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The exponentiation algorithms in this section are based on performing exponentiation by means of repeated multiplication. In a similar way, one can perform integer multiplication by means of repeated addition. The following multiplication procedure (in which it is assumed that our language can only add, not multiply) is analogous to the expt procedure: (define (* a b) (if \((=b \quad 0)\) 0 \((+a(* a(-b \quad 1)))))\)

Short Answer

Expert verified
The procedure defines integer multiplication using repeated addition, and works recursively.

Step by step solution

01

Understanding the Problem

The task is to understand the recursive procedure defined for multiplying two integers, \(a\) and \(b\), using repeated addition. The function checks if \(b\) equals 0 (base case) and returns 0 in that case. Otherwise, it adds \(a\) to the result of multiplying \(a\) with \(b-1\).
02

Identifying the Base Case

The base case of the recursion is when \(b = 0\). The function returns 0 in this case because multiplying any number by 0 results in 0.
03

Recursive Call Analysis

If \(b\) is not 0, the function calculates \(a\) times \(b\) by adding \(a\) repeatedly. Specifically, it computes \((+ a (* a (- b 1)))\), which means it adds \(a\) to the result of recursively calling the same function with \(b\) decreased by 1.
04

Example Calculation

Let's consider an example to solidify our understanding. Suppose we want to multiply \(a = 3\) and \(b = 2\). Initially, \(b\) is not 0, so it adds 3 to the recursive call \(* 3 (2-1)\). This will now calculate \(* 3 1\). In the next recursive step, \(* 3 1\) becomes \((+ 3 (* 3 0))\), and since \(b = 0\), it returns 0. So, \(* 3 1\) results in 3, making \(* 3 2\) result in 6.
05

Summarizing the Procedure

The multiplication function effectively sums the number \(a\), \(b\) times through recursive calls. Each recursion reduces \(b\) by 1 until it reaches the base case. The recursive nature of the procedure builds up the multiplication through addition.

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Key Concepts

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

Exponentiation
Exponentiation is the mathematical operation involving raising a base number to the power of an exponent. In simple terms, it's multiplying a number by itself a certain number of times. For example, if we have a base number 2 and an exponent 3, it means we multiply 2 by itself three times, resulting in 8. In terms of algorithms, exponentiation is similar to recursive procedures that help simplify complex calculations into smaller, more manageable steps.

Using recursive algorithms for exponentiation often involves both a base and a recursive case. They break down the task of multiplying the base number by itself repeatedly, reducing the problem size until it hits the base case (e.g., when the exponent equals 0). This systematic approach allows programmers to implement powerful algorithms efficiently, making use of simple operations like multiplication.
Integer Multiplication
Integer multiplication can be achieved through repeated addition, much like how exponentiation involves repeated multiplication. When we multiply two integers using recursive procedures, we repeatedly add one integer, a specified number of times. For instance, the multiplication of 3 and 4 is equivalent to adding 3, four times as in 3 + 3 + 3 + 3.

When an algorithm can only perform addition, integer multiplication is transformed into a repeated addition problem. This showcases how simple operations can build up to solve more complex problems like multiplication. Recursion helps here by decomposing the task of addition into smaller parts, easily manageable by the computer or the human mind.
Base Case
In a recursive algorithm, the base case is the condition that provides a straightforward answer to the problem, effectively stopping further recursion. It serves as an anchor that ensures the recursive algorithm doesn't run infinitely. For example, in integer multiplication by repeated addition, the base case is when the integer being decremented (usually the second operand) reaches 0.

The function checks this base case first. If it is true, it returns a simple value, often 0 or 1, depending on the problem. This base case resolves the smallest instance of the problem and allows for the cumulative build-up of the result as the recursion "unwinds.", leading to a complete solution.
Recursive Procedures
Recursive procedures are algorithms that solve complex problems by breaking them into smaller sub-problems that have the same structure. They rely on two key ideas: a base case and a recursive call that moves the problem towards the base case.

A recursive call is made within the procedure itself. This self-referencing ensures that each step of the recursion acts as a mini-loop, processing the variables until the base case is met. For integer multiplication, the recursive procedure repeatedly calls itself with one operand reduced by 1 until it reaches 0.

Recursive algorithms are especially efficient because they reduce repetitive coding and simplify the process of solving complex mathematical equations into a series of simpler, repeated steps.

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Most popular questions from this chapter

Newton's method for cube roots is based on the fact that if \(y\) is an approximation to the cube root of \(x\), then a better approximation is given by the value \(\frac{x / y^{2}+2 y}{3}\) Use this formula to implement a cube-root procedure analogous to the squareroot procedure. (In section \(1.3 .4\) we will see how to implement Newton's method in general as an abstraction of these square-root and cube- root procedures.)

The good-enough? test used in computing square roots will not be very effective for finding the square roots of very small numbers. Also, in real computers, arithmetic operations are almost always performed with limited precision. This makes our test inadequate for very large numbers. Explain these statements, with examples showing how the test fails for small and large numbers. An alternative strategy for implementing good-enough? is to watch how guess changes from one iteration to the next and to stop when the change is a very small fraction of the guess. Design a square-root procedure that uses this kind of end test. Does this work better for small and large numbers? \({ }^{24}\) Readers who are worried about the efficiency issues involved in using procedure calls to implement iteration should note the remarks on "tail recursion" in section 1.2.1.

Define a procedure double that takes a procedure of one argument as argument and returns a procedure that applies the original procedure twice. For example, if inc is a procedure that adds 1 to its argument, then (double inc) should be a procedure that adds 2 . What value is returned by (((double (double double)) inc) 5)

Several of the numerical methods described in this chapter are instances of an extremely general computational strategy known as iterative improvement. Iterative improvement says that, to compute something, we start with an initial guess for the answer, test if the guess is good enough, and otherwise improve the guess and continue the process using the improved guess as the new guess. Write a procedure iterative-improve that takes two procedures as arguments: a method for telling whether a guess is good enough and a method for improving a guess. Iterative-improve should return as its value a procedure that takes a guess as argument and keeps improving the guess until it is good enough. Rewrite the sqrt procedure of section \(1.1 .7\) and the fixed-point procedure of section \(1.3 .3\) in terms of iterative-improve.

Modify fixed-point so that it prints the sequence of approximations it generates, using the newline and display primitives shown in exercise \(1.22\). Then find a solution to \(x^{x}=1000\) by finding a fixed point of \(x \mapsto \log (1000) / \log (x)\). (Use Scheme's primitive \(\log\) procedure, which computes natural logarithms.) Compare the number of steps this takes with and without average damping. (Note that you cannot start \(\mathrm{fixed}\)-point with a guess of 1 , as this would cause division by \(\log (1)=0\).)

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