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:
- First, check if a and b are non-zero.
- If both of them are zero and c is non-zero then, no solution exists. If c is also zero then infinite solution exits.
- For given a and b, calculate the value of x1, y1, and gcd using Extended Euclidean Algorithm.
- Now, for a solution to existing gcd(a, b) should be multiple of c.
- 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 algorithmint 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 = cvoid 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 Codeint 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 approachclass GFG{static int x, y; // Function to implement the extended// euclid algorithmstatic 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 = cstatic 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 Codepublic 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 mathx, y = 0, 0 # Function to implement the extended# euclid algorithmdef 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 print the solutions of# the given equations ax + by = cdef 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 coefficientsa = 4b = 18c = 10# Function Callprint_solution(a, b, c)# This code is contributed by decode2207. |
C#
// C# program for the above approachusing System;class GFG{static int x, y; // Function to implement the extended// euclid algorithmstatic 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 = cstatic 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 Codepublic 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 algorithmfunction 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 print the solutions of// the given equations ax + by = cfunction 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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