Bisection Method Implementation

Question

I'm working on a program to program the bisection method: https://www.calculushowto.com/bisection-method/

I know there's questions similar to this one but I want to see if my own one works.

double func(double x) {
    return x * x - 3 * x - 1;
}
    
double bisect(double (*f)(double), double a, double b, double e) {
    double mid = (a + b) / 2;
    while (abs(mid) > e) {
        if (f(mid) < 0) {
            mid = a;
        } else {
            mid = b;
        }
    }
      
    return mid;       
}

func() is the function I'm using to test the bisection method. In the other function, a is the left point, b is the right point and e is the error bound.

Any mistakes that I didn't catch?

Is there something you think is wrong? Or you are just looking for a code review? (if the latter, maybe try on [codereview](https://codereview.stackexchange.com) — Alex, Feb 02 '21 at 05:19
It may look like bisection at first sight, but I very much doubt it is. And that's not how the error bound works either. [Use a debugger to step through the program and understand what is going on and compare that with your expectations.](https://stackoverflow.com/questions/25385173/what-is-a-debugger-and-how-can-it-help-me-diagnose-problems) — , Feb 02 '21 at 08:03

Bob__ · Accepted Answer · 2021-02-02T10:28:57.470

There are multiple issues with the posted implementation, many of them could be easily addressed if the function were tested in a complete program.

The midpoint is calculated only once, at the beginning. Unless this was meant to be a recursive function, it should be updated inside the loop. To be nitpicking, the evaluation itself could be done differently, by first calculating a width = right - left and then mid = left + width / 2.0.
The termination condition is wrong, it should compare the width of the current interval with the limit or even the absolute value of f(mid) with a chosen minimal bound, not the value of mid.
Inside the loop, the value of mid is changed according to the signness of f(mid), but it should be the opposite: either the left bound or the right bound should be assigned to the value mid. The assumption that the function is ascending should be also somehow documented.
The function returns mid unconditionally, but either abs(f(a)) or abs(f(b)) could be less than abs(f(mid)).

Bisection Method Implementation

1 Answers1