Flavius Josephus

February 19, 2009

Flavius Josephus was a famous Jewish historian of the first century, at the time of the destruction of the Second Temple. According to legend, during the Jewish-Roman war he was trapped in a cave with a group of forty soldiers surrounded by Romans. Preferring death to capture, the Jews decided to form a circle and, proceeding around it, to kill every third person remaining until no one was left. Josephus found the safe spot in the circle and thus stayed alive.

Write a function josephus(n,m) that returns a list of n people, numbered from 0 to n-1, in the order in which they are executed, every mth person in turn, with the sole survivor as the last person in the list. What is the value of josephus(41,3)? In what position did Josephus survive?

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18 Responses to “Flavius Josephus”

  1. Sasha B said

    Depending on your interpretation of “every third person”, this question could be done a few different ways.

    def josephus(n, m):
        """Returns a list of n people, numbered from 0 to n - 1,
        in the order in which they are executed (every mth person
        in turn). The sole surviver is the last person in the list.
        >>> josephus(40, 3) # Kill every 3rd person
        people = range(n)
        current = m - 1
        executions = []
        while len(people) > 1:
            current += m - 1 # Advance m people
            current = current % len(people) # In case we go past the first person
        return executions + [people[0]]
  2. […] For the correct Haskell solution check the first comment on the challenge page. […]

  3. kawas said

    I’ve tried two clojure versions : one using modulus, filtering “alive” vector and an other using the trick of the front and back queue.
    Both functions have the same size but the second one is clearly an order of magnitude faster.
    Seems like modulus and filtering entire vector are expensive compared to only playing with head, tail and last element of front & back vectors :)

    (defn josephus [n m]
      (loop [idx m front (range n) back [] dead []]
          (and (empty? front) (empty? back)) dead
          (empty? front) (recur idx back [] dead)
          (= idx 1) (recur m (next front) back (conj dead (first front)))
          :else (recur (dec idx) (next front) (conj back (first front)) dead))))
    (josephus 41 3)
  4. Alex said

    My solution using C
    void josephus ( int n, int m)
    int i, j, kill = m – 1;
    int reset = 0;
    int circle[n], killOrder[n];

    for (i = 0; i < n; i++)
    circle[i] = i;

    for ( i = 0; i < n; i++)
    killOrder[i] = kill;
    circle[kill] = -1;
    for ( j = 0; j n – 1)
    kill = 0;
    while (circle[kill] == -1)


    for (i = 0; i < n; i++)
    printf("%d\t%d \n", i, killOrder[i]);


  5. Alex said

    Ok i fail at posting. Not sure how to do the code block thingy. Copy pasting lost some of my code.

  6. Alex said

    OK.. sorry about the triple post… I read the HOWTO post source code section.

    And here is my solution using C:


  7. My variant using Python lambda’s and list slices:


  8. Yuri said

    Scheme solution that makes use of circular lists:

  9. j0sejuan said


    package main
    import (
    func josephus(N, k int) {
    	r := ring.New(N)
    	for n := 0; n < N; n++ {
    		r.Value = n
    	for r.Len() > 1 {
    		fmt.Printf("%d killed\n", r.Value.(int))
    func main() {
    	josephus(41, 3)
  10. kevin galkov said

    in coffeescript:

    murder = (howMany, bounce) ->
    people = [0..howMany] # people from 0 to 40
    for i in people
    people[i] = 1
    killed = Array()
    for i in [1..howMany]
    whichOne = i*bounce – howMany*count
    count++ if whichOne> (howMany-bounce)
    if people[whichOne] is 1
    killed.push whichOne
    people[whichOne] = 0;
    for i in [0..howMany]
    if people[i] is 1 then console.log i
    output = ""
    for i in killed
    output += i+", "
    console.log output
    console.log "not killed:"
    console.log killed[howMany-1]

    view raw
    hosted with ❤ by GitHub

  11. […] is another one taken from Programming Praxis #5 – Flavius Josephus, has a historical background which can be found here. For this exercise, write a function that […]

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  13. Christophe said

    Solution using Clojure. Very slow, but correct.

    (defn rotate  
      [n coll] 
      (let [c (count coll)] 
        (take c (drop (mod n c) (cycle coll))))) 
    (defn execute
      [m ps dead]
      (if (< (count ps) m)
        (reverse (concat (reverse ps) dead))
        (let [executed (nth ps (dec m))
              alive    (take (dec m) ps)
              rest     (drop m ps)
              newalive (rotate (dec m) (concat alive rest))
              newdead  (cons executed dead)]
          (execute m newalive newdead (dec limit)))))
    (defn josephus
      [n m]
      (let [people (range 0 n)]
       (execute m people '())))
  14. Bill Wood said

    Rust version

    /*  Author Bill Wood
        Order N
        "killed" people are tracked using a vec, where each entry contains the index of
        the next "living" person, and is updated to bypass each person who is "killed"
    struct Josephus {
        people: Vec<usize>,
        m: usize,
        current: usize,
        count: usize,
    impl Iterator for Josephus {
        type Item = usize;
        fn next(&mut self) -> Option<Self::Item> {
            if self.count == 0 {
                return None
            let mut prev = self.current;
            for _i in 0..self.m {
                prev = self.current;
                self.current = self.people[self.current];
            self.count -= 1;
            self.people[prev] = self.people[self.current];
    impl Josephus {
        fn new(size: usize, m: usize) -> Self {
            let mut j = Josephus{ count: size, m: m, current: size - 1, people: vec!() };
            for i in 0..size - 1 {
                j.people.push(i + 1);
    fn main() {
        let mut x = Josephus::new(41, 3).peekable();
        while let Some(p) = x.next() {
            if x.peek().is_none() {
                println!("\nfinal position = {}", p)
            } else {
                print!("{} ", p);

    Playground: https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=1cf29532b1f6ff85b5afd189cebb8ec3

  15. matthew said

    If we are doing old problems, here’s a neat recursive solution in Python (not the most efficient being n^2, we can use a similar trick to find the final position in linear time though).

    def josephus(m,k):
        if m == 1: return [0]
        else: return [k-1] + [(k+i)%m for i in josephus(m-1,k)]
    # [2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 0,
    #  4, 9, 13, 18, 22, 27, 31, 36, 40, 6, 12, 19, 25, 33,
    #  39, 7, 16, 28, 37, 10, 24, 1, 21, 3, 34, 34, 30]
  16. matthew said

    Oops, should have checked that more carefully (note repeated 34 in “solution” there). I hope this is right:

    def josephus(m,k):
        if m == 1: return [0]
        else: return [(k-1)%m] + [(k+i)%m for i in josephus(m-1,k)]

    We should be able to define a similar but more complex recurrence that takes a complete trip around the remaining circle at each step.

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