15-451 Algorithms D. Sleator Amortized analysis --- growing and shrinking a table I'm sure you all know the trick of doubling an array when it needs to grow. Then the amortized cost of all the growing operations is still O(1). Why? Because if the array ends up of size N, the total cost of all the doublings is 1+2+4+...N, which is at most 2N. And you needed the array to be size at least N/2 (or the last doubling would not have been done). Thus the cost of all the doublings is at most 4 times the number of insertions. The cost of the insertions themselves is 1 per insertion. Thus we get an amortized bound of 5. [ This can be proven using a potential, of course. Let s = number of things in the array, let n be the current size of the array. Here's the potential: Phi(n,s) = 0 if s4) then shrink() now delete the object from the table (cost of delete part is 1) Lemma: The amortized cost of an insertion or deletion are bounded by 5. Proof: Let the potential be the following: Phi(n,s) = 4|s-n/2| What is the amortized cost of an insertion? It's the actual cost plus the change in potential. A grow() may or may not happen as a result of an insertion. If a grow() occurs, what is the amortized cost of it? Before the grow() the potential is 4|n-n/2| = 2n. After the grow the potential is 0. The actual cost of the grow is 2n. Thus the amortized cost of the grow is 0. What about the rest of the insert? actual cost of insert = 1 change in potential <= 4. ====> amortized cost of insert <= 5. What about delete? If a shrink() happens, then the potential decreases by n, and the cost is n, so the amortized cost of shrink() is 0. What about the rest of the delete: actual cost of delete = 1 change in potential <= 4. ====> amortized cost of delete <= 5. This completes the proof. QED. Theorem: The total cost of a sequence of N insertions and deletions is at most 5N + 8. Proof: The amortized and real costs are related as follows: SUM (actual costs) <= SUM(amortized costs) + initial_potential - final_potential The initial potential is 8. The final potential is non-negative. The sum of the amortized costs are 5. QED. Actually, it's easy to get rid of the "+8" part. All we have to do is change the potential for n=4 to the following: [ 4|s-2| if s >= 2 Phi(4,s) = [ [ 0 otherwise Thus, we've zeroed the potential for the case when the initial array of size 4 is less than half full. The proof still goes through, because we never shrink an array of size 4, and this part of the potential is not important. This new potential has an initial value of 0, completing the proof.