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Consider a queue data structure, where the two operations of interest are enqueue (at the back of the queue) and dequeue (from the front of the queue). A queue is thus a FIFO (first in-first out) structure. Suppose we implement a queue by using two stacks, A and B, which support operations of push and pop, as follows: enqueue(x): push x onto stack A dequeue: if B is not empty return x = pop B if B is empty repeat until A is empty x = pop A push x onto stack B return x = pop B a) This implementation seems to correctly emulate a queue. Illustrate (by drawing the two stacks) how it works for the following sequence of enqueue and dequeue operations: enqueue 5 enqueue 7 dequeue enqueue 3 enqueue 6 enqueue 9 dequeue dequeue dequeue dequeue b) The actual cost of a push or a pop operation is 1 unit of work. What is the worst-case single operation possible after a sequence of n enqueue and dequeue operations, and what is its cost? c) Explain how a simplistic worst-case analysis would lead to the conclusion that after n operations, O(n³) units of work would have been done. d) Using the accounting method, assign the lowest (integer) amortized cost possible to the enqueue operation and to the dequeue operation so that after any sequence of nl enqueues and n2 dequeues, n1 2 n2, the amortized cost is 2 the actual cost. Explain why this works. e) Show that the total amortized cost, which is an upper bound for the actual cost, over a sequence of n operations is O(n).

Consider a queue data structure, where the two operations of interest are enqueue (at the back of the queue) and dequeue (from the front of the queue). A queue is thus a FIFO (first in-first out) structure. Suppose we implement a queue by using two stacks, A and B, which support operations of push and pop, as follows: enqueue(x): push x onto stack A dequeue: if B is not empty return x = pop B if B is empty repeat until A is empty x = pop A push x onto stack B return x = pop B a) This implementation seems to correctly emulate a queue. Illustrate (by drawing the two stacks) how it works for the following sequence of enqueue and dequeue operations: enqueue 5 enqueue 7 dequeue enqueue 3 enqueue 6 enqueue 9 dequeue dequeue dequeue dequeue b) The actual cost of a push or a pop operation is 1 unit of work. What is the worst-case single operation possible after a sequence of n enqueue and dequeue operations, and what is its cost? c) Explain how a simplistic worst-case analysis would lead to the conclusion that after n operations, O(n³) units of work would have been done. d) Using the accounting method, assign the lowest (integer) amortized cost possible to the enqueue operation and to the dequeue operation so that after any sequence of nl enqueues and n2 dequeues, n1 2 n2, the amortized cost is 2 the actual cost. Explain why this works. e) Show that the total amortized cost, which is an upper bound for the actual cost, over a sequence of n operations is O(n).

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Consider a queue data structure, where the two operations of interest are enqueue (at the back of the queue) and dequeue (from the front of the queue). A queue is thus a FIFO (first in-first out) structure. Suppose we implement a queue by using two stacks, A and B, which support operations of push and pop, as follows: enqueue(x): push x onto stack A if B is not empty return x = pop B if B is empty repeat until A is empty x = pop A push x onto stack B return x = pop B dequeue: a) This implementation seems to correctly emulate a queue. Illustrate (by drawing the two stacks) how it works for the following sequence of enqueue and dequeue operations: enqueue 5 enqueue 7 dequeue enqueue 3 enqueue 6 enqueue 9 dequeue dequeue dequeue dequeue b) The actual cost of a push or a pop operation is 1 unit of work. What is the worst-case single operation possible after a sequence of n enqueue and dequeue operations, and what is its cost? c) Explain how a simplistic worst-case analysis would lead to the conclusion that after n operations, O(n²) units of work would have been done. d) Using the accounting method, assign the lowest (integer) amortized cost possible to the enqueue operation and to the dequeue operation so that after any sequence of nl enqueues and n2 dequeues, n1 2 n2, the amortized cost is 2 the actual cost. Explain why this works. e) Show that the total amortized cost, which is an upper bound for the actual cost, over a sequence of n operations is O(n).
Transcribed Image Text:Consider a queue data structure, where the two operations of interest are enqueue (at the back of the queue) and dequeue (from the front of the queue). A queue is thus a FIFO (first in-first out) structure. Suppose we implement a queue by using two stacks, A and B, which support operations of push and pop, as follows: enqueue(x): push x onto stack A if B is not empty return x = pop B if B is empty repeat until A is empty x = pop A push x onto stack B return x = pop B dequeue: a) This implementation seems to correctly emulate a queue. Illustrate (by drawing the two stacks) how it works for the following sequence of enqueue and dequeue operations: enqueue 5 enqueue 7 dequeue enqueue 3 enqueue 6 enqueue 9 dequeue dequeue dequeue dequeue b) The actual cost of a push or a pop operation is 1 unit of work. What is the worst-case single operation possible after a sequence of n enqueue and dequeue operations, and what is its cost? c) Explain how a simplistic worst-case analysis would lead to the conclusion that after n operations, O(n²) units of work would have been done. d) Using the accounting method, assign the lowest (integer) amortized cost possible to the enqueue operation and to the dequeue operation so that after any sequence of nl enqueues and n2 dequeues, n1 2 n2, the amortized cost is 2 the actual cost. Explain why this works. e) Show that the total amortized cost, which is an upper bound for the actual cost, over a sequence of n operations is O(n).
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