Lecture 20 - Amortized Analysis

Question 1

What is amortized analysis?

What is the goal?

What are the 2 methods we use?

Answer 1

Analyze a sequence of operations on a data structure.
• We will talk about average cost in the worst case (i.e. not averaging over a distribution of inputs. No probability!)

Goal: Show that although some individual operations may be expensive, on average the cost per operation is small.

1. aggregate analysis
2. accounting method

Question 2

What are the running times of Push and Pop on a stack? What about a sequence of each?

Answer 2

O(1)

O(n) for a sequence

Question 3

What’s the worst case running time of a sequence of Multipop?

Answer 3

O(n^2)

Question 4

What’s the average over the n operations of push pop? why?

Answer 4

O(1).

Each object can be popped only once per time that it is pushed.
• Have ≤ n PUSHes ⇒ ≤ n POPs, including those in MULTIPOP.
• Therefore, total cost = O(n).

Question 5

What is the cost of the INCREMENT function we saw?

Answer 5

Cost of INCREMENT = Q(# of bits flipped)

Question 6

What is the total # of flips per INCREMENT? Explain how this was achieved.

What is the cost of n INCREMENTs. and Average?

Answer 6

2*n
Every bit i is flipped 1/2^i times, thus geometric.

O(n)
O(1)

Question 7

What is the amortized cost in accounting method? How is it handled when cost is higher than actual cost?

Answer 7

Amortized cost = amount we charge different operations
• When amortized cost is higher than the actual cost, store the difference on specific objects in the data structure as credit.
• Use credit later to pay for operations whose actual cost is higher than the amortized cost.

Question 8

How is the accounting method different than aggregate method?

Answer 8

Accounting method differs from aggregate analysis:
• In the accounting method, different operations can have different costs.
• In aggregate analysis, all operations have same cost.

Question 9

What’s the restriction we put on the credit in the accounting method? Explain how this impacts the sum of amortized cost and actual cost.

Answer 9

The credit cannot be negative.

Therefore, the sum of actual cost of operations has to be larger or equal to the sum of amortized cost.

Question 10

What is a good amortized cost for Push Pop Multipop? Explain the intuition behind this.

What would be the total amortized cost?

Answer 10

Push = 2
Pop = 0
Multipop = 0
Intuition: When pushing an object, pay $2.
• $1 pays for the PUSH.
• $1 is prepayment for it being popped by either POP or MULTIPOP.
• Since each object has $1, which is credit, the credit can never go negative.

Total amortized cost (= O(n)) is an upper bound on total actual cost.

Question 11

What is a good amortized cost for setting a bit in INCREMENT? Explain the intuition behind this.

What would be the total amortized cost?

Answer 11

Charge $2 to set a bit to 1.
INTUITION
• $1 pays for setting a bit to 1.
• $1 is prepayment for flipping it back to 0.
• Have $1 of credit for every 1 in the counter.
• Therefore, credit ≥ 0.

Amortized cost of INCREMENT:
• Cost of resetting bits to 0 is paid by credit.
• At most 1 bit is set to 1.
• Therefore, amortized cost ≤ $2.
• For n operations, amortized cost = O(n).

Question 12

Using aggregate analysis, what is the amortized cost per operation of inserting in a dynamic table? Explain how this was derived.

Answer 12

3.
Cost of insertion = 1
ci = actual cost of ith operation
If not full, ci = 1 
If full, have i−1 items in the table at the start of the i'th operation. Have to copy all i − 1 existing items, then insert ith item ⇒ ci = i

Sum of ci <= n + sum2^j (j0->logn) = n+2n = 3n.

Question 13

Using Accounting method, what is the amortized cost per operation of inserting in a dynamic table? Explain how this was derived.

Prove the credit never goes negative.

Answer 13

Charge $3 per insertion of x.
• $1 pays for x’s insertion.
• $1 pays for x to be moved in the future.
• $1 pays for some other item to be moved

Prove the credit never goes negative:
• size=m before and size=2m after expansion.
• Assume that the expansion used up all the credit, thus that there is no credit available after the expansion.
• We will expand again after another m insertions.
• Each insertion will put $1 on one of the m items that were in the table just after expansion and will put $1 on the item inserted.
• Have $2m of credit by next expansion, when there are 2m items to move. Just enough to pay for the expansion…