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

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 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 value of x1, y1 and gcd using Extended Euclidean Algorithm.
  4. Now, for solution to exist gcd(a, b) should be multiple of c.
  5. Calculate solution of the equation as follows:
    x = x1 * ( c / gcd )
    y = y1 * ( c / gcd )
    

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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C#

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// 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

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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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