Revision of 1.2.5 Greatest Common Divisors from 13 July 2009 - 10:03pm
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The greatest common divisor (GCD) of two integers a and b is defined to be the largest integer that divides both a and b with no remainder. For example, the GCD of 16 and 28 is 4. In chapter 2, when we investigate how to implement rational-number arithmetic, we will need to be able to compute GCDs in order to reduce rational numbers to lowest terms. (To reduce a rational number to lowest terms, we must divide both the numerator and the denominator by their GCD. For example, 16/28 reduces to 4/7.) One way to find the GCD of two integers is to factor them and search for common factors, but there is a famous algorithm that is much more efficient.
The idea of the algorithm is based on the observation that, if r is the remainder when a is divided by b, then the common divisors of a and b are precisely the same as the common divisors of b and r. Thus, we can use the equation
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to successively reduce the problem of computing a GCD to the problem of computing the GCD of smaller and smaller pairs of integers. For example,

reduces GCD(206,40) to GCD(2,0), which is 2. It is possible to show that starting with any two positive integers and performing repeated reductions will always eventually produce a pair where the second number is 0. Then the GCD is the other number in the pair. This method for computing the GCD is known as Euclid’s Algorithm.[42]
It is easy to express Euclid’s Algorithm as a procedure:
(define (gcd a b)
(if (= b 0)
a
(gcd b (remainder a b))))
This generates an iterative process, whose number of steps grows as the logarithm of the numbers involved.
The fact that the number of steps required by Euclid’s Algorithm has logarithmic growth bears an interesting relation to the Fibonacci numbers:
Lamé’s Theorem
If Euclid’s Algorithm requires k steps to compute the GCD of some pair, then the smaller number in the pair must be greater than or equal to the kth Fibonacci number.[43]
We can use this theorem to get an order-of-growth estimate for Euclid’s
Algorithm. Let n be the smaller of the two inputs to the
procedure. If the process takes k steps, then we must have
n ≥ Fib(k) ≈ φk/√5. Therefore the number of steps k grows as the logarithm (to the base
φ) of n. Hence, the order of growth is Θ(log n).
(ak, bk)
(ak-1, bk-1) are three successive pairs in the reduction process, then we must have bk+1> bk + bk-1. To verify the claim, consider that a reduction step is defined by applying the transformation ak-1 = bk, bk-1 = remainder of ak divided by bk. The second equation means that ak = qbk + bk-1 for some positive integer q. And since q must be at least 1 we have ak = qbk + bk-1 > bk + bk-1. But in the previous reduction step we have bk+1 = ak. Therefore, bk+1 = ak> bk + bk-1. This verifies the claim. Now we can prove the theorem by induction on k, the number of steps that the algorithm requires to terminate. The result is true for k = 1, since this merely requires that b be at least as large as Fib(1) = 1. Now, assume that the result is true for all integers less than or equal to k and establish the result for k + 1. Let (ak+1, bk+1)
(ak, bk)
(ak-1, bk-1) be successive pairs in the reduction process. By our induction hypotheses, we have bk-1> Fib(k - 1) and bk> Fib(k). Thus, applying the claim we just proved together with the definition of the Fibonacci numbers gives bk+1 > bk + bk-1> Fib(k) + Fib(k - 1) = Fib(k + 1), which completes the proof of Lamé’s Theorem. [back]Exercises

Comments
qWKhkAlwZvEGgxEaW
It’s much easier to undretsnad when you put it that way!
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