====== Lab 09 - Amortized Analysis======

**1.** Stacks

**2.** Binary Counter

**3.** [[https://ocw.cs.pub.ro/ppcarte/doku.php?id=aa:ammortized-analysis|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