Question

Supposing simple uniform hashing, that being, any given value is equally like to hash into any of the slots of the hash. Why is it better to use a table of size 127 and not 128? I really don’t understand what’s the problem with the power of 2 numbers. Or how it actually makes any difference at all.

When using the division method,
we usually avoid certain values
of m (table size). For example, m
should not be a power of 2, since if m
= 2^p , then h(k) is just the p lowest-order bits of k.

Let’s suppose the possible elements are only between 1 and 10000 and I picked the table size as 128. How can 127 be better?
So 128 is 2^6 (1000000) and 127 is 0111111. What difference does this make? All numbers (when hashed) are still going to be the p lowest-order bits of k for 127 too. Did I get something wrong?

I’m looking for some examples as I really can’t understand why is this bad. Thanks a lot in advance!

PS: I am aware of:
Hash table: why size should be prime?

Answer 1

All numbers (when hashed) are still going to be the p lowest-order bits of k for 127 too.

That is wrong (or I misunderstood..). k % 127 depends on all bits of k. k % 128 only depends on the 7 lowest bits.

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EDIT:

If you have a perfect distribution between 1 and 10,000. 10,000 % 127 and 10,000 % 128 both will turn this in a excellent smaller distribution. All buckets will contain 10,000 /128 = 78 (or 79) items.

If you have a distribution between 1 and 10,000 that is biased, because {x, 2x, 3x, ..} occur more often. Then a prime size will give a much, much better distribution as explained in this answer. (Unless x is exactly that prime size.)

Thus, cutting off the high bits (using a size of 128) is no problem whatsoever if the distribution in the lower bits is good enough. But, with real data and real badly designed hash functions, you will need those high bits.