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12.16 Program 7: Linear Diophantine Equations In this PROGRAM, you will be solving linear Diophantine equations recursively. A linear Diophantine equation is an equation in the form: ax + by = c where a, b, and c are all integers and the solutions will also be integers. See the following entry in Wikipedia: Linear Diophantine equations. You will be solving this using the recursive version of the Extended Euclidean algorithm for finding the integers x and y in Bezout's identity: ax + by = gcd (a,b) • You will need a Greatest Common Divisor helper function gcd(int x, int y): A recursive algorithm for GCD: the GCD of x and y is y if x mod y is 0 otherwise the GCD is the GCD of y and the remainder of x/y Required Recursive Function /* Returns true if a solution was found and false if there is no solution. x and y will contain a solution if a solution was found. This function will NOT output anything. */ bool diophantine (int a, int b, int c, int &x, int &y); Basic Algorithm • If gcd(a,b) does not divide c, then there is no solution. • If b divides a (the remainder of a / bis e), then you can just divide by b to get the solution: x = 0, y = c / b. • Otherwise (b does not divide a), through a substitution method, we can come up with a simpler version of the original problem and solve the simpler problem using recursion.

12.16 Program 7: Linear Diophantine Equations In this PROGRAM, you will be solving linear Diophantine equations recursively. A linear Diophantine equation is an equation in the form: ax + by = c where a, b, and c are all integers and the solutions will also be integers. See the following entry in Wikipedia: Linear Diophantine equations. You will be solving this using the recursive version of the Extended Euclidean algorithm for finding the integers x and y in Bezout's identity: ax + by = gcd (a,b) • You will need a Greatest Common Divisor helper function gcd(int x, int y): A recursive algorithm for GCD: the GCD of x and y is y if x mod y is 0 otherwise the GCD is the GCD of y and the remainder of x/y Required Recursive Function /* Returns true if a solution was found and false if there is no solution. x and y will contain a solution if a solution was found. This function will NOT output anything. */ bool diophantine (int a, int b, int c, int &x, int &y); Basic Algorithm • If gcd(a,b) does not divide c, then there is no solution. • If b divides a (the remainder of a / bis e), then you can just divide by b to get the solution: x = 0, y = c / b. • Otherwise (b does not divide a), through a substitution method, we can come up with a simpler version of the original problem and solve the simpler problem using recursion.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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C++ 

given main function

int main() {

int a, b, c, x, y;

cout << "Enter a b c" << endl;
cin >> a >> b >> c;
cout << endl;

cout << "Result: ";
if (diophantine(a, b, c, x, y)) {
cout << x << " " << y << endl;
}
else {
cout << "No solution!" << endl;
}

return 0;
}

12.16 Program 7: Linear Diophantine Equations In this PROGRAM, you will be solving linear Diophantine equations recursively. A linear Diophantine equation is an equation in the form: ax + by =c where a, b, and c are all integers and the solutions will also be integers. See the following entry in Wikipedia: Linear Diophantine equations. You will be solving this using the recursive version of the Extended Euclidean algorithm for finding the integers x and y in Bezout's identity: ax + by = gcd (a, b) • You will need a Greatest Common Divisor helper function gcd(int x, int y): A recursive algorithm for GCD: the GCD of x and y is y if x mod y is 0 otherwise the GCD is the GCD of y and the remainder of x/y Required Recursive Function /* Returns true if a solution was found and false if there is no solution. x and y will contain a solution if a solution was found. This function will NOT output anything. */ bool diophantine (int a, int b, int c, int &X, int &y); Basic Algorithm • If gcd(a,b) does not divide c, then there is no solution. • If b divides a (the remainder of a / bis e), then you can just divide by b to get the solution: x = 0, y = c / b. • Otherwise (b does not divide a), through a substitution method, we can come up with a simpler version of the original problem and solve the simpler problem using recursion.
Transcribed Image Text:12.16 Program 7: Linear Diophantine Equations In this PROGRAM, you will be solving linear Diophantine equations recursively. A linear Diophantine equation is an equation in the form: ax + by =c where a, b, and c are all integers and the solutions will also be integers. See the following entry in Wikipedia: Linear Diophantine equations. You will be solving this using the recursive version of the Extended Euclidean algorithm for finding the integers x and y in Bezout's identity: ax + by = gcd (a, b) • You will need a Greatest Common Divisor helper function gcd(int x, int y): A recursive algorithm for GCD: the GCD of x and y is y if x mod y is 0 otherwise the GCD is the GCD of y and the remainder of x/y Required Recursive Function /* Returns true if a solution was found and false if there is no solution. x and y will contain a solution if a solution was found. This function will NOT output anything. */ bool diophantine (int a, int b, int c, int &X, int &y); Basic Algorithm • If gcd(a,b) does not divide c, then there is no solution. • If b divides a (the remainder of a / bis e), then you can just divide by b to get the solution: x = 0, y = c / b. • Otherwise (b does not divide a), through a substitution method, we can come up with a simpler version of the original problem and solve the simpler problem using recursion.
Substitution method ax + by = c Now, we can define a as: a = bg + r where q is (a / b) (using integer division) and r is the remainder (a % b). Substituting (bq + r) in for a now: (bg + r)x + by = c which is the same as b (сх + y) + rx 3D с and now we have the equation in the same form, only with smaller coefficients: bu + rv = c with u = qx + y and v = x Finally, you recursively call your function on this simpler version of the original problem. Don't forget that this recursive call will actually solve for u and v in this case, so you still have to solve for x and y to get the solution to the original problem: x = v y = u - qx Linear Diophantine Example Example showing steps of algorithm: LINK Input/Output Test Samples Here are some examples you can test your function on: Input (a b c) | Results (x y) | 28 7 49o I0 70 | 1024 96 2048 | -64 704 | No solution! | 11 11 2010 | 1984 3070 1 | 395 252 1 | No solution! | -37 58 | -6 4 I0 -2 | 4 0 | 25 38 2 | 200 -2 4 | 25 75 100 | 25 75 1000 | 25 75 1 | 40 0 | No solution! |0 -10 | -10 -10 100 | 12 24 48 | 4 0 | 5 -29 6 | 36 6
Transcribed Image Text:Substitution method ax + by = c Now, we can define a as: a = bg + r where q is (a / b) (using integer division) and r is the remainder (a % b). Substituting (bq + r) in for a now: (bg + r)x + by = c which is the same as b (сх + y) + rx 3D с and now we have the equation in the same form, only with smaller coefficients: bu + rv = c with u = qx + y and v = x Finally, you recursively call your function on this simpler version of the original problem. Don't forget that this recursive call will actually solve for u and v in this case, so you still have to solve for x and y to get the solution to the original problem: x = v y = u - qx Linear Diophantine Example Example showing steps of algorithm: LINK Input/Output Test Samples Here are some examples you can test your function on: Input (a b c) | Results (x y) | 28 7 49o I0 70 | 1024 96 2048 | -64 704 | No solution! | 11 11 2010 | 1984 3070 1 | 395 252 1 | No solution! | -37 58 | -6 4 I0 -2 | 4 0 | 25 38 2 | 200 -2 4 | 25 75 100 | 25 75 1000 | 25 75 1 | 40 0 | No solution! |0 -10 | -10 -10 100 | 12 24 48 | 4 0 | 5 -29 6 | 36 6
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