Josephus Problem Using Bit Magic

15 Jun, 2022
The Problem
This problem is named after Flavius Josephus a Jewish historian who fought against the Romans. According to Josephus he and his group of Jewish soldiers were cornered & surrounded by the Romans inside a cave, and they choose to murder and suicide inside of surrender and capture. They decided that all the soldiers will sit in a circle and starting from the soldier sitting at the first position every soldier will kill the soldier to their sequentially. So if there are 5 soldiers sitting in a circle with positions numbered as 1, 2, 3, 4, 5. The soldier 1 kills 2, then 3 kills 4, then 5 kills 1, then 3 kills 5, and since 3 is the only one left then 3 commits suicide. Now Josephus doesn’t want to get murdered or commit suicide. He would rather be captured by the Romans and is presented with a problem. He has to figure out at which position should he sit in a circle (provided there are n men in total and the man sitting at position 1 gets the first chance to murder) so that he is the last man standing and instead of committing suicide he will surrender to the Romans. 
The Pattern
If you work this out for different values of n you will find a pattern here. If n is a true power of 2 then the answer is always 1. For every n greater than that power of 2 the answer is incremented by 2.  
Now for every n, the right position for Josephus can be found out by deducting the biggest possible power of 2 from the number and we get the answer (provided that value of n is not a pure power of 2 otherwise the answer is 1)  
N = 2a + something
where a = the biggest possible power
The TrickWhenever someone talks about the powers of 2 the first word that comes to mind is “binary”. The solution to this problem is much is easier and shorter in binary than in decimal. There is a trick to this. Since we need to deduct the biggest possible power of in binary that number is the Most Significant Bit. In the original Josephus problem, there were 40 other soldiers along with Josephus which makes n = 41. 41 in binary is 101001. If we shift the MSB i.e. the leftmost 1 to the rightmost place we get 010011 which is 19 (in decimal) which is the answer. This is true for all cases. This can be done easily using bit manipulation.
C++
C
Java
Python3
C#
PHP
Javascript
// C++ program for josephus problem#include <bits/stdc++.h>using namespace std; // function to find the position of the Most// Significant Bitint msbPos(int n){    int pos = 0;    while (n != 0) {        pos++;         // keeps shifting bits to the right        // until we are left with 0        n = n >> 1;    }    return pos;} // function to return at which place Josephus// should sit to avoid being killedint josephify(int n){    /* Getting the position of the Most Significant        Bit(MSB). The leftmost '1'. If the number is    '41' then its binary is '101001'.        So msbPos(41) = 6 */    int position = msbPos(n);     /* 'j' stores the number with which to XOR the        number 'n'. Since we need '100000'    We will do 1<<6-1 to get '100000' */    int j = 1 << (position - 1);     /* Toggling the Most Significant Bit. Changing    the leftmost '1' to '0'.    101001 ^ 100000 = 001001 (9) */    n = n ^ j;     /* Left-shifting once to add an extra '0' to        the right end of the binary number        001001 = 010010 (18) */    n = n << 1;     /* Toggling the '0' at the end to '1' which    is essentially the same as putting the    MSB at the rightmost place. 010010 | 1    = 010011 (19) */    n = n | 1;     return n;} // hard coded driver main function to run the programint main(){    int n = 41;    cout <<josephify(n);    return 0;}// This code is contributed by Mukul singh.
// C program for josephus problem #include <stdio.h> // function to find the position of the Most// Significant Bitint msbPos(int n){    int pos = 0;    while (n != 0) {        pos++;         // keeps shifting bits to the right        // until we are left with 0        n = n >> 1;    }    return pos;} // function to return at which place Josephus// should sit to avoid being killedint josephify(int n){    /*  Getting the position of the Most Significant        Bit(MSB). The leftmost '1'. If the number is       '41' then its binary is '101001'.        So msbPos(41) = 6   */    int position = msbPos(n);     /* 'j' stores the number with which to XOR the        number 'n'. Since we need '100000'       We will do 1<<6-1 to get '100000' */    int j = 1 << (position - 1);     /* Toggling the Most Significant Bit. Changing       the leftmost '1' to '0'.       101001 ^ 100000 = 001001 (9) */    n = n ^ j;     /*  Left-shifting once to add an extra '0' to        the right end of the binary number        001001 = 010010 (18) */    n = n << 1;     /* Toggling the '0' at the end to '1' which       is essentially the same as putting the       MSB at the rightmost place. 010010 | 1       = 010011 (19) */    n = n | 1;     return n;} // hard coded driver main function to run the programint main(){    int n = 41;    printf("%d\n", josephify(n));    return 0;}
// Java program for josephus problem public class GFG{    // method to find the position of the Most    // Significant Bit    static int msbPos(int n)    {        int pos = 0;        while (n != 0) {            pos++;                  // keeps shifting bits to the right            // until we are left with 0            n = n >> 1;        }        return pos;    }          // method to return at which place Josephus    // should sit to avoid being killed    static int josephify(int n)    {        /*  Getting the position of the Most Significant            Bit(MSB). The leftmost '1'. If the number is           '41' then its binary is '101001'.            So msbPos(41) = 6   */        int position = msbPos(n);              /* 'j' stores the number with which to XOR the            number 'n'. Since we need '100000'           We will do 1<<6-1 to get '100000' */        int j = 1 << (position - 1);              /* Toggling the Most Significant Bit. Changing           the leftmost '1' to '0'.           101001 ^ 100000 = 001001 (9) */        n = n ^ j;              /*  Left-shifting once to add an extra '0' to            the right end of the binary number            001001 = 010010 (18) */        n = n << 1;              /* Toggling the '0' at the end to '1' which           is essentially the same as putting the           MSB at the rightmost place. 010010 | 1           = 010011 (19) */        n = n | 1;              return n;    }         // Driver Method    public static void main(String[] args)    {        int n = 41;        System.out.println(josephify(n));    }}
# Python3 program for josephus problem # Function to find the position# of the Most Significant Bitdef msbPos(n):    pos = 0    while n != 0:        pos += 1        n = n >> 1    return pos # Function to return at which# place Josephus should sit to# avoid being killeddef josephify(n):         # Getting the position of the Most    # Significant Bit(MSB). The leftmost '1'.    # If the number is '41' then its binary    # is '101001'. So msbPos(41) = 6    position = msbPos(n)     # 'j' stores the number with which to XOR     # the number 'n'. Since we need '100000'    # We will do 1<<6-1 to get '100000'    j = 1 << (position - 1)     # Toggling the Most Significant Bit.    # Changing the leftmost '1' to '0'.    # 101001 ^ 100000 = 001001 (9)    n = n ^ j     # Left-shifting once to add an extra '0'     # to the right end of the binary number    # 001001 = 010010 (18)    n = n << 1     # Toggling the '0' at the end to '1'     # which is essentially the same as    # putting the MSB at the rightmost    # place. 010010 | 1 = 010011 (19)    n = n | 1     return n # Driver Coden = 41print (josephify(n)) # This code is contributed by Shreyanshi Arun.
// C# program for Josephus Problemusing System; public class GFG{         // Method to find the position    // of the Most Significant Bit    static int msbPos(int n)    {        int pos = 0;        while (n != 0) {            pos++;                 // keeps shifting bits to the right            // until we are left with 0            n = n >> 1;        }        return pos;    }         // method to return at which place Josephus    // should sit to avoid being killed    static int josephify(int n)    {                 // Getting the position of the Most Significant        // Bit(MSB). The leftmost '1'. If the number is        // '41' then its binary is '101001'.        // So msbPos(41) = 6        int position = msbPos(n);             // 'j' stores the number with which to XOR         // the number 'n'. Since we need '100000'        // We will do 1<<6-1 to get '100000'        int j = 1 << (position - 1);             // Toggling the Most Significant Bit.        // Changing the leftmost '1' to '0'.        // 101001 ^ 100000 = 001001 (9)        n = n ^ j;             // Left-shifting once to add an extra '0'        // to the right end of the binary number        // 001001 = 010010 (18)        n = n << 1;             // Toggling the '0' at the end to '1' which        // is essentially the same as putting the        // MSB at the rightmost place. 010010 | 1        // = 010011 (19)        n = n | 1;             return n;    }         // Driver code    public static void Main()    {        int n = 41;        Console.WriteLine(josephify(n));    }} // This code is contributed by vt_m .
<?php// PHP program for josephus problem // function to find the position of// the Most Significant Bitfunction msbPos($n){    $pos = 0;    while ($n != 0)    {        $pos++;         // keeps shifting bits to the right        // until we are left with 0        $n = $n >> 1;    }    return $pos;} // function to return at which place Josephus// should sit to avoid being killedfunction josephify($n){    /* Getting the position of the Most       Significant Bit(MSB). The leftmost '1'.       If the number is '41' then its binary       is '101001'. So msbPos(41) = 6 */    $position = msbPos($n);     /* 'j' stores the number with which to        XOR the number 'n'. Since we need    '100000'. We will do 1<<6-1 to get '100000' */    $j = 1 << ($position - 1);     /* Toggling the Most Significant Bit.    Changing the leftmost '1' to '0'.    101001 ^ 100000 = 001001 (9) */    $n = $n ^ $j;     /* Left-shifting once to add an extra '0'       to the right end of the binary number       001001 = 010010 (18) */    $n = $n << 1;     /* Toggling the '0' at the end to '1' which    is essentially the same as putting the    MSB at the rightmost place. 010010 | 1    = 010011 (19) */    $n = $n | 1;     return $n;} // Driver Code$n = 41;print(josephify($n)); // This code is contributed by mits?>
<script> // javascript program for josephus problem // method to find the position of the Most// Significant Bitfunction msbPos(n){    var pos = 0;    while (n != 0) {        pos++;          // keeps shifting bits to the right        // until we are left with 0        n = n >> 1;    }    return pos;}  // method to return at which place Josephus// should sit to avoid being killedfunction josephify(n){    /*  Getting the position of the Most Significant        Bit(MSB). The leftmost '1'. If the number is       '41' then its binary is '101001'.        So msbPos(41) = 6   */    var position = msbPos(n);      /* 'j' stores the number with which to XOR the        number 'n'. Since we need '100000'       We will do 1<<6-1 to get '100000' */    var j = 1 << (position - 1);      /* Toggling the Most Significant Bit. Changing       the leftmost '1' to '0'.       101001 ^ 100000 = 001001 (9) */    n = n ^ j;      /*  Left-shifting once to add an extra '0' to        the right end of the binary number        001001 = 010010 (18) */    n = n << 1;      /* Toggling the '0' at the end to '1' which       is essentially the same as putting the       MSB at the rightmost place. 010010 | 1       = 010011 (19) */    n = n | 1;      return n;} // Driver Methodvar n = 41;document.write(josephify(n)); // This code is contributed by Amit Katiyar</script>
Output:  
19
Time Complexity: O(log n)Space Complexity: O(1)
Previous articles on the same topic:  
Josephus problem | Set 1 (A O(n) Solution)Josephus problem | Set 2 (A Simple Solution when k = 2)
Josephus problem | Set 1 (A O(n) Solution)
Josephus problem | Set 2 (A Simple Solution when k = 2)
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