# Josephus Problem when k is 2

• Difficulty Level : Medium
• Last Updated : 02 Mar, 2023

There are n people standing in a circle waiting to be executed. The counting out begins at some point in the circle and proceeds around the circle in a fixed direction. In each step, a certain number of people are skipped and the next person is executed. The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom. Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle. The task is to choose the place in the initial circle so that you are the last one remaining and so survive.
We have discussed a generalized solution in below set 1.
Josephus problem | Set 1 (A O(n) Solution)
In this post, a special case is discussed when k = 2

Examples :

Input  : n = 5
Output : The person at position 3 survives
Explanation : Firstly, the person at position 2 is killed,
then at 4, then at 1 is killed. Finally, the person at
position 5 is killed. So the person at position 3 survives.

Input  : n = 14
Output : The person at position 13 survives

Below are some interesting facts.

• In first round all even positioned persons are killed.
• For second round two cases arise
1. If n is even : For example n = 8. In first round, first 2 is killed, then 4, then 6, then 8. In second round, we have 1, 3, 5 and 7 in positions 1st, 2nd, 3rd and 4th respectively.
2. If n is odd : For example n = 7. In first round, first 2 is killed, then 4, then 6. In second round, we have 3, 5, 7 in positions 1st, 2nd and 3rd respectively.

If n is even and a person is in position x in current round, then the person was in position 2x – 1 in previous round.
If n is odd and a person is in position x in current round, then the person was in position 2x + 1 in previous round.
From above facts, we can recursively define the formula for finding position of survivor.

Let f(n) be position of survivor for input n,
the value of f(n) can be recursively written
as below.

If n is even
f(n) = 2f(n/2) - 1
Else
f(n) = 2f((n-1)/2) + 1

Solution of above recurrence is

f(n) = 2(n - 2floor(Log2n) + 1
= 2n - 21 + floor(Log2n) + 1

Below is the implementation to find value of above formula.

## C++

 // C/C++ program to find solution of Josephus// problem when size of step is 2.#include  // Returns position of survivor among a circle// of n persons and every second person being// killedint josephus(int n){    // Find value of 2 ^ (1 + floor(Log n))    // which is a power of 2 whose value    // is just above n.    int p = 1;    while (p <= n)        p *= 2;     // Return 2n - 2^(1+floor(Logn)) + 1    return (2 * n) - p + 1;} // Driver Program to test above functionint main(){    int n = 16;    printf("The chosen place is %d", josephus(n));    return 0;}

## Java

 // Java program to find solution of Josephus// problem when size of step is 2.import java.io.*; class GFG {     // Returns position of survivor among    // a circle of n persons and every    // second person being killed    static int josephus(int n)    {         // Find value of 2 ^ (1 + floor(Log n))        // which is a power of 2 whose value        // is just above n.        int p = 1;        while (p <= n)            p *= 2;         // Return 2n - 2^(1+floor(Logn)) + 1        return (2 * n) - p + 1;    }     // Driver Program to test above function    public static void main(String[] args)    {        int n = 16;         System.out.println("The chosen place is "                           + josephus(n));    }} // This Code is Contributed by Anuj_67

## Python3

 # Python3 program to find solution of# Josephus problem when size of step is 2. # Returns position of survivor among a# circle of n persons and every second# person being killeddef josephus(n):         # Find value of 2 ^ (1 + floor(Log n))    # which is a power of 2 whose value    # is just above n.    p = 1    while p <= n:        p *= 2     # Return 2n - 2^(1 + floor(Logn)) + 1    return (2 * n) - p + 1 # Driver Coden = 16print ("The chosen place is", josephus(n)) # This code is contributed by Shreyanshi Arun.

## C#

 // C# program to find solution of Josephus// problem when size of step is 2.using System; class GFG {     // Returns position of survivor among    // a circle of n persons and every    // second person being killed    static int josephus(int n)    {         // Find value of 2 ^ (1 + floor(Log n))        // which is a power of 2 whose value        // is just above n.        int p = 1;        while (p <= n)            p *= 2;         // Return 2n - 2^(1+floor(Logn)) + 1        return (2 * n) - p + 1;    }     // Driver Program to test above function    static void Main()    {        int n = 16;         Console.Write("The chosen place is "                      + josephus(n));    }} // This Code is Contributed by Anuj_67



## Javascript



Output:

The chosen place is 1

Time complexity of above solution is O(Log n).

An another interesting solution to the problem while k=2 can be given based on an observation, that we just have to left rotate the binary representation of N to get the required answer. A working code for
the same is provided below considering number to be 64-bit number.

Below is the implementation of the above approach:

## C++

 // C++ program to find solution of Josephus// problem when size of step is 2.#include  using namespace std; // Returns position of survivor among a circle// of n persons and every second person being// killedint josephus(int n){    // An interesting observation is that    // for every number of power of two    // answer is 1 always.    if (!(n & (n - 1)) && n) {        return 1;    }     // The trick is just to right rotate the    // binary representation of n once.    // Find whether the number shed off    // during left shift is set or not     bitset<64> Arr(n);     // shifting the bitset Arr    // f will become true once leftmost    // set bit is found    bool f = false;    for (int i = 63; i >= 0; --i) {        if (Arr[i] == 1 && !f) {            f = true;            Arr[i] = Arr[i - 1];        }        if (f) {             // shifting bits            Arr[i] = Arr[i - 1];        }    }     Arr[0] = 1;     int res;     // changing bitset to int    res = (int)(Arr.to_ulong());    return res;} // Driver Program to test above functionint main(){    int n = 16;    printf("The chosen place is %d", josephus(n));    return 0;}

## Java

 public class Main {  public static int josephus(int n)  {         // An interesting observation is that    // for every number of power of two    // answer is 1 always.    if (~((n & (n - 1))) != 0 && n != 0) {      return 1;    }     // The trick is just to right rotate the    // binary representation of n once.    // Find whether the number shed off    // during left shift is set or not    int[] arr = new int[64];    char[] bin = Integer.toBinaryString(n).toCharArray();    int i = 64 - bin.length;    for (int j = 0; j < bin.length; j++) {      arr[i++] = bin[j] - '0';    }     // shifting the bitset arr    // f will become true once leftmost    // set bit is found    boolean f = false;    for (i = 63; i >= 0; i--) {      if (arr[i] == 1 && !f) {        f = true;        arr[i] = arr[i - 1];      }       if (f) {        // shifting bits        arr[i] = arr[i - 1];      }    }     arr[0] = 1;     // changing bitset to int    int res = 0;    for (i = 0; i < 64; i++) {      res += arr[i] * (1L << (63 - i));    }    return res;  }   public static void main(String[] args) {    int n = 16;    System.out.println("The chosen place is " + josephus(n));  }}

## Python3

 # Python 3 program to find solution of Josephus# problem when size of step is 2.  # Returns position of survivor among a circle# of n persons and every second person being# killeddef josephus(n):    # An interesting observation is that    # for every number of power of two    # answer is 1 always.    if (~(n & (n - 1)) and n) :        return 1          # The trick is just to right rotate the    # binary representation of n once.    # Find whether the number shed off    # during left shift is set or not     Arr=list(map(lambda x:int(x),list(bin(n)[2:])))    Arr=[0]*(64-len(Arr))+Arr     # shifting the bitset Arr    # f will become true once leftmost    # set bit is found    f = False    for i in range(63,-1,-1) :        if (Arr[i] == 1 and not f) :            f = True            Arr[i] = Arr[i - 1]                 if (f) :             # shifting bits            Arr[i] = Arr[i - 1]                   Arr[0] = 1     # changing bitset to int    res = int(''.join(Arr),2)    return res  # Driver Program to test above functionif __name__ == '__main__':    n = 16    print("The chosen place is", josephus(n))

## C#

 using System; public class Mainn{    public static long josephus(long n)    {         // An interesting observation is that        // for every number of power of two        // answer is 1 always.        if (~((n & (n - 1))) != 0 && n != 0)        {            return 1;        }         // The trick is just to right rotate the        // binary representation of n once.        // Find whether the number shed off        // during left shift is set or not        int[] arr = new int[64];        char[] bin = Convert.ToString(n, 2).ToCharArray();        int i = 64 - bin.Length;        for (int j = 0; j < bin.Length; j++)        {            arr[i++] = bin[j] - '0';        }         // shifting the bitset arr        // f will become true once leftmost        // set bit is found        bool f = false;        for (i = 63; i >= 0; i--)        {            if (arr[i] == 1 && !f)            {                f = true;                arr[i] = arr[i - 1];            }             if (f)            {                // shifting bits                arr[i] = arr[i - 1];            }        }         arr[0] = 1;         // changing bitset to long        long res = 0;        for (i = 0; i < 64; i++)        {            res += arr[i] * (1L << (63 - i));        }        return res;    }     public static void Main(string[] args)    {        long n = 16;        Console.WriteLine("The chosen place is " + josephus(n));    }}

## Javascript

 function josephus(n) {  // An interesting observation is that  // for every number of power of two  // answer is 1 always.  if (~(n & (n - 1)) && n) {    return 1;  }   // The trick is just to right rotate the  // binary representation of n once.  // Find whether the number shed off  // during left shift is set or not  let Arr = n.toString(2).split('').map(x => parseInt(x));  Arr = Array(64 - Arr.length).fill(0).concat(Arr);   // shifting the bitset Arr  // f will become true once leftmost  // set bit is found  let f = false;  for (let i = 63; i >= 0; i--) {    if (Arr[i] == 1 && !f) {      f = true;      Arr[i] = Arr[i - 1];    }     if (f) {       // shifting bits      Arr[i] = Arr[i - 1];    }  }   Arr[0] = 1;   // changing bitset to int  let res = parseInt(Arr.join(''), 2);  return res;} // Driver Program to test above functionlet n = 16;console.log("The chosen place is", josephus(n));

Output:

The chosen place is 1

Time Complexity : O(log(n))
Auxiliary Space: O(log(n))
This idea is contributed by Anukul Chand
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