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Find initial integral solution of Linear Diophantine equation if finite solution exists

  • Difficulty Level : Medium
  • Last Updated : 27 Sep, 2021
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Given three integers a, b, and c representing a linear equation of the form: ax + by = c. The task is to find the initial integral solution of the given equation if a finite solution exists.

A Linear Diophantine equation (LDE) is an equation with 2 or more integer unknowns and the integer unknowns are each to at most degree of 1. Linear Diophantine equation in two variables takes the form of ax+by=c, where x,y are integer variables and a, b, c are integer constants. x and y are unknown variables.

Examples: 
 

Input: a = 4, b = 18, c = 10 
Output: x = -20, y = 5 
 

Explanation: (-20)*4 + (5)*18 = 10



Input: a = 9, b = 12, c = 5 
Output: No Solutions exists 
 

 

Approach: 
 

  1. First, check if a and are non-zero.
  2. If both of them are zero and c is non-zero then, no solution exists. If c is also zero then infinite solution exits.
  3. For given a and b, calculate the value of x1, y1, and gcd using Extended Euclidean Algorithm.
  4. Now, for a solution to existing gcd(a, b) should be multiple of c.
  5. Calculate the solution of the equation as follows:
x = x1 * ( c / gcd )
y = y1 * ( c / gcd )

Below is the implementation of the above approach:

 

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to implement the extended
// euclid algorithm
int gcd_extend(int a, int b,
               int& x, int& y)
{
    // Base Case
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
 
    // Recursively find the gcd
    else {
        int g = gcd_extend(b,
                           a % b, x, y);
        int x1 = x, y1 = y;
        x = y1;
        y = x1 - (a / b) * y1;
        return g;
    }
}
 
// Function to print the solutions of
// the given equations ax + by = c
void print_solution(int a, int b, int c)
{
    int x, y;
    if (a == 0 && b == 0) {
 
        // Condition for infinite solutions
        if (c == 0) {
            cout
                << "Infinite Solutions Exist"
                << endl;
        }
 
        // Condition for no solutions exist
        else {
            cout
                << "No Solution exists"
                << endl;
        }
    }
    int gcd = gcd_extend(a, b, x, y);
 
    // Condition for no solutions exist
    if (c % gcd != 0) {
        cout
            << "No Solution exists"
            << endl;
    }
    else {
 
        // Print the solution
        cout << "x = " << x * (c / gcd)
             << ", y = " << y * (c / gcd)
             << endl;
    }
}
 
// Driver Code
int main(void)
{
    int a, b, c;
 
    // Given coefficients
    a = 4;
    b = 18;
    c = 10;
 
    // Function Call
    print_solution(a, b, c);
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
static int x, y;
 
// Function to implement the extended
// euclid algorithm
static int gcd_extend(int a, int b)
{
     
    // Base Case
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
 
    // Recursively find the gcd
    else
    {
        int g = gcd_extend(b, a % b);
        int x1 = x, y1 = y;
        x = y1;
        y = x1 - (a / b) * y1;
        return g;
    }
}
 
// Function to print the solutions of
// the given equations ax + by = c
static void print_solution(int a, int b, int c)
{
    if (a == 0 && b == 0)
    {
         
        // Condition for infinite solutions
        if (c == 0)
        {
            System.out.print("Infinite Solutions " +
                             "Exist" + "\n");
        }
 
        // Condition for no solutions exist
        else
        {
            System.out.print("No Solution exists" +
                             "\n");
        }
    }
    int gcd = gcd_extend(a, b);
 
    // Condition for no solutions exist
    if (c % gcd != 0)
    {
        System.out.print("No Solution exists" + "\n");
    }
    else
    {
         
        // Print the solution
        System.out.print("x = " + x * (c / gcd) +
                       ", y = " + y * (c / gcd) + "\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int a, b, c;
 
    // Given coefficients
    a = 4;
    b = 18;
    c = 10;
 
    // Function Call
    print_solution(a, b, c);
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program for the above approach
import math
 
x, y = 0, 0
   
# Function to implement the extended
# euclid algorithm
def gcd_extend(a, b):
    global x, y
     
    # Base Case
    if (b == 0):
        x = 1
        y = 0
        return a
   
    # Recursively find the gcd
    else:
        g = gcd_extend(b, a % b)
        x1, y1 = x, y
        x = y1
        y = x1 - math.floor(a / b) * y1
        return g
   
# Function to prlet the solutions of
# the given equations ax + by = c
def print_solution(a, b, c):
    if (a == 0 and b == 0):
       
        # Condition for infinite solutions
        if (c == 0):
            print("Infinite Solutions Exist")
   
        # Condition for no solutions exist
        else :
            print("No Solution exists")
    gcd = gcd_extend(a, b)
   
    # Condition for no solutions exist
    if (c % gcd != 0):
        print("No Solution exists")
    else:
        # Print the solution
        print("x = ", int(x * (c / gcd)), ", y = ", int(y * (c / gcd)), sep = "")
         
# Given coefficients
a = 4
b = 18
c = 10
 
# Function Call
print_solution(a, b, c)
 
# This code is contributed by decode2207.

C#




// C# program for the above approach
using System;
 
class GFG{
 
static int x, y;
 
// Function to implement the extended
// euclid algorithm
static int gcd_extend(int a, int b)
{
     
    // Base Case
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
 
    // Recursively find the gcd
    else
    {
        int g = gcd_extend(b, a % b);
        int x1 = x, y1 = y;
        x = y1;
        y = x1 - (a / b) * y1;
        return g;
    }
}
 
// Function to print the solutions of
// the given equations ax + by = c
static void print_solution(int a, int b, int c)
{
    if (a == 0 && b == 0)
    {
         
        // Condition for infinite solutions
        if (c == 0)
        {
            Console.Write("Infinite Solutions " +
                          "Exist" + "\n");
        }
 
        // Condition for no solutions exist
        else
        {
            Console.Write("No Solution exists" +
                          "\n");
        }
    }
    int gcd = gcd_extend(a, b);
 
    // Condition for no solutions exist
    if (c % gcd != 0)
    {
        Console.Write("No Solution exists" + "\n");
    }
    else
    {
         
        // Print the solution
        Console.Write("x = " + x * (c / gcd) +
                    ", y = " + y * (c / gcd) + "\n");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int a, b, c;
 
    // Given coefficients
    a = 4;
    b = 18;
    c = 10;
 
    // Function call
    print_solution(a, b, c);
}
}
 
// This code contributed by amal kumar choubey

Javascript




<script>
 
// Javascript program for the above approach
  
let x, y;
 
// Function to implement the extended
// euclid algorithm
function gcd_extend(a, b)
{
     
    // Base Case
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
 
    // Recursively find the gcd
    else
    {
        let g = gcd_extend(b, a % b);
        let x1 = x, y1 = y;
        x = y1;
        y = x1 - Math.floor(a / b) * y1;
        return g;
    }
}
 
// Function to prlet the solutions of
// the given equations ax + by = c
function print_solution(a, b, c)
{
    if (a == 0 && b == 0)
    {
         
        // Condition for infinite solutions
        if (c == 0)
        {
            document.write("Infinite Solutions " +
                             "Exist" + "\n");
        }
 
        // Condition for no solutions exist
        else
        {
            document.write("No Solution exists" +
                             "\n");
        }
    }
    let gcd = gcd_extend(a, b);
 
    // Condition for no solutions exist
    if (c % gcd != 0)
    {
        document.write("No Solution exists" + "\n");
    }
    else
    {
         
        // Print the solution
        document.write("x = " + x * (c / gcd) +
                       ", y = " + y * (c / gcd) + "<br/>");
    }
}
 
// Driver Code
     
   let a, b, c;
 
    // Given coefficients
    a = 4;
    b = 18;
    c = 10;
 
    // Function Call
    print_solution(a, b, c);
         
</script>
Output: 
x = -20, y = 5

 

Time Complexity: O(log(max(A, B))), where A and B are the coefficient of x and y in the given linear equation. 
Auxiliary Space: O(1)
 

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