Complexity of the Euclidean Algorithm (2/7) The complexity of an algorithm is the approximate number of steps necessary for the algorithm to finish as a function of the size of the input. Most commonly we are interested in either average case complexity or worst case complexity. These mean exactly what they sound like. We shall now analyze the worst case complexity of the Euclidean Algorithm (EA).

EA Worst Case The EA will take the longest if each remainder is not much smaller than the previous remainder. Said another way, EA will take the longest if the quotients q k are relatively small. In particular, the worst case is going to occur when every quotient is 1 ! So, what does the EA sequence look like if every quotient is 1: (Recall: r -1 = a and r 0 = b.) r -1 = r 0 + r 1 r 0 = r 1 + r 2 r 1 = r 2 + r r n-2 = r n-1 + r n

EA Worst Case Continued For simplicity, let’s assume n is even (easy to adjust if odd), where r n is the last non-zero remainder. From above we have then: b = r 1 + r 2 > 2r 2, r 2 = r 3 + r 4 > 2r 4, r 4 = r 5 + r 6 > 2r 6, etc., and finally r n-2 = r n-1 + r n > 2r n. Combining all these, we get b > 2 n/2 r n  2 n/2. Now solve this equation for n. (How do you solve equations for things in the exponent?) Finally, change the base-2 log to a base-10 log, since base10 logs reveal the number of decimal digits. (Discuss.)

Conclusion: Theorem. The worst case complexity for the Euclidean Algorithm is that it will complete in a number of steps which is at most about 6.7 times the number of decimal digits in b. Finally, are there pairs of numbers which have the property that this worst case occurs when you apply the EA to them to get their GCD? Answer: Yes! Consider the Fibonacci Numbers. For Monday, absorb these ideas!!