Paul Crowley

I used log(ab) (ie log(a) + log(b)) as my estimate of the cost. That led me to believe that calculating (ab)(cd) would be much faster than calculating ((ab)c)d.

Consider the classic dynamic programming example of reordering matrix multiplication for potentially enormous speedups (eg http://www.ics.uci.edu/~dan/class/161/notes/6/Dynamic.html) - that's where a strategy like this works.