ECJ-INTEREST-L@LISTSERV.GMU.EDU

View:

 Message: [ First | Previous | Next | Last ] By Topic: [ First | Previous | Next | Last ] By Author: [ First | Previous | Next | Last ] Font: Monospaced Font

Subject:

Re: RMHC random mutation hill climber

From:

Date:

Fri, 15 Jul 2005 15:26:06 -0400

Content-Type:

text/plain

Parts/Attachments:

 text/plain (42 lines)
 Dear Sean, Thank you for the guidance. It certainly looks easier than I had originally thought. w00t! I have a couple questions/clarifications (see below). Thanks again, Steve On Jul 12, 2005, at 3:57 PM, Sean Luke wrote: > On Jul 12, 2005, at 3:32 PM, Steve Butcher (Steve-O) wrote: > > >> Conceptually, a hill climber is a population of 1 who at each step is >> compared to the performance of a clone of itself that has undergone a >> random mutation. >> > > I think generally a hill-climber has a population of two. I've had a > spirited debate with Paul Wiegand about the difference between > hill-climbing and the 1+1 ES, and it comes down to the assumption that > a hill-climber's mutator is "local" -- it cannot mutate to every > possible spot in space in a single jump. > > Here's a 1+1 mutation-only ES for the ec/app/sum problem. It's using > VectorMutationPipeline which could in theory mutate to any point in > space, but you could replace it with a true "local" hill-climbing > mutator if you so desire. > I'm a bit confused on this point. The two most common ways to implement mutation in a vector-based individual are either (1) for each individual, on some probability, flip a random bit (clearly local) or (2) for each individual, for each bit, on some probability, flip the bit (probably not "local"). Which method does VectorMutationPipeline use? I can't think of why it would be any other way but is the child always a mutation of the parent (even at initialization)?