A Practical Introduction to Data Structures and Algorithm Analysis

Sec. 9.4 Hashing 345found. Alternatively, if the hash table size is a power of two and the probe functionis p(K, i) = (i 2 + i)/2, then every slot in the table will be visited by the probefunction.Both pseudo-random probing and quadratic probing eliminate primary clustering,which is the problem of keys sharing substantial segments of a probe sequence.If two keys hash to the same home position, however, then they will always followthe same probe sequence for every collision resolution method that we have seen sofar. The probe sequences generated by pseudo-random and quadratic probing (forexample) are entirely a function of the home position, not the original key value.This is because function p ignores its input parameter K for these collision resolutionmethods. If the hash function generates a cluster at a particular home position,then the cluster remains under pseudo-random and quadratic probing. This problemis called secondary clustering.To avoid secondary clustering, we need to have the probe sequence make use ofthe original key value in its decision-making process. A simple technique for doingthis is to return to linear probing by a constant step size for the probe function, butto have that constant be determined by a second hash function, h 2 . Thus, the probesequence would be of the form p(K, i) = i ∗ h 2 (K). This method is called doublehashing.Example 9.11 Assume a hash table has size M = 101, and that thereare three keys k 1 , k 2 , and k 3 with h(k 1 ) = 30, h(k 2 ) = 28, h(k 3 ) = 30,h 2 (k 1 ) = 2, h 2 (k 2 ) = 5, and h 2 (k 3 ) = 5. Then, the probe sequencefor k 1 will be 30, 32, 34, 36, and so on. The probe sequence for k 2 willbe 28, 33, 38, 43, and so on. The probe sequence for k 3 will be 30, 35,40, 45, and so on. Thus, none of the keys share substantial portions of thesame probe sequence. Of course, if a fourth key k 4 has h(k 4 ) = 28 andh 2 (k 4 ) = 2, then it will follow the same probe sequence as k 1 . Pseudorandomor quadratic probing can be combined with double hashing to solvethis problem.A good implementation of double hashing should ensure that all of the probesequence constants are relatively prime to the table size M. This can be achievedeasily. One way is to select M to be a prime number, and have h 2 return a value inthe range 1 ≤ h 2 (K) ≤ M − 1. Another way is to set M = 2 m for some value mand have h 2 return an odd value between 1 and 2 m .Figure 9.7 shows an implementation of the dictionary ADT by means of a hashtable. The simplest hash function is used, with collision resolution by linear probing,as the basis for the structure of a hash table implementation. A suggested