About this lecture. Three Methods for the Same Purpose (1) Aggregate Method (2) Accounting Method (3) Potential Method.

Transcription

1 About this lecture Given a data structure, amortized analysis studies in a sequence of operations, the average time to perform an operation Introduce amortized cost of an operation Three Methods for the Same Purpose (1) Aggregate Method (2) Accounting Method (3) Potential Method This Lecture 2

2 Super Stack Your friend has created a super stack, which, apart from PUSH/POP, supports: SUPER-POP(k): pop top k items Suppose SUPER-POP never pops more items than current stack size ThetimeforSUPER-POP is O(k) ThetimeforPUSH/POP is O(1) 3

3 Super Stack Suppose we start with an empty stack, and we have performed n operations But we don t knowtheorder Questions: Worst-case time of a SUPER-POP? Ans. O(n) time [why?] Total time of n operations in worst case? Ans. O(n 2 ) time [correct, but not tight] 4

4 Super Stack Though we don t know the order of the operations, we still know that: There are at most n PUSH/POP Time spent on PUSH/POP = O(n) # items popped by all SUPER-POP cannot exceed total # items ever pushed into stack Time spent on SUPER-POP = O(n) So, total time of n operations = O(n)!!! 5

5 Amortized Cost So far, there are no assumptions on n and the order of operations. Thus, we have: For any n and any sequence of n operations, worst-case total time = O(n) We can think of each operation performs in average O(n) / n = O(1) time We say amortized cost = O(1) per operation (or, each runs in amortized O(1) time) 6

6 Amortized Cost In general, we can say something like: OP 1 runs in amortized O(x) time OP 2 runs in amortized O(y) time OP 3 runs in amortized O(z) time Meaning: For any sequence of operations with #OP 1 = n 1,#OP 2 = n 2,#OP 3 = n 3, worst-case total time = O(n 1 x + n 2 y + n 3 z) 7

7 Binary Counter Let us see another example of implementing a k-bit binary counter At the beginning, count is, and the counter will be like (assume k=5): which is the binary representation of the count 8

8 Binary Counter When the counter is incremented, the content will change Example: content of counter when: count = 5 cost = 2 count = 6 The cost of the increment is equal to the number of bits flipped 9

9 Special case: Binary Counter When all bits in the counter are 1, an increment resets all bits to count = MAX cost = k count = The cost of the corresponding increment is equal to k, the number of bits flipped 1

10 Binary Counter Suppose we have performed n increments Questions: Worst-case time of an increment? Ans. O(k) time Total time of n operations in worst case? Ans. O(nk) time [correct, but not tight] 11

11 Binary Counter Let us denote the bits in the counter by b, b 1, b 2,, b k-1, starting from the right b 4 b 3 b 2 b 1 b Observation: b i is flipped only once in every 2 i increments Precisely, b i is flipped at x th increment x is divisible by 2 i 12

12 Amortized Cost So, for n increments, the total cost is: i= to k b n /2 i c i= to k ( n /2 i ) 2n By dividing total cost with #increments, amortized cost of increment = O(1) 13

13 Aggregate Method The computation of amortized cost of an operation in super stack or binary counter follows similar steps: 1. Find total cost (thus, an aggregation ) 2. Divide total cost by #operations This method is called Aggregate Method 14

14 About this lecture Previous lecture shows Aggregate Method This lecture shows two more methods: (2) Accounting Method (3) Potential Method 2

15 Accounting Method In real life, a bank account allows us to save our excess money, and the money can be used later when needed We also have an easy way to check the savings In amortized analysis, the accounting method is very similar 3

16 Accounting Method Each operation pays an amortized cost if amortized cost actual cost, wesavethe excess in the bank Else, we use savings to help the payment Often, savings can be checked easily based on the objects in the current data structure Lemma: For a sequence of operations, if we have enough to pay for each operation, total actual cost total amortized cost 4

17 Super Stack (Take 2) Recall that apart from PUSH/POP, a super stack, supports: SUPER-POP(k): pop top k items in k time Let us now assign the amortized cost for each operation as follows: PUSH =$2 POP or SUPER-POP =$ 5

18 Super Stack (Take 2) Questions: Which operation saves money to the bank when performed? Which operation needs money from the bank when performed? How to check the savings in the bank? 6

19 Super Stack (Take 2) Does our bank have enough to pay for each SUPER-POP operation? Ans. When SUPER-POP is performed, each popped item donates its corresponding $1 to help the payment Enough $$ to pay for each SUPER-POP 7

20 Super Stack (Take 2) Conclusion: Amortized cost of PUSH =2 Amortized cost of POP/SUPER-POP = Meaning: For any sequence of operations with #PUSH = n 1,#POP = n 2,#SUPER-POP = n 3, total actual cost 2n 1 8

21 Binary Counter (Take 2) Let us use accounting method to analyze increment operation in a binary counter, whose initial count = We assign amortized cost for each increment = $2 Recall: actual cost = #bits flipped 9

22 Binary Counter (Take 2) Observation: In each increment operation, at most one bit is set from to 1 (whereas the remaining bits are set from 1 to ). E.g., count = 4 count = count = 5 count = 6 1

23 Binary Counter (Take 2) Lemma: Savings = # of 1 s inthecounter Proof: By induction To show amortized cost =$2isenough, we use $1 to pay for flipping some bit x from to 1, and store the excess $1 For other bits being flipped (from 1 to ), each donates its corresponding $1 to help in paying the operation Enough to pay for each increment 11

24 Binary Counter (Take 2) Conclusion: Amortized cost of increment =2 Meaning: For n increments (with initial count = ) total actual cost 2n Question: What s wrong if initial count? 12

25 Accounting Method (Remarks) In contrast to the aggregate method, the accounting method may assign different amortized costs to different operations Another thing: To help the analysis, we usually link each excess $ to a specific object in the data structure (such as an item in a stack, or a bit in a binary counter) called the credit stored in the object 13

26 Potential Method Inphysics,anobjectatahigherplace has more potential energy (due to gravity) than an object at a lower place More potential Less potential 14

27 Potential Method The potential energy can usually be measured by some function of the status of the object (in fact, its height) In amortized analysis, the potential method is very similar It uses a potential function to measure the potential of a data structure, based on its current status 15

28 Potential Method Thus, potential of a data structure may increase or decrease after an operation The potential is similar to the $ in the accounting method, which can be used to help in paying an operation 16

29 Potential Method Each operation pays an amortized cost, and If potential increases by d after an operation, we need: amortized cost actual cost + d If potential decreases by d after an operation, we need: amortized cost + d actual cost 17

30 Potential Method To combine the above, we let = potential function D i = data structure after i th operation c i i = actual cost of i th operation = amortized cost of i th operation Then, we always need: i c i + D i D i-1 18

31 Potential Method Because smaller amortized cost gives better (tighter) analysis, so in general, we set: i = c i + D i D i-1 Consequently, after n operations, total amortized cost = total actual cost + D n D 19

32 Potential Method Any such that D i D for all i should work, as it implies total amortized cost at any time total actual cost at any time Our target is to find the best such so that amortized cost can be minimized 2

33 Super Stack (Take 3) Let us now use potential method to analyze the operations on a super stack Define such that for a super stack S Thus we have: S = #itemsins D =, and D i D for all i 21

34 Super Stack (Take 3) PUSH increases potential by 1 amortized cost of PUSH =1+1=2 POP decreases potential by 1 amortized cost of POP =1+(-1)= SUPER-POP(k) decreases potential by k amortized cost of SUPER-POP = k +(-k) = [Assume: Stack has enough items before POP/SUPER-POP] 22

35 Super Stack (Take 3) Conclusion: Because D =,and D i D for all i, total amortized cost total actual cost Then, by setting amortized cost for each operation accordingly (according to what??): amortized cost = O(1) 23

36 Binary Counter (Take 3) Let us now use potential method to analyze the increment in a binary counter Define such that for a binary counter B Thus we have: B = #bitsinb which are 1 D =, and D i D for all i Assume: initial count = 24

37 Binary Counter (Take 3) From our previous observation, at most 1 bit is set from to 1, the corresponding increase in potential is at most 1 Now, suppose the i th operation resets t i bits from 1 to actual cost c i = t i +1 potential change = (-t i )+1 amortized cost i = c i + potential change = 2 25

38 Binary Counter (Take 3) Conclusion: Because D =,and D i D for all i, total amortized cost total actual cost Then, by setting amortized cost for each operation accordingly: amortized cost = 2 = O(1) 26

39 Potential Method (Remarks) Potential method is very similar to the accounting method: we can save something ($/potential) now, which can be used later It usually gives a neat analysis, as the cost of each operation is very specific However, finding a good potential function can be extremely difficult (like magic) Analyzing Union-Find data structure 27