I'm not sure if this belongs in this forum, but it is more an algorithm question than a programming, or language specific question.

Given a function that takes 2 integer parameters, f(a, b), it should return a number representing how much you have to add to a to make it a multiple of b.

Edit: If a is already a multiple of b then my formula returns b and your formula returns 0. You might think yours is "righter" than mine in that case but both satisfy the stated requirements.

Paul Clapham wrote:
Edit: If a is already a multiple of b then my formula returns b and your formula returns 0. You might think yours is "righter" than mine in that case but both satisfy the stated requirements.

I would call Garrett's solution "righter" because you "have to" add 0 to 6 to make it a multiple of 3, but you don't "have to" add 3. I interpreted the original problem statement to say we're looking for the minimum non-negative integer that satisfies the condition.

How about...

b - 1 - ((a -1 ) % b)

That formula replaces the second MOD operation from Garrett's version with an extra +1 and -1. However, it has a minor fail when a=0, in which case a-1 == -1 and -1 % b == -1. We could modify it a bit so that the left side of the % doesn't go negative for non-negative values of a:

(b-1) - ((a + (b-1)) % b)

That looks like a total of four addition and subtractions. However, you could cache the value for b-1:

c = b-1
answer = c - ((a + c) % b)

So... three addition/subtractions and one MOD.

Ryan McGuire
Ranch Hand

Joined: Feb 18, 2005
Posts: 1007

3

posted

0

...or just use Paul's formula with an if() for the special case:

One MOD, one subtraction, one comparison and one extra assignment one time out of b on average. The comparison and possibly extra assignment are probably cheaper computationally than a second MOD.

Ryan McGuire wrote:...or just use Paul's formula with an if() for the special case...

With two variables you could save yourself a subtraction, viz:I suspect also that comparison with 0 is quicker, but it might all be offset by having an extra variable.

Ain't optimization fun?

Winston

Isn't it funny how there's always time and money enough to do it WRONG?
Articles by Winston can be found here