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Check if it is possible to reach (X, Y) from (1, 0) by given steps

  • Last Updated : 07 Apr, 2021

Given two positive integers X and Y, the task is to check if it is possible to reach (X, Y) from (1, 0) by the given steps. In each step, possible moves from any cell (a, b) are (a, b + a) or (a + b, b). Print “Yes” if possible. Otherwise, print “No”.

Examples:

Input: X = 2, Y = 7
Output: Yes
Explanation: Sequence of moves to reach (2, 7) are: (1, 0) -> (1, 1) -> (2, 1) -> (2, 3) -> (2, 5) -> (2, 7).

Input: X = 30, Y = 24
Output: No

Naive Approach: The simplest approach is to try to move from points (X, Y) to (1, 0) recursively by using the operation (X – Y, Y) or (X, Y – X) until it becomes equals to (1, 0). If the X-coordinate becomes less than 1 or Y-coordinate becomes less than 0, then it is not possible to reach (1, 0). Therefore, print “No”. Otherwise, if (1, 0) is reached, print “Yes”



Time Complexity: O(2log (min(X, Y)))
Auxiliary Space: O(1)

Efficient Approach: The idea is to observe the following properties:

  • Try to solve the problem in reverse order i.e., it is possible to move from (X, Y) to (1, 0) by reclusively moving to points (X – Y, Y) or (X, Y – X).
  • The above property can be represented as follows:

GCD(X, Y) = GCD(X, Y – X) or GCD(X – Y, Y)
where,  
Base Case is GCD(X, 0) = X
Now, notice that gcd of 1 and 0 i.e., gcd(1, 0) is 1.
Therefore, gcd of X and Y must also be 1 to reach (1, 0).

Therefore, from the above observations, the path from (1, 0) to (X, Y) always exists if GCD(X, Y) = 1. Print “Yes” if the GCD of the given two numbers is 1. Otherwise, print “No”.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <iostream>
using namespace std;
 
// Function to find the GCD of two
// numbers a and b
int GCD(int a, int b)
{
    // Base Case
    if (b == 0)
        return a;
 
    // Recursively find the GCD
    else
        return GCD(b, a % b);
}
 
// Function to check if (x, y) can be
// reached from (1, 0) from given moves
void check(int x, int y)
{
    // If GCD is 1, then print "Yes"
    if (GCD(x, y) == 1) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    // Given X and Y
    int X = 2, Y = 7;
 
    // Function call
    check(X, Y);
 
    return 0;
}

Java




// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function to find the
// GCD of two numbers a
// and b
static int GCD(int a,
               int b)
{
  // Base Case
  if (b == 0)
    return a;
 
  // Recursively find
  // the GCD
  else
    return GCD(b, a % b);
}
 
// Function to check if
// (x, y) can be reached
// from (1, 0) from given
// moves
static void check(int x,
                  int y)
{
  // If GCD is 1, then
  // print "Yes"
  if (GCD(x, y) == 1)
  {
    System.out.print("Yes");
  }
  else
  {
    System.out.print("No");
  }
}
 
// Driver Code
public static void main(String[] args)
{
  // Given X and Y
  int X = 2, Y = 7;
 
  // Function call
  check(X, Y);
}
}
 
// This code is contributed by shikhasingrajput

Python3




# Python3 program for the above approach
 
# Function to find the GCD of two
# numbers a and b
def GCD(a, b):
     
    # Base Case
    if (b == 0):
        return a
         
    # Recursively find the GCD
    else:
        return GCD(b, a % b)
 
# Function to check if (x, y) can be
# reached from (1, 0) from given moves
def check(x, y):
     
    # If GCD is 1, then pr"Yes"
    if (GCD(x, y) == 1):
        print("Yes")
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
     
    # Given X and Y
    X = 2
    Y = 7
 
    # Function call
    check(X, Y)
 
# This code is contributed by mohit kumar 29

C#




// C# program for the
// above approach
using System;
 
class GFG{
   
// Function to find the
// GCD of two numbers a
// and b
static int GCD(int a, int b)
{
   
  // Base Case
  if (b == 0)
    return a;
 
  // Recursively find
  // the GCD
  else
    return GCD(b, a % b);
}
 
// Function to check if
// (x, y) can be reached
// from (1, 0) from given
// moves
static void check(int x, int y)
{
   
  // If GCD is 1, then
  // print "Yes"
  if (GCD(x, y) == 1)
  {
    Console.WriteLine("Yes");
  }
  else
  {
    Console.WriteLine("No");
  }
}
 
// Driver Code
public static void Main()
{
   
  // Given X and Y
  int X = 2, Y = 7;
 
  // Function call
  check(X, Y);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript




<script>
// JavaScript program for the above approach
 
// Function to find the GCD of two
// numbers a and b
function GCD(a, b)
{
 
    // Base Case
    if (b == 0)
        return a;
 
    // Recursively find the GCD
    else
        return GCD(b, a % b);
}
 
// Function to check if (x, y) can be
// reached from (1, 0) from given moves
function check(x, y)
{
 
    // If GCD is 1, then print "Yes"
    if (GCD(x, y) == 1) {
        document.write("Yes");
    }
    else {
        document.write("No");
    }
}
 
// Driver Code
 
    // Given X and Y
    let X = 2, Y = 7;
 
    // Function call
    check(X, Y);
 
// This code is contributed by Surbhi Tyagi.
</script>
Output: 
Yes

 

Time Complexity: O(log(min(X, Y))
Auxiliary Space: O(1)

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