Largest odd divisor Game to check which player wins
Last Updated :
09 Jan, 2023
Two players are playing a game starting with a number n. In each turn, a player can make any one of the subsequent moves:
- Divide n by any of its odd divisors greater than 1. Divisors of a number include the number itself.
- Subtract 1 from n if n > k where k < n.
Player 1 makes the primary move, print “yes” if player 1 wins otherwise print “no” if both play optimally. The player who is unable to make a move loses the game.
Examples:
Input: n = 12, k = 1
Output: Yes
Explanation:
Player 1 first move = 12 / 3 = 4
Player 2 first move = 4 – 1 = 3
Player 1 second move = 3 / 3 = 1
Player 2 second move can be done and hence he loses.
Input: n = 1, k = 1
Output: No
Explanation:
Player 1 first move is not possible because n = k and hence player 1 loses.
Approach: The idea is to analyze the problem for the following 3 cases:
- When integer n is odd, player 1 can divide n by itself, since it is odd and hence n / n = 1, and player 2 loses. Note that here n = 1 is an exception.
- When integer n is even and has no odd divisors greater than 1 then n is of the form 2x. Player 1 is bound to subtract it by 1 making n odd. So if x > 1, player 2 wins. Note that for x = 1, n – 1 is equal to 1, so Player 1 wins.
- When integer n is even and has odd divisors, the task remains to check if n is divisible by 4 then player 1 can divide n by its largest odd factor after which n becomes of the form 2x where x > 1, so again player 1 wins.
- Otherwise, n must be of form 2 * p, where p is odd. If p is prime, player 1 loses since he can either reduce n by 1 or divide it by p both of which would be losing for him. If p is not prime then p must be of the form p1 * p2 where p1 is prime and p2 is any odd number > 1, for which player 1 can win by dividing n by p2.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void findWinner( int n, int k)
{
int cnt = 0;
if (n == 1)
cout << "No" << endl;
else if ((n & 1) or n == 2)
cout << "Yes" << endl;
else {
int tmp = n;
int val = 1;
while (tmp > k and tmp % 2 == 0) {
tmp /= 2;
val *= 2;
}
for ( int i = 3; i <= sqrt (tmp); i++) {
while (tmp % i == 0) {
cnt++;
tmp /= i;
}
}
if (tmp > 1)
cnt++;
if (val == n)
cout << "No" << endl;
else if (n / tmp == 2 and cnt == 1)
cout << "No" << endl;
else
cout << "Yes" << endl;
}
}
int main()
{
long long n = 1, k = 1;
findWinner(n, k);
return 0;
}
|
Java
import java.util.*;
class GFG{
public static void findWinner( int n, int k)
{
int cnt = 0 ;
if (n == 1 )
System.out.println( "No" );
else if ((n & 1 ) != 0 || n == 2 )
System.out.println( "Yes" );
else
{
int tmp = n;
int val = 1 ;
while (tmp > k && tmp % 2 == 0 )
{
tmp /= 2 ;
val *= 2 ;
}
for ( int i = 3 ;
i <= Math.sqrt(tmp); i++)
{
while (tmp % i == 0 )
{
cnt++;
tmp /= i;
}
}
if (tmp > 1 )
cnt++;
if (val == n)
System.out.println( "No" );
else if (n / tmp == 2 && cnt == 1 )
System.out.println( "No" );
else
System.out.println( "Yes" );
}
}
public static void main(String[] args)
{
int n = 1 , k = 1 ;
findWinner(n, k);
}
}
|
Python3
import math
def findWinner(n, k):
cnt = 0 ;
if (n = = 1 ):
print ( "No" );
elif ((n & 1 ) or n = = 2 ):
print ( "Yes" );
else :
tmp = n;
val = 1 ;
while (tmp > k and tmp % 2 = = 0 ):
tmp / / = 2 ;
val * = 2 ;
for i in range ( 3 , int (math.sqrt(tmp)) + 1 ):
while (tmp % i = = 0 ):
cnt + = 1 ;
tmp / / = i;
if (tmp > 1 ):
cnt + = 1 ;
if (val = = n):
print ( "No" );
elif (n / tmp = = 2 and cnt = = 1 ):
print ( "No" );
else :
print ( "Yes" );
if __name__ = = "__main__" :
n = 1 ; k = 1 ;
findWinner(n, k);
|
C#
using System;
class GFG{
public static void findWinner( int n, int k)
{
int cnt = 0;
if (n == 1)
Console.Write( "No" );
else if ((n & 1) != 0 || n == 2)
Console.Write( "Yes" );
else
{
int tmp = n;
int val = 1;
while (tmp > k && tmp % 2 == 0)
{
tmp /= 2;
val *= 2;
}
for ( int i = 3;
i <= Math.Sqrt(tmp); i++)
{
while (tmp % i == 0)
{
cnt++;
tmp /= i;
}
}
if (tmp > 1)
cnt++;
if (val == n)
Console.Write( "No" );
else if (n / tmp == 2 && cnt == 1)
Console.Write( "No" );
else
Console.Write( "Yes" );
}
}
public static void Main( string [] args)
{
int n = 1, k = 1;
findWinner(n, k);
}
}
|
Javascript
<script>
function findWinner(n, k)
{
let cnt = 0;
if (n == 1)
document.write( "No" );
else if ((n & 1) != 0 || n == 2)
document.write( "Yes" );
else
{
let tmp = n;
let val = 1;
while (tmp > k && tmp % 2 == 0)
{
tmp /= 2;
val *= 2;
}
for (let i = 3;
i <= Math.sqrt(tmp); i++)
{
while (tmp % i == 0)
{
cnt++;
tmp /= i;
}
}
if (tmp > 1)
cnt++;
if (val == n)
document.write( "No" );
else if (n / tmp == 2 && cnt == 1)
document.write( "No" );
else
document.write( "Yes" );
}
}
let n = 1, k = 1;
findWinner(n, k);
</script>
|
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
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