Lecture 8 handout question:

def exp3(a,b):
if b == 1:
return a
if (b%2)*2 == b:
return exp3(a*a, b/2)
else: return a*exp3(a,b-1)

line 4 tests whether or not b is even. b%2 is used which will return either a 0 or a 1. Shouldn't it be b/2 instead?

Question

Lecture 8 handout question: 

def exp3(a,b):
    if b == 1:
        return a
    if (b%2)*2 == b:
        return exp3(a*a, b/2)
    else: return a*exp3(a,b-1)

line 4 tests whether or not b is even. b%2 is used which will return either a 0 or a 1. Shouldn't it be b/2 instead?

vaboro · Answer

If b is not integer b%2 will not be 0 if b is even; whereas (b%2)*2 will be b

anonymous · Answer

Both ways should work correctly. They actually do the same thing (test b to see if it's even).

anonymous · Answer

Sorry-but I still don't see it... if b is 6 (even), then 6%2 is 0, right? 0 * 2 is 0, making 0==b  false. So an even number will skip this loop and the else loop (for odd numbers) will run. The if loop never runs. I'm missing something obvious, aren't I?

vaboro · Answer

Right, if it were division instead of modulus, (b/2)*b then it would make perfect sense.
Anyway, I should probably need to refresh the lecture in order to be able to tell for sure what it means.