Function as parameter of a function, using bisection method C++ - c++

Question

// Function as parameter of a function
#include <iostream>
#include <cmath>
#include <cassert>
using namespace std;
const double PI = 4 * atan(1.0); // tan^(-1)(1) == pi/4 then 4*(pi/4)== pi
typedef double(*FD2D)(double);
double root(FD2D, double, double); //abscissae of 2-points,
//For the algorithm to work, the function must assume values of opposite sign in these two points, check
// at point 8
double polyn(double x) { return 3 - x*(1 + x*(27 - x * 9)); }
int main() {
double r;
cout.precision(15);
r = root(sin, 3, 4);
cout << "sin: " << r << endl
<< "exactly: " << PI << endl << endl;
r = root(cos, -2, -1.5);
cout << "cos: " << r << endl
<< "exactly: " << -PI/2 << endl << endl;
r = root(polyn, 0, 1);
cout << "polyn: " << r << endl
<< "exactly: " << 1./3 << endl << endl;
/*
we look for the root of the function equivalent to function polyn
but this time defined as a lambda function
*/
r = root([](double x) -> double {
return 3 - x*(1 + x*(27 - x * 9));
}, 0, 1);
cout << "lambda: " << r << endl
<< "exactly: " << 1. / 3 << endl << endl;
return 0;
}
// Finding root of function using bisection.
// fun(a) and fun(b) must be of opposite sign
double root(FD2D fun, double a, double b) {
static const double EPS = 1e-15; // 1×10^(-15)
double f, s, h = b - a, f1 = fun(a), f2 = fun(b);
if (f1 == 0) return a;
if (f2 == 0) return b;
assert(f1*f2<0); // 8.- macro assert from header cassert.
do {
if ((f = fun((s = (a + b) / 2))) == 0) break;
if (f1*f2 < 0) {
f2 = f;
b = s;
}
else {
f1 = f;
a = s;
}
} while ((h /= 2) > EPS);
return (a + b) / 2;
}
Could somebody explain me how the loop in double root function works? I don't seem to understand 100%, I checked this bisection method online and try on paper, but I can't manage to figure it out from this example.
Thanks in advance!

Answer 1

It's a lot easier to understand if you split the "clever" line into multiple lines. Here's some modifications, plus comments:
double root(FD2D fun, double a, double b) {
static const double EPS = 1e-15; // 1×10^(-15)
double fa = fun(a), fb = fun(b);
// if either f(a) or f(b) are the root, return that
// nothing else to do
if (fa == 0) return a;
if (fb == 0) return b;
// this method only works if the signs of f(a) and f(b)
// are different. so just assert that
assert(fa * fb < 0); // 8.- macro assert from header cassert.
do {
// calculate fun at the midpoint of a,b
// if that's the root, we're done
// this line is awful, never write code like this...
//if ((f = fun((s = (a + b) / 2))) == 0) break;
// prefer:
double midpt = (a + b) / 2;
double fmid = fun(midpt);
if (fmid == 0) return midpt;
// adjust our bounds to either [a,midpt] or [midpt,b]
// based on where fmid ends up being. I'm pretty
// sure the code in the question is wrong, so I fixed it
if (fa * fmid < 0) { // fmid, not f1
fb = fmid;
b = midpt;
}
else {
fa = fmid;
a = midpt;
}
} while (b-a > EPS); // only loop while
// a and b are sufficiently far
// apart
return (a + b) / 2; // approximation
}