Verify the correctness of the following program segment to exchange the values ofx and y:temp = xy = tempAt the beginning of this program segment, x and y have certain values. Thuswe may express the actual precondition as x = a and y = b. The desired postcondi-tion is then x = b and y = a. Using the assignment rule, we can work backwardfrom the postcondition to find the earlier assertions (read the following from thebottom to the top).{y = b, x = a}temp = x{y = b, temp = a}{x = b, temp = a}y = temp{x = b, y = a}The first assertion agrees with the precondition; the assignment rule, appliedrepeatedly, assures us that the program segment is correct.

#include <iostream> using namespace std; int main(void) { int x = 1, y = 2; int…

Answered: Verify the correctness of the following…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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The two most recent operands are popped in the postfix expression evaluation example whenever an operator is met so that the subexpression may be evaluated. In the subexpression, the first operand popped is regarded as the second operand, and the second operand popped as the first. Give a case study that exemplifies the significance of this component of the solution.
Use DrRacket. Define the procedures below. Write a procedure reverse-digits (n) which returns a number with the digits of n in reverse order. For example, (reverse-digits 397164) → 461793. Define a predicate same-digits? (a b) which returns #t if the integers a and b contain the same digits. If a digit occurs n times in a, it must also occur n times in b. For example, (same-digits? 133042 420313) → #t, but (same-digits? 133042 42013) → #f. To solve the problem, one technique would be to add some extra procedures for manipulating digits inside a number. Here are some hints: To find the number of digits in a number, count how many times you need to divide it by 10 until you reach 0. To extract the second digit of 5784: floor( 5784/102 ) mod 10=7. Note that mod in racket is modulo or remainder To assemble the digits 3, 4, and 5 into a number: 3×102+4×101+5×100 = 345 The standard Scheme procedures modulo and floor may prove useful. The solution only needs to consider non-negative…
A problem with returning truthy values is that code may be introducedwhich depends upon the value not being a boolean and subsequently becomesdifficult to understand.code to example of how that approach can confuse things:
A problem with returning truthy values is that code may be introducedwhich depends upon the value not being a boolean and subsequently becomesdifficult to understand.code to example of how that approach can confuse things:
Write a tail-call recursive function in the Racket language called circles. This function should take two numbers, x and size, and a Scene scn. If size is less-than or equal to 2, then your function returns the scn unchanged (base case). Otherwise, your function should place a circle in the Scene that results from a recursive call to circles with x shifted by (+size (next-size size)), a new size of (next-size size), and using the original Scene. After evaluating such as (circles 100 48 ES), your function should output an image of five blue circles, with the first circle at (x,y) coordinate position (100, 80):
An expression of the form a = a, +az + a, ++ a,+** is called an infinite series. The letterk is called the index of summation and takes on values from the lower bound (which is 1 here) to the upper bound (which is infinity here). The dots mean that we are to continue the addition indefinitely. However, it is not practical to keep adding the terms indefinitely; we should be able to stop at some point. Hence, we need a criterion to stop the summation. The sum of part of the sequence is a partial sum. Usually, we either have to calculate the sum of the first n terms of a series or stop the summation when a term becomes less than some pre-defined tolerance value. The Maclaurin series expansion for f(x) =on an intervalfrom -1
Implement the function integrate_3(func, xa, xb, ya, yb, za, zb). xa and xb are the endpoints in the x-axis. ya and yb are the endpoints in the y-axis. za and zb are the endpoints in the z-axis. For example, if we have def h(x, y, z): returnmath.sin(x)+math.cos(y)+math.tan(z) r = integrate_3(h, 0, 3, 0, 2, 0, 1) print("{:.3f}".format(r)) This should print 10.453. Set Δx, Δy, and Δz equal to 0.01. The code above corresponds to the following: (see the attached picture below) The functions should return values, NOT print them. Use python, only import math library nothing else.
The following argument is valid. ((A ∨ D) → W) (R ↔ (C ∧ T)) (← R → A) ← (B ∨ C) ∴ ← (W → C) Below is a natural deduction proof, but with some of the lines and justifications missing. Using any of the proof rules (from Ch. 16 & 18 of the text) in the natural deduction system, fill in the missing lines and justifications to complete the proof (3 points). 1 ((A ∨ D) → W) Pr. 2 (R ↔ (C ∧ T)) Pr. 3 (- R → A) Pr. 4 -(B ∨ C) Pr. /∴← (W → C) 5 (- B ∧ - C) _________ 6 (W → C) _________ 7 _________ ∧E, 5 8 - W MT, 6, 7 9 - (A ∨ D) _________ 10 _________ DM, 9 11 - A ∧E, 10 12 _________ MT, 4 13 R _________ 14 (C ∧ T) _________ 15 C ∧E, 14 16 _________ -E, 7, 15 17 - (W → C) _________
Write a MATLAB function sumOddInts(). sumOddInts should return the sum of all odd integers between the values provided to the two parameters, inclusive, e.g., sumOddInts(5,9) should return 21. start and finish do not have to be odd, e.g, sumOddInts(4,8) should return 12. start and finish do not have to be positive, e.g., sumOddInts(-5,7) should return 7, However, finish may be assumed to be greater or equal to start. whereas Python has a modulo operator "%", MATLAB has a built-in mod() function that returns the remainder. For instance, >> mod (13,5) ans = 3>> mod (13,2) ans = 1
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2. Define the function O(k), with k > 2, as 22 + 4? + 62 + ·..+ m? where m is the largest even number that is < k. Please implement the function in MIPS and make sure your implementation indeed works on the simulator, by printing the value O(15) to the terminal.
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