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27.3. Amortized Analysis

This module presents the concept of amortized analysis, which is the analysis for a series of operations taken as a whole. In particular, amortized analysis allows us to deal with the situation where the worst-case cost for \(n\) operations is less than \(n\) times the worst-case cost of any one operation. Rather than focusing on the individual cost of each operation independently and summing them, amortized analysis looks at the cost of the entire series and "charges" each individual operation with a share of the total cost.

We can apply the technique of amortized analysis in the case of a series of sequential searches in an unsorted array. For \(n\) random searches, the average-case cost for each search is \(n/2\), and so the expected total cost for the series is \(n^2/2\). Unfortunately, in the worst case all of the searches would be to the last item in the array. In this case, each search costs \(n\) for a total worst-case cost of \(n^2\). Compare this to the cost for a series of \(n\) searches such that each item in the array is searched for precisely once. In this situation, some of the searches must be expensive, but also some searches must be cheap. The total number of searches, in the best, average, and worst case, for this problem must be \(\sum_{i=i}^n i \approx n^2/2\). This is a factor of two better than the more pessimistic analysis that charges each operation in the series with its worst-case cost.

As another example of amortized analysis, consider the process of incrementing a binary counter. The algorithm is to move from the lower-order (rightmost) bit toward the high-order (leftmost) bit, changing 1s to 0s until the first 0 is encountered. This 0 is changed to a 1, and the increment operation is done. Below is an implementation for the increment operation, assuming that a binary number of length \(n\) is stored in array A of length \(n\).

for (i=0; ((i<A.length) && (A[i] == 1)); i++)
  A[i] = 0;
if (i < A.length)
  A[i] = 1;

If we count from 0 through \(2^n - 1\), (requiring a counter with at least \(n\) bits), what is the average cost for an increment operation in terms of the number of bits processed? Naive worst-case analysis says that if all \(n\) bits are 1 (except for the high-order bit), then \(n\) bits need to be processed. Thus, if there are \(2^n\) increments, then the cost is \(n 2^n\). However, this is much too high, because it is rare for so many bits to be processed. In fact, half of the time the low-order bit is 0, and so only that bit is processed. One quarter of the time, the low-order two bits are 01, and so only the low-order two bits are processed. Another way to view this is that the low-order bit is always flipped, the bit to its left is flipped half the time, the next bit one quarter of the time, and so on. We can capture this with the summation (charging costs to bits going from right to left)

\[\begin{split}\sum_{i=0}^{n-1} \frac{1}{2^i} < 2.\end{split}\]

In other words, the average number of bits flipped on each increment is 2, leading to a total cost of only \(2 \cdot 2^n\) for a series of \(2^n\) increments.

A useful concept for amortized analysis is illustrated by a simple variation on the stack data structure, where the pop function is slightly modified to take a second parameter \(k\) indicating that \(k\) pop operations are to be performed.

The "local" worst-case analysis for multipop is \(\Theta(n)\) for \(n\) elements in the stack. Thus, if there are \(m_1\) calls to push and \(m_2\) calls to multipop, then the naive worst-case cost for the series of operation is \(m_1 + m_2\cdot n = m_1 + m_2 \cdot m_1\). This analysis is unreasonably pessimistic. Clearly it is not really possible to pop \(m_1\) elements each time multipop is called. Analysis that focuses on single operations cannot deal with this global limit, and so we turn to amortized analysis to model the entire series of operations.

The key to an amortized analysis of this problem lies in the concept of potential. At any given time, a certain number of items may be on the stack. The cost for multipop can be no more than this number of items. Each call to push places another item on the stack, which can be removed by only a single multipop operation. Thus, each call to push raises the potential of the stack by one item. The sum of costs for all calls to multipop can never be more than the total potential of the stack (aside from a constant time cost associated with each call to multipop itself).

The amortized cost for any series of push and multipop operations is the sum of three costs. First, each of the push operations takes constant time. Second, each multipop operation takes a constant time in overhead, regardless of the number of items popped on that call. Finally, we count the sum of the potentials expended by all multipop operations, which is at most \(m_1\), the number of push operations. This total cost can therefore be expressed as

\[m_1 + (m_2 + m_1) = \Theta(m_1 + m_2).\]

A similar argument was used in our analysis for the partition function in the Quicksort algorithm. While on any given pass through the while loop the left or right pointers might move all the way through the remainder of the partition, doing so would reduce the number of times that the while loop can be further executed.

Our final example uses amortized analysis to prove a relationship between the cost of the move-to-front self-organizing list heuristic and the cost for the optimal static ordering of the list.

Recall that, for a series of search operations, the minimum cost for a static list results when the list is sorted by frequency of access to its records. This is the optimal ordering for the records if we never allow the positions of records to change, because the most-frequently accessed record is first (and thus has least cost), followed by the next most frequently accessed record, and so on.

Theorem 27.3.1

Theorem: The total number of comparisons required by any series \(S\) of \(n\) or more searches on a self-organizing list of length \(n\) using the move-to-front heuristic is never more than twice the total number of comparisons required when series \(S\) is applied to the list stored in its optimal static order.

Proof: Each comparison of the search key with a record in the list is either successful or unsuccessful. For \(m\) searches, there must be exactly \(m\) successful comparisons for both the self-organizing list and the static list. The total number of unsuccessful comparisons in the self-organizing list is the sum, over all pairs of distinct keys, of the number of unsuccessful comparisons made between that pair.

Consider a particular pair of keys: \(A\) and \(B\). For any sequence of searches \(S\), the total number of (unsuccessful) comparisons between \(A\) and \(B\) is identical to the number of comparisons between \(A\) and \(B\) required for the subsequence of \(S\) made up only of searches for \(A\) or \(B\). Call this subsequence \(S_{AB}\). In other words, including searches for other keys does not affect the relative position of \(A\) and \(B\) and so does not affect the relative contribution to the total cost of the unsuccessful comparisons between \(A\) and \(B\).

The number of unsuccessful comparisons between \(A\) and \(B\) made by the move-to-front heuristic on subsequence \(S_{AB}\) is at most twice the number of unsuccessful comparisons between \(A\) and \(B\) required when \(S_{AB}\) is applied to the optimal static ordering for the list. To see this, assume that \(S_{AB}\) contains \(i\) \(A\) s and \(j\) \(B\) s, with \(i \leq j\). Under the optimal static ordering, \(i\) unsuccessful comparisons are required because \(B\) must appear before \(A\) in the list (because its access frequency is higher). Move-to-front will yield an unsuccessful comparison whenever the request sequence changes from \(A\) to \(B\) or from \(B\) to \(A\). The total number of such changes possible is \(2i\) because each change involves an \(A\) and each \(A\) can be part of at most two changes.

Because the total number of unsuccessful comparisons required by move-to-front for any given pair of keys is at most twice that required by the optimal static ordering, the total number of unsuccessful comparisons required by move-to-front for all pairs of keys is also at most twice as high. Because the number of successful comparisons is the same for both methods, the total number of comparisons required by move-to-front is less than twice the number of comparisons required by the optimal static ordering.

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