5.2

The `ChainedHashTable` data structure uses an array of lists, where the
th
list stores all elements
such that
. An alternative,
called open addressing is to store the elements directly in an
array,
, with each array location in
storing at most one value.
This is the approach taken by the `LinearHashTable` described in this
section. In some places, this data structure is described as open
addressing with linear probing.

The main idea behind a `LinearHashTable` is that we would, ideally, like
to store the element
with hash value
in the table location
. If we can't do this (because some element is already stored
there) then we try to store it at location
;
if that's not possible, then we try
, and
so on, until we find a place for
.

There are three types of entries stored in :

- data values: actual values in the
`USet`that we are representing; - values: at array locations where no data has ever been stored; and
- values: at array locations where data was once stored but that has since been deleted.

To summarize, a `LinearHashTable` contains an array,
, that stores
data elements, and integers
and
that keep track of the number of
data elements and non-
values of
, respectively. Because many
hash functions only work for table sizes that are a power of 2, we also
keep an integer
and maintain the invariant that
.

T[] t; // the table int n; // the size int d; // t.length = 2^d int q; // number of non-null entries in t

The
operation in a `LinearHashTable` is simple. We start
at array entry
where
and search entries
,
,
, and so on,
until we find an index
such that, either,
, or
.
In the former case we return
. In the latter case, we conclude
that
is not contained in the hash table and return
.

T find(T x) { int i = hash(x); while (t[i] != null) { if (t[i] != del && x.equals(t[i])) return t[i]; i = (i == t.length-1) ? 0 : i + 1; // increment i (mod t.length) } return null; }

The operation is also fairly easy to implement. After checking that is not already stored in the table (using ), we search , , , and so on, until we find a or and store at that location, increment , and , if appropriate.:

boolean add(T x) { if (find(x) != null) return false; if (2*(q+1) > t.length) resize(); // max 50% occupancy int i = hash(x); while (t[i] != null && t[i] != del) i = (i == t.length-1) ? 0 : i + 1; // increment i (mod t.length) if (t[i] == null) q++; n++; t[i] = x; return true; }

By now, the implementation of the operation should be obvious. We search , , , and so on until we find an index such that or . In the former case, we set and return . In the latter case we conclude that was not stored in the table (and therefore cannot be deleted) and return .

T remove(T x) { int i = hash(x); while (t[i] != null) { T y = t[i]; if (y != del && x.equals(y)) { t[i] = del; n--; if (8*n < t.length) resize(); // min 12.5% occupancy return y; } i = (i == t.length-1) ? 0 : i + 1; // increment i (mod t.length) } return null; }

The correctness of the , , and methods is easy to verify, though it relies on the use of values. Notice that none of these operations ever sets a non- entry to . Therefore, when we reach an index such that , this is a proof that the element, , that we are searching for is not stored in the table; has always been , so there is no reason that a previous operation would have proceeded beyond index .

The method is called by when the number of non- entries exceeds or by when the number of data entries is less than . The method works like the methods in other array-based data structures. We find the smallest non-negative integer such that . We reallocate the array so that it has size and then we insert all the elements in the old version of into the newly-resized copy of . While doing this we reset equal to since the newly-allocated has no values.

void resize() { d = 1; while ((1<<d) < 3*n) d++; T[] told = t; t = newArray(1<<d); q = n; // insert everything in told for (int k = 0; k < told.length; k++) { if (told[k] != null && told[k] != del) { int i = hash(told[k]); while (t[i] != null) i = (i == t.length-1) ? 0 : i + 1; t[i] = told[k]; } } }

Notice that each operation, , , or , finishes as soon as (or before) it discovers the first entry in . The intuition behind the analysis of linear probing is that, since at least half the elements in are equal to , an operation should not take long to complete because it will very quickly come across a entry. We shouldn't rely too heavily on this intuition though, because it would lead us to (the incorrect) conclusion that the expected number of locations in examined by an operation is at most 2.

For the rest of this section, we will assume that all hash values are independently and uniformly distributed in . This is not a realistic assumption, but it will make it possible for us to analyze linear probing. Later in this section we will describe a method, called tabulation hashing, that produces a hash function that is ``good enough'' for linear probing. We will also assume that all indices into the positions of are taken modulo , so that is really a shorthand for .

We say that a run of length that starts at occurs when are all non- and . The number of non- elements of is exactly and the method ensures that, at all times, . There are elements that have been inserted into since the last operation. By our assumption, each of these has a hash value, , that is uniform and independent of the rest. With this setup, we can prove the main lemma required to analyze linear probing.

In the following derivation we will cheat a little and replace with . Stirling's Approximation (Section 1.2.2) shows that this is only a factor of from the truth. This is just done to make the derivation simpler; Exercise 5.4 asks the reader to redo the calculation more rigorously using Stirling's Approximation in its entirety.

The value of is maximized when is minimum, and the data structure maintains the invariant that , so

[Stirling's approximation] | ||||

(In the last step, we use the inequality , which holds for all .) Since , this completes the proof.

Using Lemma 5.4 to prove upper-bounds on the expected
running time of
,
, and
is now fairly
straight-forward. Consider the simplest case, where we execute
for some value
that has never been stored in the
`LinearHashTable`. In this case,
is a random value in
independent of the contents of
. If
is part of a run of length then the time it takes to execute the
operation is at most . Thus, the expected running time
can be upper-bounded by

The last step in this derivation comes from the fact that is an exponentially decreasing series.

If we ignore the cost of the
operation, the above
analysis gives us all we need to analyze the cost of operations on a
`LinearHashTable`.

First of all, the analysis of given above applies to the operation when is not contained in the table. To analyze the operation when is contained in the table, we need only note that this is the same as the cost of the operation that previously added to the table. Finally, the cost of a operation is the same as the cost of a operation.

In summary, if we ignore the cost of calls to
, all operations
on a `LinearHashTable` run in expected time. Accounting for the
cost of resize can be done using the same type of amortized analysis
performed for the `ArrayStack` data structure in Section 2.1.

The following theorem summarizes the performance of the `LinearHashTable`
data structure:

*Furthermore, beginning with an empty LinearHashTable, any
sequence of
and
operations results in a total
of time spent during all calls to
.*

5.2.3 Tabulation Hashing

While analyzing the `LinearHashTable` structure, we made a very strong
assumption: That for any set of elements,
,
the hash values
are independently
and uniformly distributed over
. One way
to imagine getting this is to have a giant array,
, of length
, where each entry is a random
-bit integer, independent
of all the other entries. In this way, we could implement
by extracting a
-bit integer from
:

int idealHash(T x) { return tab[x.hashCode() >>> w-d]; }

Unfortunately, storing an array of size is prohibitive in terms of memory usage. The approach used by tabulation hashing is to, instead, treat -bit integers as being comprised of integers, each having only bits. In this way, tabulation hashing only needs arrays each of length . All the entries in these arrays are independent -bit integers. To obtain the value of we split up into -bit integers and use these as indices into these arrays. We then combine all these values with the bitwise exclusive-or operator to obtain . The following code shows how this works when and :

int hash(T x) { int h = x.hashCode(); return (tab[0][h&0xff] ^ tab[1][(h>>>8)&0xff] ^ tab[2][(h>>>16)&0xff] ^ tab[3][(h>>>24)&0xff]) >>> (w-d); }In this case, is a 2-dimensional array with 4 columns and rows.

One can easily verify that, for any
,
is uniformly
distributed over
. With a little work, one
can even verify that any pair of values have independent hash values.
This implies tabulation hashing could be used in place of multiplicative
hashing for the `ChainedHashTable` implementation.

However, it is not true that any set of distinct values gives a set of independent hash values. Nevertheless, when tabulation hashing is used, the bound of Theorem 5.2 still holds. References for this are provided at the end of this chapter.

- ... locations.
^{5.1} - Note that is greater than the probability that a run of length starts at , since the definition of does not include the requirement .
- ...
series.
^{5.2} - In the terminology of many calculus texts, this sum passes the ratio test: There exists a positive integer such that, for all , .