CPSC 320: Intermediate Algorithm Design and Analysis August 6, 2014 - PowerPoint PPT Presentation

CPSC 320: Intermediate Algorithm Design and Analysis August 6, 2014 1

Schedule Monday: BC Day, no classes • Wednesday: • Quiz 5 (NP and NP completeness) • Review: Amortized Analysis • Friday: • Survey (bonus marks) • Review: probably Graph Theory and Dynamic Programming • 2

Amortized Analysis 3

Amortized cost Although the worst case for one call has a large bound, we are interested in the • total cost of a sequence of calls In other words, 𝑙 calls to a 𝑃 𝑔 𝑜 function may be better than 𝑃 𝑙 ⋅ 𝑔 𝑜 • The amortized cost of the function takes the total amount in consideration • 4

Aggregate Method Evaluate a sequence of calls, and calculate the total cost • Usually changes in data structure are considered instead of operations • 5

Potential Method Each operation receives a potential cost, which if larger than actual cost can be • “saved” towards future operations Operations that take longer can use “saved” value • Total cost of sequence of operations is the sum of potential costs • Properties: • Assume 𝐸 𝑗 is state of algorithm after 𝑗 iterations • Φ 𝐸 𝑗 ≥ 0 (you cannot use more than what you saved) • Φ 𝐸 0 = 0 (you start with no saved cost – usually) • 𝑑𝑝𝑡𝑢 𝑏𝑛 𝑝𝑞 𝑗 = Φ 𝐸 𝑗 − Φ 𝐸 𝑗−1 + 𝑑𝑝𝑡𝑢 𝑠𝑓𝑏𝑚 (𝑝𝑞 𝑗 ) • Alternative notation: • 𝑑 𝑗 for real cost, 𝑑 𝑗 for amortized cost • 6

Garbage Collection Assume a memory management process where: • allocating memory takes a constant time 𝑏 in general case • there is no freeing process, but memory reference is taken into account • if there is no memory left (or memory runs over a limit), unreferenced • memory space is freed each freed element takes constant time 𝑔 • For simplicity, assume there is trivial way to retrieve unreferenced memory space • 7

Garbage Collection Real cost of allocate: • If no garbage collection is needed: 𝑏 • If garbage collection is needed: 𝑏 + 𝑔𝑠 ( 𝑠 is number of freeable objects) • Amortized cost of allocate: • Assuming initially no elements are allocated • Each freed element in garbage collection requires a previous allocation, and • every allocation has a single free Sequence of 𝑙 calls to allocate costs less than 𝑏𝑙 + 𝑔𝑙 • Each allocate can be assumed to have amortized cost 𝑏 + 𝑔 ∈ Θ(1) • 8

Garbage Collection – Potential Method What information in the data structure determines an expensive operation? • Allocate is costly if number of elements is large • In worst-case, allocate takes time linear to size of memory • Potential function: number of allocated words × 𝑔 • If no GC is needed: 𝑑𝑝𝑡𝑢 𝑏𝑛 𝑏𝑚𝑚𝑝𝑑 = 𝑑𝑝𝑡𝑢 𝑠𝑓𝑏𝑚 𝑏𝑚𝑚𝑝𝑑 + Δ Φ = 𝑏 + 𝑔 • If GC is needed: 𝑑𝑝𝑡𝑢 𝑏𝑛 𝑏𝑚𝑚𝑝𝑑 = 𝑑𝑝𝑡𝑢 𝑠𝑓𝑏𝑚 𝑏𝑚𝑚𝑝𝑑 + Δ Φ = 𝑏 + 𝑔𝑠 + 𝑔 − 𝑔𝑠 = 𝑏 + 𝑔 • 9

More examples Consider an operation that, when called the 𝑜 th time, costs 𝑜 if 𝑜 is a power of 2, • and costs 1 otherwise Aggregate analysis: • A sequence of 𝑜 operations will cost 1 every time except lg 𝑜 + 1 times • ( 2 0 , 2 1 , … 2 lg 𝑜 ) 𝑜 operations will cost: • lg 𝑜 2 𝑗 ≤ 𝑜 + 2 lg 𝑜+1 − 1 𝐷 ≤ 𝑜 + = 𝑜 + 2𝑜 − 1 ≤ 3𝑜 ∈ 𝑃 𝑜 2 − 1 𝑗=0 Each operation has an amortized constant cost • 10

More examples Consider an operation that, when called the 𝑜 th time, costs 𝑜 2 if 𝑜 is a power of 2, • and costs 1 otherwise Aggregate analysis: • A sequence of 𝑜 operations will cost 1 every time except lg 𝑜 + 1 times • ( 2 0 , 2 1 , … 2 lg 𝑜 ) 𝑜 operations will cost: • lg 𝑜 2 2𝑗 ≤ 𝑜 + 4 lg 𝑜+1 − 1 = 𝑜 + 4𝑜 2 − 1 ∈ 𝑃 𝑜 2 𝐷 ≤ 𝑜 + 4 − 1 3 𝑗=0 Each operation has an amortized linear cost • 11

Exercise Georgia Street has many tall buildings, but only some of them have a clear view of Stanley Park. Suppose we are given an array 𝐵 1. . 𝑜 that stores the height of 𝑜 buildings on a city block, indexed from South to North. Building 𝑗 has a good view of Stanley Park if and only if every building to the North of 𝑗 is shorter than 𝑗 . Here is an algorithm that computes which buildings have a good view of Stanley Park. What is the running time of this algorithm? Algorithm Goodview(A[1 .. n]): Initialize a stack S For i ← 1 to n Do While (S not empty and A[i] > A[Top(S)]) Do Pop(S) Push(S, i) Return S 12

Good view – Aggregate Analysis Consider each different building • How many times is it pushed? • How many times is it popped? • How many times is Top called? • For each building: • Push is called exactly once • Pop is called no more than once • Top is called once for every pop and once for every push • So, potentially, each building triggers up to 4 queue operations • Since there are 𝑜 buildings, this algorithm takes 𝑃(𝑜) total time • 13

Good view – Potential Method Potential method corresponds to twice the size of the stack • Push: • 𝑑𝑝𝑡𝑢 𝑏𝑛 𝑞𝑣𝑡ℎ = 𝑑𝑝𝑡𝑢 𝑠𝑓𝑏𝑚 𝑞𝑣𝑡ℎ + Δ Φ = 1 + 2 = 3 Pop: • 𝑑𝑝𝑡𝑢 𝑏𝑛 𝑞𝑝𝑞 = 𝑑𝑝𝑡𝑢 𝑠𝑓𝑏𝑚 𝑞𝑝𝑞 + Δ Φ = 1 − 2 = −1 Top: • 𝑑𝑝𝑡𝑢 𝑏𝑛 𝑢𝑝𝑞 = 𝑑𝑝𝑡𝑢 𝑠𝑓𝑏𝑚 𝑢𝑝𝑞 + Δ Φ = 1 + 0 = 1 In each iteration of the for loop, if there are 𝑙 calls to Pop, there are 𝑙 + 1 calls to Top • Cost for each iteration: 𝑙 ⋅ −1 + 𝑙 + 1 ⋅ 1 + 3 = 4 • Since the loop iterates 𝑜 times, we have a total cost of 4𝑜 ∈ 𝑃 𝑜 • 14