array, generate a stream by inserting n at different positions:
def insert(m;n):
if m >= 0 then (.[0:m] + [n] + .[m:]), insert(m-1;n) else empty end;
if n==0 then []
elif n == 1 then [1]
else
permutations(n-1) | insert(n-1; n)
end;
# Count the number of items in a stream
def count(f): reduce f as $_ (0; .+1);
```
'''Part 2: The no-connections puzzle for N pegs and holes'''
```jq
# Generate a stream of solutions.
# Input should be the connections array, i.e. an array of [i,j] pairs;
# N is the number of pegs and holds.
def solutions(N):
def abs: if . < 0 then -. else . end;
# Is the proposed permutation (the input) ok?
def ok(connections):
. as $p
| forall( connections;
(($p[.[0]] - $p[.[1]])|abs) != 1 );
. as $connections | permutations(N) | select(ok($connections);
```
'''Part 3: The 8-peg no-connection puzzle'''
```jq
# The connectedness matrix:
# In this table, 0 represents "A", etc, and an entry [i,j]
# signifies that the holes with indices i and j are connected.
def connections:
[[0, 2], [0, 3], [0, 4],
[1, 3], [1, 4], [1, 5],
[6, 2], [6, 3], [6, 4],
[7, 3], [7, 4], [7, 5],
[2, 3], [3, 4], [4, 5]]
;
def solve:
connections | solutions(8);
# pretty-print a solution for the 8-peg puzzle
def pp:
def pegs: ["A", "B", "C", "D", "E", "F", "G", "H"];
. as $in
| ("
A B
/|\\ /|\\
/ | X | \\
/ |/ \\| \\
C - D - E - F
\\ |\\ /| /
\\ | X | /
\\|/ \\|/
G H
" | explode) as $board
| (pegs | map(explode)) as $letters
| $letters
| reduce range(0;length) as $i ($board; index($letters[$i]) as $ix | .[$ix] = $in[$i] + 48)
| implode;
```
'''Examples''':
```jq
# To print all the solutions:
# solve | pp
# To count the number of solutions:
# count(solve)
# jq 1.4 lacks facilities for harnessing generators,
# but the following will suffice here:
def one(f): reduce f as $s
(null; if . == null then $s else . end);
one(solve) | pp
```
{{out}}
```sh
$ jq -n -r -f no_connection.jq
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4
```
## Julia
```julia
using Combinatorics
const HOLES = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H']
const PEGS = [1, 2, 3, 4, 5, 6, 7, 8]
const EDGES = [('A', 'C'), ('A', 'D'), ('A', 'E'),
('B', 'D'), ('B', 'E'), ('B', 'F'),
('C', 'G'), ('C', 'D'), ('D', 'G'),
('D', 'E'), ('D', 'H'), ('E', 'F'),
('E', 'G'), ('E', 'H'), ('F', 'H')]
goodperm(p) = all(e->abs(p[e[1]-'A'+1] - p[e[2]-'A'+1]) > 1, EDGES)
goodplacements() = [p for p in permutations(PEGS) if goodperm(p)]
const BOARD = raw"""
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
"""
function printsolutions()
solutions = goodplacements()
println("Found $(length(solutions)) solutions.")
for soln in solutions
board = BOARD
for (i, n) in enumerate(soln)
board = replace(board, string('A' + i - 1) => string(n))
end
println(board); exit(1) # remove this exit for all solutions
end
end
printsolutions()
```
{{output}}
```txt
Found 16 solutions.
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
```
## Kotlin
{{trans|Go}}
```scala
// version 1.2.0
import kotlin.math.abs
// Holes A=0, B=1, …, H=7
// With connections:
const val conn = """
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
"""
val connections = listOf(
0 to 2, 0 to 3, 0 to 4, // A to C, D, E
1 to 3, 1 to 4, 1 to 5, // B to D, E, F
6 to 2, 6 to 3, 6 to 4, // G to C, D, E
7 to 3, 7 to 4, 7 to 5, // H to D, E, F
2 to 3, 3 to 4, 4 to 5 // C-D, D-E, E-F
)
// 'isValid' checks if the pegs are a valid solution.
// If the absolute difference between any pair of connected pegs is
// greater than one it is a valid solution.
fun isValid(pegs: IntArray): Boolean {
for ((a, b) in connections) {
if (abs(pegs[a] - pegs[b]) <= 1) return false
}
return true
}
fun swap(pegs: IntArray, i: Int, j: Int) {
val tmp = pegs[i]
pegs[i] = pegs[j]
pegs[j] = tmp
}
// 'solve' is a simple recursive brute force solver,
// it stops at the first found solution.
// It returns the solution, the number of positions tested,
// and the number of pegs swapped.
fun solve(): Triple {
val pegs = IntArray(8) { it + 1 }
var tests = 0
var swaps = 0
fun recurse(i: Int): Boolean {
if (i >= pegs.size - 1) {
tests++
return isValid(pegs)
}
// Try each remaining peg from pegs[i] onwards
for (j in i until pegs.size) {
swaps++
swap(pegs, i, j)
if (recurse(i + 1)) return true
swap(pegs, i, j)
}
return false
}
recurse(0)
return Triple(pegs, tests, swaps)
}
fun pegsAsString(pegs: IntArray): String {
val ca = conn.toCharArray()
for ((i, c) in ca.withIndex()) {
if (c in 'A'..'H') ca[i] = '0' + pegs[c - 'A']
}
return String(ca)
}
fun main(args: Array) {
val (p, tests, swaps) = solve()
println(pegsAsString(p))
println("Tested $tests positions and did $swaps swaps.")
}
```
{{out}}
```txt
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.
```
## M2000 Interpreter
Final Version, print all solutions (16 from 40320 permutations)
Press space bar to see solutions so far.
```M2000 Interpreter
Module no_connection_puzzle {
\\ Holes
Inventory Connections="A":="CDE","B":="DEF","C":="ADG", "D":="ABCEGH"
Append Connections, "E":="ABDFGH","F":="HEB", "G":="CDE","H":="DEF"
Inventory ToDelete, Solutions
\\ eliminate double connnections
con=each(Connections)
While con {
m$=eval$(con, con^)
c$=eval$(con)
If c$="*" Then continue
For i=1 to len(C$) {
d$=mid$(c$,i,1)
r$=Filter$(Connections$(d$), m$)
If r$<>"" Then {
Return connections, d$:=r$
} else {
If m$=connections$(d$) Then {
Return connections, d$:="*" : If not exist(todelete, d$) Then Append todelete, d$
}
}
}
}
con=each(todelete)
While con {
Delete Connections, eval$(con)
}
Inventory Holes
For i=0 to 7 : Append Holes, Chr$(65+i):=i : Next i
CheckValid=lambda Holes, Connections (a$, arr) -> {
val=Array(arr, Holes(a$))
con$=Connections$(a$)
res=True
For i=1 to Len(con$) {
If Abs(Array(Arr, Holes(mid$(con$,i,1)))-val)<2 Then res=False: Exit
}
=res
}
a=(1,2,3,4,5,6,7,8)
h=(,)
solution=(,)
done=False
counter=0
Print "Wait..."
P(h, a)
sol=Each(Solutions)
While sol {
Print "Solution:";sol^+1
Disp(Eval(Solutions))
aa$=Key$
}
Sub P(h, a)
If len(a)<=1 Then process(cons(h, a)) : Exit Sub
local b=cons(a)
For i=1 to len(b) {
b=cons(cdr(b),car(b))
P(cons(h,car(b)), cdr(b))
}
End Sub
Sub Process(a)
counter++
Print counter
If keypress(32) Then {
local sol=Each(Solutions)
aa$=Key$
While sol {
Print "Solution:";sol^+1
Disp(Eval(Solutions))
aa$=Key$
}
}
hole=each(Connections)
done=True
While hole {
If not CheckValid(Eval$(hole, hole^), a) Then done=False : Exit
}
If done Then Append Solutions, Len(Solutions):=a : Print a
End Sub
Sub Disp(a)
Print format$(" {0} {1}", array(a), array(a,1))
Print " /|\ /|\"
Print " / | X | \"
Print " / |/ \| \"
Print Format$("{0} - {1} - {2} - {3}", array(a,2),array(a,3), array(a,4), array(a,5))
Print " \ |\ /| /"
Print " \ | X | /"
Print " \|/ \|/"
Print Format$(" {0} {1}", array(a,6), array(a,7))
End Sub
}
no_connection_puzzle
```
{{out}}
```txt
3 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 6
## Mathematica
This one simply takes all permutations of the pegs and filters out invalid solutions.
```Mathematica
sol = Fold[
Select[#,
Function[perm, Abs[perm[[#2[[1]]]] - perm[[#2[[2]]]]] > 1]] &,
Permutations[
Range[8]], {{1, 3}, {1, 4}, {1, 5}, {2, 4}, {2, 5}, {2, 6}, {3,
4}, {3, 7}, {4, 5}, {4, 7}, {4, 8}, {5, 6}, {5, 7}, {5, 8}, {6,
8}}][[1]];
Print[StringForm[
" `` ``\n /|\\ /|\\\n / | X | \\\n / |/ \\| \\\n`` - `` \
- `` - ``\n \\ |\\ /| /\n \\ | X | /\n \\|/ \\|/\n `` ``",
Sequence @@ sol]];
```
{{out}}
```txt
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
```
## Perl
```perl
#!/usr/bin/perl
use strict;
use warnings;
my $gap = qr/.{3}/s;
find( <Solution 1
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Solution 2
3 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 6
Solution 3
3 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 5
Solution 4
3 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 4
Solution 5
4 3
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
6 5
Solution 6
4 5
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
6 3
Solution 7
4 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 6
Solution 8
4 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 5
Solution 9
5 3
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
6 4
Solution 10
5 4
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
6 3
Solution 11
5 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 6
Solution 12
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4
Solution 13
6 3
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
4 5
Solution 14
6 3
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
5 4
Solution 15
6 4
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
5 3
Solution 16
6 5
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
4 3
```
## Racket
```racket
#lang racket
;; Solve the no connection puzzle. Tim Brown Oct. 2014
;; absolute difference of a and b if they are both true
(define (and- a b) (and a b (abs (- a b))))
;; Finds the differences of all established connections in the network
(define (network-diffs (A #f) (B #f) (C #f) (D #f) (E #f) (F #f) (G #f) (H #f))
(list (and- A C) (and- A D) (and- A E)
(and- B D) (and- B E) (and- B F)
(and- C D) (and- C G)
(and- D E) (and- D G) (and- D H)
(and- E F) (and- E G) (and- E H)
(and- F G)))
;; Make sure there is “no connection” in the network N; return N if good
(define (good-network? N)
(and (for/and ((d (filter values (apply network-diffs N)))) (> d 1)) N))
;; possible optimisation is to reverse the arguments to network-diffs, reverse the return value from
;; this function and make this a cons but we're pretty quick here as it is.
(define (find-good-network pegs (n/w null))
(if (null? pegs) n/w
(for*/or ((p pegs))
(define n/w+ (append n/w (list p)))
(and (good-network? n/w+)
(find-good-network (remove p pegs =) n/w+)))))
(define (render-puzzle pzl)
(apply printf (regexp-replace* "O" #<_ then iterate
subs=subs + 1; !._.subs=__
end /*#*/
!._.0=subs /*assign the number of the node paths. */
end /*pegs*/
pegs=pegs-1 /*the number of pegs to be seated. */
_=' ' /*_ is used for indenting the output.*/
do a=1 for pegs; if ?('A') then iterate
do b=1 for pegs; if ?('B') then iterate
do c=1 for pegs; if ?('C') then iterate
do d=1 for pegs; if ?('D') then iterate
do e=1 for pegs; if ?('E') then iterate
do f=1 for pegs; if ?('F') then iterate
do g=1 for pegs; if ?('G') then iterate
do h=1 for pegs; if ?('H') then iterate
say _ 'a='a _ 'b='||b _ 'c='c _ 'd='d _ 'e='e _ 'f='f _ 'g='g _ 'h='h
cnt=cnt+1; if cnt==limit then leave a
end /*h*/
end /*g*/
end /*f*/
end /*e*/
end /*d*/
end /*c*/
end /*b*/
end /*a*/
say /*display a blank line to the terminal.*/
s= left('s', cnt\==1) /*handle the case of plurals (or not).*/
say 'found ' cnt " solution"s'.' /*display the number of solutions found*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
?: parse arg node; nn=value(node)
nH=nn+1
do cn=c2d('A') to c2d(node) - 1; if value( d2c(cn) )==nn then return 1
end /*cn*/ /* [↑] see if there any are duplicates.*/
nL=nn-1
do ch=1 for !.node.0 /* [↓] see if there any ¬= ±1 values.*/
$=!.node.ch; fn=value($) /*the node name and its current peg #.*/
if nL==fn | nH==fn then return 1 /*if ≡ ±1, then the node can't be used.*/
end /*ch*/ /* [↑] looking for suitable number. */
return 0 /*the subroutine arg value passed is OK.*/
```
'''output''' when using the default input:
```txt
a=3 b=4 c=7 d=1 e=8 f=2 g=5 h=6
found 1 solution.
```
'''output''' when using the input of: 999
```txt
a=3 b=4 c=7 d=1 e=8 f=2 g=5 h=6
a=3 b=5 c=7 d=1 e=8 f=2 g=4 h=6
a=3 b=6 c=7 d=1 e=8 f=2 g=4 h=5
a=3 b=6 c=7 d=1 e=8 f=2 g=5 h=4
a=4 b=3 c=2 d=8 e=1 f=7 g=6 h=5
a=4 b=5 c=2 d=8 e=1 f=7 g=6 h=3
a=4 b=5 c=7 d=1 e=8 f=2 g=3 h=6
a=4 b=6 c=7 d=1 e=8 f=2 g=3 h=5
a=5 b=3 c=2 d=8 e=1 f=7 g=6 h=4
a=5 b=4 c=2 d=8 e=1 f=7 g=6 h=3
a=5 b=4 c=7 d=1 e=8 f=2 g=3 h=6
a=5 b=6 c=7 d=1 e=8 f=2 g=3 h=4
a=6 b=3 c=2 d=8 e=1 f=7 g=4 h=5
a=6 b=3 c=2 d=8 e=1 f=7 g=5 h=4
a=6 b=4 c=2 d=8 e=1 f=7 g=5 h=3
a=6 b=5 c=2 d=8 e=1 f=7 g=4 h=3
found 16 solutions.
```
### annotated solutions
Usage note: if the '''limit''' (the 1st argument) is negative, a diagram (node graph) is shown.
```rexx
/*REXX program solves the "no-connection" puzzle (the puzzle has eight pegs). */
@abc='ABCDEFGHIJKLMNOPQRSTUVWXYZ'
parse arg limit . /*number of solutions wanted.*/ /* ╔═══════════════════════════╗ */
if limit=='' | limit=="." then limit=1 /* ║ A B ║ */
oLimit=limit; limit=abs(limit) /* ║ /│\ /│\ ║ */
@. = /* ║ / │ \/ │ \ ║ */
@.1 = 'A C D E' /* ║ / │ /\ │ \ ║ */
@.2 = 'B D E F' /* ║ / │/ \│ \ ║ */
@.3 = 'C A D G' /* ║ C────D────E────F ║ */
@.4 = 'D A B C E G' /* ║ \ │\ /│ / ║ */
@.5 = 'E A B D F H' /* ║ \ │ \/ │ / ║ */
@.6 = 'F B E H' /* ║ \ │ /\ │ / ║ */
@.7 = 'G C D E' /* ║ \│/ \│/ ║ */
@.8 = 'H D E F' /* ║ G H ║ */
cnt=0 /* ╚═══════════════════════════╝ */
do pegs=1 while @.pegs\==''; _=word(@.pegs, 1)
subs=0
do #=1 for words(@.pegs) -1 /*create list of node paths.*/
__=word(@.pegs, #+1); if __>_ then iterate
subs=subs + 1; !._.subs=__
end /*#*/
!._.0=subs /*assign the number of the node paths. */
end /*pegs*/
pegs=pegs - 1 /*the number of pegs to be seated. */
_=' ' /*_ is used for indenting the output. */
do a=1 for pegs; if ?('A') then iterate
do b=1 for pegs; if ?('B') then iterate
do c=1 for pegs; if ?('C') then iterate
do d=1 for pegs; if ?('D') then iterate
do e=1 for pegs; if ?('E') then iterate
do f=1 for pegs; if ?('F') then iterate
do g=1 for pegs; if ?('G') then iterate
do h=1 for pegs; if ?('H') then iterate
call showNodes
cnt=cnt+1; if cnt==limit then leave a
end /*h*/
end /*g*/
end /*f*/
end /*e*/
end /*d*/
end /*c*/
end /*b*/
end /*a*/
say /*display a blank line to the terminal.*/
s=left('s', cnt\==1) /*handle the case of plurals (or not).*/
say 'found ' cnt " solution"s'.' /*display the number of solutions found*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
?: parse arg node; nn=value(node)
nH=nn+1
do cn=c2d('A') to c2d(node)-1; if value( d2c(cn) )==nn then return 1
end /*cn*/ /* [↑] see if there're any duplicates.*/
nL=nn-1
do ch=1 for !.node.0 /* [↓] see if there any ¬= ±1 values.*/
$=!.node.ch; fn=value($) /*the node name and its current peg #.*/
if nL==fn | nH==fn then return 1 /*if ≡ ±1, then the node can't be used.*/
end /*ch*/ /* [↑] looking for suitable number. */
return 0 /*the subroutine arg value passed is OK*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
showNodes: _=left('', 5) /*_ is used for padding the output. */
show=0 /*indicates no graph has been found yet*/
do box=1 for sourceline() while oLimit<0 /*Negative? Then display the diagram. */
xw=sourceline(box) /*get a source line of this program. */
p2=lastpos('*', xw) /*the position of last asterisk.*/
p1=lastpos('*', xw, max(1, p2-1) ) /* " " " penultimate " */
if pos('╔', xw)\==0 then show=1 /*Have found the top-left box corner ? */
if \show then iterate /*Not found? Then skip this line. */
xb=substr(xw, p1+1, p2-p1-2) /*extract the "box" part of line. */
xt=xb /*get a working copy of the box. */
do jx=1 for pegs /*do a substitution for all the pegs. */
@=substr(@abc, jx, 1) /*get the name of the peg (A ──► Z). */
xt=translate(xt,value(@),@) /*substitute the peg name with a value.*/
end /*jx*/ /* [↑] graph is limited to 26 nodes.*/
say _ xb _ _ xt /*display one line of the graph. */
if pos('╝', xw)\==0 then return /*Is this last line of graph? Then stop*/
end /*box*/
say _ 'a='a _ 'b='||b _ 'c='c _ 'd='d _ ' e='e _ 'f='f _ 'g='g _ 'h='h
return
```
'''output''' when using the input of: -3
```txt
╔═══════════════════════════╗ ╔═══════════════════════════╗
║ A B ║ ║ 3 4 ║
║ /│\ /│\ ║ ║ /│\ /│\ ║
║ / │ \/ │ \ ║ ║ / │ \/ │ \ ║
║ / │ /\ │ \ ║ ║ / │ /\ │ \ ║
║ / │/ \│ \ ║ ║ / │/ \│ \ ║
║ C────D────E────F ║ ║ 7────1────8────2 ║
║ \ │\ /│ / ║ ║ \ │\ /│ / ║
║ \ │ \/ │ / ║ ║ \ │ \/ │ / ║
║ \ │ /\ │ / ║ ║ \ │ /\ │ / ║
║ \│/ \│/ ║ ║ \│/ \│/ ║
║ G H ║ ║ 5 6 ║
╚═══════════════════════════╝ ╚═══════════════════════════╝
╔═══════════════════════════╗ ╔═══════════════════════════╗
║ A B ║ ║ 3 5 ║
║ /│\ /│\ ║ ║ /│\ /│\ ║
║ / │ \/ │ \ ║ ║ / │ \/ │ \ ║
║ / │ /\ │ \ ║ ║ / │ /\ │ \ ║
║ / │/ \│ \ ║ ║ / │/ \│ \ ║
║ C────D────E────F ║ ║ 7────1────8────2 ║
║ \ │\ /│ / ║ ║ \ │\ /│ / ║
║ \ │ \/ │ / ║ ║ \ │ \/ │ / ║
║ \ │ /\ │ / ║ ║ \ │ /\ │ / ║
║ \│/ \│/ ║ ║ \│/ \│/ ║
║ G H ║ ║ 4 6 ║
╚═══════════════════════════╝ ╚═══════════════════════════╝
╔═══════════════════════════╗ ╔═══════════════════════════╗
║ A B ║ ║ 3 6 ║
║ /│\ /│\ ║ ║ /│\ /│\ ║
║ / │ \/ │ \ ║ ║ / │ \/ │ \ ║
║ / │ /\ │ \ ║ ║ / │ /\ │ \ ║
║ / │/ \│ \ ║ ║ / │/ \│ \ ║
║ C────D────E────F ║ ║ 7────1────8────2 ║
║ \ │\ /│ / ║ ║ \ │\ /│ / ║
║ \ │ \/ │ / ║ ║ \ │ \/ │ / ║
║ \ │ /\ │ / ║ ║ \ │ /\ │ / ║
║ \│/ \│/ ║ ║ \│/ \│/ ║
║ G H ║ ║ 4 5 ║
╚═══════════════════════════╝ ╚═══════════════════════════╝
found 3 solutions.
```
## Ruby
Be it Golden Frogs jumping on trancendental lilly pads, or a Knight on a board, or square pegs into round holes this is essentially a Hidato Like Problem, so I use [http://rosettacode.org/wiki/Solve_a_Hidato_puzzle#With_Warnsdorff HLPSolver]:
```ruby
# Solve No Connection Puzzle
#
# Nigel_Galloway
# October 6th., 2014
require 'HLPSolver'
ADJACENT = [[0,0]]
A,B,C,D,E,F,G,H = [0,1],[0,2],[1,0],[1,1],[1,2],[1,3],[2,1],[2,2]
board1 = < !links(i).forall(peg => math.abs(pegs(i) - peg) == 1))
private def printResult(pegs: Seq[Int]) = {
println(f"${pegs(0)}%3d${pegs(1)}%2d")
println(f"${pegs(2)}%1d${pegs(3)}%2d${pegs(4)}%2d${pegs(5)}%2d")
println(f"${pegs(6)}%3d${pegs(7)}%2d")
}
printResult(genRandom.dropWhile(!notSolved(links, _)).head)
}
```
## Tcl
{{tcllib|struct::list}}
```tcl
package require Tcl 8.6
package require struct::list
proc haveAdjacent {a b c d e f g h} {
expr {
[edge $a $c] ||
[edge $a $d] ||
[edge $a $e] ||
[edge $b $d] ||
[edge $b $e] ||
[edge $b $f] ||
[edge $c $d] ||
[edge $c $g] ||
[edge $d $e] ||
[edge $d $g] ||
[edge $d $h] ||
[edge $e $f] ||
[edge $e $g] ||
[edge $e $h] ||
[edge $f $h]
}
}
proc edge {x y} {
expr {abs($x-$y) == 1}
}
set layout [string trim {
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
} \n]
struct::list foreachperm p {1 2 3 4 5 6 7 8} {
if {![haveAdjacent {*}$p]} {
puts [string map [join [
lmap name {A B C D E F G H} val $p {list $name $val}
]] $layout]
break
}
}
```
{{out}}
```txt
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
```
## XPL0
```XPL0
include c:\cxpl\codes;
int Hole, Max, I;
char Box(8), Str;
def A, B, C, D, E, F, G, H;
[for Hole:= 0 to 7 do Box(Hole):= Hole+1;
Max:= 7;
while abs(Box(D)-Box(A)) < 2 or abs(Box(D)-Box(C)) < 2 or
abs(Box(D)-Box(G)) < 2 or abs(Box(D)-Box(E)) < 2 or
abs(Box(A)-Box(C)) < 2 or abs(Box(C)-Box(G)) < 2 or
abs(Box(G)-Box(E)) < 2 or abs(Box(E)-Box(A)) < 2 or
abs(Box(E)-Box(B)) < 2 or abs(Box(E)-Box(H)) < 2 or
abs(Box(E)-Box(F)) < 2 or abs(Box(B)-Box(D)) < 2 or
abs(Box(D)-Box(H)) < 2 or abs(Box(H)-Box(F)) < 2 or
abs(Box(F)-Box(B)) < 2 do
loop [I:= Box(0); \next permutation
for Hole:= 0 to Max-1 do Box(Hole):= Box(Hole+1);
Box(Max):= I;
if I # Max+1 then [Max:= 7; quit]
else Max:= Max-1];
Str:= "
# #
/|\ /|\
/ | X | \
/ |/ \| \
# - # - # - #
\ |\ /| /
\ | X | /
\|/ \|/
# #
";
Hole:= 0; I:= 0;
repeat if Str(I)=^# then [Str(I):= Box(Hole)+^0; Hole:= Hole+1];
I:= I+1;
until Hole = 8;
Text(0, Str);
]
```
{{out}}
```txt
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4
```
## zkl
{{trans|D}}
```zkl
const PegA=0, PegB=1, PegC=2, PegD=3, PegE=4, PegF=5, PegG=6, PegH=7;
connections:=T(
T(PegA, PegC), T(PegA, PegD), T(PegA, PegE),
T(PegB, PegD), T(PegB, PegE), T(PegB, PegF),
T(PegC, PegD), T(PegD, PegE), T(PegE, PegF),
T(PegG, PegC), T(PegG, PegD), T(PegG, PegE),
T(PegH, PegD), T(PegH, PegE), T(PegH, PegF) );
CZ:=connections.len();
#<<< // Use "raw" string in a "here doc" so \ isn't a quote char
board:=
0'$ A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H$;
#<<< // end "here doc"
perm:=T(PegA,PegB,PegC,PegD,PegE,PegF,PegG,PegH); // Peg[8]
foreach p in (Utils.Helpers.permuteW(perm)){ // permutation iterator
if(connections.filter1('wrap([(a,b)]){ (p[a] - p[b]).abs()<=1 })) continue;
board.translate("ABCDEFGH",p.apply('+(1)).concat()).println();
break; // comment out to see all 16 solutions
}
```
The filter1 method stops on the first True, so it acts like a conditional or.
{{out}}
```txt
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4
```