Unit 6: Amortized Analysis

Transcription

1 : Amortized Analysis Course contents: Aggregate method Accounting method Potential method Reading: Chapter 17 Y.-W. Chang 1 Amortized Analysis Why Amortized Analysis? Find a tight bound of a sequence of data structure operations. No probability involved, guarantees the average performance of each operation in the worst case. Three popular methods Aggregate method Accounting method Potential method Y.-W. Chang 2 1

2 Methods for Amortized Analysis Aggregate method n operations take T(n) time. Average cost of an operation is T(n)/n time. Accounting method Charge each operation an amortized cost. Store the amount not used in bank. Use the stored amount for later operations. Must guarantee nonnegative balance!! Potential method View stored amount as potential energy. Y.-W. Chang 3 Aggregate method: MULTIPOP n operations take T(n) time average cost of an operation is T(n)/n time. Consider a sequence of n PUSH, POP, and MULTIPOP operations on an initially empty stack. Worst-case analysis: a MULTIPOP operation takes O(n). Aggregate method: Any sequence of n PUSH, POP, MULTIPOP costs at most O(n) time (why?) amortized cost of an operation: O(n)/n=O(1). MULTIPOP(S, k) 1.while not Stack-Empty(S) and k > 0 2. Pop(S); 3. k k-1; Y.-W. Chang 4 2

3 Incrementing a Binary Counter Increment an initially zero k-bit binary counter. Increment(A) 1. i = 0 2. while i < length[a] and A[I] == 1 3. A[i] = 0 4. i = i if i < length[a] 6. A[i] = 1 Y.-W. Chang 5 Aggregate method: Incrementing a Binary Counter Worst case: an INCREMENT operation takes O(k) time. Amortized cost: O(1) time. A[0], A[1], A[2], : flips each time, every other time, every fourth time,... that INCREMENT is called. lg n n # Flips = i0 i < 2n Amortized Cost = O(n)/n = O(1). 2 Y.-W. Chang 6 3

4 Accounting method Stack operations (s: stack size): For any sequence of n operations, total actual cost total amortized cost = O(n). Incrementing a binary counter Charge an amortized cost of $2 to set a bit to 1. 1 actual cost 1 credit Don't charge anything to reset a bit to 0 (-$1 from credit). For n increment operations, total credit = # of 1's in the counter 0, Total actual cost total amortized cost = O(n) Y.-W. Chang 7 The Potential Method View the prepaid work as potential that can be released to pay for future operations. Potential is associated with the whole data structure, not with specific items in the data structure. The potential method: D 0 : initial data structure D i : data structure after applying the i-th operation to D i-1 c i : actual cost of the i-th operation Define the potential function : D i R. c ˆi Amortized cost, = c i + (D i ) - (D i-1 ). Pick (D n ) (D 0 ) to make i1 i i1 Often define (D 0 ) = 0 and then show that (D i ) 0, i. n cˆ n c i Y.-W. Chang 8 4

5 The Potential Method: Stack Operations Amortized cost of each operation = O(1). (D) = # of objects in the stack D; (D 0 )=0, (D i ) 0. c c PUSH: = c i + (D i ) - (D i-1 ) = 1 + (s+1) - s = 2. ˆi POP: ˆi = c i + (D i ) - (D i-1 ) = 1 + (-1) = 0. MULTIPOP(S, k): k' = min(k, s) objects are popped off. = c i + (D i ) - (D i-1 ) = k' - k' = 0. Y.-W. Chang 9 Potential Method: Incrementing a Binary Counter Amortized cost of each operation = O(1). (D)= # of 1's in the counter D; let b i = (D i ) 0. Suppose the i-th increment operation resets t i bits. c i t i + 1 b i b i-1 - t i +1. c Amortized cost ˆi = c i + (D i ) - (D i-1 ) (t i + 1) - t i + 1 = 2. For b 0 k, let n = (k). c i = 5, t i = 4 b i-1 = 4, b i = 1 Y.-W. Chang 10 5

6 Dynamic Table Expansion Insertion only for the time being. Goal: Try to make table as small as possible. Idea: Allocate more memory when needed 1. Initialize table size m = Insert elements until the # of elements > m. 3. Generate a new table of size 2m. 4. Copy old elements into a new table. 5. Goto Step 2. Actual costs: c i = ith insertion. One Insertion can be costly, but in total? Worst-case cost of an insertion = O(n) total time for n insertions =O(n 2 ). Not tight!! Y.-W. Chang 11 Expansion: Aggregate and Accounting Analyses Aggregate analysis: amortized cost of an operation < 3. Accounting analysis: amortized cost of an operation < 3. Charge each operation $3 (amortized cost): $1 for immediate insertion and store $2. When table doubles, $1 for a re-inserting item and $1 for re-inserting another old item. Total runtime = # of dollars spent # of dollars entered table = 3n. $2 $2 $2 $0 $1 $1 $2 $2 $0 $1 $2 $0 $0 $0 $1 Y.-W. Chang 12 6

7 Expansion: Potential Analyses num[t]: # of elements in T; size[t]: size of the table T. (T)= 2 num[t] - size[t] Right before expansion: (T) = num[t]. Right after expansion: (T) = 0. 0 = 0; i 0. Table is always at least half full: num[t] size[t]/2 (T) 0. If i-th operation does not trigger an expansion (size i = size i-1 ): = c i + i - i-1 = 1 + (2 num i -size i ) - (2 num i-1 -size i-1 ) = 1 + (2 num i -size i ) - (2 (num i -1)-size i ) = 3. If i-th operation triggers an expansion (size i /2 = size i-1 = num i -1): = c i + i - i-1 = num i + (2 num i -size i ) - (2 num i-1 -size i-1 ) = num i + (2 num i -(2 num i -2))-(2(num i -1)-(num i -1)) = 3. Y.-W. Chang 13 Effect of Expansion (T)= 2 num[t] - size[t] Right before expansion: (T) = num[t]. Right after expansion: (T) = 0. 0 = 0; i 0. Table is always at least half full: num[t] size[t]/2 (T) 0. Y.-W. Chang 14 7

8 Expansion and Contraction: Accounting Analysis Bad idea: Double the table when overflow (as before), halve it when table < full /2. Cause thrashing when repeatedly halve and double it if repeatedly insert and delete 2 items. n = 2 k : insert n/2 items and then Insert/Delete I D D I I D D amortized cost of an operation = (n). Better idea: Double the table when overflow and halve it when table < full /4. Accounting analysis Charge $3 for each insertion (as before). Charge $2 for deletion: Store extra $1 in emptied slot; use later to pay to copy remaining items to a new table while shrinking the table. Assume we halve the table when size <= 1/4 Y.-W. Chang 15 Potential Analysis: Expansion and Contraction Load factor: (T) = num[t]/size[t] if num[t] > 0; (T) = 1 if num [T] = 0. Define the potential function (T): num 0 = 0, size 0 = 0, 0 = 1, 0 = 0, i 0. = 1 (T) = num[t]: sufficient potential for expansion. = 1/2 (T) = 0. = 1/4 (T) = num[t]: sufficient potential for contraction. Y.-W. Chang 16 8

9 Potential Analysis: Insertion i-th operation is an insertion: num i = num i i-1 1/2: 3 (as before) i-1 < 1/2, i < 1/2: = c i + i - i-1 = 1 + (size i /2 - num i ) - (size i-1 /2 - num i-1 ) = 1 + (size i /2 - num i ) - (size i /2 - (num i -1)) = 0. i-1 < 1/2, i 1/2: = c i + i - i-1 = 1 + (2 num i - size i ) - (size i-1 /2 - num i-1 ) = 1 + (2 (num i-1 + 1) - size i-1 ) - (size i-1 /2 - num i-1 ) = 3 num i-1-3size i-1 /2 + 3 = 3 i-1 size i-1-3size i-1 /2 + 3 < 3size i-1 /2-3size i-1 /2 + 3 = 3. Y.-W. Chang 17 Potential Analysis: Deletion i-th operation is a deletion: num i = num i-1-1 i-1 < 1/2, i 1/4: no contraction size i =size i-1 = c i + i - i-1 = 1 + (size i /2 - num i ) - (size i-1 /2 - num i-1 ) = 1 + (size i /2 - num i ) - (size i /2 - (num i + 1)) = 2. i-1 < 1/2, i < 1/4: with contraction size i /2 =size i-1 /4 = num i + 1 = c i + i - i-1 = (num i + 1) + (size i /2 - num i ) - (size i-1 /2 - num i-1 ) = (num i + 1) + ((num i + 1) - num i ) ((2num i + 2) - (num i + 1)) = 1. i-1 1/2 : c some constant. ˆi Y.-W. Chang 18 9

10 Appendix: B*-tree Packing Revisited x-coordinates can be determined by the tree structure. Left child: the lowest, adjacent block on the right (x j = x i + w i ). Right child: the first block above, with the same x-coordinate (x j = x i ). y-coordinates? b 5 b 6 n 0 b 1 b 3 b 2 b 4 b 10 n 7 n 1 x 1 =x 0 n 8 n 2 n 5 b 0 b 9 b 8 b 11 n 11 n 9 n 3 n 6 (x 0,y 0 ) b 7 x 7 =x 0 +w 0 n 10 n 4 w 0 Y.-W. Chang 19 Computing y-coordinates Contour: Use a doubly linked list to record the current max y-coordinate for each x-range With the B*-tree topology and block width, the y-coordinate is the max y value within the x-ranges Amortized O(1) time for computing a y-coordinate (i.e., linear time for all blocks) n 3 contour p p B*-tree topology b b 3 1 b 4 b 0 b 10 b 9 b 8 b 11 b 7 Y.-W. Chang 20 10