Author: J. Wesley Cleveland
Date: 09:40:53 10/04/01
Take an iterative function F(a), i.e. F1(a) = a, F2(a) = F(a), F3(a) = F(F(a)), ... Assume the function is complicated enough to use all the available registers on your machine and requires the same number of machine cycles for each iteration. Now find the first iteration for which F(b) > F(a). A serial implementation will spend a significant amount of time saving and restoring state as it goes between the two functions, while the two processor solution will only have to spend a minimal amount of time syncronizing at the end of each iteration (due to my second condition).
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