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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 \$\require{cancel}1 \, 2 \, \cancel{3} \, 4 \, 5 \, \cancel{6} \, 7 \, 1 \, \cancel{2} \, 4 \, 5 \, \cancel{7} \, 1 \, 4 \, \cancel{5} \, 1 \, 4 \, \cancel{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
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31 Answers 31

Reset to default
1
2
14
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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: Josephus problem as Minsky register machine

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.

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4
  • \$\begingroup\$ I think you have to adjust your machine slightly. If I read it correctly the output is zero-indexed, isn't it? \$\endgroup\$
    – Howard
    May 21, 2012 at 13:45
  • \$\begingroup\$ I was thinking the other way: if k=1 then r=0. Hmm, I have to think about this one again... \$\endgroup\$
    – Howard
    May 21, 2012 at 14:40
  • \$\begingroup\$ As I read your diagram, i is simply counting from 2 to n while r is the register which accumulates the result. \$\endgroup\$
    – Howard
    May 21, 2012 at 15:39
  • \$\begingroup\$ @Howard, I looked up the comments I made when I first wrote this and you're right. Whoops. Now corrected (I believe - will test more thoroughly later). \$\endgroup\$
    – Peter Taylor
    May 21, 2012 at 16:19
7
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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
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6
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Mathematica, 38 36 bytes

Same Wikipedia formula:

1~f~_=1
n_~f~k_:=Mod[f[n-1,k]+k,n,1]
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1
  • 1
    \$\begingroup\$ If[#<2,1,Mod[#0[#-1,#2]+#2,#,1]]& \$\endgroup\$
    – alephalpha
    May 7, 2015 at 5:27
5
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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;
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6
  • \$\begingroup\$ Can you add an explanation, please? \$\endgroup\$
    – Sherlock9
    Nov 26, 2015 at 18:49
  • \$\begingroup\$ @Sherlock9, I managed to figure out what I was doing despite almost 3.5 years having passed. Who says that GolfScript is read-only? ;) \$\endgroup\$
    – Peter Taylor
    Nov 26, 2015 at 21:45
  • 1
    \$\begingroup\$ Ahem. s/read/write/ \$\endgroup\$
    – Peter Taylor
    Nov 26, 2015 at 22:40
  • \$\begingroup\$ Sorry. I've only started learning Golfscript two or three days ago and I every time I read your code, I kept thinking I'd missed something. ... Ok, I'm still missing something, how does folding {k block} over [0..n-1] get you g(0,k) 0 k to start with? Sorry, if I'm posting these questions in the wrong place. \$\endgroup\$
    – Sherlock9
    Nov 27, 2015 at 5:30
  • \$\begingroup\$ @Sherlock9, fold works pairwise, so the first thing it does is evaluate 0 1 block. Very conveniently, that happens to be g(1, k) (2-1) block. So it's starting at g(1,k) 1 rather than g(0,k) 0. Then after executing the block, it pushes the next item from the array (2) and executes the block again, etc. \$\endgroup\$
    – Peter Taylor
    Nov 27, 2015 at 6:47
5
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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;}
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1
  • 4
    \$\begingroup\$ Save a character: j(n,k){return--n?(j(n,k)+k-1)%-~n+1:1;}. \$\endgroup\$
    – ugoren
    May 20, 2012 at 10:06
5
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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)
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4
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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
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4
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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.

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14
  • 1
    \$\begingroup\$ 1- has special opcode (. Similarly 1+ is ). You don't have to use alphabetic characters for storage, so you could use e.g. ^ instead of J and not need a space after it. You have far more $s than are usual in a well-golfed program: consider whether you can reduce them using some combination of \@.. \$\endgroup\$
    – Peter Taylor
    May 21, 2012 at 7:27
  • \$\begingroup\$ @PeterTaylor Thanks a lot for these great tips! It's pretty hard to grasp all the Golfscript operators and I overlooked these two very straightforward one. Only by applying the first two suggestions I manage to shorten the code by 5 chars. I'll also try to remove the $ references. \$\endgroup\$
    – Cristian Lupascu
    May 21, 2012 at 8:23
  • 1
    \$\begingroup\$ Also, recursion isn't really GolfScript's thing. Try flipping it round and doing a loop. I can get it down to 19 chars (albeit untested code) that way. Hint: unroll the function g from the Wikipedia article, and use , and /. \$\endgroup\$
    – Peter Taylor
    May 21, 2012 at 9:21
  • 1
    \$\begingroup\$ {,{\2$+\)%}*)\;}:f; Make sure you understand why it works ;) \$\endgroup\$
    – Peter Taylor
    May 21, 2012 at 17:12
  • 1
    \$\begingroup\$ One final trick: rather than using 2 characters to access k inside the loop and then 2 more to discard it at the end, we can pull it inside using + to get down to 17 characters: {{@+\)%}+\,*)}:f; I doubt that can be improved. \$\endgroup\$
    – Peter Taylor
    May 21, 2012 at 18:13
3
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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

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3
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Groovy, 36 bytes

def j(n,k){n>1?(j(n-1,k)+k-1)%n+1:1}
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2
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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

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2
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Scala, 53 bytes

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

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1
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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;
}
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3
  • 2
    \$\begingroup\$ You could save bytes on the J function, by using the ternary operator. \$\endgroup\$
    – Yytsi
    Jun 26, 2016 at 9:52
  • \$\begingroup\$ intn in your golfed version won't compile \$\endgroup\$
    – Felipe Nardi Batista
    Oct 24, 2017 at 10:56
  • \$\begingroup\$ you can remove space in #include <list> \$\endgroup\$
    – Felipe Nardi Batista
    Oct 24, 2017 at 10:56
1
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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

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1
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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
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1
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Ruby, 34 bytes

J=->n,k{n<2?1:(J(n-1,k)+k-1)%n+1}
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1
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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
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1
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Nibbles, 6 bytes

+~/\,@0%+_@

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Uses the formula as described on Wikipedia: \$J(n,k) = g(n,k) + 1\$, where

$$\begin{align}g(1,k) &= 0 \\ g(n,k) &= (g(n-1,k)+k) \bmod n\end{align}$$

+~/\,@0%+_@
+~          Add 1
  /          fold right
   \          reverse
    ,          range from 1 to
     @          n
      0       with initial value 0
       %      modulo
        +      add
         _      k
          @     accumutalor
               item
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0
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Haskell, 29

Using the formula from wikipedia.

1#_=1
n#k=mod((n-1)#k+k-1)n+1
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0
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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.

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0
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05AB1E, 11 bytes

L[Dg#²<FÀ}¦

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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
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0
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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
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0
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J, 24 bytes

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

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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
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0
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J, 19 bytes

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

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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
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0
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Dart, 33 bytes

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

Try it online!

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0
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Japt, 15 bytes

_é1-V Å}h[Uõ] Ì

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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
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0
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Japt -h, 10 bytes

õ
£=éVn1¹v

Try it

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0
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APL (Dyalog Unicode), 17 bytes

{¯1↓⍺⌽⍵}⍣{1=≢⍺}∘⍳

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Takes input as k f n.

Explanation

{¯1↓⍺⌽⍵}⍣{2=≢⍵}∘⍳
               ∘⍳ list from 1..n
        ⍣         (f⍣g) → apply f repeatedly till g is true
{¯1↓⍺⌽⍵}          f: rotate list through k elements, drop last
         {1=≢⍺}   g: is the length = 1, for the previous iteration?

APL (Dyalog Unicode), 36 bytes

1+{⍺>1:⍺|⍵+⍵∇⍨⍺-1⋄0}

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Recursive function.

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0
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Jelly, 7 bytes

RṙṖ¥ƬṖṪ

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How it works

RṙṖ¥ƬṖṪ - Main link. Takes n on the left and k on the right
R       - Range; Yield [1, 2, ..., n]
   ¥Ƭ   - Do the following to a fixed point and collect intermediate steps:
 ṙ      -   Rotate k steps to the left
  Ṗ     -   Remove the last element
     Ṗ  - Remove the empty list (the fixed point)
      Ṫ - Return the single element left over
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1
2

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