# 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.