Munchausen numbers
You are encouraged to solve this task according to the task description, using any language you may know.
A 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: A046253
- The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
360 Assembly
<lang 360asm>* 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
BR R14 return to caller
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 </lang>
- Output:
1
3435
ALGOL 68
<lang algol68># 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 </lang>
- Output:
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. <lang algol68># 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</lang>
- Output:
+0
+1
+3435
+438579088
ALGOL W
<lang algolw>% 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.</lang>
- Output:
1 3435
AppleScript
<lang 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</lang>
- Output:
<lang AppleScript>{1, 3435}</lang>
AWK
<lang 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)
} </lang>
- Output:
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.
<lang basic>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</lang>
- Output:
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.
<lang basic> 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</lang>
- Output:
1 3435
BBC BASIC
<lang bbcbasic>REM >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%</lang>
- Output:
1
3435
C
Adapted from Zack Denton's code posted on Munchausen Numbers and How to Find Them. <lang C>#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;
}</lang>
- Output:
1 3435
C#
<lang csharp>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);</lang>
- Output:
1 3435
Faster version
<lang csharp>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);
}
}
Console.ReadLine();
}
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;
}
}
}</lang>
0 1 3435 438579088
Faster version alternate
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. <lang csharp>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;
}
}
}</lang>
- Output:
0 1 3435 438579088
C++
<lang cpp>
- 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;
} </lang>
- Output:
Munchausen Numbers ================== 1 3435
Clojure
<lang lisp>(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))
</lang>
- Output:
(1 3435)
Common Lisp
<lang 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)))))
</lang>
- Output:
> (check4munch 5000) (1 3435) > (munchp 438579088) T
D
<lang D>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);
}
}
}</lang>
- Output:
1 3435
Dc
Needs a modern Dc due to ~.
Use S1S2l2l1/L2L1% instead of ~ to run it in older Dcs.
<lang dc>[ O ~ S! d 0!=M L! d ^ + ] sM
[p] sp
[z d d lM x =p z 5001>L ] sL
lL x</lang>
Cosmetic: The stack is dirty after execution. The loop L needs a fix if that is a problem.
Elixir
<lang 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)</lang>
- Output:
1 3435
Factor
<lang 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 </lang>
- Output:
V{ 1 3435 }
Fōrmulæ
In 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 (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.
Fortran
Fortran IV
<lang fortran>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 </lang>
- Output:
1
3435
Fortran 77
<lang fortran>! 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 </lang>
- Output:
1
3435
FreeBASIC
Version 1
<lang freebasic>' 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</lang>
- Output:
The Munchausen numbers between 0 and 500000000 are : 0 1 3435 438579088
Version 2
<lang freebasic>' 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</lang>
- Output:
0 1 3435 438579088
F#
<lang fsharp>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)</lang>
- Output:
[1; 3435]
Go
<lang go>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()
}</lang>
- Output:
0 1 3435 438579088
Haskell
<lang haskell>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]</lang>
- Output:
[1,3435]
The Haskell libraries provide a lot of flexibility – we could also rework the sum, map, and unfold above to a single fold:
<lang haskell>import Control.Monad (join) import Data.Char (digitToInt)
isMunchausen :: Int -> Bool isMunchausen =
let go = (+) . join (^) . digitToInt in (==) <*> foldr go 0 . show
main :: IO () main = print $ filter isMunchausen [1 .. 5000]</lang>
- Output:
[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:
<lang J> munch=: +/@(^~@(10&#.inv))
(#~ ] = munch"0) 1+i.5000
1 3435</lang>
Note that 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:
<lang J> munch=: +/@((**^~)@(10&#.inv))
(#~ ] = munch"0) 1+i.5000
1 3435</lang>
Java
Adapted from Zack Denton's code posted on Munchausen Numbers and How to Find Them. <lang Java> 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)");
}
}
}
}
</lang>
- Output:
1 (munchausen) 3435 (munchausen)
Faster version
<lang java>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; }
}</lang>
0 1 3435 438579088
JavaScript
ES6
<lang javascript>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);</lang>
- Output:
1 3435
Or, composing reusable primitives:
<lang JavaScript>(() => {
'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();
})();</lang>
- Output:
<lang JavaScript>[1, 3435]</lang>
jq
<lang jq>def sigma( stream ): reduce stream as $x (0; . + $x ) ;
def ismunchausen:
def digits: tostring | split("")[] | tonumber;
. == sigma(digits | pow(.;.));
- Munchausen numbers from 1 to 5000 inclusive:
range(1;5001) | select(ismunchausen)</lang>
- Output:
<lang jq>1 3435</lang>
Julia
<lang julia>println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])</lang>
- Output:
[1, 3435]
Kotlin
As it doesn't take long to find all 4 known Munchausen numbers, we will test numbers up to 500 million here rather than just 5000: <lang scala>// version 1.0.6
val powers = IntArray(10)
fun isMunchausen(n: Int): Boolean {
if (n < 0) return false
var sum = 0L
var nn = n
while (nn > 0) {
sum += powers[nn % 10]
if (sum > n.toLong()) return false
nn /= 10
}
return sum == n.toLong()
}
fun main(args: Array<String>) {
// cache n ^ n for n in 0..9, defining 0 ^ 0 = 0 for this purpose for (i in 1..9) powers[i] = Math.pow(i.toDouble(), i.toDouble()).toInt()
// check numbers 0 to 500 million
println("The Munchausen numbers between 0 and 500 million are:")
for (i in 0..500000000) if (isMunchausen(i))print ("$i ")
println()
}</lang>
- Output:
The Munchausen numbers between 0 and 500 million are: 0 1 3435 438579088
Langur
sort of copying from the C# example... <lang Langur># sum power of digits val .spod = f(.n) fold f .x + .y, map(f (.x-'0') ^ (.x-'0'), stringToCp toString .n)
- Munchausen
writeln "Answers: ", where f(.n) .n == .spod(.n), series 1..5000</lang>
- Output:
Answers: [1, 3435]
Lua
<lang 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</lang>
- Output:
1 3435
Nim
<lang 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</lang>
- Output:
1 3435
M2000 Interpreter
<lang 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 </lang> Using Array instead of Inventory <lang M2000 Interpreter> 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 </lang>
- Output:
1 3435
Mathematica
<lang Mathematica>Off[Power::indet];(*Supress 0^0 warnings*) Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]</lang>
- Output:
{1,3435}
min
<lang min>(dup string "" split (int dup pow) (+) map-reduce ==) :munchausen? 1 :i (i 5000 <=) ((i munchausen?) (i puts!) when i succ @i) while</lang>
- Output:
1 3435
Modula-2
<lang modula2>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;
ReadChar;
END MunchausenNumbers.</lang>
Pascal
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 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. <lang pascal>{$IFDEF FPC}{$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. </lang>
- Output:
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
<lang perl>use List::Util "sum"; for my $n (1..5000) {
print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) );
}</lang>
- Output:
1 3435
Perl 6
<lang perl6>sub is_munchausen ( Int $n ) {
constant @powers = 0, |map { $_ ** $_ }, 1..9;
$n == @powers[$n.comb].sum;
} .say if .&is_munchausen for 1..5000;</lang>
- Output:
1 3435
Phix
<lang 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</lang>
- Output:
1 3435
PicoLisp
<lang PicoLisp>(for N 5000
(and
(=
N
(sum
'((N) (** N N))
(mapcar format (chop N)) ) )
(println N) ) )</lang>
- Output:
1 3435
PowerBASIC
(Translated from the FreeBasic Version 2 example.)
<lang powerbasic>#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</lang>
- Output:
0 1 3435 438579088 execution time: 78 ms; hit any key to end program
Pure
<lang 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;</lang>
- Output:
[1,3435]
PureBasic
<lang PureBasic>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</lang>
- Output:
1 3435
Python
<lang python>for i in range(5000):
if i == sum(int(x) ** int(x) for x in str(i)):
print(i)</lang>
- Output:
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():
<lang python>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()</lang>
[1, 3435]
Racket
<lang>#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"))</lang>
- Output:
1 3435
REXX
version 1
<lang rexx>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</lang>
- Output:
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. <lang rexx>/*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.*/</lang>
output
1 is a Münchhausen number.
3435 is a Münchhausen number.
version 3
It is about 3 times faster than version 1. <lang rexx>/*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; $=@.a+@.b+@.c+@.d+@.e /*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*/</lang>
output is the same as the 2nd REXX version.
Ring
<lang 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 </lang> Output:
1 3435
Ruby
<lang ruby>class Integer
def munchausen?
self.digits.map{|d| d**d}.sum == self
end
end
puts (1..5000).select(&:munchausen?)</lang>
- Output:
1 3435
Rust
<lang 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);
}</lang>
- Output:
Munchausen numbers below 5_000 : [1, 3435]
Scala
Adapted from Zack Denton's code posted on Munchausen Numbers and How to Find Them. <lang Scala> 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)")
}
}
} </lang>
- Output:
1 (munchausen) 3435 (munchausen)
Sidef
<lang ruby>func is_munchausen(n) {
n.digits.map{|d| d**d }.sum == n
}
say (1..5000 -> grep(is_munchausen))</lang>
- Output:
[1, 3435]
SuperCollider
<lang supercollider>(1..5000).select { |n| n == n.asDigits.sum { |x| pow(x, x) } }</lang>
[1, 3435]
Swift
<lang 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)
}</lang>
- Output:
1 3435
TI-83 BASIC
<lang ti83b> 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 </lang>
- Output:
1
3435
Execution time: 15 min
VBA
<lang vb> 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 </lang>
- Output:
1 is a munchausen number. 3435 is a munchausen number.
VBScript
<lang 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 </lang>
- Output:
1 is a Munchausen number 3435 is a Munchausen number
Visual Basic
(Translated from the FreeBasic Version 2 example.)
<lang vb>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</lang>
- Output:
0 1 3435 438579088 execution time: 156 ms
Visual Basic .NET
(Translated from the FreeBasic Version 2 example.)
Computation time is under 4 seconds on tio.run. <lang vbnet>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</lang>
- Output:
0 1 3435 438579088
zkl
<lang zkl>[1..5000].filter(fcn(n){ n==n.split().reduce(fcn(s,n){ s + n.pow(n) },0) }) .println();</lang>
- Output:
L(1,3435)
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