/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Verify that the program segment ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Verify that the program segment $$ \begin{array}{c}{\text { if } xy \wedge \min =y)\).

Short Answer

Expert verified
The program segment is correct as it satisfies \((x \leq y \wedge \min = x) \vee (x > y \wedge \min = y)\) in all cases.

Step by step solution

01

Understand the Problem

The program segment needs to be verified against the initial and final assertions. The initial assertion is true (represented by \(\mathbf{T}\)), and the final assertion is \((x \leq y \wedge \min =x) \vee (x > y \wedge \min =y)\).
02

Identify Conditions

The program has a conditional statement that checks if \(x < y\). There are two main branches to evaluate: one for \(x < y\) and one for \(x \geq y\).
03

Evaluate the True Branch (x < y)

If \(x < y\), then the program sets \(\min := x\). After this assignment, the goal is to check if \((x \leq y \wedge \min =x) \vee (x > y \wedge \min =y)\) holds. Since \(x < y\) implies \(x \leq y\), and \(\min = x\) is set, the condition \(x \leq y \wedge \min = x\) holds true.
04

Evaluate the False Branch (x >= y)

If \(x \geq y\), then the program sets \(\min := y\). Given \(x < y\) is false, \(x \geq y\) implies that \(x > y\) or \(x = y\). The important case is for \(x > y\), where \(\min = y\), maintaining \(x > y \wedge \min = y\). When \(x = y\), since both \(\min = y\) and \(x = y\) hold, the overall condition \( \min = y \) satisfies the final assertion.
05

Combine the Results

Both branches of the conditional statement lead to conditions that satisfy the final assertion: \((x \leq y \wedge \min = x)\) for the true branch, and \((x > y \wedge \min = y)\) for the false branch. As such, the entire program segment is correct with respect to the initial and final assertions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Assertion
When verifying a program's correctness, the initial assertion is the starting point. It sets the condition before any program execution. In this particular exercise, the initial assertion is denoted by the symbol \(\textbf{T}\).
This symbol simply means that the condition is 'True' universally. In other words, no specific constraints are set at the beginning.
Understanding the initial assertion helps identify any assumptions made before the code runs. Here, the lack of specific conditions means we assume nothing about the values of x and y initially.
Final Assertion
The final assertion is the condition we expect to hold after the program runs. It helps determine if the program's output is correct.
In this case, the final assertion is: \((x \leq y \wedge \min = x) \vee (x > y \wedge \min = y)\).
This means that upon completion, one of the two following conditions should be true:
  • If x is less than or equal to y, then min should be equal to x.
  • Otherwise, if x is greater than y, min should be equal to y.
Ensuring that the program satisfies this assertion confirms that it works correctly for all possible inputs.
Conditional Statement
A conditional statement determines the flow of execution based on a condition. In our program segment, the conditional statement is: \(\text{if } x < y\).
This creates two possible paths:
  • If the condition \(x < y\) is true, the program executes one branch.
  • If the condition is false, it executes another.
Conditional statements help decision-making within programs. Understanding how they split program logic into different branches is crucial for verifying that all possible scenarios are handled correctly.
Branch Evaluation
Branch evaluation involves examining different paths the program can take and ensuring each meets the final assertion. For this exercise, we'll evaluate both branches:
  • **True Branch:** If \(x < y\), then \(\text{min := x}\). Here, \(\text{x}\text{ is less than y}\), and \(\text{min is set to x}\). Therefore, \((x \leq y \wedge \min = x)\) holds true.
  • **False Branch:** If \(x >= y\), then \(\text{min := y}\). This implies \(\text{x}\) is greater than or equal to y, and \(\text{min}\) is y. So, \((x > y \wedge \min = y)\) also holds true.
By individually verifying each branch, we confirm that the program always satisfies the final assertion, demonstrating its correctness.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use a merge sort to sort 4, 3, 2, 5, 1, 8, 7, 6 into increasing order. Show all the steps used by the algorithm.

Suppose there are \(n\) people in a group, each aware of a scandal no one else in the group knows about. These people communicate by telephone; when two people in the group talk, they share information about all scandals each knows about. For example, on the first call, two people share information, so by the end of the call, each of these people knows about two scandals. The gossip problem asks for \(G(n),\) the minimum number of telephone calls that are needed for all \(n\) people to learn about all the scandals. Exercises \(69-71\) deal with the gossip problem. Use mathematical induction to prove that \(G(n) \leq 2 n-4\) for \(n \geq 4 .\) [Hint: In the inductive step, have a new person call a particular person at the start and at the end. \(]\)

Suppose there are \(n\) people in a group, each aware of a scandal no one else in the group knows about. These people communicate by telephone; when two people in the group talk, they share information about all scandals each knows about. For example, on the first call, two people share information, so by the end of the call, each of these people knows about two scandals. The gossip problem asks for \(G(n),\) the minimum number of telephone calls that are needed for all \(n\) people to learn about all the scandals. Exercises \(69-71\) deal with the gossip problem. Find \(G(1), G(2), G(3),\) and \(G(4)\)

Devise a recursive algorithm to find the \(n\) th term of the sequence defined by \(a_{0}=1, a_{1}=2, a_{2}=3,\) and \(a_{n}=\) \(a_{n-1}+a_{n-2}+a_{n-3},\) for \(n=3,4,5, \ldots\)

Give a recursive definition of \(w^{i},\) where \(w\) is a string and \(i\) is a nonnegative integer. (Here \(w^{i}\) represents the concatenation of \(i\) copies of the string \(w . )\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.