CS333 / Cutler Amortized Analysis 1 Amortized Analysis The average cost of a sequence of n operations on a given Data Structure. Aggregate Analysis Accounting Method

CS333 / Cutler Amortized Analysis 2 Amortized Analysis Amortized analysis computes the average time required to perform a sequence of n operations on a data structure Often worst case analysis is not tight and the amortized cost of an operation is less that its worst case. Analogy (making coffee)

CS333 / Cutler Amortized Analysis 3 Applications of amortized analysis Vectors/ tables Disjoint sets Priority queues Heaps, Binomial heaps, Fibonacci heaps Hashing

CS333 / Cutler Amortized Analysis 4 Difference between amortized and average cost To do averages we need to use probability For amortized analysis no such assumptions are needed We compute the average cost per operation for any mix of n operations

CS333 / Cutler Amortized Analysis 5 Operations on Data Structures A data structure has a set of operations associated with it. Example: A stack with –push(), pop() and MultiPop(k). Often some operations may be slow while others are fast. push() and pop() are fast. MultiPop(k) may be slow. Sometimes the time of a single operation can vary

CS333 / Cutler Amortized Analysis 6 Methods Aggregate analysis- the total amount of time needed for the n operations is computed and divided by n Accounting - operations are assigned an amortized cost. Objects of the data structure are assigned a credit Potential – The prepaid work (money in the “bank”) is represented as “potential” energy that can be released to pay for future operations

CS333 / Cutler Amortized Analysis 7 Aggregate analysis n operations take T(n) time Amortized cost of an operation is T(n)/n

CS333 / Cutler Amortized Analysis 8 Stack - aggregate analysis A stack with operations Push, Pop and Multipop. Multipop(S, k) while not empty(S) and k>0 do Pop(S); k:=k-1 end while

CS333 / Cutler Amortized Analysis 9 Stack - aggregate analysis Push and Pop are O(1) (to move 1 data element) Multipop is O(min(s, k)) where s is the size of the stack and k the number of elements to pop. Assume a sequence of n Push, Pop and Multipop operations

CS333 / Cutler Amortized Analysis 10 Stack - aggregate analysis Each object can be popped only once for each time it is pushed So the total number of times Pop can be called ( directly or from Multipop) is bound by the number of Pushes <=n.

CS333 / Cutler Amortized Analysis 11 Stack - aggregate analysis A sequence of n Push and Pop operations is therefore O(n) and the amortized cost of each is O(n)/n=O(1)

CS333 / Cutler Amortized Analysis 12 Stack Example: Op/Moves <=2 Start push a a push b a b push c a b c Multipop(3) a b pop b a pop a OperationStack a a b a b c pop c 6 Operation: 6 Moves 4 Operation: 6 Moves

CS333 / Cutler Amortized Analysis 13 Accounting Method Charge each operation an (invented) amortized cost. –Often different from actual run time cost. Some operations may have an amortized cost larger than runtime, others may have less. –Unlike businesses we do not want to make a profit –We want to cover the actual cost Amount charged but not used in performing an operation is stored with objects of the data structure

CS333 / Cutler Amortized Analysis 14 Accounting method Later operations can use stored amount to pay for their actual cost Credit balance must not go negative (always enough to pay for performance of future operations)

CS333 / Cutler Amortized Analysis 15 Stack - amortized analysis We assign the amortized costs: $2 for Push $0 for both Pop and Multipop For a sequence of n Push and Pop operations the total amortized cost is at most 2n or O(n)

CS333 / Cutler Amortized Analysis 16 Stack - amortized analysis Each time we do a Push we pay $1 for the actual cost of the Push and the element has a credit of $1. Each time an element is popped we take the $1 credit to pay for it Thus the balance is always nonnegative

CS333 / Cutler Amortized Analysis 17 k bit binary counter Increment (A) i = 0 while i < length[A] and A[i] = 1 A[i] = 0 i ++ if i < length[A] A[i] = 1 Initially the counter contains 0 Eventually it becomes 2 k -1 Next it is reset to k -1 A

CS333 / Cutler Amortized Analysis 18 Aggregate analysis Count number of times a bit is flipped. Let number increments n = 2 k (if n < 2 k analysis similar) A[0] flipped n times A[1] flipped n/2 1 times … A[k - 1] flipped n/2 k-1 times Bit D No Flips bit counter k=3

CS333 / Cutler Amortized Analysis 19 Accounting method Charge amortized cost of $2 to set a bit to 1 When a bit is set to 1, pay $1 for actual cost and store $1 with bit Note: at all times a bit with value 1 has $1 When a bit is reset to 0 use $1 to pay for actual cost $2 per Increment operation

CS333 / Cutler Amortized Analysis 20 Accounting method Let the value stored in the counter be: After increment and a payment of $2 = $1 +$1: $1 $1 $0 $0 $1 $0 $0 $1 $1 $1 $1 $1 $0 $0 $1 $0 $1 $0 $0 $

CS333 / Cutler Amortized Analysis 21 Dynamic table (object table, hash table, vector, etc) The table is dynamic We can’t predict its maximum size We would like to avoid allocating a lot of unused space (reasonable load balance) May not be able to avoid table overflow. Overflow should not cause run time failure

CS333 / Cutler Amortized Analysis 22 java.util.Vector A built in class which is a “growable” array. The user can set: –initialCapacity - the initial capacity of the vector. –capacityIncrement - the amount by which the capacity is increased when the vector overflows. The default for capacityIncrement is to double the size

CS333 / Cutler Amortized Analysis 23 Dynamic table Idea: Allocate more memory as needed. Duplicate the size of the table after each overflow. After each duplication must copy elements from old to new table We assume for now the only operation is insert and calculate amortized cost Table sizes: 1, 2, 4, 8, …, 2 k

CS333 / Cutler Amortized Analysis 24 Aggregate method Op. Size Cost The table before after / Copy

CS333 / Cutler Amortized Analysis 25 Let the number of inserts be 2 k < n  2 k+1 (Note: 2*2 k = 2 k+1 < 2n) At this point the size of the table is 2 k+1 After n inserts: –The total cost for copy operations only is k = 2 k < 2n –The total cost for n inserts (without copy) is n. Total < 3n Therefore the amortized time is O(1). Aggregate Analysis 2k2k 2k2k n

CS333 / Cutler Amortized Analysis 26 Accounting analysis Charge each insert $3. When the table is not full, use $1 for the cost of insert, and store $2 with element When the table doubles from m to 2m: –m/2 elements that never moved before have $2 credit, –m/2 elements which already moved have $0 –After copy all m elements have $0 credit

CS333 / Cutler Amortized Analysis 27 Accounting analysis $0 $2 $0 $2 Size = 4 2 copied elements with $0 2 new elements with $2 $0 Size =8 4 copied elements with $0 4 new elements with $2 Size = 8 4 copied elements with $0

CS333 / Cutler Amortized Analysis 28 java.util.Vector fixed increment Let the number of inserts n satisfy c 0 +(m-1)c < n  c 0 +mc So (n- c 0 )/c  m <1+ (n- c 0 )/c and m =  (n) At this point the size of the array is c 0 +mc After the nth insert: –The total time for copy operations is ? –The total time for n inserts (without copy) is n.

CS333 / Cutler Amortized Analysis 29 java.util.Vector fixed increment m-1 Cost for m vector copies = mc 0 + c  ( … +(m-1)) = mc 0 + c  m(m-1)/2=  (m 2 ) =  (n 2 ) c0c0 c0c0 c0c0 c0c0 ccc c c  Initial capacity c 0 Capacity increment c

CS333 / Cutler Amortized Analysis 30 java.util.Vector fixed increment After the nth insert: –The total time for copy operations is  (n 2 ) –The total time for n inserts (without copy) is n. Average insert time is  (n)