Amortized Analysis performed An individual operation may itself be - PDF document

4/18/2013 Amortized Analysis  Amortized analysis averages the time required to perform a sequence of operations on a data structure over all the operations Amortized Analysis performed  An individual operation may itself be relatively expensive, but its average cost (amortized Chapter 14 cost) may be small CPTR 318 1 Some Simple Examples Augmented Stack A normal stack with an additional operation: multipop()  Two simple applications which illustrate how  push(x) — pushes element x onto the stack, as usual the principles of amortized analysis apply:  pop() — pops the top element off of the stack and  Augmented stack returns the popped element, as usual  Binary counter  multipop(k) pops the top k elements off of the stack  Returns a vector containing the popped elements  Notice, cannot just adjust the top-of-stack pointer; each element must be inserted into the result vector Augmented Stack Implementation Analysis of Stack Operations template <typename T> T pop() class MultiStack { vector<T> stack; // Vector of elements {  push(x) — O(1) int top; // Top of stack T result; public: if ( top >= 0 ) { result = stack[top--]; MultiStack(int capacity): stack(vector<T>(capacity)), }  pop() — O(1) top(-1) {} return result; bool isEmpty() const } { return top < 0; vector<T> multipop(int k)  multipop(k) — min( n , k ) , where n is } { int size() const vector<T> result; { while ( !isEmpty() && k != 0 ) the size of the stack { return top + 1; } result.push_back(pop()); void push(const T& newElement) k--; { }  worst case is O( n ) int size = stack.size(); return result; if ( top < size - 1 ) { } stack[++top] = newElement; }; } } 1

4/18/2013 Analysis of Stack Operations Analysis of Stack Operations  push(x) O(1)  push(x) O(1)  pop() — O(1)  pop() O(1)  multipop(k) — min( n , k ) , where n is  multipop(k) — min( n , k ) , where n is the size of the stack the size of the stack  worst case is O( n )  worst case is O( n ) Analysis of Stack Operations Analysis of Stack Operations  push(x) O(1)  push(x) O(1)  pop() O(1)  pop() O(1)  multipop(k) O(min( n , k )) , where n  multipop(k) O(min( n , k )) , where n is the size of the stack is the size of the stack  worst case is O( n )  worst case is O( n ) Analysis of Stack Operations Accounting Method  push(x) — O(1)  The accounting method charges differing amounts to  pop() — O(1) different operations  multipop(k) — O(min( n , k )) , where n is the size of the  Some operations may be charged more or less than they stack — worst case is O( n ) actually cost  Consider a sequence of n stack operations on an initially  The amount charged to an operation is its amortized cost empty stack  Operations charged more than their actual cost build  The worst-case cost of a multipop is O( n ) , since the stack may credit stored within a specific object contain n elements  This credit can be used to help pay for operations that  The worst-case time for any stack operation is thus O( n ) are charged less than their actual cost  A sequence on n stack operations, worst-case, would be O( n 2 ) → n multipop operations each costing O( n )  This bound is correct but not tight (Why?) 2

4/18/2013 Stack Charges Stack Example (cont.) Operation Actual Cost Amortized Cost Every push operation is not only charged its actual cost (1), but an  push 1 2 additional cost of 1 for prepayment for any future pop operation, pop whether it be from a single pop , or multiple pops from within 1 0 multipop multipop min(n, k) 0 This credit is associated with the object pushed onto the stack  The operations pop and multipop need be charged nothing, as  All three operations have effective costs  their cost has already been paid of O(1) When an object is popped off, the credit associated with that object   Can we pay for any sequence of stack pays for the pop operation Thus for any sequence of n push , pop , and multipop operations,  operations by charging simply the amortized the total amortized cost is an upper bound on the total actual cost costs? The total amortized cost is O( n ) , and so is the total actual cost  Binary Counter Operation Actual Amortized Credit Cost Cost (Empty stack) - - 0 push(5) 1 2 1 push(3) 1 2 2 push(11) 1 2 3 push(12) 1 2 4 push(3) 1 2 5 pop() 1 0 4 pop() 1 0 3 push(7) 1 2 4 push(10) 1 2 5 push(8) 1 2 6 multipop(5) 5 5 0 1 Total 15 16 Counter Complexity Counter Amortization 3

4/18/2013 Counter Amortized Analysis 4