On this page:
16.1 A First Example
16.2 The New Form of Analysis
16.3 An Example:   Queues from Lists
16.3.1 List Representations
16.3.2 A First Analysis
16.3.3 More Liberal Sequences of Operations
16.3.4 A Second Analysis
16.3.5 Amortization Versus Individual Operations
16.4 Reading More

16 Halloween Analysis

    16.1 A First Example

    16.2 The New Form of Analysis

    16.3 An Example: Queues from Lists

      16.3.1 List Representations

      16.3.2 A First Analysis

      16.3.3 More Liberal Sequences of Operations

      16.3.4 A Second Analysis

      16.3.5 Amortization Versus Individual Operations

    16.4 Reading More

In Predicting Growth, we introduced the idea of big-Oh complexity to measure the worst-case time of a computation. As we saw in Choosing Between Representations, however, this is sometimes too coarse a bound when the complexity is heavily dependent on the exact sequence of operations run. Now, we will consider a different style of complexity analysis that better accommodates operation sequences.

16.1 A First Example

Consider, for instance, a set that starts out empty, followed by a sequence of \(k\) insertions and then \(k\) membership tests, and suppose we are using the representation without duplicates. Insertion time is proportional to the size of the set (and list); this is initially \(0\), then \(1\), and so on, until it reaches size \(k\). Therefore, the total cost of the sequence of insertions is \(k \cdot (k+1) / 2\). The membership tests cost \(k\) each in the worst case, because we’ve inserted up to \(k\) distinct elements into the set. The total time is then

\begin{equation*}k^2 / 2 + k / 2 + k^2\end{equation*}

for a total of \(2k\) operations, yielding an average of

\begin{equation*}\frac{3}{4} k + \frac{1}{4}\end{equation*}

steps per operation in the worst case.

16.2 The New Form of Analysis

What have we computed? We are still computing a worst case cost, because we have taken the cost of each operation in the sequence in the worst case. We are then computing the average cost per operation. Therefore, this is a average of worst cases.Importantly, this is different from what is known as average-case analysis, which uses probability theory to compute the estimated cost of the computation. We have not used any probability here. Note that because this is an average per operation, it does not say anything about how bad any one operation can be (which, as we will see [Amortization Versus Individual Operations], can be quite a bit worse); it only says what their average is.

In the above case, this new analysis did not yield any big surprises. We have found that on average we spend about \(k\) steps per operation; a big-Oh analysis would have told us that we’re performing \(2k\) operations with a cost of \(O([k \rightarrow k])\) each in the number of distinct elements; per operation, then, we are performing roughly linear work in the worst-case number of set elements.

As we will soon see, however, this won’t always be the case: this new analysis can cough up pleasant surprises.

Before we proceed, we should give this analysis its name. Formally, it is called amortized analysis. Amortization is the process of spreading a payment out over an extended but fixed term. In the same way, we spread out the cost of a computation over a fixed sequence, then determine how much each payment will be.We have given it a whimsical name because Halloween is a(n American) holiday devoted to ghosts, ghouls, and other symbols of death. Amortization comes from the Latin root mort-, which means death, because an amortized analysis is one conducted “at the death”, i.e., at the end of a fixed sequence of operations.

16.3 An Example: Queues from Lists

We have already seen lists [From Tables to Lists] and sets [Sets Appeal]. Now let’s consider another fundamental computer science data structure: the queue. A queue is a linear, ordered data structure, just like a list; however, the set of operations they offer is different. In a list, the traditional operations follow a last-in, first-out discipline: .first returns the element most recently linked. In contrast, a queue follows a first-in, first-out discipline. That is, a list can be visualized as a stack, while a queue can be visualized as a conveyer belt.

16.3.1 List Representations

We can define queues using lists in the natural way: every enqueue is implemented with link, while every dequeue requires traversing the whole list until its end. Conversely, we could make enqueuing traverse to the end, and dequeuing correspond to .rest. Either way, one of these operations will take constant time while the other will be linear in the length of the list representing the queue.

In fact, however, the above paragraph contains a key insight that will let us do better.

Observe that if we store the queue in a list with most-recently-enqueued element first, enqueuing is cheap (constant time). In contrast, if we store the queue in the reverse order, then dequeuing is constant time. It would be wonderful if we could have both, but once we pick an order we must give up one or the other. Unless, that is, we pick...both.

One half of this is easy. We simply enqueue elements into a list with the most recent addition first. Now for the (first) crucial insight: when we need to dequeue, we reverse the list. Now, dequeuing also takes constant time.

16.3.2 A First Analysis

Of course, to fully analyze the complexity of this data structure, we must also account for the reversal. In the worst case, we might argue that any operation might reverse (because it might be the first dequeue); therefore, the worst-case time of any operation is the time it takes to reverse, which is linear in the length of the list (which corresponds to the elements of the queue).

However, this answer should be unsatisfying. If we perform \(k\) enqueues followed by \(k\) dequeues, then each of the enqueues takes one step; each of the last \(k-1\) dequeues takes one step; and only the first dequeue requires a reversal, which takes steps proportional to the number of elements in the list, which at that point is \(k\). Thus, the total cost of operations for this sequence is \(k \cdot 1 + k + (k-1) \cdot 1 = 3k-1\) for a total of \(2k\) operations, giving an amortized complexity of effectively constant time per operation!

16.3.3 More Liberal Sequences of Operations

In the process of this, however, I’ve quietly glossed over something you’ve probably noticed: in our candidate sequence all dequeues followed all enqueues. What happens on the next enqueue? Because the list is now reversed, it will have to take a linear amount of time! So we have only partially solved the problem.

Now we can introduce the second insight: have two lists instead of one. One of them will be the tail of the queue, where new elements get enqueued; the other will be the head of the queue, where they get dequeued:

data Queue<T>: | queue(tail :: List<T>, head :: List<T>) end mt-q :: Queue = queue(empty, empty)

Provided the tail is stored so that the most recent element is the first, then enqueuing takes constant time:

fun enqueue<T>(q :: Queue<T>, e :: T) -> Queue<T>: queue(link(e, q.tail), q.head) end

For dequeuing to take constant time, the head of the queue must be stored in the reverse direction. However, how does any element ever get from the tail to the head? Easy: when we try to dequeue and find no elements in the head, we reverse the (entire) tail into the head (resulting in an empty tail). We will first define a datatype to represent the response from dequeuing:

data Response<T>: | elt-and-q(e :: T, r :: Queue<T>) end

Now for the implementation of dequeue:

fun dequeue<T>(q :: Queue<T>) -> Response<T>: cases (List) q.head: | empty => new-head = q.tail.reverse() elt-and-q(new-head.first, queue(empty, new-head.rest)) | link(f, r) => elt-and-q(f, queue(q.tail, r)) end end

16.3.4 A Second Analysis

We can now reason about sequences of operations as we did before, by adding up costs and averaging. However, another way to think of it is this. Let’s give each element in the queue three “credits”. Each credit can be used for one constant-time operation.

One credit gets used up in enqueuing. So long as the element stays in the tail list, it still has two credits to spare. When it needs to be moved to the head list, it spends one more credit in the link step of reversal. Finally, the dequeuing operation performs one operation too.

Because the element does not run out of credits, we know it must have had enough. These credits reflect the cost of operations on that element. From this (very informal) analysis, we can conclude that in the worst case, any permutation of enqueues and dequeues will still cost only a constant amount of amortized time.

16.3.5 Amortization Versus Individual Operations

Note, however, that the constant represents an average across the sequence of operations. It does not put a bound on the cost of any one operation. Indeed, as we have seen above, when dequeue finds the head list empty it reverses the tail, which takes time linear in the size of the tail—not constant at all! Therefore, we should be careful to not assume that every step in the sequence will is bounded. Nevertheless, an amortized analysis sometimes gives us a much more nuanced understanding of the real behavior of a data structure than a worst-case analysis does on its own.

16.4 Reading More

At this point we have only briefly touched on the subject of amortized analysis. A very nice tutorial by Rebecca Fiebrink provides much more information. The authoritative book on algorithms, Introduction to Algorithms by Cormen, Leiserson, Rivest, and Stein, covers amortized analysis in extensive detail.