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Problem Statement: We can store k stacks in a single array if we use the data structure suggested in Figure 1 shown below, for the case k = 3. We push and pop from each stack as suggested in connection with Figure 2 below. However, if pushing onto stack i causes TOP(i) to equal BOTTOM(i – 1), we first move all the stacks so that there is an appropriate size gap between each adjacent pair of stacks. For example, we might make the gaps above all stacks equal, or we might make the gap above stack i proportional to the current size of stack i (on the theory that larger stacks are likely to grow sooner, and we want to postpone as long as possible the next reorganization).

Problem Statement: We can store k stacks in a single array if we use the data structure suggested in Figure 1 shown below, for the case k = 3. We push and pop from each stack as suggested in connection with Figure 2 below. However, if pushing onto stack i causes TOP(i) to equal BOTTOM(i – 1), we first move all the stacks so that there is an appropriate size gap between each adjacent pair of stacks. For example, we might make the gaps above all stacks equal, or we might make the gap above stack i proportional to the current size of stack i (on the theory that larger stacks are likely to grow sooner, and we want to postpone as long as possible the next reorganization).

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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Can the problem be solved without using push operation in excessive or with different approach?

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Problem Statement: We can store k stacks in a single array if we use the data structure suggested in Figure 1 shown below, for the case k = 3. We push and pop from each stack as suggested in connection with Figure 2 below. However, if pushing onto stack i causes TOP(i) to equal BOTTOM(i – 1), we first move all the stacks so that there is an appropriate size gap between each adjacent pair of stacks. For example, we might make the gaps above all stacks equal, or we might make the gap above stack i proportional to the current size of stack i (on the theory that larger stacks are likely to grow sooner, and we want to postpone as long as possible the next reorganization). 1 stack 1 3 stack 2 bottom 1 2 stack 3 3 top stackspace Figure 1 top first element second element maxlength last element elements Figure 2 1. On the assumption that there is a procedure reorganize to call when stacks collide, write code for the five stack operations. 2. On the assumption that there is a procedure MakeNewTops that computes newtop[i], the "appropriate" position for the top of stack i, for 1 <i<k, write the procedure reorganize. Hint. Note that stack i could move up or down, and it is necessary to move stack i before stack j if the new position of stack j overlaps the old position of stack i. Consider stacks 1, 2,..., k in order, but keep a stack of "goals," each goal being to move a particular stack. If on considering stack i, we can move it safely, do so, and then reconsider the stack whose number is on top of the goal stack. If we cannot safely move stack i, push i onto the goal stack. 3. What is an appropriate implementation for the goal stack in (2)? Do we really need to keep it as a list of integers, or will a more succinct representation do? 4. Implement MakeNewTops in such a way that space above each stack is proportional to the current size of that stack. The implementation of this stack management software should be as structured as possible.
Transcribed Image Text:Problem Statement: We can store k stacks in a single array if we use the data structure suggested in Figure 1 shown below, for the case k = 3. We push and pop from each stack as suggested in connection with Figure 2 below. However, if pushing onto stack i causes TOP(i) to equal BOTTOM(i – 1), we first move all the stacks so that there is an appropriate size gap between each adjacent pair of stacks. For example, we might make the gaps above all stacks equal, or we might make the gap above stack i proportional to the current size of stack i (on the theory that larger stacks are likely to grow sooner, and we want to postpone as long as possible the next reorganization). 1 stack 1 3 stack 2 bottom 1 2 stack 3 3 top stackspace Figure 1 top first element second element maxlength last element elements Figure 2 1. On the assumption that there is a procedure reorganize to call when stacks collide, write code for the five stack operations. 2. On the assumption that there is a procedure MakeNewTops that computes newtop[i], the "appropriate" position for the top of stack i, for 1 <i<k, write the procedure reorganize. Hint. Note that stack i could move up or down, and it is necessary to move stack i before stack j if the new position of stack j overlaps the old position of stack i. Consider stacks 1, 2,..., k in order, but keep a stack of "goals," each goal being to move a particular stack. If on considering stack i, we can move it safely, do so, and then reconsider the stack whose number is on top of the goal stack. If we cannot safely move stack i, push i onto the goal stack. 3. What is an appropriate implementation for the goal stack in (2)? Do we really need to keep it as a list of integers, or will a more succinct representation do? 4. Implement MakeNewTops in such a way that space above each stack is proportional to the current size of that stack. The implementation of this stack management software should be as structured as possible.
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