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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

Suppose that \(P\) is a simple polygon with vertices \(v_{1}, v_{2}, \ldots, v_{n}\) listed so that consecutive vertices are con- nected by an edge, and \(v_{1}\) and \(v_{n}\) are connected by an edge. A vertex \(v_{i}\) is called an ear if the line segment connecting the two vertices adjacent to \(v_{i}\) is an interior diagonal of the simple polygon. Two ears \(v_{i}\) and \(v_{j}\) are called nonoverlapping if the interiors of the triangles with vertices \(v_{i}\) and its two adjacent vertices and \(v_{j}\) and its two adjacent vertices do not intersect. Prove that every simple polygon with at least four vertices has at least two nonoverlapping ears.

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