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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 \( \text{TOP}(i) \) to equal \( \text{BOTTOM}(i+1) \), 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). ### Figure Explanations **Figure 1:** - Illustrates the top and bottom pointers for each of the three stacks (stack 1, stack 2, stack 3) within the array. - Shows a structure where stacks are separated by a stack space, which can be adjusted. **Figure 2:** - Depicts a single stack’s elements from top to bottom. - Lists elements from the top element through to the last element indicated by the `maxlength`. - Contains labels for first element, second element, etc. ### Instructions 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 \( 1 \leq i \leq k \), write the procedure `reorganize`. - **Hint:** Note that stack \( i \) could move up or down, and it is necessary to move stack \( j \) before stack \( j \) if the new position of stack \( j \) overlaps the old position of stack \( i \). Consider stacks \( 1, 2, \ldots, k \) 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
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 \( \text{TOP}(i) \) to equal \( \text{BOTTOM}(i+1) \), 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). ### Figure Explanations **Figure 1:** - Illustrates the top and bottom pointers for each of the three stacks (stack 1, stack 2, stack 3) within the array. - Shows a structure where stacks are separated by a stack space, which can be adjusted. **Figure 2:** - Depicts a single stack’s elements from top to bottom. - Lists elements from the top element through to the last element indicated by the `maxlength`. - Contains labels for first element, second element, etc. ### Instructions 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 \( 1 \leq i \leq k \), write the procedure `reorganize`. - **Hint:** Note that stack \( i \) could move up or down, and it is necessary to move stack \( j \) before stack \( j \) if the new position of stack \( j \) overlaps the old position of stack \( i \). Consider stacks \( 1, 2, \ldots, k \) 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
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