8.7.1 Linear Probing by Steps

How can we avoid primary clustering? One possible improvement might be to use linear probing, but to skip slots by some constant $c$ other than 1. This would make the probe function $\textbf{p}(K, i) = ci$, and so the $i$ th slot in the probe sequence will be $(\textbf{h}(K) + ic) \mod M$. In this way, records with adjacent home positions will not follow the same probe sequence.

One quality of a good probe sequence is that it will cycle through all slots in the hash table before returning to the home position. Clearly linear probing (which “skips” slots by one each time) does this. Unfortunately, not all values for $c$ will make this happen. For example, if $c = 2$ and the table contains an even number of slots, then any key whose home position is in an even slot will have a probe sequence that cycles through only the even slots. Likewise, the probe sequence for a key whose home position is in an odd slot will cycle through the odd slots. Thus, this combination of table size and linear probing constant effectively divides the records into two sets stored in two disjoint sections of the hash table. So long as both sections of the table contain the same number of records, this is not really important. However, just from chance it is likely that one section will become fuller than the other, leading to more collisions and poorer performance for those records. The other section would have fewer records, and thus better performance. But the overall system performance will be degraded, as the additional cost to the side that is more full outweighs the improved performance of the less-full side.

Constant $c$ must be relatively prime to $M$ to generate a linear probing sequence that visits all slots in the table (that is, $c$ and $M$ must share no factors). For a hash table of size $M = 10$, if $c$ is any one of 1, 3, 7, or 9, then the probe sequence will visit all slots for any key. When $M = 11$, any value for $c$ between 1 and 10 generates a probe sequence that visits all slots for every key.

Now you can practice linear probing by different step sizes.

8.7.2 Pseudo-Random Probing

Consider the situation where $c = 2$ and we wish to insert a record with key $k_1$ such that $\textbf{h}(k_1) = 3$. The probe sequence for $k_1$ is 3, 5, 7, 9, and so on. If another key $k_2$ has home position at slot 5, then its probe sequence will be 5, 7, 9, and so on. The probe sequences of $k_1$ and $k_2$ are linked together in a manner that contributes to clustering. In other words, linear probing with a value of $c > 1$ does not solve the problem of primary clustering. We would like to find a probe function that does not link keys together in this way. We would prefer that the probe sequence for $k_1$ after the first step on the sequence should not be identical to the probe sequence of $k_2$. Instead, their probe sequences should diverge.

The ideal probe function would select the next position on the probe sequence at random from among the unvisited slots; that is, the probe sequence should be a random permutation of the hash table positions. Unfortunately, we cannot actually select the next position in the probe sequence at random, because we would not be able to duplicate this same probe sequence when searching for the key. However, we can do something similar called pseudo-random probing. In pseudo-random probing, the $i$ th slot in the probe sequence is $(\textbf{h}(K) + r_i) \mod M$ where $r_i$ is the $i$ th value in a random permutation of the numbers from 1 to $M-1$. All inserts and searches must use the same sequence of random numbers. The probe function would be $\textbf{p}(K, i) = \textbf{Permutation}[i]$ where Permutation is an array of length $M$ that stores a value of 0 in position Permutation[0], and stores a random permutation of the values from 1 to $M - 1$ in slots 1 to $M - 1$.

Here is a practice exercise for pseudo-random probing.

Pseudo-random probing exhibits another desirable feature in a hash function.

8.7.3 Quadratic Probing

Another probe function that eliminates primary clustering is called quadratic probing. Here the probe function is some quadratic function $\textbf{p}(K, i) = c_1 i^2 + c_{2}i + c_3$ for some choice of constants $c_1$, $c_2$, and $c_3$.

The simplest variation is $\textbf{p}(K, i) = i^2$ (i.e., $c_1 = 1$, $c_2 = 0$, and $c_3 = 0$). Then the $i$ th value in the probe sequence would be $(\textbf{h}(K) + i^2) \mod M$.

Now you can practice quadratic probing.

There is one problem with quadratic probing: Its probe sequence typically will not visit all slots in the hash table.

For many hash table sizes, this probe function will cycle through a relatively small number of slots. If all slots on that cycle happen to be full, this means that the record cannot be inserted at all! A more realistic example is a table with 105 slots. The probe sequence starting from any given slot will only visit 23 other slots in the table. If all 24 of these slots should happen to be full, even if other slots in the table are empty, then the record cannot be inserted because the probe sequence will continually hit only those same 24 slots.

Fortunately, it is possible to get good results from quadratic probing at low cost. The right combination of probe function and table size will visit many slots in the table. In particular, if the hash table size is a prime number and the probe function is $\textbf{p}(K, i) = i^2$, then at least half the slots in the table will be visited. Thus, if the table is less than half full, we can be certain that a free slot will be found. Alternatively, if the hash table size is a power of two and the probe function is $\textbf{p}(K, i) = (i^2 + i)/2$, then every slot in the table will be visited by the probe function.

8.7.4 Double Hashing

Both pseudo-random probing and quadratic probing eliminate primary clustering, which is the name given to the the situation when keys share substantial segments of a probe sequence. If two keys hash to the same home position, however, then they will always follow the same probe sequence for every collision resolution method that we have seen so far. The probe sequences generated by pseudo-random and quadratic probing (for example) 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 resolution methods. If the hash function generates a cluster at a particular home position, then the cluster remains under pseudo-random and quadratic probing. This problem is called secondary clustering.

To avoid secondary clustering, we need to have the probe sequence make use of the original key value in its decision-making process. A simple technique for doing this is to return to linear probing by a constant step size for the probe function, but to have that constant be determined by a second hash function, $\textbf{h}_2$. Thus, the probe sequence would be of the form $\textbf{p}(K, i) = i * \textbf{h}_2(K)$. This method is called double hashing.

There are important restrictions on $h_2$. Most importantly, the value returned by $h_2$ must never be zero (or $M$) because that will immediately lead to an infinite loop as the probe sequence makes no progress. However, a good implementation of double hashing should also ensure that all of the probe sequence constants are relatively prime to the table size $M$. For example, if the hash table size were 100 and the step size for linear probing (as generated by function $h_2$) were 50, then there would be only one slot on the probe sequence. If instead the hash table size is 101 (a prime number), than any step size less than 101 will visit every slot in the table.

This can be achieved easily. One way is to select $M$ to be a prime number, and have $\textbf{h}_2$ return a value in the range $1 <= \textbf{h}_2(k) <= M - 1$. We can do this by using this secondary hash function: $\textbf{h}_2(k) = 1 + (k \mod (M-1))$. An alternative is to set $M = 2^m$ for some value $m$ and have $\textbf{h}_2$ return an odd value between 1 and $2^m$. We can get that result with this secondary hash function: $\textbf{h}_2(k) = (((k/M) \mod (M/2)) * 2) + 1$.

Now you can try it.