5 Is This Fair to Hill-climbers?

The results of Section 5 are of this sort. One sometimes hears the question: Is a GA more powerful than an equal population of hill-climbers, and if so, why? The answer seems to be yes, as noted. But why? Perhaps the basic reason is that the hill-climber can get tripped up by an individual case, whereas on a problem that is GA-easy but not SAO, the GA is carried over such cases by the statistical power of the schema theorem. Note also that each of a population of hill-climbers is likely to get stuck somewhere, but in different places than its neighbors. In a problem like the one constructed in this paper, each hill-climber will at the same time be correct in a number of places (in the string), again at different places than its neighbors. Crossover would permit the good parts to be communicated and accumulated. But only the GA has crossover. One might suggest that the genetic algorithm has a built-in unfair advantage over a steepest-ascender since the GA works from a population of N initial strings whereas the steepest-ascender uses just one. What if there were N steepest-ascenders each starting with a random string and searching independently? The probability of finding the optimum somewhere among N such steepest-ascenders certainly increases with N. How does this compare with a population of size N under the GA? Here we are on somewhat more complicated ground. Let us start by assuming we are trying to optimize the function F using N steepest-ascenders. Since a steepest-ascender starting at four of the eight possible starting strings for f 2 will reach the optimum of f 2 , the probability of reaching the optimum of F, starting with a random 3m-bit string, is (4/8) m. The probability of not finding the optimum in such a string is then [1-(.5) m ]. If we have N random strings, the probability that the steepest-ascender will not find the optimum in any of the strings is [1-(.5) m ] N. Consequently, the probability of finding at least one optimum in a population of N strings, written in terms of string length n = 3m, is 1-[1-(.5) n/3 ] N. Let us go one step further and suppose this probability equals some criterion value, say one-half. We could then set the above expression equal to one-half and solve for N. The result N(n) would be the minimum …