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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.

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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.

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Most popular questions from this chapter

Exercises \(49-51\) present incorrect proofs using mathematical induction. You will need to identify an error in reasoning in each exercise. What is wrong with this "proof" that all horses are the same color? Let \(P(n)\) be the proposition that all the horses in a set of \(n\) horses are the same color. Basis Step: Clearly, \(P(1)\) is true. Inductive Step: Assume that \(P(k)\) is true, so that all the horses in any set of \(k\) horses are the same color. Consider any \(k+1\) horses; number these as horses \(1,2,3, \ldots, k, k+1 .\) Now the first \(k\) of these horses all must have the same color, and the last \(k\) of these must also have the same color. Because the set of the first \(k\) horses and the set of the last \(k\) horses overlap, all \(k+1\) must be the same color. This shows that \(P(k+1)\) is true and finishes the proof by induction.

Sometimes we cannot use mathematical induction to prove a result we believe to be true, but we can use mathematical induction to prove a stronger result. Because the inductive hypothesis of the stronger result provides more to work with, this process is called inductive loading. We use inductive loading in Exercise \(74-76\) . Suppose that we want to prove that $$ \sum_{j=1}^{n} j /(j+1) !<1 $$ for all positive integers \(n .\) a) Show that if we try to prove this inequality using mathematical induction, the basis step works, but the inductive step fails. b) Show that mathematical induction can be used to prove the stronger inequality $$ \sum_{j=1}^{n} j /(j+1) ! \leq 1-1 /(n+1) ! $$ for all positive integers \(n,\) implying that the weaker inequality is also true.

Use strong induction to show that every positive integer can be written as a sum of distinct powers of two, that is, as a sum of a subset of the integers \(2^{0}=1,2^{1}=2,2^{2}=4\) and so on. [Hint: For the inductive step, separately con- sider the case where \(k+1\) is even and where it is odd. When it is even, note that \((k+1) / 2\) is an integer. \(]\)

Give a recursive definition of a) the set of odd positive integers. b) the set of positive integer powers of 3 . c) the set of polynomials with integer coefficients.

Use mathematical induction in Exercises \(38-46\) to prove results about sets. Prove that a set with \(n\) elements has \(n(n-1) / 2\) subsets containing exactly two elements whenever \(n\) is an integer greater than or equal to \(2 .\)
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