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 = 5Explanation: (-20)*4 + (5)*18 = 10
Input: a = 9, b = 12, c = 5
Output: No Solutions exists
Approach:
- First check if a and 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 value of x1, y1 and gcd using Extended Euclidean Algorithm.
- Now, for solution to exist gcd(a, b) should be multiple of c.
- Calculate 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 |
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 |
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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