3.3. Search in Sorted Arrays¶

3.3.1. Analysis¶

For large collections of records that are searched repeatedly, sequential search is unacceptably slow. One way to reduce search time is to preprocess the records by sorting them. Given a sorted array, an obvious improvement over simple linear search is to test if the current element in L is greater than $K$. If it is, then we know that $K$ cannot appear later in the array, and we can quit the search early. But this still does not improve the worst-case cost of the algorithm.

3.3.1.1. Jump Search¶

We can also observe that if we look first at position 1 in sorted array L and find that K is bigger, then we rule out position 0 as well as position 1. Because more is often better, what if we look at position 2 in L and find that $K$ is bigger yet? This rules out positions 0, 1, and 2 with one comparison. What if we carry this to the extreme and look first at the last position in L and find that $K$ is bigger? Then we know in one comparison that $K$ is not in L. This is useful to know, but what is wrong with the conclusion that we should always start by looking at the last position? The problem is that, while we learn a lot sometimes (in one comparison we might learn that $K$ is not in the list), usually we learn only a little bit (that the last element is not $K$).

The question then becomes: What is the right amount to jump? This leads us to an algorithm known as Jump Search. For some value $j$, we check every $j$ ‘th element in L, that is, we check elements $\mathbf{L}[j]$, $\mathbf{L}[2j]$, and so on. So long as $K$ is greater than the values we are checking, we continue on. But when we reach a value in L greater than $K$, we do a linear search on the piece of length $j-1$ that we know brackets $K$ if it is in the list.

If we define $m$ such that $mj \leq n < (m+1)j$, then the total cost of this algorithm is at most $m + j - 1$ 3-way comparisons. (They are 3-way because at each comparison of $K$ with some $\mathbf{L}[i]$ we need to know if $K$ is less than, equal to, or greater than $\mathbf{L}[i]$.) Therefore, the cost to run the algorithm on $n$ items with a jump of size $j$ is

\[\mathbf{T}(n, j) = m + j - 1 = \left\lfloor \frac{n}{j} \right\rfloor + j - 1.\]

What is the best value that we can pick for $j$? We want to minimize the cost:

\[\min_{1 \leq j \leq n} \left\{\left\lfloor\frac{n}{j}\right\rfloor + j - 1\right\}\]

Take the derivative and solve for $f'(j) = 0$ to find the minimum, which is $j = \sqrt{n}$. In this case, the worst case cost will be roughly $2\sqrt{n}$.

This example invokes a basic principle of algorithm design. We want to balance the work done while selecting a sublist with the work done while searching a sublist. In general, it is a good strategy to make subproblems of equal effort. This is an example of a divide and conquer algorithm.

What if we extend this idea to three levels? We would first make jumps of some size $j$ to find a sublist of size $j-1$ whose end values bracket value $K$. We would then work through this sublist by making jumps of some smaller size, say $j_1$. Finally, once we find a bracketed sublist of size $j_1 - 1$, we would do sequential search to complete the process.

This probably sounds convoluted to do two levels of jumping to be followed by a sequential search. While it might make sense to do a two-level algorithm (that is, jump search jumps to find a sublist and then does sequential search on the sublist), it almost never seems to make sense to do a three-level algorithm. Instead, when we go beyond two levels, we nearly always generalize by using recursion. This leads us to the most commonly used search algorithm for sorted arrays, the binary search.

3.3.1.2. Binary Search¶

You are probably pretty familiar with Binary Search already. So that we have a concrete example to discuss, here is an implementation.

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Of couse you know that Binary Search is far better than Sequential Why would that be? Because we have additional information to work with that we do not have when the list is unsorted. You probably already “know” that the standard binary search algorithm has a worst case cost of $O(\log n)$. Let’s do the math to make sure that it really is in $O(\log n$, and see how to handle the nasty details of modeling the exact behavior of a recursive algorithm. After that, we can deal with proving that Binary Search is indeed optimal (at least in the worst case) for solving the problem of search in a sorted list.

If we are willing to be casual about our analysis, we can reason that we look at one element (for a cost of one), and then repeat the process on half of the array. This would give us a recurrence that looks like $f(n) = 1 + f(n/2)$. But if we want to be more precise, then we need to think carefully about what is going on in the worst case. First, we should notice that we are doing a little more than cutting the array in half. We never look again at a particular position that we test. For example, if the input size is nine, then we actually look at position 4 (since $(9-0)/2 = 4$ when rounded down), and we then either continue to consider four positions to the left (positions 0 to 3) or four positions to the right (positions 5 to 8). But what if there are ten element? Then we actually look at position 5 (since $(10-0)/2 = 5$). We will then either need to continue dealing with five positions to the left (positions 0 to 4), or four positions to the right. Which means that in the worst case, we are looking at a little less than half when the array size is odd, or exactly half when the array size is even. To capture this, we can use the floor function, to get an exact worst case model as follows:

\[\begin{split}f(n) = \left\{ \begin{array}{ll} 1 & n=1\\ f(\lfloor n/2 \rfloor) + 1 & n > 1 \end{array} \right.\end{split}\]

Since $n/2 \geq \lfloor n/2 \rfloor$, and since $f(n)$ is assumed to be non-decreasing (since adding more elements won’t decrease the work) we can estimate the upper bound with the simplification $f(n) = f(n/2) + 1$.

This recurrence is fairly easy to solve via expansion:

\[\begin{split}\begin{eqnarray*} f(n) &=& f(n/2) + 1\\ &=& \{f(n/4) + 1\} + 1\\ &=& \{\{f(n/8) + 1\} + 1\} + 1 \end{eqnarray*}\end{split}\]

Then, collapse to

\[f(n) = f(n/2^i) + i = f(1) + \log n = \log n + 1\]

Now, we can prove that this is correct with induction.

By the IH, $f(n/2) = \log(n/2) + 1$.

\[\begin{split}\begin{eqnarray*} f(n/2) + 1 &=& (\log(n/2) + 1) + 1\\ &=& (\log n - 1 + 1) + 1\\ &=& \log n + 1 = f(n). \end{eqnarray*}\end{split}\]

How do we calculate the average cost for Binary Search? This requires some modeling, because we need to know things about the probabilities of the various inputs. We will estimate given these assumptions:

$X$ is in L.
$X$ is equally likely to be in any position.
$n = 2^k - 1$ for some non-negative integer $k$.

What is the cost?

There is one chance to hit in one probe.
There are two chances to hit in two probes.
There are $2^{i-1}$ chances to hit in $i$ probes.
$i \leq k$.

What is the resulting equation?

\[\frac{1\times 1 + 2\times 2 + 3 \times 4 + ... + \log n 2^{\log n-1}}{n} = \frac{1}{n}\sum_{i=1}^{\log n}i 2^{i-1}\]

Note that $2^{\log n-1} = n/2$.

To solve the summation:

\[\begin{split}\begin{eqnarray*} \sum_{i=1}^k i2^{i-1} &=& \sum_{i=0}^{k-1}(i+1)2^i = \sum_{i=0}^{k-1} i 2^i + \sum_{i=0}^{k-1} 2^i\\ &=& 2 \sum_{i=0}^{k-1} i 2^{i-1} + 2^k - 1\\ &=& 2 \sum_{i=1}^{k} i 2^{i-1} - k 2^k + 2^k - 1 \end{eqnarray*}\end{split}\]

Note that in the above series of equations, we change variables: $i \rightarrow i+1$.

Now what? Subtract from the original!

\[\sum_{i=1}^{k} i 2^{i-1} = k 2^k - 2^k + 1 = (k - 1)2^k + 1.\]

Note that

\[\sum_{i=1}^k i 2^{i-1} = 2 \sum_{i=1}^k i 2^{i-1} - k 2^k + 2^k -1\]

So,

\[\begin{split}\begin{eqnarray*} \sum_{i=1}^k i 2^{i-1} &=& k2^k - 2^k +1\\ &=& (k-1)2^k +1 \end{eqnarray*}\end{split}\]

Now we come back to solving the original equation. Since we have a closed-form solution for the summation in hand, we can restate the equation with the appropriate variable substitutions.

\[\begin{split}\begin{eqnarray*} \frac{1}{n}\sum_{i=1}^{\log n}i 2^{i-1} &=& \frac{(\log n - 1)2^{\log n} + 1}{n}\\ &=& \frac{n (\log n -1) + 1}{n}\\ &\approx& \log n - 1 \end{eqnarray*}\end{split}\]

So the average cost is only about one or two comparisons less than the worst cost.

If we want to relax the assumption that $n = 2^k - 1$, we get this as the exact cost:

\[\begin{split}f(n) = \left\{ \begin{array}{ll} 0 & n=0\\ 1 & n=1\\ \frac{\lceil \frac{n}{2} \rceil - 1}{n}f(\lceil \frac{n}{2} \rceil - 1) + \frac{1}{n} 0\ + \\ \frac{\lfloor \frac{n}{2} \rfloor}{n}f(\lfloor \frac{n}{2} \rfloor) + 1& n > 1 \end{array} \right.\end{split}\]

Identify each of the components of this equation as follows:

Left side: $X < L[i]$
$L(i) == X$ has no additional cost, with chance $1/n$
Right side: $X > L[i]$

3.3.1.3. Lower Bounds Proof¶

So, $O(\log n)$ time for Binary Search seems pretty good. Can we do better than this? We can prove that this is the best possible algorithm in the worst case for searching in a sorted list by using a proof similar to that used to show the lower bound on sorting.

We use the decision tree to model our algorithm. Unlike when searching an unsorted list, comparisons between elements of L tell us nothing new about their relative order (since L is already sorted), so we consider only comparisons between $K$ and an element in L. At the root of the decision tree, our knowledge rules out no positions in L, so all are potential candidates. As we take branches in the decision tree based on the result of comparing $K$ to an element in L, we gradually rule out potential candidates. Eventually we reach a leaf node in the tree representing the single position in L that can contain $K$. There must be at least $n+1$ nodes in the tree because we have $n+1$ distinct positions that $K$ can be in (any position in L, plus not in L at all). Some path in the tree must be at least $\log n$ levels deep, and the deepest node in the tree represents the worst case for that algorithm. Thus, any algorithm on a sorted array requires at least $\Omega(\log n)$ comparisons in the worst case.

We can modify this proof to find the average cost lower bound. Again, we model algorithms using decision trees. Except now we are interested not in the depth of the deepest node (the worst case) and therefore the tree with the least-deepest node. Instead, we are interested in knowing what the minimum possible is for the “average depth” of the leaf nodes. Define the total path length as the sum of the levels for each node. The cost of an outcome is the level of the corresponding node plus 1. The average cost of the algorithm is the average cost of the outcomes (total path length / $n$). What is the tree with the least average depth? This is equivalent to the tree that corresponds to binary search. Thus, binary search is optimal in the average case.

While binary search is indeed an optimal algorithm for a sorted list in the worst and average cases when searching a sorted array, there are a number of circumstances that might lead us to select another algorithm instead. One possibility is that we know something about the distribution of the data in the array. If each position in L is equally likely to hold $K$ (equivalently, the data are well distributed along the full key range), then an interpolation search is $\Theta(\log \log n)$ in the average case. If the data are not sorted, then using binary search requires us to pay the cost of sorting the list in advance, which is only worthwhile if many (at least $O(\log n)$ searches will be performed on the list. Binary search also requires that the list (even if sorted) be implemented using an array or some other structure that supports random access to all elements with equal cost. Finally, if we know all search requests in advance, we might prefer to sort the list by frequency and do linear search in extreme search distributions, or use a self-organizing list.

3.3.1.4. Interpolation and Quadratic Binary Search¶

If we know nothing about the distribution of key values, then we have just proved that binary search is the best algorithm available for searching a sorted array. However, sometimes we do know something about the expected key distribution. Consider the typical behavior of a person looking up a word in a large dictionary. Most people certainly do not use sequential search! Typically, people use a modified form of binary search, at least until they get close to the word that they are looking for. The search generally does not start at the middle of the dictionary. People looking for a word starting with ‘S’ generally assume that entries beginning with ‘S’ start about three quarters of the way through the dictionary. Thus, they will first open the dictionary about three quarters of the way through and then make a decision based on what is found as to where to look next. In other words, people typically use some knowledge about the expected distribution of key values to “compute” where to look next. This form of “computed” binary search is called a dictionary search or interpolation search. In a dictionary search, we search L at a position $p$ that is appropriate to the value of $K$ as follows.

\[p = \frac{K - \mathbf{L}[1]}{\mathbf{L}[n] - \mathbf{L}[1]}\]

This equation is computing the position of $K$ as a fraction of the distance between the smallest and largest key values. This will next be translated into that position which is the same fraction of the way through the array, and this position is checked first. As with binary search, the value of the key found eliminates all records either above or below that position. The actual value of the key found can then be used to compute a new position within the remaining range of the array. The next check is made based on the new computation. This proceeds until either the desired record is found, or the array is narrowed until no records are left.

A variation on dictionary search is known as $Quadratic Binary Search$ (QBS), and we will analyze this in detail because its analysis is easier than that of the general dictionary search. QBS will first compute (p) and then examine $\mathbf{L}[\lceil pn\rceil]$. If $K < \mathbf{L}[\lceil pn\rceil]$ then QBS will sequentially probe to the left by steps of size $\sqrt{n}$, that is, we step through

\[\mathbf{L}[\lceil pn - i\sqrt{n}\rceil], i = 1, 2, 3, ...\]

until we reach a value less than or equal to $K$. Similarly for $K > \mathbf{L}[\lceil pn\rceil]$ we will step to the right by $\sqrt{n}$ until we reach a value in L that is greater than $K$. We are now within $\sqrt{n}$ positions of $K$. Assume (for now) that it takes a constant number of comparisons to bracket $K$ within a sublist of size $\sqrt{n}$. We then take this sublist and repeat the process recursively. That is, at the next level we compute an interpolation to start somewhere in the subarray. We then step to the left or right (as appropriate) by steps of size $\sqrt{\sqrt{n}}$.

What is the cost for QBS? Note that $\sqrt{c^n} =c^{n/2}$, and we will be repeatedly taking square roots of the current sublist size until we find the item that we are looking for. Because $n = 2^{\log n}$ and we can cut $\log n$ in half only $\log \log n$ times, the cost is $\Theta(\log \log n)$ if the number of probes on jump search is constant.

Say that the number of comparisons needed is $i$, in which case the cost is $i$ (since we have to do $i$ comparisons). If $\mathbf{P}_i$ is the probability of needing exactly $i$ probes, then

\[\begin{split}\sum_{i=1}^{\sqrt{n}} i \mathbf{P}(\mbox{need exactly $i$ probes})\\ = 1 \mathbf{P}_1 + 2 \mathbf{P}_2 + 3 \mathbf{P}_3 + \cdots + \sqrt{n} \mathbf{P}_{\sqrt{n}}\end{split}\]

We now show that this is the same as

\[\sum_{i=1}^{\sqrt{n}} \mathbf{P}(\mbox{need at least $i$ probes})\]

\[\begin{split}&=& 1 + (1-\mathbf{P}_1) + (1-\mathbf{P}_1-\mathbf{P}_2) + \cdots + \mathbf{P}_{\sqrt{n}}\\ &=& (\mathbf{P}_1 + ... + \mathbf{P}_{\sqrt{n}}) + (\mathbf{P}_2 + ... + \mathbf{P}_{\sqrt{n}}) +\\ && \qquad (\mathbf{P}_3 + ... + \mathbf{P}_{\sqrt{n}}) + \cdots\\ &=& 1 \mathbf{P}_1 + 2 \mathbf{P}_2 + 3 \mathbf{P}_3 + \cdots + \sqrt{n} \mathbf{P}_{\sqrt{n}}\end{split}\]

We require at least two probes to set the bounds, so the cost is

\[2 + \sum_{i=3}^{\sqrt{n}} \mathbf{P}(\mbox{need at least $i$ probes}).\]

We now make take advantage of a useful fact known as Chebyshev’s Inequality. Chebyshev’s inequality states that $\mathbf{P}(\mbox{need exactly}\ i\ \mbox{probes})$, or $\mathbf{P}_i$, is

\[\mathbf{P}_i \leq \frac{p(1 - p)n}{(i - 2)^2 n} \leq \frac{1}{4(i-2)^2}\]

because $p(1-p) \leq 1/4$ for any probability $p$. This assumes uniformly distributed data. Thus, the expected number of probes is

\[2 + \sum_{i=3}^{\sqrt{n}} \frac{1}{4(i-2)^2} < 2 + \frac{1}{4}\sum_{i=1}^\infty \frac{1}{i^2} = 2 + \frac{1}{4}\frac{\pi}{6} \approx 2.4112\]

Is QBS better than binary search? Theoretically yes, because $O(\log \log n)$ grows slower than $O(\log n)$. However, we have a situation here which illustrates the limits to the model of asymptotic complexity in some practical situations. Yes, $c_1 \log n$ does grow faster than $c_2 \log \log n$. In fact, it is exponentially faster! But even so, for practical input sizes, the absolute cost difference is fairly small. Thus, the constant factors might play a role. First we compare $\log \log n$ to $\log n$.

\[\begin{split}\begin{array}{llll} &&&{\rm Factor}\\ n &\log n&\log \log n&{\rm Difference}\\ \hline 16 &4 &2 &2\\ 256&8 &3 &2.7\\ 2^{16}&16 &4 &4\\ 2^{32}&32 &5 &6.4\\ \end{array}\end{split}\]

It is not always practical to reduce an algorithm’s growth rate. There is a “practicality window” for every problem, in that we have a practical limit to how big an input we wish to solve for. If our problem size never grows too big, it might not matter if we can reduce the cost by an extra log factor, because the constant factors in the two algorithms might differ by more than the log of the log of the input size.

For our two algorithms, let us look further and check the actual number of comparisons used. For binary search, we need about $\log n-1$ total comparisons. Quadratic binary search requires about $2.4 \log \log n$ comparisons. If we incorporate this observation into our table, we get a different picture about the relative differences.

\[\begin{split}\begin{array}{llll} &&&{\rm Factor}\\ n &\log n -1&2.4 \log \log n&{\rm Difference}\\ \hline 16&3&4.8&{\rm worse}\\ 256&7&7.2&\approx {\rm same}\\ 64K&15&9.6&1.6\\ 2^{32}&31&12&2.6 \end{array}\end{split}\]

But we still are not done. This is only a count of raw comparisons. Binary search is inherently much simpler than QBS, because binary search only needs to calculate the midpoint position of the array before each comparison, while quadratic binary search must calculate an interpolation point which is more expensive. So the constant factors for QBS are even higher.

Not only are the constant factors worse on average, but QBS is far more dependent than binary search on good data distribution to perform well. For example, imagine that you are searching a telephone directory for the name “Young”. Normally you would look near the back of the book. If you found a name beginning with ‘Z’, you might look just a little ways toward the front. If the next name you find also begins with ‘Z’ you would look a little further toward the front. If this particular telephone directory were unusual in that half of the entries begin with ‘Z’, then you would need to move toward the front many times, each time eliminating relatively few records from the search. In the extreme, the performance of interpolation search might not be much better than sequential search if the distribution of key values is badly calculated.

While it turns out that QBS is not a practical algorithm, this is not a typical situation. Fortunately, algorithm growth rates are usually well behaved, so that asymptotic algorithm analysis nearly always gives us a practical indication for which of two algorithms is better.

Senior Algorithms

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