## 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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### 13 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:
executions.append(people.pop(current))
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 []]
(cond
(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)
kill++;
}

}

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:

http://pastebin.com/T08Jpj7G

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

http://pastebin.com/pVgE9hF2

8. Yuri said

Scheme solution that makes use of circular lists:
http://pastebin.com/faTX0j6q

9. j0sejuan said

golang

```package main

import (
"fmt"
"container/ring"
)

func josephus(N, k int) {
r := ring.New(N)
for n := 0; n < N; n++ {
r.Value = n
r.Next()
}
for r.Len() > 1 {
fmt.Printf("%d killed\n", r.Value.(int))