# 8.2. Amortized Analysis¶

Our analysis of the efficiency of hash table operations concluded that find runs in expected constant time, where the modifier “expected” is needed to express the fact the performance is on average and depends on the hash function satisfying certain properties.

We also concluded that insert would usually run in expected constant time, but that in the worst case it would require linear time because of needing to rehash the entire table. That kind of defeats the goal of a hash table, which is to offer constant-time performance, or at least as close to it as we can get.

It turns out there is another way of looking at this analysis that allows us to conclude that insert does have “amortized” expected constant time performance—that is, for excusing the occasional worst-case linear performance. Right away, we have to acknowledge this technique is just a change in perspective. We’re not going to change the underlying algorithms. The insert algorithm will still have worst-case linear performance. That’s a fact.

But the change in perspective we now undertake is to recognize that if it’s very rare for insert to require linear time, then maybe we can “spread out” that cost over all the other calls to insert. It’s a creative accounting trick!

Sushi vs. Ramen. Let’s amuse ourselves with a real-world example for a moment. Suppose that you have $20 to spend on lunches for the week. You like to eat sushi, but you can’t afford to have sushi every day. So instead you eat as follows: • Monday:$1 ramen

• Tuesday: $1 ramen • Wednesday:$1 ramen

• Thursday: $1 ramen • Friday:$16 sushi

Most of the time, your lunch was cheap. On a rare occasion, it was expensive. So you could look at it in one of two ways:

• My worst-case lunch cost was $16. • My average lunch cost was$4.

Both are true statements, but maybe the latter is more helpful in understanding your spending habits.

Back to Hash Tables. It’s the same with hash tables. Even though insert is occasionally expensive, it’s so rarely expensive that the average cost of an operation is actually constant time! But, we need to do more complicated math (or more complicated than our lunch budgeting anyway) to actually demonstrate that’s true.

## 8.2.1. Amortized Analysis of Hash Tables¶

“Amortization” is a financial term. One of its meanings is to pay off a debt over time. In algorithmic analysis, we use it to refer to paying off the cost of an expensive operation by inflating the cost of inexpensive operations. In effect, we pre-pay the cost of a later expensive operation by adding some additional cost to earlier cheap operations.

The amortized complexity or amortized running time of a sequence of operations that each have cost $$T_1, T_2, \ldots, T_n$$, is just the average cost of each operation:

$\frac{T_1 + T_2 + ... + T_n}{n}.$

Thus, even if one operation is especially expensive, we could average that out over a bunch of inexpensive operations.

Applying that idea to a hash table, suppose the table has 8 bindings and 8 buckets. Then 8 more inserts are made. The first 7 are (expected) constant-time, but the 8th insert is linear time: it increases the load factor to 2, causing a resize, thus causing rehashing of all 16 bindings into a new table. The total cost over that series of operations is therefore the cost of 8+16 inserts. For simplicity of calculation, we could grossly round that up to 16+16 = 32 inserts. So the average cost of each operation in the sequence is 32/8 = 4 inserts.

In other words, if we just pretended each insert cost four times its normal price, the final operation in the sequence would have been “pre-paid” by the extra price we paid for earlier inserts. And all of them would be constant-time, since four times a constant is still a constant.

Generalizing from the example above, let’s suppose that the the number of buckets currently in a hash table is $$2^n$$, and that the load factor is currently 1. Therefore, there are currently $$2^n$$ bindings in the table. Next:

• A series of $$2^n - 1$$ inserts occurs. There are now $$2^n + 2^n - 1$$ bindings in the table.

• One more insert occurs. That brings the number of bindings up to $$2^n + 2^n$$, which is $$2^{n+1}$$. But the number of buckets is $$2^n$$, so the the load factor just reached 2. A resize is necessary.

• The resize occurs. That doubles the number of buckets. All $$2^{n+1}$$ bindings have to be reinserted into the new table, which is of size $$2^{n+1}$$. The load factor is back down to 1.

So in total we did $$2^n + 2^{n+1}$$ inserts, which included $$2^n$$ inserts of bindings and $$2^{n+1}$$ re-insertions after the resize. We could grossly round that quantity up to $$2^{n+2}$$. Over a series of $$2^n$$ insert operations, that’s an average cost of $$\frac{2^{n+2}}{2^n}$$, which equals 4. So if we just pretend each insert costs four times its normal price, every operation in the sequence is amortized (and expected) constant time.

Doubling vs. Constant-size Increasing. Notice that it is crucial that the array size grows by doubling (or at least geometrically). A bad mistake would be to instead grow the array by a fixed increment—for example, 100 buckets at time. Then we’d be in real trouble as the number of bindings continued to grow:

• Round 1. Insert 100 bindings. There are now 200 bindings and 100 buckets. The load factor is 2.

• Increase the number of buckets by 100 and rehash. That’s 200 more insertions. The load factor is back down to 1.

• The average cost of each insert is so far just 3x the cost of an actual insert (100+200 insertions / 100 bindings inserted). So far so good.

• Round 2. Insert 200 more bindings. There are now 400 bindings and 200 buckets. The load factor is 2.

• Increase the number of buckets by 100 and rehash. That’s 400 more insertions. There are now 400 bindings and 300 buckets. The load factor is 400/300 = 4/3, not 1.

• The average cost of each insert is now 100+200+200+400 / 300 = 3. That’s still okay.

• Round 3. Insert 200 more bindings. There are now 600 bindings and 300 buckets. The load factor is 2.

• Increase the number of buckets by 100 and rehash. That’s 600 more insertions. There are now 600 bindings and 400 buckets. The load factor is 3/2, not 1.

• The average cost of each insert is now 100+200+200+400+200+600 / 500 = 3.2. It’s going up.

• Round 4. Insert 200 more bindings. There are now 800 bindings and 400 buckets. The load factor is 2.

• Increase the number of buckets by 100 and rehash. That’s 800 more insertions. There are now 800 bindings and 500 buckets. The load factor is 8/5, not 1.

• The average cost of each insert is now 100+200+200+400+200+600+200+800 / 700 = 3.7. It’s continuing to go up, not staying constant.

After $$k$$ rounds we have $$200k$$ bindings and $$100k$$ buckets. We have called insert to insert $$100+200k$$ bindings, but all the rehashing has caused us to do $$100+200(k-1)+\sum_{i=1}^{k} 200i$$ actual insertions. That last term is the real problem. It’s quadratic:

$\sum_{i=1}^{k} 200i \quad = \quad \frac{200k (200 (k+1))}{2} \quad = \quad 20,000 (k^2 + k)$

So over a series of $$n$$ calls to insert, we do $$O(n^2)$$ actual inserts. That makes the amortized cost of insert be $$O(n)$$, which is linear! Not constant.

That’s why it’s so important to double the size of the array at each rehash. It’s what gives us the amortized constant-time performance.

## 8.2.2. Amortized Analysis of Batched Queues¶

The implementation of batched queues with two lists was in a way more efficient than the implementation with just one list, because it managed to achieve a constant time enqueue operation. But, that came at the tradeoff of making the dequeue operation sometimes take more than constant time: whenever the front became empty, the back had to be reversed, which required an additional linear-time operation.

As we observed then, the reversal is relatively rare. It happens only when the front gets exhausted. Amortized analysis gives us a way to account for that. We can actually show that the dequeue operation is amortized constant time.

To keep the analysis simple at first, let’s assume the queue starts off with exactly one element 1 already enqueued, and that we do three enqueue operations of 2, 3, then 4, followed by a single dequeue. The single initial element had to be in the front of the queue. All three enqueue operations will cons an element onto the back. So just before the dequeue, the queue looks like:

{front = ; back = [4; 3; 2]}


and after the dequeue:

{front = [2; 3; 4]; back = []}


It required

• 3 cons operations to do the 3 enqueues, and

• another 3 cons operations to finish the dequeue by reversing the list.

That’s a total of 6 cons operations to do the 4 enqueue and dequeue operations. The average cost is therefore 1.5 cons operations per queue operation. There were other pattern matching operations and record constructions, but those all took only constant time, so we’ll ignore them.

What about a more complicated situation, where there are enqueues and dequeues interspersed with one another? Trying to take averages over the series is going to be tricky to analyze. But, inspired by our analysis of hash tables, suppose we pretend that the cost of each enqueue is twice its actual cost, as measured in cons operations? Then at the time an element is enqueued, we could “prepay” the later cost that will be incurred when that element is cons’d onto the reversed list.

The enqueue operation is still constant time, because even though we’re now pretending its cost is 2 instead of 1, it’s still the case that 2 is a constant. And the dequeue operation is amortized constant time:

• If dequeue doesn’t need to reverse the back, it really does just constant work, and

• If dequeue does need to reverse a back list with $$n$$ elements, it already has $$n$$ units of work “saved up” from each of the enqueues of those $$n$$ elements.

So if we just pretend each enqueue costs twice its normal price, every operation in a sequence is amortized constant time. Is this just a bookkeeping trick? Absolutely. But it also reveals the deeper truth that on average we get constant-time performance, even though some operations might rarely have worst-case linear-time performance.

## 8.2.3. Bankers and Physicists¶

Conceptually, amortized analysis can be understood in three ways:

1. Taking the average cost over a series of operations. This is what we’ve done so far.

2. Keeping a “bank account” at each individual element of a data structure. Some operations deposit credits, and others withdraw them. The goal is for account totals to never be negative. The amortized cost of any operation is the actual cost, plus any credits deposited, minus any credits spent. So if an operation actually costs $$n$$ but spends $$n-1$$ credits, then its amortized cost is just $$1$$. This is called the banker’s method of amortized analysis.

3. Regarding the entire data structure as having an amount of “potential energy” stored up. Some operations increase the energy, some decrease it. The energy should never be negative. The amortized cost of any operation is its actual cost, plus the change in potential energy. So if an operation actually costs $$n$$, and before the operation the potential energy is $$n$$, and after the operation the potential energy is $$0$$, then the amortized cost is $$n + (0 - n)$$, which is just $$0$$. This is called the physicist’s method of amortized analysis.

The banker’s and physicist’s methods can be easier to use in many situations than a complicated analysis of a series of operations. Let’s revisit our examples so far to illustrate their use:

• Banker’s method, hash tables: The table starts off empty. When a binding is added to the table, save up 1 credit in its account. When a rehash becomes necessary, every binding is guaranteed to have 1 credit. Use that credit to pay for the rehash. Now all bindings have 0 credits. From now on, when a binding is added to the table, save up 1 credit in its account and 1 credit in the account of any one of the bindings that has 0 credits. At the time the next rehash becomes necessary, the number of bindings has doubled. But since we’ve saved 2 credits at each insertion, every binding now has 1 credit in its account again. So we can pay for the rehash. The accounts never go negative, because they always have either 0 or 1 credit.

• Banker’s method, two-list queues: When an element is added to the queue, save up 1 credit in its account. When the back must be reversed, use the credit in each element to pay for the cons onto the front. Since elements enter at the back and transition at most once to the front, every element will have 0 or 1 credits. So the accounts never go negative.

• Physicist’s method, hash tables: At first, define the potential energy of the table to be the number of bindings inserted. That energy will therefore never be negative. Each insertion increases the energy by 1 unit. When the first rehash is needed after inserting $$n$$ bindings, the potential energy is $$n$$. The potential goes back down to $$0$$ at the rehash. So the actual cost is $$n$$, but the change in potential is $$n$$, which makes the amortized cost $$0$$, or constant. From now on, define the potential energy to be twice the number of bindings inserted since the last rehash. Again, the energy will never be negative. Each insertion increases the energy by 2 units. When the next rehash is needed after inserting $$n$$ bindings, there will be $$2n$$ bindings that need to be rehashed. Again, the amortized cost will be constant, because the actual cost of $$2n$$ re-insertions is offset by the $$2n$$ change in potential.

• Physicist’s method, two-list queues: Define the potential energy of the queue to be the length of the back. It therefore will never be negative. When a dequeue has to reverse a back list of length $$n$$, there is an actual cost of $$n$$ but a change in potential of $$n$$ too, which offsets the cost and makes it constant.

The two methods are equivalent in their analytical power:

• To convert a banker’s analysis into a physicist’s, just make the potential be the sum of all the credits in the individual accounts.

• To convert a physicist’s analysis into a banker’s, just designate one distinguished element of the data structure to be the only one that will ever hold any credits, and have each operation deposit or withdraw the change in potential into that element’s account.

So, the choice of which to use really just depends on which is easier for the data structure being analyzed, or which is easier for you to wrap your head around. You might find one or the other of the methods easier to understand for the data structures above, and your friend might have a different opinion.

## 8.2.4. Amortized Analysis and Persistence¶

Amortized analysis breaks down as a technique when data structures are used persistently. For example, suppose we have a two-list queue q into which we’ve inserted $$n+1$$ elements. One elements will be in the front, and the other $$n$$ will be in the back. Now we do the following:

# let q1 = dequeue q
# let q2 = dequeue q
...
# let qn = dequeue q


Each one of those $$n$$ dequeue operations requires an actual cost of $$O(n)$$ to reverse the back list. So the entire series has an actual cost of $$O(n^2)$$. But the amortized analysis techniques only apply to the first dequeue. After that, all the the accounts are empty (banker’s method), or the potential is zero (physicist’s), which means the remaining operations can’t use them to pay for the expensive list reversal. The total cost of the series is therefore $$O(n^2 - n)$$, which is $$O(n^2)$$.

The problem with persistence is that it violates the assumption built-in to amortized analysis that credits (or energy units) are spent only once. Every persistent copy of the data structure instead tries to spend them itself, not being aware of all the other copies.

So we must be careful not to use functional data structures persistently if we want amortized analysis bounds to hold. There are more advanced techniques for amortized analysis that can account for persistence. Those techniques are based on the idea of accumulating debt that is later paid off, rather than accumulating savings that are later spent. The difference is that although our banks would never (financially speaking) allow us to spend money twice, they would be fine with us paying off our debt multiple times.