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vb.netmodulus
David BS
David BS

Reputation: 53

How fast calculate MODULUS for divisor greater than 8, 16, 32

I´m trying to make very fast MOD operations. I saw in several pages we can do alternatively calculate the MOD using the AND operator with (Divisor-1). Eg.:

result = (100 mod 8) is the same as result = (100 and 7)

It functions perfectly if the divisor is less than 8 bits, but if we calculate (1245 mod 67) we can see the result is different of (1245 and 66).

So, how can I calculate this faster than using the MOD operator provided by VB.NET language?

Thanks!

Upvotes: 0

Views: 185

Answers (2)

user1889116
user1889116

Reputation:

Well, 100 mod 8 = 100 and 7 works, because 7 is binary 111b.

Whatever number you have (100 decimal is 1100100b), the last 3 binary digits (bits) are kept due to the 111b. For 100, that would be 100b = 4.

Now consider what you did attempt with 1245 mod 67.

1245 is binary 10011011101b.

67 is binary 1000011b.

Can you see, why it's not working?

Upvotes: 0

Douglas Barbin
Douglas Barbin

Reputation: 3615

Using a bitwise AND only works for modulus powers of 2 (and only positive ones, at that). It does not work for other numbers. See this link

I think that the modulus operator built into the framework is fast and you probably won't be able to improve on it.

Upvotes: 2

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