Steven D'Aprano wrote: >>With two numbers, 15 and 30, all you really need is five test cases:
My solution assumed integers also, but after I posted it, I thought: "What about floating points?" On Tue, Oct 7, 2014 at 1:48 AM, Steven D'Aprano <st...@pearwood.info> wrote: > On Sun, 05 Oct 2014 20:18:13 -0400, Seymore4Head wrote: > >> I think I get it now. You are using a sample of answers. So you could >> actually just run through them all. (I haven't tried this yet) >> >> for x in range(lo,hi) >> print((15 <= x < 30) == (15<= x and x <30)) > > Yes, except using print is probably not the best idea, since you might > have dozens of True True True True ... printed, one per line, and if you > blink the odd False might have scrolled off screen before you notice. > > With two numbers, 15 and 30, all you really need is five test cases: > > - a number lower than the smaller of the two numbers (say, 7); > - a number equal to the smaller of the two numbers (that is, 15); > - a number between the two numbers (say, 21); > - a number equal to the larger of the two numbers (that is, 30); > - a number higher than the larger of the two numbers (say, 999); > > > The exact numbers don't matter, so long as you test all five cases. And > rather than printing True True True... let's use assert instead: > > > for x in (7, 15, 21, 30, 999): > assert (15 <= x < 30) == (15<= x and x <30) > > > If the two cases are equal, assert will do nothing. But if they are > unequal, assert will raise an exception and stop, and you know that the > two cases are not equivalent and can go on to the next possibility. > > > -- > Steven > -- > https://mail.python.org/mailman/listinfo/python-list -- https://mail.python.org/mailman/listinfo/python-list
