Munchausen numbers: Difference between revisions
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=={{header|Pure}}== |
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<lang Pure>// split numer into digits |
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digits n::number = loop n [] with |
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loop n l = loop (n div 10) ((n mod 10):l) if n > 0; |
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= l otherwise; end; |
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munchausen n::int = (filter isMunchausen list) when |
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list = 1..n; end with |
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isMunchausen n = n == foldl (+) 0 |
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(map (\d -> d^d) |
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(digits n)); end; |
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munchausen 5000;</lang> |
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{{out}} |
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<pre>[1,3435]</pre> |
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=={{header|Python}}== |
=={{header|Python}}== |
||
Revision as of 18:56, 11 October 2017
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.
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
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
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
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)
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
FreeBASIC
<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
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]
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 Data.Char (digitToInt)
isMunchausen :: Int -> Bool isMunchausen n =
n ==
foldr
(\c n ->
let v = digitToInt c
in n + v ^ v)
0
(show n)
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>(function () {
'use strict';
// isMunchausen :: Int -> Bool
let isMunchausen = n =>
!isNaN(n) && (
n.toString()
.split()
.reduce((a, c) => {
let d = parseInt(c, 10);
return a + Math.pow(d, d);
}, 0) === n
),
// range(intFrom, intTo, intStep?)
// Int -> Int -> Maybe Int -> [Int]
range = (m, n, step) => {
let d = (step || 1) * (n >= m ? 1 : -1);
return Array.from({
length: Math.floor((n - m) / d) + 1
}, (_, i) => m + (i * d));
};
return range(1, 5000)
.filter(isMunchausen);
})();</lang>
- Output:
<lang JavaScript>[1, 3435]</lang>
Julia
<lang julia>ismunchausen(n) = sum(d ^ d for d in digits(n)) == n println(filter(ismunchausen, 1:5000))</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 ("$i ")
println()
}</lang>
- Output:
The Munchausen numbers between 0 and 500 million are: 0 1 3435 438579088
Lua
<lang Lua>function isMunchausen (n)
local sum, nStr, digit = 0, tostring(n)
for pos = 1, #nStr do
digit = tonumber(nStr:sub(pos, pos))
sum = sum + digit ^ digit
end
return sum == n
end
for i = 1, 5000 do
if isMunchausen(i) then print(i) end
end</lang>
- Output:
1 3435
Mathematica
<lang Mathematica>Off[Power::indet];(*Supress 0^0 warnings*) Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]</lang>
- Output:
{1,3435}
Pascal
tried to speed things up.Only checking one arrangement of 123456789 instead of all 9! = 362880 permutations.This ist possible, because summing up is commutative. So I only have to create Combinations_with_repetitions and need to check, that the number and the sum of power of digits have the same amount in every possible digit. This means, that a combination of the digits of number leads to the sum of power of digits. Therefore I need leading zero's. <lang pascal>{$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
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]
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
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 Munchhausen 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 Munchhausen 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
- Date : 2017/09/30
- Author : Gal Zsolt (~ CalmoSoft ~)
- Email : <calmosoft@gmail.com>
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>POW = [0] + (1..9).map{|i| i**i}
def munchausen_number?(n)
digits(n).inject(0){|sum,i| sum + POW[i]} == n
end
def digits(n)
ary = [] while n > 0 n,mod = n.divmod(10) ary << mod end ary
end
(1..5000).each do |i|
puts i if munchausen_number?(i)
end</lang>
- Output:
1 3435
Rust
<lang rust> use std::f64;
fn main () {
let mut solutions = Vec::new();
for i in 1..5_000 {
if i.to_string()
.chars()
.map(|c| (c.to_digit(10).unwrap() as f64).powi(c.to_digit(10).unwrap() as i32) as i32)
.fold(0, |n, i| n + i) == i { solutions.push(i); }
}
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]
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
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)