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
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Joined: Feb 18, 2005
Posts: 1040

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...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?
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I’ve looked at a lot of different solutions, and in my humble opinion Aspose is the way to go. Here’s the link: http://aspose.com

subject: Algorithm help: minimum distance to multiple