Exponentiation

Consider the problem of computing the exponential of a given number. We would like a procedure that takes as arguments a base b and a positive integer exponent n and computes bⁿ. One way to do this is via the recursive definition

which translates readily into the procedure

(define (expt b n)
  (if (= n 0)
      1
      (* b (expt b (- n 1)))))

This is a linear recursive process, which requires (n) steps and (n) space. Just as with factorial, we can readily formulate an equivalent linear iteration:

(define (expt b n)
  (expt-iter b n 1))

(define (expt-iter b counter product)
  (if (= counter 0)
      product
      (expt-iter b
                (- counter 1)
                (* b product))))

This version requires (n) steps and (1) space.

We can compute exponentials in fewer steps by using successive squaring. For instance, rather than computing b⁸ as

we can compute it using three multiplications:

This method works fine for exponents that are powers of 2. We can also take advantage of successive squaring in computing exponentials in general if we use the rule

We can express this method as a procedure:

(define (fast-expt b n)
  (cond ((= n 0) 1)
        ((even? n) (square (fast-expt b (/ n 2))))
        (else (* b (fast-expt b (- n 1))))))

where the predicate to test whether an integer is even is defined in terms of the primitive procedure remainder by

(define (even? n)
  (= (remainder n 2) 0))

The process evolved by fast-expt grows logarithmically with n in both space and number of steps. To see this, observe that computing b²ⁿ using fast-expt requires only one more multiplication than computing bⁿ. The size of the exponent we can compute therefore doubles (approximately) with every new multiplication we are allowed. Thus, the number of multiplications required for an exponent of n grows about as fast as the logarithm of n to the base 2. The process has (log n) growth.³⁶

The difference between (log n) growth and (n) growth becomes striking as n becomes large. For example, fast-expt for n = 1000 requires only 14 multiplications.³⁷ It is also possible to use the idea of successive squaring to devise an iterative algorithm that computes exponentials with a logarithmic number of steps (see exercise 1.16), although, as is often the case with iterative algorithms, this is not written down so straightforwardly as the recursive algorithm.³⁸

Design a procedure that evolves an iterative exponentiation process that uses successive squaring and uses a logarithmic number of steps, as does fast-expt. (Hint: Using the observation that (b^n/2)² = (b²)^n/2, keep, along with the exponent n and the base b, an additional state variable a, and define the state transformation in such a way that the product a bⁿ is unchanged from state to state. At the beginning of the process a is taken to be 1, and the answer is given by the value of a at the end of the process. In general, the technique of defining an invariant quantity that remains unchanged from state to state is a powerful way to think about the design of iterative algorithms.)

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 0)
      0
      (+ a (* a (- b 1)))))

This algorithm takes a number of steps that is linear in b. Now suppose we include, together with addition, operations double, which doubles an integer, and halve, which divides an (even) integer by 2. Using these, design a multiplication procedure analogous to fast-expt that uses a logarithmic number of steps.

Using the results of exercises 1.16 and 1.17, devise a procedure that generates an iterative process for multiplying two integers in terms of adding, doubling, and halving and uses a logarithmic number of steps.³⁹

There is a clever algorithm for computing the Fibonacci numbers in a logarithmic number of steps. Recall the transformation of the state variables a and b in the fib-iter process of section 1.2.2: a a + b and b a. Call this transformation T, and observe that applying T over and over again n times, starting with 1 and 0, produces the pair Fib(n + 1) and Fib(n). In other words, the Fibonacci numbers are produced by applying Tⁿ, the nth power of the transformation T, starting with the pair (1,0). Now consider T to be the special case of p = 0 and q = 1 in a family of transformations T_pq, where T_pq transforms the pair (a,b) according to a bq + aq + ap and b bp + aq. Show that if we apply such a transformation T_pq twice, the effect is the same as using a single transformation T_p'q' of the same form, and compute p' and q' in terms of p and q. This gives us an explicit way to square these transformations, and thus we can compute Tⁿ using successive squaring, as in the fast-expt procedure. Put this all together to complete the following procedure, which runs in a logarithmic number of steps:⁴⁰

(define (fib n)
  (fib-iter 1 0 0 1 n))
(define (fib-iter a b p q count)
  (cond ((= count 0) b)
        ((even? count)
         (fib-iter a
                   b
                   <??>      ; compute p'
                   <??>      ; compute q'
                   (/ count 2)))
        (else (fib-iter (+ (* b q) (* a q) (* a p))
                        (+ (* b p) (* a q))
                        p
                        q
                        (- count 1)))))