} n ops take time T(n) PUSH(S,x) --- O(1) Avg cost per op = - PDF document

Stack Example Stack Example Amortized Analysis: 3 Methods Amortized Analysis: 3 Methods • Aggregate Consider a stack data structure with 3 ops } • n ops take time T(n) • PUSH(S,x) --- O(1) • Avg cost per op = T(n)/n (Amortized Cost) • Imprecise – no seperate cost for each item • POP(S) --- O(1) s • Accounting --- (Book-Keeping scheme for counting ops • MULTIPOP(S,k) --- Min(s,k) • Charge each op an amortized cost Stack S • Store in “bank” amount not used • Later ops can used “banked” work • Potential • “Stored work” viewed as “potential energy” • Most flexible & powerful point of view Stack Example (cont.) Aggregate Method Stack Example (cont.) Aggregate Method • If the stack S is initially empty, Non-Tight Bound then an object can be Popped only after it has been pushed. So if Assume O(n) ops. Then the worst there are n ops, then case is a MultiPop of n objects. ≤ ≤ ≤ ≤ Hence, T(n) = O(n 2 ). # Pops # Pushes n • Moreover, counts all Pops, # Pops even those executed by all MultiPop Ergo, T(n)/n =O(n) cost per op. ops. So the total work T(n) for n ops is at most n+n = O(n). Hence, the amortized cost per operation is = T n ( )/ n O (1) 1

Incrementing a Counter: Aggregate Analysis Incrementing a Counter: Aggregate Analysis Incrementing a Counter: Aggregate Analysis Incrementing a Counter: Aggregate Analysis We have a k-bit register which counts up from 0. [ [ ] ] [ ] [ ] [ ] [ ] [ ] [ ] Least Sig A k − 1 A 2 A 1 A 0 bit Most Sig � bit − = ∑ k 1 [ ] [ ] Value of counter i A i 2 = i 0 There are two types of bit flip ops: • Reset to O • Set to 1 Cursory Analysis: Non- -Tight Bound Tight Bound Better Analysis: Tighter Bound Cursory Analysis: Non Better Analysis: Tighter Bound Observation: If register is initially • The computation work for a single bit = set to 0, then after a seq of n ops, O(k), since worst case is k-1 resets and bit A[i] flips exactly: 1 set. { ≤  times, if    / 2 i   i lg n  n  • Hence, work for n ops = O(nk) >  0 times, if    i lg n ∞   lg n ∑ n ∑ Hence, = = < < − = i T n ( ) n 2 2 n   • Ergo, Amortized cost =T(n)/n =O(k)   i 2 = = = = i 0 i 0 = ⇒ = So, T n ( ) O n ( ) T n ( )/ n O (1) 2

Accounting Method Accounting Method Stack Example: Accounting Method Stack Example: Accounting Method Assign Amortized Costs in such a way that Actual Costs: you overcharge on some ops, and undercharge Push 1 on others. Pop 1 Overcharge = Actual Cost + Credit MultiPop Min(k,s) Undercharge = Actual Cost –Credit Amortized Costs: ± Amortized Cost = Actual Cost Credit Push 2 But always, Pop 0 ≤ Total Actual Cost Total Amortized Cost MultiPop 0 Stack Example: Accounting Method Incrementing Counter: Accounting Method Stack Example: Accounting Method Incrementing Counter: Accounting Method Amortized Costs: Associate Credit with Data Structure Set Bit to 1 --- $2 Push Push Push Pop Push Push MultiP Reset Bit to 0 -- $0 P=$ P=$ P=$ P=$ P=$ P=$ P=$ P=$ Note. A bit cannot be reset unless it has P=$ P=$ P=$ P=$ P=$ P=$ P=$ been set sometime in the past. So if we pay $1 $2 $3 $2 $3 $4 $0 Total Credit $2 for each Set (which is $1 too much), $4 $1 $2 $3 $5 $6 $10 Tot Act. Cost then there will be $1 left to pay for a Reset, should it occur. Therefore, the total $2 $4 $6 $6 $8 $10 $10 Total ≥ Amortized available credit will always be 0 Cost ≤ ≤ Actual Cost Total Amortized Cost = Total Actual Cost Total Amortized Cost ≤ ≤ = = 2#Sets ( ( ) ) 2 n O n ( ) = = = = ⇒ = 2 # Push O n ( ) T n ( ) O n ( ) 3

The Potential Method The Potential Method Example: A Dynamic Table Example: A Dynamic Table To be Completed Example: A Dynamic Table(Cont Table(Cont.) .) Example: A Dynamic Table(Cont Table(Cont.) .) Example: A Dynamic Example: A Dynamic 4