⚠️ Warning: This is a draft ⚠️

This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.

If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.

A [[wp:Munchausen number|Munchausen number]] is a natural number ''n'' the sum of whose digits (in base 10), each raised to the power of itself, equals ''n''.

('''Munchausen''' is also spelled: '''Münchhausen'''.)

For instance: 3435 = 33 + 44 + 33 + 55

;Task Find all Munchausen numbers between 1 and 5000

;Also see: :* The OEIS entry: [[oeis:A046253| A046253]] :* The Wikipedia entry: [[wp:Perfect_digit-to-digit_invariant| Perfect digit-to-digit invariant, redirected from ''Munchausen Number'']]

## 360 Assembly

*        Munchausen numbers        16/03/2019
MUNCHAU  CSECT
USING  MUNCHAU,R12        base register
LR     R12,R15            set addressability
L      R3,=F'5000'        for do i=1 to 5000
LA     R6,1               i=1
LOOPI    SR     R10,R10              s=0
LR     R0,R6                ii=i
LA     R11,4                for do j=1 to 4
LA     R7,P10               j=1
LOOPJ    L      R8,0(R7)               d=p10(j)
LR     R4,R0                  ii
SRDA   R4,32                  ~
DR     R4,R8                  (n,r)=ii/d
SLA    R5,2                   ~
L      R1,POW(R5)             pow(n+1)
AR     R10,R1                 s=s+pow(n+1)
LR     R0,R4                  ii=r
LA     R7,4(R7)               j++
BCT    R11,LOOPJ            enddo j
CR     R10,R6               if s=i
BNE    SKIP                 then
XDECO  R6,PG                  edit i
XPRNT  PG,L'PG                print i
SKIP     LA     R6,1(R6)             i++
BCT    R3,LOOPI           enddo i
POW      DC     F'0',F'1',F'4',F'27',F'256',F'3125',4F'0'
P10      DC     F'1000',F'100',F'10',F'1'
PG       DC     CL12' '            buffer
REGEQU
END    MUNCHAU

{{out}}

1
3435

## ALGOL 68

# Find Munchausen Numbers between 1 and 5000                                        #
# note that 6^6 is 46 656 so we only need to consider numbers consisting of 0 to 5   #

# table of Nth powers - note 0^0 is 0 for Munchausen numbers, not 1                 #
[]INT nth power = ([]INT( 0, 1, 2 * 2, 3 * 3 * 3, 4 * 4 * 4 * 4, 5 * 5 * 5 * 5 * 5 ))[ AT 0 ];

INT d1 := 0; INT d1 part := 0;
INT d2 := 0; INT d2 part := 0;
INT d3 := 0; INT d3 part := 0;
INT d4 := 1;
WHILE d1 < 6 DO
INT number           = d1 part + d2 part + d3 part + d4;
INT digit power sum := nth power[ d1 ]
+ nth power[ d2 ]
+ nth power[ d3 ]
+ nth power[ d4 ];
IF digit power sum = number THEN
print( ( whole( number, 0 ), newline ) )
FI;
d4 +:= 1;
IF d4 > 5 THEN
d4       := 0;
d3      +:= 1;
d3 part +:= 10;
IF d3 > 5 THEN
d3       := 0;
d3 part  := 0;
d2      +:= 1;
d2 part +:= 100;
IF d2 > 5 THEN
d2       := 0;
d2 part  := 0;
d1      +:= 1;
d1 part +:= 1000;
FI
FI
FI
OD

{{out}}

1
3435

Alternative that finds all 4 Munchausen numbers. As noted by the Pascal sample, we only need to consider one arrangement of the digits of each number (e.g. we only need to consider 3345, not 3435, 3453, etc.). This also relies on the non-standard 0^0 = 0.

# Find all Munchausen numbers - note 11*(9^9) has only 10 digits so there are no    #
# Munchausen numbers with 11+ digits                                                #
# table of Nth powers - note 0^0 is 0 for Munchausen numbers, not 1                 #
[]INT nth power = ([]INT( 0, 1, 2 ^ 2, 3 ^ 3, 4 ^ 4, 5 ^ 5, 6 ^ 6, 7 ^ 7, 8 ^ 8, 9 ^ 9 ) )[ AT 0 ];

[       ]INT z count  = []INT( ( 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ) )[ AT 0 ];
[ 0 : 9 ]INT d count := z count;

# as the digit power sum is independent of the order of the digits, we need only    #
# consider one arrangement of each possible combination of digits                   #
FOR d1 FROM 0 TO 9 DO
FOR d2 FROM 0 TO d1 DO
FOR d3 FROM 0 TO d2 DO
FOR d4 FROM 0 TO d3 DO
FOR d5 FROM 0 TO d4 DO
FOR d6 FROM 0 TO d5 DO
FOR d7 FROM 0 TO d6 DO
FOR d8 FROM 0 TO d7 DO
FOR d9 FROM 0 TO d8 DO
FOR da FROM 0 TO d9 DO
LONG INT digit power sum  := nth power[ d1 ] + nth power[ d2 ];
digit power sum          +:= nth power[ d3 ] + nth power[ d4 ];
digit power sum          +:= nth power[ d5 ] + nth power[ d6 ];
digit power sum          +:= nth power[ d7 ] + nth power[ d8 ];
digit power sum          +:= nth power[ d9 ] + nth power[ da ];
# count the occurrences of each digit (including leading zeros #
d count        := z count;
d count[ d1 ] +:= 1; d count[ d2 ] +:= 1; d count[ d3 ] +:= 1;
d count[ d4 ] +:= 1; d count[ d5 ] +:= 1; d count[ d6 ] +:= 1;
d count[ d7 ] +:= 1; d count[ d8 ] +:= 1; d count[ d9 ] +:= 1;
d count[ da ] +:= 1;
# subtract the occurrences of each digit in the power sum      #
# (also including leading zeros) - if all counts drop to 0 we  #
# have a Munchausen number                                     #
LONG INT number        := digit power sum;
INT      leading zeros := 10;
WHILE number > 0 DO
d count[ SHORTEN ( number MOD 10 ) ] -:= 1;
leading zeros -:= 1;
number OVERAB 10
OD;
d count[ 0 ] -:= leading zeros;
IF  d count[ 0 ] = 0 AND d count[ 1 ] = 0 AND d count[ 2 ] = 0
AND d count[ 3 ] = 0 AND d count[ 4 ] = 0 AND d count[ 5 ] = 0
AND d count[ 6 ] = 0 AND d count[ 7 ] = 0 AND d count[ 8 ] = 0
AND d count[ 9 ] = 0
THEN
print( ( digit power sum, newline ) )
FI
OD
OD
OD
OD
OD
OD
OD
OD
OD
OD

{{out}}

+0
+1
+3435
+438579088

## ALGOL W

{{Trans|ALGOL 68}}

% Find Munchausen Numbers between 1 and 5000                                         %
% note that 6^6 is 46 656 so we only need to consider numbers consisting of 0 to 5   %
begin

% table of nth Powers - note 0^0 is 0 for Munchausen numbers, not 1              %
integer array nthPower( 0 :: 5 );
integer d1, d2, d3, d4, d1Part, d2Part, d3Part;
nthPower( 0 ) := 0;             nthPower( 1 ) := 1;
nthPower( 2 ) := 2 * 2;         nthPower( 3 ) := 3 * 3 * 3;
nthPower( 4 ) := 4 * 4 * 4 * 4; nthPower( 5 ) := 5 * 5 * 5 * 5 * 5;
d1 := d2 := d3 := d1Part := d2Part := d3Part := 0;
d4 := 1;
while d1 < 6 do begin
integer number, digitPowerSum;
number        := d1Part + d2Part + d3Part + d4;
digitPowerSum := nthPower( d1 )
+ nthPower( d2 )
+ nthPower( d3 )
+ nthPower( d4 );
if digitPowerSum = number then begin
write( i_w := 1, number )
end;
d4 := d4 + 1;
if d4 > 5 then begin
d4     := 0;
d3     := d3 + 1;
d3Part := d3Part + 10;
if d3 > 5 then begin
d3     := 0;
d3Part := 0;
d2     := d2 + 1;
d2Part := d2Part + 100;
if d2 > 5 then begin
d2     := 0;
d2Part := 0;
d1     := d1 + 1;
d1Part := d1Part + 1000;
end
end
end
end

end.

{{out}}

1
3435

## AppleScript

-- MUNCHAUSEN NUMBER ? -------------------------------------------------------

-- isMunchausen :: Int -> Bool
on isMunchausen(n)

-- digitPowerSum :: Int -> Character -> Int
script digitPowerSum
on |λ|(a, c)
set d to c as integer
a + (d ^ d)
end |λ|
end script

(class of n is integer) and ¬
foldl(digitPowerSum, 0, characters of (n as string)) = n

end isMunchausen

-- TEST ----------------------------------------------------------------------
on run

filter(isMunchausen, enumFromTo(1, 5000))

--> {1, 3435}

end run

-- GENERIC FUNCTIONS ---------------------------------------------------------

-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m > n then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo

-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter

-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn

{{Out}}

{1, 3435}

## AWK

# syntax: GAWK -f MUNCHAUSEN_NUMBERS.AWK
BEGIN {
for (i=1; i<=5000; i++) {
sum = 0
for (j=1; j<=length(i); j++) {
digit = substr(i,j,1)
sum += digit ^ digit
}
if (i == sum) {
printf("%d\n",i)
}
}
exit(0)
}

{{out}}

1
3435

## BASIC

This should need only minimal modification to work with any old-style BASIC that supports user-defined functions. The call to INT in line 10 is needed because the exponentiation operator may return a (floating-point) value that is slightly too large.

10 DEF FN P(X)=INT(X^X*SGN(X))
20 FOR I=0 TO 5
30 FOR J=0 TO 5
40 FOR K=0 TO 5
50 FOR L=0 TO 5
60 M=FN P(I)+FN P(J)+FN P(K)+FN P(L)
70 N=1000*I+100*J+10*K+L
80 IF M=N AND M>0 THEN PRINT M
90 NEXT L
100 NEXT K
110 NEXT J
120 NEXT I

{{out}}

1
3435

=

## Sinclair ZX81 BASIC

= Works with 1k of RAM. The word FAST in line 10 shouldn't be taken too literally. We don't have DEF FN, so the expression for exponentiation-where-zero-to-the-power-zero-equals-zero is written out inline.

10 FAST
20 FOR I=0 TO 5
30 FOR J=0 TO 5
40 FOR K=0 TO 5
50 FOR L=0 TO 5
60 LET M=INT (I**I*SGN I+J**J*SGN J+K**K*SGN K+L**L*SGN L)
70 LET N=1000*I+100*J+10*K+L
80 IF M=N AND M>0 THEN PRINT M
90 NEXT L
100 NEXT K
110 NEXT J
120 NEXT I
130 SLOW

{{out}}

1
3435

## BBC BASIC

munchausen
FOR i% = 0 TO 5
FOR j% = 0 TO 5
FOR k% = 0 TO 5
FOR l% = 0 TO 5
m% = FNexp(i%) + FNexp(j%) + FNexp(k%) + FNexp(l%)
n% = 1000 * i% + 100 * j% + 10 * k% + l%
IF m% = n% AND m% > 0 THEN PRINT m%
NEXT
NEXT
NEXT
NEXT
END
:
DEF FNexp(x%)
IF x% = 0 THEN
= 0
ELSE
= x% ^ x%

{{out}}

1
3435

## C

Adapted from Zack Denton's code posted on [https://zach.se/munchausen-numbers-and-how-to-find-them/ Munchausen Numbers and How to Find Them].

#include <stdio.h>
#include <math.h>

int main() {
for (int i = 1; i < 5000; i++) {
// loop through each digit in i
// e.g. for 1000 we get 0, 0, 0, 1.
int sum = 0;
for (int number = i; number > 0; number /= 10) {
int digit = number % 10;
// find the sum of the digits
// raised to themselves
sum += pow(digit, digit);
}
if (sum == i) {
// the sum is equal to the number
// itself; thus it is a
// munchausen number
printf("%i\n", i);
}
}
return 0;
}

{{out}}

1
3435

## C#

Func<char, int> toInt = c => c-'0';

foreach (var i in Enumerable.Range(1,5000)
.Where(n => n == n.ToString()
.Sum(x => Math.Pow(toInt(x), toInt(x)))))
Console.WriteLine(i);

{{out}}

1
3435

### Faster version

{{Trans|Kotlin}}

using System;

namespace Munchhausen
{
class Program
{
static readonly long[] cache = new long[10];

static void Main()
{
// Allow for 0 ^ 0 to be 0
for (int i = 1; i < 10; i++)
{
cache[i] = (long)Math.Pow(i, i);
}

for (long i = 0L; i <= 500_000_000L; i++)
{
if (IsMunchhausen(i))
{
Console.WriteLine(i);
}
}
}

private static bool IsMunchhausen(long n)
{
long sum = 0, nn = n;
do
{
sum += cache[(int)(nn % 10)];
if (sum > n)
{
return false;
}
nn /= 10;
} while (nn > 0);

return sum == n;
}
}
}
0
1
3435
438579088

### Faster version alternate

{{trans|Visual Basic .NET}} Search covers all 11 digit numbers (as pointed out elsewhere, 11*(9^9) has only 10 digits, so there are no Munchausen numbers with 11+ digits), not just the first half of the 9 digit numbers. Computation time is under 1.5 seconds.

using System;

static class Program
{
public static void Main()
{
long sum, ten1 = 0, ten2 = 10; byte [] num; int [] pow = new int[10];
int i, j, n, n1, n2, n3, n4, n5, n6, n7, n8, n9, s2, s3, s4, s5, s6, s7, s8;
for (i = 1; i <= 9; i++) { pow[i] = i; for (j = 2; j <= i; j++) pow[i] *= i; }
for (n = 1; n <= 11; n++) { for (n9 = 0; n9 <= n; n9++) { for (n8 = 0; n8 <= n - n9; n8++) {
for (n7 = 0; n7 <= n - (s8 = n9 + n8); n7++) { for (n6 = 0; n6 <= n - (s7 = s8 + n7); n6++) {
for (n5 = 0; n5 <= n - (s6 = s7 + n6); n5++) { for (n4 = 0; n4 <= n - (s5 = s6 + n5); n4++) {
for (n3 = 0; n3 <= n - (s4 = s5 + n4); n3++) { for (n2 = 0; n2 <= n - (s3 = s4 + n3); n2++) {
for (n1 = 0; n1 <= n - (s2 = s3 + n2); n1++) {
sum = n1 * pow[1] + n2 * pow[2] + n3 * pow[3] + n4 * pow[4] +
n5 * pow[5] + n6 * pow[6] + n7 * pow[7] + n8 * pow[8] + n9 * pow[9];
if (sum < ten1 || sum >= ten2) continue;
num = new byte[10]; foreach (char ch in sum.ToString()) num[Convert.ToByte(ch) - 48] += 1;
if (n - (s2 + n1) == num[0] && n1 == num[1] && n2 == num[2]
&& n3 == num[3] && n4 == num[4] && n5 == num[5] && n6 == num[6]
&& n7 == num[7] && n8 == num[8] && n9 == num[9]) Console.WriteLine(sum);
} } } } } } } } }
ten1 = ten2; ten2 *= 10;
}
}
}

{{out}}

0
1
3435
438579088

## C++

#include <math.h>
#include <iostream>

unsigned pwr[10];

unsigned munch( unsigned i ) {
unsigned sum = 0;
while( i ) {
sum += pwr[(i % 10)];
i /= 10;
}
return sum;
}

int main( int argc, char* argv[] ) {
for( int i = 0; i < 10; i++ )
pwr[i] = (unsigned)pow( (float)i, (float)i );
std::cout << "Munchausen Numbers\n
### ============
\n";
for( unsigned i = 1; i < 5000; i++ )
if( i == munch( i ) ) std::cout << i << "\n";
return 0;
}

{{out}}

Munchausen Numbers

### ============

1
3435

## Clojure

(ns async-example.core
(:require [clojure.math.numeric-tower :as math])
(:use [criterium.core])
(:gen-class))

(defn get-digits [n]
" Convert number of a list of digits  (e.g. 545 -> ((5), (4), (5)) "
(map #(Integer/valueOf (str %)) (String/valueOf n)))

(defn sum-power [digits]
" Convert digits such as abc... to a^a + b^b + c^c ..."
(let [digits-pwr (fn [n]
(apply + (map #(math/expt % %) digits)))]
(digits-pwr digits)))

(defn find-numbers [max-range]
" Filters for Munchausen numbers "
(->>
(range 1 (inc max-range))
(filter #(= (sum-power (get-digits %)) %))))

(println (find-numbers 5000))

{{Output}}

(1 3435)

## Common Lisp

;;; check4munch maximum &optional b
;;; Return a list with all Munchausen numbers less then or equal to maximum.
;;; Checks are done in base b (<=10, dpower is the limiting factor here).
(defun check4munch (maximum &optional (base 10))
(do ((n 1 (1+ n))
(result NIL (if (munchp n base) (cons n result) result)))
((> n maximum)
(nreverse result))))

;;;
;;; munchp n &optional b
;;; Return T if n is a Munchausen number in base b.
(defun munchp (n &optional (base 10))
(if (= n (apply #'+ (mapcar #'dpower (n2base n base)))) T NIL))

;;; dpower d
;;; Returns d^d. I.e. the digit to the power of itself.
;;; 0^0 is set to 0. For discussion see e.g. the wikipedia entry.
;;; This function is mainly performance optimization.
(defun dpower (d)
(aref #(0 1 4 27 256 3125 45556 823543 16777216 387420489) d))

;;; divmod a b
;;; Return (q,k) such that a = b*q + k and k>=0.
(defun divmod (a b)
(let ((foo (mod a b)))
(list (/ (- a foo) b) foo)))

;;; n2base n &optional b
;;; Return a list with the digits of n in base b representation.
(defun n2base (n &optional (base 10) (digits NIL))
(if (zerop n) digits
(let ((dm (divmod n base)))
(n2base (car dm) base (cons (cadr dm) digits)))))

{{Out}}

> (check4munch 5000)
(1 3435)
> (munchp 438579088)
T

## D

{{trans|C}}

import std.stdio;

void main() {
for (int i=1; i<5000; i++) {
// loop through each digit in i
// e.g. for 1000 we get 0, 0, 0, 1.
int sum = 0;
for (int number=i; number>0; number/=10) {
int digit = number % 10;
// find the sum of the digits
// raised to themselves
sum += digit ^^ digit;
}
if (sum == i) {
// the sum is equal to the number
// itself; thus it is a
// munchausen number
writeln(i);
}
}
}

{{out}}

1
3435

## Dc

Needs a modern Dc due to ~. Use S1S2l2l1/L2L1% instead of ~ to run it in older Dcs.

[ O ~ S! d 0!=M L! d ^ + ] sM
[p] sp
[z d d lM x =p z 5001>L ] sL
lL x

Cosmetic: The stack is dirty after execution. The loop L needs a fix if that is a problem.

## Elixir

defmodule Munchausen do
@pow  for i <- 0..9, into: %{}, do: {i, :math.pow(i,i) |> round}

def number?(n) do
n == Integer.digits(n) |> Enum.reduce(0, fn d,acc -> @pow[d] + acc end)
end
end

Enum.each(1..5000, fn i ->
if Munchausen.number?(i), do: IO.puts i
end)

{{out}}

1
3435

## Factor

USING: kernel math.functions math.ranges math.text.utils
prettyprint sequences ;
IN: rosetta-code.munchausen

: munchausen? ( n -- ? )
dup 1 digit-groups dup [ ^ ] 2map sum = ;

: main ( -- ) 5000 [1,b] [ munchausen? ] filter . ;

MAIN: main

{{out}}

V{ 1 3435 }

In [http://wiki.formulae.org/Munchausen_numbers this] page you can see the solution of this task.

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text ([http://wiki.formulae.org/Editing_F%C5%8Drmul%C3%A6_expressions more info]). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.

The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.

## Forth

{{works with|GNU Forth|0.7.0}}

: dig.num                                       \ returns input number and the number of its digits ( n -- n n1 )
dup
0 swap
begin
swap 1 + swap
dup 10 >= while
10 /
repeat
drop ;

: to.self                                        \ returns input number raised to the power of itself ( n -- n^n  )
dup 1 = if drop 1 else                   \ positive numbers only, zero and negative returns zero
dup 0 <= if drop 0 else
dup
1 do
dup
loop
dup
1 do
*
loop
then then ;

: ten.to			                    \ ( n -- 10^n ) returns 1 for zero and negative
dup 0 <= if drop 1 else
dup 1 = if drop 10 else
10 swap
1 do
10 *
loop then then ;

: zero.divmod                                       \ /mod that returns zero if number is zero
dup
0 = if drop 0
else /mod
then ;

: split.div                                         \ returns input number and its digits ( n -- n n1 n2 n3....)
dup 10 < if dup 0 else		     \ duplicates single digit numbers adds 0 for add.pow
dig.num			             \ provides number of digits
swap dup rot dup 1 - ten.to swap           \ stack juggling, ten raised to number of digits - 1...
1 do                                       \ ... is the needed divisor, counter on top and ...
dup rot swap zero.divmod swap rot 10 /     \ ...division loop
loop drop then ;

: add.pow	  				     \ raises each number on the stack except last one to ...
to.self                                    \ ...the power of itself and adds them
depth					     \ needs at least 3 numbers on the stack
2 do
swap to.self +
loop ;

: check.num

: munch.num                                         \ ( n -- ) displays Munchausen numbers between 1 and n
1 +
page
1 do
i check.num = if i . cr
then loop ;

{{out}}

1
3435
ok

## Fortran

{{trans|360 Assembly}}

### Fortran IV

C MUNCHAUSEN NUMBERS - FORTRAN IV
DO 2 I=1,5000
IS=0
II=I
DO 1 J=1,4
ID=10**(4-J)
N=II/ID
IR=MOD(II,ID)
IF(N.NE.0) IS=IS+N**N
1       II=IR
2     IF(IS.EQ.I) WRITE(*,*) I
END

{{out}}

1
3435

### Fortran 77

! MUNCHAUSEN NUMBERS - FORTRAN 77
DO I=1,5000
IS=0
II=I
DO J=1,4
ID=10**(4-J)
N=II/ID
IR=MOD(II,ID)
IF(N.NE.0) IS=IS+N**N
II=IR
END DO
IF(IS.EQ.I) WRITE(*,*) I
END DO
END

{{out}}

1
3435

## FreeBASIC

### Version 1

' FB 1.05.0 Win64
' Cache n ^ n for the digits 1 to 9
' Note than 0 ^ 0 specially treated as 0 (not 1) for this purpose
Dim Shared powers(1 To 9) As UInteger
For i As UInteger = 1 To 9
Dim power As UInteger = i
For j As UInteger = 2 To i
power *= i
Next j
powers(i) = power
Next i

Function isMunchausen(n As UInteger) As Boolean
Dim p As UInteger = n
Dim As UInteger digit, sum
While p > 0
digit = p Mod 10
If digit > 0 Then sum += powers(digit)
p \= 10
Wend
Return n = sum
End Function

Print "The Munchausen numbers between 0 and 500000000 are : "
For i As UInteger = 0 To 500000000
If isMunchausen(i) Then Print i
Next

Print
Print "Press any key to quit"

Sleep

{{out}}

The Munchausen numbers between 0 and 500000000 are :
0
1
3435
438579088

### Version 2

' version 12-10-2017
' compile with: fbc -s console

Dim As UInteger i, j, n, sum, ten1, ten2 = 10
Dim As UInteger n0, n1, n2, n3, n4, n5, n6, n7, n8, n9
Dim As UInteger     s1, s2, s3, s4, s5, s6, s7, s8
Dim As UInteger pow(9), num()
Dim As String number

For i = 1 To 9
pow(i) = i
For j = 2 To i
pow(i) *= i
Next
Next

For n = 1 To 11
For n9 = 0 To n
For n8 = 0 To n - n9
s8 = n9 + n8
For n7 = 0 To n - s8
s7 = s8 + n7
For n6 = 0 To n - s7
s6 = s7 + n6
For n5 = 0 To n - s6
s5 = s6 + n5
For n4 = 0 To n - s5
s4 = s5 + n4
For n3 = 0 To n - s4
s3 = s4 + n3
For n2 = 0 To n - s3
s2 = s3 + n2
For n1 = 0 To n - s2
n0 = n - (s2 + n1)
sum = n1 * pow(1) + n2 * pow(2) + n3 * pow(3) + _
n4 * pow(4) + n5 * pow(5) + n6 * pow(6) + _
n7 * pow(7) + n8 * pow(8) + n9 * pow(9)
If sum < ten1 Or sum >= ten2 Then Continue For
ReDim num(9) : number = Str(sum)
For i = 0 To n -1
j = number[i] -48
num(j) += 1
Next i
If n0 = num(0) AndAlso n1 = num(1) AndAlso n2 = num(2) AndAlso _
n3 = num(3) AndAlso n4 = num(4) AndAlso n5 = num(5) AndAlso _
n6 = num(6) AndAlso n7 = num(7) AndAlso n8 = num(8) AndAlso _
n9 = num(9) Then Print sum
Next n1
Next n2
Next n3
Next n4
Next n5
Next n6
Next n7
Next n8
Next n9
ten1 = ten2
ten2 *= 10
Next n

' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End

{{out}}

0
1
3435
438579088

let toFloat x = x |> int |> fun n -> n - 48 |> float
let power x = toFloat x ** toFloat x |> int
let isMunchausen n = n = (string n |> Seq.map char |> Seq.map power |> Seq.sum)

printfn "%A" ([1..5000] |> List.filter isMunchausen)

{{out}}

[1; 3435]

## Go

{{trans|Kotlin}}

package main

import(
"fmt"
"math"
)

var powers [10]int

func isMunchausen(n int) bool {
if n < 0 { return false }
n64 := int64(n)
nn  := n64
var sum int64 = 0
for nn > 0 {
sum += int64(powers[nn % 10])
if sum > n64 { return false }
nn /= 10
}
return sum == n64
}

func main() {
// cache n ^ n for n in 0..9, defining 0 ^ 0 = 0 for this purpose
for i := 1; i <= 9; i++ {
d := float64(i)
powers[i] = int(math.Pow(d, d))
}

// check numbers 0 to 500 million
fmt.Println("The Munchausen numbers between 0 and 500 million are:")
for i := 0; i <= 500000000; i++ {
if isMunchausen(i) { fmt.Printf("%d ", i) }
}
fmt.Println()
}

{{out}}

0 1 3435 438579088

import Data.List (unfoldr)

isMunchausen :: Integer -> Bool
isMunchausen n = (n ==) \$ sum \$ map (\x -> x^x) \$ unfoldr digit n where
digit 0 = Nothing
digit n = Just (r,q) where (q,r) = n `divMod` 10

main :: IO ()
main = print \$ filter isMunchausen [1..5000]

{{out}}

[1,3435]

The Haskell libraries provide a lot of flexibility – we could also reduce the sum and map (above) down to a single foldr:

import Data.Char (digitToInt)

isMunchausen :: Int -> Bool
isMunchausen = (==) <*> foldr ((+) . (id >>=) (^) . digitToInt) 0 . show

main :: IO ()
main = print \$ filter isMunchausen [1 .. 5000]

{{Out}}

[1,3435]

## J

Here, it would be useful to have a function which sums the powers of the digits of a number. Once we have that we can use it with an equality test to filter those integers:

munch=: +/@(^~@(10&#.inv))
(#~ ] = munch"0) 1+i.5000
1 3435

Note that [[wp:Munchausen_number|wikipedia]] claims that 0=0^0 in the context of Munchausen numbers. It's not clear why this should be (1 is the multiplicative identity and if you do not multiply it by zero it should still be 1), but it's easy enough to implement. Note also that this does not change the result for this task:

munch=: +/@((**^~)@(10&#.inv))
(#~ ] = munch"0) 1+i.5000
1 3435

## Java

Adapted from Zack Denton's code posted on [https://zach.se/munchausen-numbers-and-how-to-find-them/ Munchausen Numbers and How to Find Them].

public class Main {
public static void main(String[] args) {
for(int i = 0 ; i <= 5000 ; i++ ){
int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum();
if( i == val){
System.out.println( i + " (munchausen)");
}
}
}
}

{{out}}

1 (munchausen)
3435 (munchausen)

### Faster version

{{trans|Kotlin}}

public class Munchhausen {

static final long[] cache = new long[10];

public static void main(String[] args) {
// Allowing 0 ^ 0 to be 0
for (int i = 1; i < 10; i++) {
cache[i] = (long) Math.pow(i, i);
}
for (long i = 0L; i <= 500_000_000L; i++) {
if (isMunchhausen(i)) {
System.out.println(i);
}
}
}

private static boolean isMunchhausen(long n) {
long sum = 0, nn = n;
do {
sum += cache[(int)(nn % 10)];
if (sum > n) {
return false;
}
nn /= 10;
} while (nn > 0);

return sum == n;
}
}
0
1
3435
438579088

## JavaScript

### ES6

for (let i of [...Array(5000).keys()]
.filter(n => n == n.toString().split('')
.reduce((a, b) => a+Math.pow(parseInt(b),parseInt(b)), 0)))
console.log(i);

{{out}}

1
3435

Or, composing reusable primitives:

(() => {
'use strict';

const main = () =>
filter(isMunchausen, enumFromTo(1, 5000));

// isMunchausen :: Int -> Bool
const isMunchausen = n =>
n.toString()
.split('')
.reduce(
(a, c) => (
d => a + Math.pow(d, d)
)(parseInt(c, 10)),
0
) === n;

// GENERIC ---------------------------

// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: 1 + n - m
}, (_, i) => m + i);

// filter :: (a -> Bool) -> [a] -> [a]
const filter = (f, xs) => xs.filter(f);

// MAIN ---
return main();
})();

{{Out}}

[1, 3435]

## jq

{{works with|jq|1.5}}

def sigma( stream ): reduce stream as \$x (0; . + \$x ) ;

def ismunchausen:
def digits: tostring | split("")[] | tonumber;
. == sigma(digits | pow(.;.));

# Munchausen numbers from 1 to 5000 inclusive:
range(1;5001) | select(ismunchausen)

{{out}}

1
3435

## Julia

{{works with|Julia|1.0}}

println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])

{{out}}

[1, 3435]

## Kotlin

As it doesn't take long to find all 4 known Munchausen numbers, we will test numbers up to 500 million here rather than just 5000:

// version 1.0.6

val powers = IntArray(10)

fun isMunchausen(n: Int): Boolean {
if (n < 0) return false
var sum = 0L
var nn = n
while (nn > 0) {
sum += powers[nn % 10]
if (sum > n.toLong()) return false
nn /= 10
}
return sum == n.toLong()
}

fun main(args: Array<String>) {
// cache n ^ n for n in 0..9, defining 0 ^ 0 = 0 for this purpose
for (i in 1..9) powers[i] = Math.pow(i.toDouble(), i.toDouble()).toInt()

// check numbers 0 to 500 million
println("The Munchausen numbers between 0 and 500 million are:")
for (i in 0..500000000) if (isMunchausen(i))print ("\$i ")
println()
}

{{out}}

The Munchausen numbers between 0 and 500 million are:
0 1 3435 438579088

## Langur

{{trans|C#}}

# sum power of digits
val .spod = f(.n) fold f .x + .y, map(f (.x-'0') ^ (.x-'0'), s2cp toString .n)

# Munchausen
writeln "Answers: ", where f(.n) .n == .spod(.n), series 0..5000

{{out}}

## Lua

function isMunchausen (n)
local sum, nStr, digit = 0, tostring(n)
for pos = 1, #nStr do
digit = tonumber(nStr:sub(pos, pos))
sum = sum + digit ^ digit
end
return sum == n
end

for i = 1, 5000 do
if isMunchausen(i) then print(i) end
end

{{out}}

1
3435

## Nim

import math

for i in 1..<5000:
var sum: int64 = 0
var number = i
while number > 0:
var digit = number mod 10
sum += digit ^ digit
number = number div 10
if sum == i:
echo i

{{out}}

1
3435

## M2000 Interpreter

Module Munchausen {
Inventory p=0:=0,1:=1
for i=2 to 9 {Append p, i:=i**i}
Munchausen=lambda p (x)-> {
m=0
t=x
do {
m+=p(x mod 10)
x=x div 10
} until x=0
=m=t
}
For i=1 to 5000
If Munchausen(i) then print i,
Next i
Print
}
Munchausen

Using Array instead of Inventory

Module Münchhausen {
Dim p(0 to 9)
p(0)=0, 1
for i=2 to 9 {p(i)=i**i}
Münchhausen=lambda p() (x)-> {
m=0
t=x
do {
m+=p(x mod 10)
x=x div 10
} until x=0
=m=t
}
For i=1 to 5000
If Münchhausen(i) then print i,
Next i
Print
}
Münchhausen

{{out}}

1     3435

## Maple

isMunchausen := proc(n::posint)
local num_digits;
num_digits := map(x -> StringTools:-Ord(x) - 48, StringTools:-Explode(convert(n, string)));
return evalb(n = convert(map(x -> x^x, num_digits), `+`));
end proc;

Munchausen_upto := proc(n::posint) local k, count, list_num;
list_num := [];
for k to n do
if isMunchausen(k) then
list_num := [op(list_num), k];
end if;
end do;
return list_num;
end proc;

Munchausen_upto(5000);

{{out}}

[1, 3435]

## Mathematica

Off[Power::indet];(*Supress 0^0 warnings*)
Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]

{{out}}

{1,3435}

## min

{{works with|min|0.19.3}}

(dup string "" split (int dup pow) (+) map-reduce ==) :munchausen?
1 :i
(i 5000 <=) ((i munchausen?) (i puts!) when i succ @i) while

{{out}}

1
3435

MODULE MunchausenNumbers;
FROM FormatString IMPORT FormatString;
FROM Terminal IMPORT WriteString,ReadChar;

(* Simple power function, does not handle negatives *)
PROCEDURE Pow(b,e : INTEGER) : INTEGER;
VAR result : INTEGER;
BEGIN
IF e=0 THEN
RETURN 1;
END;
IF b=0 THEN
RETURN 0;
END;

result := b;
DEC(e);
WHILE e>0 DO
result := result * b;
DEC(e);
END;
RETURN result;
END Pow;

VAR
buf : ARRAY[0..31] OF CHAR;
i,sum,number,digit : INTEGER;
BEGIN
FOR i:=1 TO 5000 DO
(* Loop through each digit in i
e.g. for 1000 we get 0, 0, 0, 1. *)
sum := 0;
number := i;
WHILE number>0 DO
digit := number MOD 10;
sum := sum + Pow(digit, digit);
number := number DIV 10;
END;
IF sum=i THEN
FormatString("%i\n", buf, i);
WriteString(buf);
END;
END;

END MunchausenNumbers.

## Pascal

{{works with|Free Pascal}} {{works with|Delphi}} tried to speed things up.Only checking one arrangement of 123456789 instead of all 9! = 362880 permutations.This ist possible, because summing up is commutative. So I only have to create [http://rosettacode.org/wiki/Combinations_with_repetitions Combinations_with_repetitions] and need to check, that the number and the sum of power of digits have the same amount in every possible digit. This means, that a combination of the digits of number leads to the sum of power of digits. Therefore I need leading zero's.

{\$IFDEF FPC}{\$MODE objFPC}{\$ELSE}{\$APPTYPE CONSOLE}{\$ENDIF}
uses
sysutils;
type
tdigit  = byte;
const
base = 10;
maxDigits = base-1;// set for 32-compilation otherwise overflow.

var
DgtPotDgt : array[0..base-1] of NativeUint;
cnt: NativeUint;

function CheckSameDigits(n1,n2:NativeUInt):boolean;
var
dgtCnt : array[0..Base-1] of NativeInt;
i : NativeUInt;
Begin
fillchar(dgtCnt,SizeOf(dgtCnt),#0);
repeat
//increment digit of n1
i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]);
//decrement digit of n2
i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]);
until (n1=0) AND (n2= 0 );
result := true;
For i := 0 to Base-1 do
result := result AND (dgtCnt[i]=0);
end;

procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt);
var
i: NativeUint;
begin
inc(cnt);
number := number*base;
IF digits > 1 then
Begin
For i := minDigit to base-1 do
Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1);
end
else
For i := minDigit to base-1 do
//number is always the arrangement of the digits leading to smallest number
IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then
IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then
iF number+i>0 then
writeln(Format('%*d  %.*d',
[maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i]));
end;

procedure InitDgtPotDgt;
var
i,k,dgtpow: NativeUint;
Begin
// digit ^ digit ,special case 0^0 here 0
DgtPotDgt[0]:= 0;
For i := 1 to Base-1 do
Begin
dgtpow := i;
For k := 2 to i do
dgtpow := dgtpow*i;
DgtPotDgt[i] := dgtpow;
end;
end;

begin
cnt := 0;
InitDgtPotDgt;
Munch(0,0,0,maxDigits);
writeln('Check Count ',cnt);
end.

{{Out}}

1  000000001
3435  000003345
438579088  034578889

Check Count 43758 ==
n= maxdigits = 9,k = 10;CombWithRep = (10+9-1))!/(10!*(9-1)!)=43758

real    0m0.002s

## Perl

use List::Util "sum";
for my \$n (1..5000) {
print "\$n\n" if \$n == sum( map { \$_**\$_ } split(//,\$n) );
}

{{out}}

1
3435

## Perl 6

sub is_munchausen ( Int \$n ) {
constant @powers = 0, |map { \$_ ** \$_ }, 1..9;
\$n == @powers[\$n.comb].sum;
}
.say if .&is_munchausen for 1..5000;

{{out}}

1
3435

## Phix

sequence powers = 0&sq_power(tagset(9),tagset(9))

function munchausen(integer n)
integer n0 = n
atom summ = 0
while n!=0 do
summ += powers[remainder(n,10)+1]
n = floor(n/10)
end while
return summ=n0
end function

for i=1 to 5000 do
if munchausen(i) then ?i end if
end for

{{out}}

1
3435

## PicoLisp

(for N 5000
(and
(=
N
(sum
'((N) (** N N))
(mapcar format (chop N)) ) )
(println N) ) )

{{out}}

1
3435

## PowerBASIC

{{trans|FreeBASIC}}(Translated from the FreeBasic Version 2 example.)

#COMPILE EXE
#DIM ALL
#COMPILER PBCC 6

DECLARE FUNCTION GetTickCount LIB "kernel32.dll" ALIAS "GetTickCount"() AS DWORD

FUNCTION PBMAIN () AS LONG
LOCAL i, j, n, sum, ten1, ten2, t AS DWORD
LOCAL n0, n1, n2, n3, n4, n5, n6, n7, n8, n9 AS DWORD
LOCAL s1, s2, s3, s4, s5, s6, s7, s8 AS DWORD
DIM pow(9) AS DWORD, num(9) AS DWORD
LOCAL pb AS BYTE PTR
LOCAL number AS STRING

t = GetTickCount()
ten2 = 10
FOR i = 1 TO 9
pow(i) = i
FOR j = 2 TO i
pow(i) *= i
NEXT j
NEXT i
FOR n = 1 TO 11
FOR n9 = 0 TO n
FOR n8 = 0 TO n - n9
s8 = n9 + n8
FOR n7 = 0 TO n - s8
s7 = s8 + n7
FOR n6 = 0 TO n - s7
s6 = s7 + n6
FOR n5 = 0 TO n - s6
s5 = s6 + n5
FOR n4 = 0 TO n - s5
s4 = s5 + n4
FOR n3 = 0 TO n - s4
s3 = s4 + n3
FOR n2 = 0 TO n - s3
s2 = s3 + n2
FOR n1 = 0 TO n - s2
n0 = n - (s2 + n1)
sum = n1 * pow(1) + n2 * pow(2) + n3 * pow(3) + _
n4 * pow(4) + n5 * pow(5) + n6 * pow(6) + _
n7 * pow(7) + n8 * pow(8) + n9 * pow(9)
SELECT CASE AS LONG sum
CASE ten1 TO ten2 - 1
number = LTRIM\$(STR\$(sum))
pb = STRPTR(number)
MAT num() = ZER
FOR i = 0 TO n -1
j = @pb[i] - 48
INCR num(j)
NEXT i
IF n0 = num(0) AND n1 = num(1) AND n2 = num(2) AND _
n3 = num(3) AND n4 = num(4) AND n5 = num(5) AND _
n6 = num(6) AND n7 = num(7) AND n8 = num(8) AND _
n9 = num(9) THEN CON.PRINT STR\$(sum)
END SELECT
NEXT n1
NEXT n2
NEXT n3
NEXT n4
NEXT n5
NEXT n6
NEXT n7
NEXT n8
NEXT n9
ten1 = ten2
ten2 *= 10
NEXT n
t = GetTickCount() - t
CON.PRINT "execution time:" & STR\$(t) & " ms; hit any key to end program"
CON.WAITKEY\$
END FUNCTION

{{out}}

0
1
3435
438579088
execution time: 78 ms; hit any key to end program

## Pure

// split numer into digits
digits n::number = loop n [] with
loop n l = loop (n div 10) ((n mod 10):l) if n > 0;
= l otherwise; end;

munchausen n::int = (filter isMunchausen list) when
list = 1..n; end with
isMunchausen n = n == foldl (+) 0
(map (\d -> d^d)
(digits n)); end;
munchausen 5000;

{{out}}

[1,3435]

## PureBasic

{{trans|C}}

EnableExplicit
Declare main()

If OpenConsole("Munchausen_numbers")
main() : Input() : End
EndIf

Procedure main()
Define i.i,
sum.i,
number.i,
digit.i
For i = 1 To 5000
sum = 0
number = i
While number > 0
digit = number % 10
sum + Pow(digit, digit)
number / 10
Wend
If sum = i
PrintN(Str(i))
EndIf
Next
EndProcedure

{{out}}

1
3435

## Python

for i in range(5000):
if i == sum(int(x) ** int(x) for x in str(i)):
print(i)

{{out}}

1
3435

Or, defining an '''isMunchausen''' predicate in terms of a single fold – rather than a two-pass ''sum'' after ''map'' (or comprehension) –

and reaching for a specialised '''digitToInt''', which turns out to be a little faster than type coercion with the more general built-in '''int()''':

{{Works with|Python|3}}

'''Munchausen numbers'''

from functools import (reduce)

# isMunchausen :: Int -> Bool
def isMunchausen(n):
'''True if n equals the sum of
each of its digits raised to
the power of itself.'''
def powerOfSelf(d):
i = digitToInt(d)
return i**i
return n == reduce(
lambda n, c: n + powerOfSelf(c),
str(n), 0
)

# main :: IO ()
def main():
'''Test'''
print(list(filter(
isMunchausen,
enumFromTo(1)(5000)
)))

# GENERIC -------------------------------------------------

# digitToInt :: Char -> Int
def digitToInt(c):
'''The integer value of any digit character
drawn from the 0-9, A-F or a-f ranges.'''
oc = ord(c)
if 48 > oc or 102 < oc:
return None
else:
dec = oc - 48   # ord('0')
hexu = oc - 65  # ord('A')
hexl = oc - 97  # ord('a')
return dec if 9 >= dec else (
10 + hexu if 0 <= hexu <= 5 else (
10 + hexl if 0 <= hexl <= 5 else None
)
)

# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))

if __name__ == '__main__':
main()
[1, 3435]

## Racket

#lang racket

(define (expt:0^0=1 r p) (if (zero? r) 0 (expt r p)))

(define (munchausen-number? n (t n)) (if (zero? n) (zero? t) (let-values (([q r] (quotient/remainder n 10))) (munchausen-number? q (- t (expt:0^0=1 r r))))))

(module+ main (for-each displayln (filter munchausen-number? (range 1 (add1 5000)))))

(module+ test (require rackunit) ;; this is why we have the (if (zero? r)...) test (check-equal? (expt 0 0) 1) (check-equal? (expt:0^0=1 0 0) 0) (check-equal? (expt:0^0=1 0 4) 0) (check-equal? (expt:0^0=1 3 4) (expt 3 4)) ;; given examples (check-true (munchausen-number? 1)) (check-true (munchausen-number? 3435)) (check-false (munchausen-number? 3)) (check-false (munchausen-number? -45) "no recursion on -ve numbers"))

{{out}}

```txt
1
3435

## REXX

### version 1

Do n=0 To 10000
If n=m(n) Then
Say n
End
Exit
m: Parse Arg z
res=0
Do While z>''
Parse Var z c +1 z
res=res+c**c
End
Return res

{{out}}

D:\mau>rexx munch
1
3435

### version 2

This REXX version uses the requirement that '''0**0''' equals zero.

It is about 2.5 times faster than version 1.

For the high limit of '''5,000''', optimization isn't needed. But for much higher limits, optimization becomes significant.

/*REXX program finds and displays Münchhausen numbers from one to a specified number (Z)*/
@.=0;          do i=1  for 9;  @.i=i**i;  end    /*precompute powers for non-zero digits*/
parse arg z .                                    /*obtain optional argument from the CL.*/
if z=='' | z==","  then z=5000                   /*Not specified?  Then use the default.*/
@is='is a Münchhausen number.';   do j=1  for z  /* [↓]  traipse through all the numbers*/
if isMunch(j)  then say  right(j, 11)    @is
end   /*j*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isMunch: parse arg x 1 ox;  \$=0;  do  until  x==''  |  \$>ox         /*stop if too large.*/
parse var x _ +1 x;  \$=\$ + @._    /*add the next power*/
end   /*while*/                   /* [↑]  get a digit.*/
return \$==ox                                               /*it is or it ain't.*/

'''output'''

1 is a Münchhausen number.
3435 is a Münchhausen number.

### version 3

It is about 3 times faster than version 1.

/*REXX program finds and displays Münchhausen numbers from one to a specified number (Z)*/
@.=0;          do i=1  for 9;  @.i=i**i;  end    /*precompute powers for non-zero digits*/
parse arg z .                                    /*obtain optional argument from the CL.*/
if z=='' | z==","  then z=5000                   /*Not specified?  Then use the default.*/
@is='is a Münchhausen number.';   do j=1  for z  /* [↓]  traipse through all the numbers*/
if isMunch(j)  then say  right(j, 11)    @is
end   /*j*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isMunch: parse arg a 2 b 3 c 4 d 5 e 6 x 1 ox; [email protected][email protected][email protected][email protected][email protected] /*sum 1st 5 digits.*/
if \$>ox  then return 0                                      /*is sum too large?*/
do  while  x\==''  &  \$<=ox        /*any more digits ?*/
parse var x _ +1 x;   \$=\$ + @._    /*sum 6th & up digs*/
end   /*while*/
return \$==ox                                                /*it is or it ain't*/

'''output''' is the same as the 2nd REXX version.

## Ring

# Project : Munchausen numbers

limit = 5000

for n=1 to limit
sum = 0
msum = string(n)
for m=1 to len(msum)
ms = number(msum[m])
sum = sum + pow(ms, ms)
next
if sum = n
see n + nl
ok
next

Output:

1
3435

## Ruby

class Integer

def munchausen?
self.digits.map{|d| d**d}.sum == self
end

end

puts (1..5000).select(&:munchausen?)

{{out}}

1
3435

## Rust

fn main() {
let mut solutions = Vec::new();

for num in 1..5_000 {
let power_sum = num.to_string()
.chars()
.map(|c| {
let digit = c.to_digit(10).unwrap();
(digit as f64).powi(digit as i32) as usize
})
.sum::<usize>();

if power_sum == num {
solutions.push(num);
}
}

println!("Munchausen numbers below 5_000 : {:?}", solutions);
}

{{out}}

Munchausen numbers below 5_000 : [1, 3435]

## Scala

Adapted from Zack Denton's code posted on [https://zach.se/munchausen-numbers-and-how-to-find-them/ Munchausen Numbers and How to Find Them].

object Munch {
def main(args: Array[String]): Unit = {
import scala.math.pow
(1 to 5000).foreach {
i => if (i == (i.toString.toCharArray.map(d => pow(d.asDigit,d.asDigit))).sum)
println( i + " (munchausen)")
}
}
}

{{out}}

1 (munchausen)
3435 (munchausen)

## Sidef

func is_munchausen(n) {
n.digits.map{|d| d**d }.sum == n
}

say (1..5000 -> grep(is_munchausen))

{{out}}

[1, 3435]

## SuperCollider

(1..5000).select { |n| n == n.asDigits.sum { |x| pow(x, x) } }

[1, 3435]

## Swift

import Foundation

func isMünchhausen(_ n: Int) -> Bool {
let nums = String(n).map(String.init).compactMap(Int.init)

return Int(nums.map({ pow(Double(\$0), Double(\$0)) }).reduce(0, +)) == n
}

for i in 1...5000 where isMünchhausen(i) {
print(i)
}

{{out}}

1
3435

=={{header|TI-83 BASIC}}== {{works with|TI-83 BASIC|TI-84Plus 2.55MP}} {{trans|Fortran}}

For(I,1,5000)
0→S:I→K
For(J,1,4)
10^(4-J)→D
iPart(K/D)→N
remainder(K,D)→R
If N≠0:S+N^N→S
R→K
End
If S=I:Disp I
End

{{out}}

1
3435

Execution time: 15 min

## VBA

Option Explicit

Sub Main_Munchausen_numbers()
Dim i&

For i = 1 To 5000
If IsMunchausen(i) Then Debug.Print i & " is a munchausen number."
Next i
End Sub

Function IsMunchausen(Number As Long) As Boolean
Dim Digits, i As Byte, Tot As Long

Digits = Split(StrConv(Number, vbUnicode), Chr(0))
For i = 0 To UBound(Digits) - 1
Tot = (Digits(i) ^ Digits(i)) + Tot
Next i
IsMunchausen = (Tot = Number)
End Function

{{out}}

1 is a munchausen number.
3435 is a munchausen number.

## VBScript

for i = 1 to 5000
if Munch(i) Then
Wscript.Echo i, "is a Munchausen number"
end if
next

'Returns True if num is a Munchausen number. This is true if the sum of
'each digit raised to that digit's power is equal to the given number.
'Example: 3435 = 3^3 + 4^4 + 3^3 + 5^5

Function Munch (num)

dim str: str = Cstr(num)    'input num as a string
dim sum: sum = 0            'running sum of n^n
dim i                       'loop index
dim n                       'extracted digit

for i = 1 to len(str)
n = CInt(Mid(str,i,1))
sum = sum + n^n
next

Munch = (sum = num)

End Function

{{out}}

1 is a Munchausen number
3435 is a Munchausen number

## Visual Basic

{{trans|FreeBASIC}}(Translated from the FreeBasic Version 2 example.)

Option Explicit

Declare Function GetTickCount Lib "kernel32.dll" () As Long
Declare Sub ZeroMemory Lib "kernel32.dll" Alias "RtlZeroMemory" (ByRef Destination As Any, ByVal Length As Long)

Sub Main()
Dim i As Long, j As Long, n As Long, t As Long
Dim sum As Double
Dim n0 As Double
Dim n1 As Double
Dim n2 As Double
Dim n3 As Double
Dim n4 As Double
Dim n5 As Double
Dim n6 As Double
Dim n7 As Double
Dim n8 As Double
Dim n9 As Double
Dim ten1 As Double
Dim ten2 As Double
Dim s1 As Long
Dim s2 As Long
Dim s3 As Long
Dim s4 As Long
Dim s5 As Long
Dim s6 As Long
Dim s7 As Long
Dim s8 As Long
Dim pow(9) As Long, num(9) As Long
Dim number As String, res As String

t = GetTickCount()
ten2 = 10
For i = 1 To 9
pow(i) = i
For j = 2 To i
pow(i) = i * pow(i)
Next j
Next i
For n = 1 To 11
For n9 = 0 To n
For n8 = 0 To n - n9
s8 = n9 + n8
For n7 = 0 To n - s8
s7 = s8 + n7
For n6 = 0 To n - s7
s6 = s7 + n6
For n5 = 0 To n - s6
s5 = s6 + n5
For n4 = 0 To n - s5
s4 = s5 + n4
For n3 = 0 To n - s4
s3 = s4 + n3
For n2 = 0 To n - s3
s2 = s3 + n2
For n1 = 0 To n - s2
n0 = n - (s2 + n1)
sum = n1 * pow(1) + n2 * pow(2) + n3 * pow(3) + _
n4 * pow(4) + n5 * pow(5) + n6 * pow(6) + _
n7 * pow(7) + n8 * pow(8) + n9 * pow(9)
Select Case sum
Case ten1 To ten2 - 1
number = CStr(sum)
ZeroMemory num(0), 40
For i = 1 To n
j = Asc(Mid\$(number, i, 1)) - 48
num(j) = num(j) + 1
Next i
If n0 = num(0) Then
If n1 = num(1) Then
If n2 = num(2) Then
If n3 = num(3) Then
If n4 = num(4) Then
If n5 = num(5) Then
If n6 = num(6) Then
If n7 = num(7) Then
If n8 = num(8) Then
If n9 = num(9) Then
res = res & CStr(sum) & vbNewLine
End If
End If
End If
End If
End If
End If
End If
End If
End If
End If
End Select
Next n1
Next n2
Next n3
Next n4
Next n5
Next n6
Next n7
Next n8
Next n9
ten1 = ten2
ten2 = ten2 * 10
Next n
t = GetTickCount() - t
res = res & "execution time:" & Str\$(t) & " ms"
MsgBox res
End Sub

{{out}}

0
1
3435
438579088
execution time: 156 ms

## Visual Basic .NET

{{trans|FreeBASIC}}(Translated from the FreeBasic Version 2 example.)
Computation time is under 4 seconds on tio.run.

Imports System

Module Program
Sub Main()
Dim i, j, n, n1, n2, n3, n4, n5, n6, n7, n8, n9, s2, s3, s4, s5, s6, s7, s8 As Integer,
sum, ten1 As Long, ten2 As Long = 10
Dim pow(9) As Long, num() As Byte
For i = 1 To 9 : pow(i) = i : For j = 2 To i : pow(i) *= i : Next : Next
For n = 1 To 11 : For n9 = 0 To n : For n8 = 0 To n - n9 : s8 = n9 + n8 : For n7 = 0 To n - s8
s7 = s8 + n7 : For n6 = 0 To n - s7 : s6 = s7 + n6 : For n5 = 0 To n - s6
s5 = s6 + n5 : For n4 = 0 To n - s5 : s4 = s5 + n4 : For n3 = 0 To n - s4
s3 = s4 + n3 : For n2 = 0 To n - s3 : s2 = s3 + n2 : For n1 = 0 To n - s2
sum = n1 * pow(1) + n2 * pow(2) + n3 * pow(3) + n4 * pow(4) +
n5 * pow(5) + n6 * pow(6) + n7 * pow(7) + n8 * pow(8) + n9 * pow(9)
If sum < ten1 OrElse sum >= ten2 Then Continue For
redim num(9)
For Each ch As Char In sum.ToString() : num(Convert.ToByte(ch) - 48) += 1 : Next
If n - (s2 + n1) = num(0) AndAlso n1 = num(1) AndAlso n2 = num(2) AndAlso
n3 = num(3) AndAlso n4 = num(4) AndAlso n5 = num(5) AndAlso n6 = num(6) AndAlso
n7 = num(7) AndAlso n8 = num(8) AndAlso n9 = num(9) Then Console.WriteLine(sum)
Next : Next : Next : Next : Next : Next : Next : Next : Next
ten1 = ten2 : ten2 *= 10
Next
End Sub
End Module

{{out}}

0
1
3435
438579088

## zkl

[1..5000].filter(fcn(n){ n==n.split().reduce(fcn(s,n){ s + n.pow(n) },0) })
.println();

{{out}}

L(1,3435)