Why is the size 127 (prime) better than 128 for a hash-table?

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?

Answer 1

Division Method

"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 = 2p , then h(k) is just the p lowest-order bits of k."

--CLRS

To understand why m = 2p uses only the p lowest bits of k, you must first understand the modulo hash function h(k) = k % m.

The key can be written in terms of a quotient q, and remainder r.

k = nq + r

Choosing the quotient to be q = m allows us to write k % m simply as the remainder in the above equation:

k % m = r = k - nm,  where r < m

Therefore, k % m is equivalent to continuously subtracting m a total of n times (until r < m):

k % m = k - m - m - ... - m,  until r < m

Lets try hashing the key k = 91 with m = 24 = 16.

  91 = 0101 1011
- 16 = 0001 0000
----------------
  75 = 0100 1011
- 16 = 0001 0000
----------------
  59 = 0011 1011
- 16 = 0001 0000
----------------
  43 = 0010 1011
- 16 = 0001 0000
----------------
  27 = 0001 1011
- 16 = 0001 0000
----------------
  11 = 0000 1011

Thus, 91 % 24 = 11 is just the binary form of 91 with only the p=4 lowest bits remaining.

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Important Distinction:

This pertains specifically to the division method of hashing. In fact, the converse is true for the multiplication method as stated in CLRS:

"An advantage of the multiplication method is that the value of m is not critical... We typically choose [m] to be a power of 2 since we can then easily implement the function on most computers."