Integral solution to equation `a + bx = c + dy`

问题

In the equation a + bx = c + dy, all variables are integers. a, b, c, and d are known. How do I find integral solutions for x and y? If I'm thinking right, there will be an infinite number of solutions, separated by the lowest common multiple of b and d, but all I need is one solution, and I can calculate the rest. Here's an example:

a = 2
b = 3
c = 4
d = 5
a + bx: (2, 5,  8, 11, 14)
c + dy: (4, 9, 14, 19, 24)

a + bx intersects c + dy at 14, so:
x = 4
y = 2

Right now, I'm looping through integral values of x until I find an integral value for y (pseudocode):

function integral_solution(int a, int b, int c, int d) {
    // a + bx == c + dy
    // (a + bx - c) / d == y

    // Some parameters may have no integral solution,
    // for example if b == d and (a - c) % b != 0
    // If we don't have a solution before x == d, there is none.

    int x = 0;
    while (x < d)
    {
        if ((a + bx - c) % d == 0)
        {
            return [x, (a + bx - c) / d];
        }
        x++;
    }
    return false;
}

I feel like there's a better way to do this. Is there any way to find x and y without a loop? I'm using C++, if that's of any importance.

回答1:

Linear Diophantine equations take the form ax + by = c. If c is the greatest common divisor of a and b this means a=z'c and b=z''c then this is Bézout's identity of the form

with a=z' and b=z'' and the equation has an infinite number of solutions. So instead of trial searching method you can check if c is the greatest common divisor (GCD) of a and b (in your case this translates into bx - dy = c - a)

If indeed a and b are multiples of c then x and y can be computed using extended Euclidean algorithm which finds integers x and y (one of which is typically negative) that satisfy Bézout's identity

and your answer is:

a = k*x, b = k*y, c - a = k * gcd(a,b) for any integer k.

(as a side note: this holds also for any other Euclidean domain, i.e. polynomial ring & every Euclidean domain is unique factorization domain). You can use Iterative Method to find these solutions:

Iterative method

By routine algebra of expanding and grouping like terms (refer to last section of wikipedia article mentioned before), the following algorithm for iterative method is obtained:

1 . Apply Euclidean algorithm, and let qn (n starts from 1) be a finite list of quotients in the division.
2 . Initialize x0, x1 as 1, 0, and y0, y1 as 0,1 respectively.
- 2.1 Then for each i so long as qi is defined,
- 2.2 Compute xi+1 = xi−1 − qixi
- 2.3 Compute yi+1 = yi−1 − qiyi
- 2.4 Repeat the above after incrementing i by 1.
3 . The answers are the second-to-last of xn and yn.

pseudocode:

function extended_gcd(a, b)
    x := 0    lastx := 1
    y := 1    lasty := 0
    while b ≠ 0
        quotient := a div b
        (a, b) := (b, a mod b)
        (x, lastx) := (lastx - quotient*x, x)
        (y, lasty) := (lasty - quotient*y, y)       
    return (lastx, lasty)

So I have written example algorithm which calculates greatest common divisor using Euclidean Algorithm iterative method for non-negative a and b (for negative - these extra steps are needed), it returns GCD and stores solutions for x and y in variables passed to it by reference:

int gcd_iterative(int a, int b, int& x, int& y) {
    int c;
    std::vector<int> r, q, x_coeff, y_coeff;
    x_coeff.push_back(1); y_coeff.push_back(0);
    x_coeff.push_back(0); y_coeff.push_back(1);

    if ( b == 0 ) return a;
    while ( b != 0 ) {
            c = b;
            q.push_back(a/b);
            r.push_back(b = a % b);
            a = c;
            x_coeff.push_back( *(x_coeff.end()-2) -(q.back())*x_coeff.back());
            y_coeff.push_back( *(y_coeff.end()-2) -(q.back())*y_coeff.back());
    }
    if(r.size()==1) {
        x = x_coeff.back();
        y = y_coeff.back();
    } else {
        x = *(x_coeff.end()-2);
        y = *(y_coeff.end()-2);
    }
    std::vector<int>::iterator it;
    std::cout << "r: ";
    for(it = r.begin(); it != r.end(); it++) { std::cout << *it << "," ; }
    std::cout << "\nq: ";
    for(it = q.begin(); it != q.end(); it++) { std::cout << *it << "," ; }
    std::cout << "\nx: ";
    for(it = x_coeff.begin(); it != x_coeff.end(); it++){ std::cout << *it<<",";}
    std::cout << "\ny: ";
    for(it = y_coeff.begin(); it != y_coeff.end(); it++){ std::cout << *it<<",";}
    return a;
}

by passing to it an example from wikipedia for a = 120 and b = 23 we obtain:

int main(int argc, char** argv) {   
    // 120x + 23y = gcd(120,23)
    int x_solution, y_solution;
    int greatestCommonDivisor =  gcd_iterative(120, 23, x_solution, y_solution);
    return 0;
}

r: 5,3,2,1,0,

q: 5,4,1,1,2,

x: 1,0,1,-4,5,-9,23,

y: 0,1,-5,21,-26,47,-120,

what is in accordance with the given table for this example: