Josephus problem (counting out)

Question

The challenge

Write a function that takes two positive integers n and k as arguments and returns the number of the last person remaining out of n after counting out each k-th person.

This is a code-golf challenge, so the shortest code wins.

The problem

n people (numbered from 1 to n) are standing in a circle and each k-th is counted out until a single person is remaining (see the corresponding wikipedia article). Determine the number of this last person.

E.g. for k=3 two people will be skipped and the third will be counted out. I.e. for n=7 the numbers will be counted out in the order 3 6 2 7 5 1 (in detail 1 2 3 4 5 6 7 1 2 4 5 7 1 4 5 1 4 1 4) and thus the answer is 4.

Examples

J(7,1) = 7      // people are counted out in order 1 2 3 4 5 6 [7]
J(7,2) = 7      // people are counted out in order 2 4 6 1 5 3 [7]
J(7,3) = 4      // see above
J(7,11) = 1
J(77,8) = 1
J(123,12) = 21

Answer 1

GolfScript, 17 bytes

{{@+\)%}+\,*)}:f;

Takes n k on the stack, and leaves the result on the stack.

Dissection

This uses the recurrence g(n,k) = (g(n-1,k) + k) % n with g(1, k) = 0 (as described in the Wikipedia article) with the recursion replaced by a fold.

{          # Function declaration
           # Stack: n k
  {        # Stack: g(i-1,k) i-1 k
    @+\)%  # Stack: g(i,k)
  }+       # Add, giving stack: n {k block}
  \,*      # Fold {k block} over [0 1 ... n-1]
  )        # Increment to move from 0-based to 1-based indexing
}:f;

Answer 2

Minsky Register Machine (25 non-halt states)

Not technically a function, but it's in a computing paradigm which doesn't have functions per se...

This is a slight variation on the main test case of my first MRM interpretation challenge:

Input in registers n and k; output in register r; it is assumed that r=i=t=0 on entry. The first two halt instructions are error cases.

Answer 3

Python, 36

I also used the formula from wikipedia:

J=lambda n,k:n<2or(J(n-1,k)+k-1)%n+1

Examples:

>>> J(7,3)
4
>>> J(77,8)
1
>>> J(123,12)
21

Answer 4

Mathematica, 38 36 bytes

Same Wikipedia formula:

1~f~_=1
n_~f~k_:=Mod[f[n-1,k]+k,n,1]

Answer 5

C, 40 chars

This is pretty much just the formula that the above-linked wikipedia article gives:

j(n,k){return n>1?(j(n-1,k)+k-1)%n+1:1;}

For variety, here's an implementation that actually runs the simulation (99 chars):

j(n,k,c,i,r){char o[999];memset(o,1,n);for(c=k,i=0;r;++i)(c-=o[i%=n])||(o[i]=!r--,c=k);
return i;}

Answer 6

dc, 27 bytes

[d1-d1<glk+r%1+]dsg?1-skrxp

Uses the recurrence from the Wikipedia article. Explanation:

# comment shows what is on the stack and any other effect the instructions have
[   # n
d   # n, n
1-  # n-1, n
d   # n-1, n-1, n
1<g # g(n-1), n ; g is executed only if n > 1, conveniently g(1) = 1
lk+ # g(n-1)+(k-1), n; remember, k register holds k-1
r%  # g(n-1)+(k-1) mod n
1+  # (g(n-1)+(k-1) mod n)+1
]
dsg # code for g; code also stored in g
?   # read user input => k, n, code for g
1-  # k-1, n, code for g
sk  # n, code for g; k-1 stored in register k
r   # code for g, n
x   # g(n)
p   # prints g(n)

Answer 7

J, 45 characters

j=.[:{.@{:]([:}.]|.~<:@[|~#@])^:(<:@#)>:@i.@[

Runs the simulation.

Alternatively, using the formula (31 characters):

j=.1:`(1+[|1-~]+<:@[$:])@.(1<[)

I hope Howard doesn't mind that I've adjusted the input format slightly to suit a dyadic verb in J.

Usage:

   7 j 3
4
   123 j 12
21

Answer 8

GolfScript, 32 24 bytes

:k;0:^;(,{))^k+\%:^;}/^)

Usage: Expects the two parameters n and k to be in the stack and leaves the output value.

(thanks to Peter Taylor for suggesting an iterative approach and many other tips)

The old (recursive) approach of 32 chars:

{1$1>{1$(1$^1$(+2$%)}1if@@;;}:^;

This is my first GolfScript, so please let me know your criticisms.

Answer 9

Groovy, 36 bytes

def j(n,k){n>1?(j(n-1,k)+k-1)%n+1:1}

Answer 10

R, 48

J=function(n,k)ifelse(n<2,1,(J(n-1,k)+k-1)%%n+1)

Running Version with examples: http://ideone.com/i7wae

Answer 11

Scala, 53 bytes

def?(n:Int,k:Int):Int=if(n<2)1 else(?(n-1,k)+k-1)%n+1

Answer 12

Haskell, 68

j n=q$cycle[0..n]
q l@(i:h:_)k|h/=i=q(drop(k-1)$filter(/=i)l)k|1>0=i

Does the actual simulation. Demonstration:

GHCi> j 7 1
7
GHCi> j 7 2
7
GHCi> j 7 3
4
GHCi> j 7 11
1
GHCi> j 77 8
1
GHCi> j 123 12
21

Answer 13

J, 8 bytes

1&|.&.#:

       1&|.&.#: 10
    5

       1&|.&.#: 69
    11

        1&|.&.#: 123456
    115841

        1&|.&.#: 123245678901234567890x NB. x keeps input integral
    98917405212792722853

All credit to Roger Hui, co-inventor of J and all-round uber-genius
www.jsoftware.com for free j software across many platforms

Explanation
    (J works right-to-left)
     #:       convert input to binary
     &.       a&.b <=> perform b, perform a, perform reverse of b
     1&|.     rotate bitwise one bit left

So
    1&|.&.#: 10

    a. #:            convert input (10) TO binary -> 1 0 1 0
    b. 1&|.          rotate result 1 bit left -> 0 1 0 1
    c. due to the &. perform convert FROM binary -> 5 (answer)

Answer 14

Ruby, 39 bytes

def J(n,k)
n<2?1:(J(n-1,k)+k-1)%n+1
end

Running version with test cases: http://ideone.com/pXOUc

Answer 15

C, 88 chars

Does the simulation, doesn't calculate the formula.
Much longer than the formula, but shorter than the other C simulation.

j(n,k){
    int i=0,c=k,r=n,*p=calloc(n,8);
    for(;p[i=i%n+1]||--c?1:--r?p[i]=c=k:0;);
    return i;
}

Notes:
1. Allocates memory and never releases.
2. Allocates n*8 instead of n*4, because I use p[n]. Could allocate (n+1)*4, but it's more characters.

Answer 16

Q, 34 bytes

f:{$[x=1;1;1+mod[;x]f[x-1;y]+y-1]}

Usage:

q)f .'(7 1;7 2;7 3;7 11;77 8;123 12)
7 7 4 1 1 21

Answer 17

Ruby, 34 bytes

J=->n,k{n<2?1:(J(n-1,k)+k-1)%n+1}

Answer 18

C++, 166 bytes

Golfed:

#include<iostream>
#include <list>
int j(int n,int k){if(n>1){return(j(n-1,k)+k-1)%n+1;}else{return 1;}}
int main(){intn,k;std::cin>>n>>k;std::cout<<j(n,k);return 0;}

Ungolfed:

#include<iostream>
#include <list>
int j(int n,int k){
    if (n > 1){
        return (j(n-1,k)+k-1)%n+1;
    } else {
        return 1;
    }
}
int main()
{
    int n, k;
    std::cin>>n>>k;
    std::cout<<j(n,k);
    return 0;
}

Answer 19

JavaScript (ECMAScript 5), 48 bytes

Using ECMAScript 5 since that was the latest version of JavaScript at the time this question was asked.

function f(a,b){return a<2?1:(f(a-1,b)+b-1)%a+1}

ES6 version (non-competing), 33 bytes

f=(a,b)=>a<2?1:(f(a-1,b)+b-1)%a+1

Explanation

Not much to say here. I'm just implementing the function the Wikipedia article gives me.

Answer 20

05AB1E, 11 bytes

L[Dg#²<FÀ}¦

Try it online!

L           # Range 1 .. n
 [Dg#       # Until the array has a length of 1:
     ²<F }  #   k - 1 times do:
        À   #     Rotate the array
          ¦ #   remove the first element

Answer 21

8th, 82 bytes

Code

: j >r >r a:new ( a:push ) 1 r> loop ( r@ n:+ swap n:mod ) 0 a:reduce n:1+ rdrop ;

SED (Stack Effect Diagram) is: n k -- m

Usage and explanation

The algorithm uses an array of integers like this: if people value is 5 then the array will be [1,2,3,4,5]

: j \ n k -- m
    >r                               \ save k
    >r a:new ( a:push ) 1 r> loop    \ make array[1:n]
    ( r@ n:+ swap n:mod ) 0 a:reduce \ translation of recursive formula with folding using an array with values ranging from 1 to n
    n:1+                             \ increment to move from 0-based to 1-based indexing
    rdrop                            \ clean r-stack
;

ok> 7 1 j . cr
7
ok> 7 2 j . cr
7
ok> 7 3 j . cr
4
ok> 7 11 j . cr
1
ok> 77 8 j . cr
1
ok> 123 12 j . cr
21

Answer 22

J, 24 bytes

1+1{1([:|/\+)/@,.1|.!.0#

Try it online!

An iterative approach based on the dynamic programming solution.

Explanation

1+1{1([:|/\+)/@,.1|.!.0#  Input: n (LHS), k (RHS)
                       #  Make n copies of k
                 1|.!.0   Shift left by 1 and fill with zero
    1          ,.         Interleave with 1
             /@           Reduce using
           +                Addition
        |/\                 Cumulative reduce using modulo
  1{                      Select value at index 1
1+                        Add 1

Answer 23

J, 19 bytes

1+(}:@|.^:({:@])i.)

Try it online!

How it works

1+(}:@|.^:({:@])i.)   Left: k, Right: n
                i.    Generate [0..n-1]
        ^:            Repeat:
   }:@|.                Rotate left k items, and remove the last item
          ({:@])        n-1 (tail of [0..n-1]) times
1+                    Add one to make the result one-based

Answer 24

Dart, 33 bytes

f(n,k)=>n<2?1:(f(n-1,k)+k-1)%n+1;

Try it online!

Answer 25

Japt, 15 bytes

_é1-V Å}h[Uõ] Ì

Try it online!

A byte could be saved by 0-indexing k, but it isn't actually an index so I decided against that.

Explanation:

         [Uõ]      :Starting with the array [1...n]
_      }h          :Repeat n-1 times:
 é1-V              : Rotate the array right 1-k times (i.e. rotate left k-1 times)
      Å            : Remove the new first element
              Ì    :Get the last value remaining

Answer 26

Powershell, 56 bytes

param($n,$k)if($n-lt2){1}else{((.\f($n-1)$k)+$k-1)%$n+1}

Important! The script calls itself recursively. So save the script as f.ps1 file in the current directory. Also you can call a script block variable instead script file (see the test script below). That calls has same length.

Test script:

$f = {

param($n,$k)if($n-lt2){1}else{((&$f($n-1)$k)+$k-1)%$n+1}

}

@(
    ,(7, 1, 7)
    ,(7, 2, 7)
    ,(7, 3, 4)
    ,(7, 11, 1)
    ,(77, 8, 1)
    ,(123,12, 21)
) | % {
    $n,$k,$expected = $_
    $result = &$f $n $k
    "$($result-eq$expected): $result"
}

Output:

True: 7
True: 7
True: 4
True: 1
True: 1
True: 21

Answer 27

Japt -h, 10 bytes

õ
£=éVn1¹v

Try it

Answer 28

Haskell, 29

Using the formula from wikipedia.

1#_=1
n#k=mod((n-1)#k+k-1)n+1