Lab 09 - Amortized Analysis

1. Stacks

2. Binary Counter

3. FIFO

4. Heaps

4.1 Aggregate analysis:

Let S be a sequence of n insertions into an initially empty heap.

At the end of the sequence, we will have an almost-complete binary tree. To make the analysis simpler, suppose it is complete.

On the level 1 (base level) of the tree, we have n/2 elements which have been either swapped-down, or remained there upon insertion

On level 2, n/4 elements must have climbed-up exactly once, after insertion

On the last level, log(n), the root must have climbed-up (via swapping) ceil(log(n)) times.

The aggregate climbing-up costs (the down-climbing is not included as no element can go down without one coming-up), per tree level, are given below:

 level 1        n/2 elems,    0 swaps      
 level 2        n/4 elems,    n/4 swaps   
 level 3        n/8 elems,    2*n/8 swaps 
 level log(n)   1 elem        log(n) * 1 swaps

cost(S) = ins_cost(S) + swap_cost(S)

      = n + complicated_sum

                log(n)

complicated_sum = sum n/2^i (elems per level) * (i-2) (swaps)

                i = 1

                 log(n)
              <= sum    n/2^i * i = n sum ...
                 i = 1

The sum i/2^i can be solved via derivation. Finally, n * sum … ~= n/2 - 1 - log(n)

Thus, cost(S) ⇐ 2n hence is done in linear time.

Banking?

Potential?

5. ArrayList with removal

Further reading: https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:ammortized-analysis