Cormen et al.'s "Introduction to Algorithms" says the following about the division method hash function $h(k)=k \text{ mod } m$:

A prime not too close to an exact power of 2 is often a good choice for $m$. For example, suppose we wish to allocate a hash table, with collisions resolved by chaining, to hold roughly $n=2000$ character strings, where a character has 8 bits. We don't mind examining an average of 3 elements in an unsuccesful search, so we allocate a table of size $m=701$. The number $701$ is chose because it is a prime near $2000/3$ but not near any power of 2.

The book, however, does not explain what it means by "near". Is $701$ preferrable over, say, $661$ because $661-512<701-512$? I do not understand how this is relevant to the modulo function involved in the calculation of $h(k)$.


2 Answers 2


If a string of 8 bit bytes is interpreted as a large integer, and you calculate the value modulo $2^{24}-1$ then the first, fourth, seventh byte etc. are all hashed to the same value, which means the hash will not be well distributed.


This answer possibly explains why choosing M(table size) equal to a power of 2 should be avoided.

Prime numbers that are too close to a power of 2 will provide the same kind of biasing as a power of 2 for the keys which differs by $+a$ or $-a$ if $2^k=a(modulo)M$.

In division method we simply use remainder modulo M: h(K) = K mod M

In this case some values of M are obviously much better than others.

Case 1: If M is an even number h(K) will be even when K is even and odd when K is odd, and this will lead to a substantial bias in many files.

Case 2: It would be even worse to let M be a power of 2 (more generally the radix of the computer), since K mod M would then be simply the least significant digits of K (independent of the other digits).

Case 3: Similarly we can argue that M probably shouldn't be a multiple of 3: for if the keys are alphabetic, two keys that differ only by the permutation of letters, would then differ in numeric value by a multiple of 3. (This occurs because $2^{2n}$ mod 3 = 1 and $10^n$ mod 3 = 1).

Case 4: In general, we want to avoid values of M that divide $r^k+a$ or $r^k-a$, where k and a are small numbers and r is the radix of the alphabetic charecter set (usually $r=64, \ 256$ or $100$), since a remainder modulo such a value of M tends to be largely a simple superposition of key digits. Such considerations suggest that we choose M to be a prime number such that $r^k!=a(modulo)M$ or $r^k!=-a(modulo)M$ for small k & a.

--Donald E. Knuth (The art of Computer Programming Vol. 3)

  • $\begingroup$ There are very few primes that are powers of two, and the question was already based on the assumption that the table size is a prime. $\endgroup$
    – gnasher729
    Commented Jan 6, 2018 at 14:45
  • $\begingroup$ @gnasher729 Please take a look, I've edited my answer. $\endgroup$ Commented Jan 6, 2018 at 20:59
  • $\begingroup$ Goo you this quote from Knuth, but question why still remains. $\endgroup$
    – Yola
    Commented Jan 17, 2018 at 16:27
  • $\begingroup$ @Yola I have mentioned that it gives the same kind of biasing as a power of 2, if you have got the case with powers of 2 then its easy to understand this. $\endgroup$ Commented Jan 17, 2018 at 16:31

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