On 30 May 2013 22:03, Carlos Nepomuceno <carlosnepomuc...@outlook.com> wrote: >> Here's another way, mathematically equivalent (although not necessarily >> equivalent using floating point computations!) which avoids the divide-by- >> zero problem: >> >> abs(a - b) < epsilon*a > > That's wrong! If abs(a) < abs(a-b)/epsilon you will break the commutative law.
There is no commutative law for relative tolerance floating point comparisons. If you want to compare with a relative tolerance then you you should choose carefully what your tolerance is to be relative to (and how big your relative tolerance should be). In some applications it's obvious which of a or b you should use to scale the tolerance but in others it is not or you should compare with something more complex. For an example where it is obvious, when testing numerical code I might write something like: eps = 1e-7 true_answer = 123.4567879 estimate = myfunc(5) assert abs(estimate - true_answer) < eps * abs(true_answer) Oscar -- http://mail.python.org/mailman/listinfo/python-list
