/*! 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 17 Describe a recursive algorithm f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 17

Describe a recursive algorithm for multiplying two non- negative integers \(x\) and \(y\) based on the fact that \(x y=2(x \text { . }\) \((y / 2) )\) when \(y\) is even and \(x y=2(x \cdot\lfloor y / 2\rfloor)+x\) when \(y\) is odd, together with the initial condition \(x y=0\) when \(y=0\)

Short Answer

Expert verified
The recursive algorithm multiplies x and y by halving y and doubling x, returning the result or adding x based on whether y is even or odd.

Step by step solution

01

Understand the Base Case

Identify the base case for the recursion. The base case occurs when one of the integers, specifically y, is 0. When y is 0, the product of x and y is 0. Therefore, the recursive function should return 0 in this case.
02

Handle the Case When y is Even

When y is even, the product of x and y can be represented as twice the product of x and half of y. Mathematically, this is expressed as: \[ xy = 2(x \times \frac{y}{2}) \] Write the recursive step in the algorithm to handle this case.
03

Handle the Case When y is Odd

When y is odd, the product of x and y can be represented as twice the product of x and the floor value of half of y, plus x. Mathematically, this is expressed as: \[ xy = 2(x \times \text{floor}(\frac{y}{2})) + x \] Write the recursive step in the algorithm to handle this case.
04

Combine the Steps into One Recursive Algorithm

Combine all the steps to form the complete recursive algorithm. The algorithm checks if y is 0, even, or odd, and performs the appropriate recursive call based on the conditions.

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 91影视!

Key Concepts

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

recursion
Recursion is a programming technique where a function calls itself to solve a problem. In the context of multiplying two non-negative integers, recursion helps break the problem into smaller, more manageable parts. The idea is to repeatedly reduce one of the integers, applying the multiplication equation for specific cases (even or odd) until reaching the simplest form, known as the base case.
By handling smaller instances of the problem within the function, recursion achieves the final result through a series of repeated function calls. This might seem complex at first, but it provides a clear and elegant way to solve problems that have repetitive, sub-problem structures.
base case
The base case is a crucial component of recursion. It defines the simplest instance of the problem, which can be solved without further recursive calls. In our algorithm for multiplying two non-negative integers, the base case occurs when one of the integers, specifically y, is 0.
When y is 0, the product of x and y is 0. Therefore, the recursive function should return 0 in this case. This prevents the recursion from continuing indefinitely and allows it to terminate with a definitive answer. Understanding the base case is essential to implementing and ensuring the correctness of any recursive algorithm.
even and odd cases
To handle the multiplication of two non-negative integers using recursion, we must consider two specific scenarios: when the second integer (y) is even and when it is odd.
  • **When y is even**: The multiplication can be simplified to twice the product of x and half of y. Mathematically, this is expressed as:
    \( xy = 2(x \times \frac{y}{2})\)
  • **When y is odd**: The multiplication can be simplified to twice the product of x and the floor value of half of y, plus x. This is expressed as:
    \( xy = 2(x \times \text{floor}(\frac{y}{2})) + x\).

Handling these cases correctly in the algorithm ensures that the recursive function reduces the problem size at each step, eventually reaching the base case. Combining the correct handling of even and odd values within the recursive calls creates a complete and efficient multiplication algorithm.

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 \(b, d, a, f, g, h, z, p, o, k\) into alphabetic order. Show all the steps used by the algorithm.

Show that we can prove the well-ordering property when we take strong induction as an axiom instead of taking the well-ordering property as an axiom.

A knight on a chessboard can move one space horizontally (in either direction) and two spaces vertically (in either direction) or two spaces horizontally (in either direction) and one space vertically (in either direction). Suppose that we have an infinite chessboard, made up of all squares \((m, n),\) where \(m\) and \(n\) are nonnegative integers that denote the row number and the column number of the square, respectively. Use mathematical induction to show that a knight starting at \((0,0)\) can visit every square using a finite sequence of moves. [Hint: Use induction on the variable \(s=m+n . ]\)

Deal with iterations of the logarithm function. Let log \(n\) denote the logarithm of \(n\) to the base \(2,\) as usual. The function \(\log ^{(k)} n\) is defined recursively by $$\log ^{(k)} n=\left\\{\begin{array}{cl}{n} & {\text { if } k=0} \\ {\log \left(\log ^{(k-1)} n\right)} & {\text { if } \log ^{(k-1)} n \text { is defined }} \\ {} & {\text { and positive }} \\ {\text { undefined }} & {\text { otherwise. }}\end{array}\right.$$ The iterated logarithm is the function log" \(n\) whose value at \(n\) is the smallest nonnegative integer \(k\) such that \(\log ^{(k)} n \leq 1\) Find the value of \(\log ^{*} n\) for these values of \(n\) a) 2 b) 4 c) 8 d) 16 e) 256 f) 65536 g) \(2^{2048}\)

Sort 3, 5, 7, 8, 1, 9, 2, 4, 6 using the quick sort.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation
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.