Data Structures and Algorithm Analysis - Computer Science at ...

524 Chap. 16 Patterns of AlgorithmsNow, for any value n there exists k and l such thatn = km + l where m > l ≥ 0.From the definition of the mod function, we can derive the fact thatn = ⌊n/m⌋m + n mod m.Since the LCF is a factor of both n and m, and since n = km + l, the LCF musttherefore be a factor of both km and l, and also the largest common factor of eachof these terms. As a consequence, LCF (n, m) = LCF (m, l) = LCF (m, n modm).This observation leads to a simple algorithm. We will assume that n ≥ m. Ateach iteration we replace n with m and m with n mod m until we have driven mto zero.int LCF(int n, int m) {if (m == 0) return n;return LCF(m, n % m);}To determine how expensive this algorithm is, we need to know how muchprogress we are making at each step. Note that after two iterations, we have replacedn with n mod m. So the key question becomes: How big is n mod mrelative to n?n ≥ m ⇒ n/m ≥ 1⇒2⌊n/m⌋ > n/m⇒ m⌊n/m⌋ > n/2⇒⇒n − n/2 > n − m⌊n/m⌋ = n mod mn/2 > n mod mThus, function LCF will halve its first parameter in no more than 2 iterations.The total cost is then O(log n).16.3.3 Matrix MultiplicationThe standard algorithm for multiplying two n × n matrices requires Θ(n 3 ) time.It is possible to do better than this by rearranging and grouping the multiplicationsin various ways. One example of this is known as Strassen’s matrix multiplicationalgorithm.For simplicity, we will assume that n is a power of two. In the following, Aand B are n × n arrays, while A ij and B ij refer to arrays of size n/2 × n/2. Using