# Josephus Problem Using Bit Magic

**The Problem**

This problem is named after Flavius Josephus a Jewish historian who fought against the Romans. According to Josephus him and his group of Jewish soldiers were cornered & surrounded by the Romans inside a cave and they chose 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 in a sequential way. 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 the 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.

n soldiers | 2^{a}+ l |
Survivor W(n) = 2l + 1 |
---|---|---|

1 | 1 + 0 | 2 * 0 + 1 = 1 |

2 | 2 + 0 | 2 * 0 + 1 = 1 |

3 | 2 + 1 | 2 * 1 + 1 = 3 |

4 | 4 + 0 | 2 * 0 + 1 = 1 |

5 | 4 + 1 | 2 * 1 + 1 = 3 |

6 | 4 + 2 | 2 * 2 + 1 = 5 |

7 | 4 + 3 | 2 * 3 + 1 = 7 |

8 | 8 + 0 | 2 * 0 + 1 = 1 |

9 | 8 + 1 | 2 * 1 + 1 = 3 |

10 | 8 + 2 | 2 * 2 + 1 = 5 |

11 | 8 + 3 | 2 * 3 + 1 = 7 |

12 | 8 + 4 | 2 * 4 + 1 = 9 |

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 = 2 ^{a} + something**

Where, a = biggest possible power

**The Trick**

Whenever 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++ program for josephus problem ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to find the position of the Most ` `// Significant Bit ` `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; ` `} ` ` ` `// function to return at which place Josephus ` `// should sit to avoid being killed ` `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; ` `} ` ` ` `// hard coded driver main function to run the program ` `int` `main() ` `{ ` ` ` `int` `n = 41; ` ` ` `cout <<josephify(n); ` ` ` `return` `0; ` `}` `// This code is contributed by Mukul singh. ` |

*chevron_right*

*filter_none*

## C

`// C program for josephus problem ` ` ` `#include <stdio.h> ` ` ` `// function to find the position of the Most ` `// Significant Bit ` `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; ` `} ` ` ` `// function to return at which place Josephus ` `// should sit to avoid being killed ` `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; ` `} ` ` ` `// hard coded driver main function to run the program ` `int` `main() ` `{ ` ` ` `int` `n = 41; ` ` ` `printf` `(` `"%d\n"` `, josephify(n)); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// 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)); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 program for josephus problem ` ` ` `# Function to find the position ` `# of the Most Significant Bit ` `def` `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 killed ` `def` `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 Shreyanshi Arun. ` |

*chevron_right*

*filter_none*

## C#

`// C# program for Josephus Problem ` `using` `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 . ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP program for josephus problem ` ` ` `// function to find the position of ` `// the Most Significant Bit ` `function` `msbPos(` `$n` `) ` `{ ` ` ` `$pos` `= 0; ` ` ` `while` `(` `$n` `!= 0) ` ` ` `{ ` ` ` `$pos` `++; ` ` ` ` ` `// keeps shifting bits to the right ` ` ` `// until we are left with 0 ` ` ` `$n` `= ` `$n` `>> 1; ` ` ` `} ` ` ` `return` `$pos` `; ` `} ` ` ` `// function to return at which place Josephus ` `// should sit to avoid being killed ` `function` `josephify(` `$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 ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

19

**References:**

**Previous articles on the same topic:**

This article is contributed by Palash Nigam . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Josephus problem | Set 1 (A O(n) Solution)
- Secretary Problem (A Optimal Stopping Problem)
- Magic Square
- Find nth Magic Number
- Magic Square | Even Order
- The Magic of Fibonacci Numbers
- Count Magic squares in a grid
- Check if a number is magic (Recursive sum of digits is 1)
- Fibonacci problem (Value of Fib(N)*Fib(N) - Fib(N-1) * Fib(N+1))
- Tiling Problem
- 21 Matchsticks Problem
- Frobenius coin problem
- The Lazy Caterer's Problem
- Problem of 8 Neighbours of an element in a 2-D Matrix
- The Knight's tour problem | Backtracking-1