On Freitag 02 Juli 2010 Lionel Guy wrote: > [...] It's a difficult problem. Imagine this: Ref ....caaaag..... Adding a read with some "junk" you my get: Ref ....c**aaaag..... R1 ....ccaaaaag..... Now comes a couple of reads with the terminal "a" Ref ....c**aaaag..... R1 ....ccaaaaag..... R2a ....ca R2b ....ca R2c ....ca Which is the most optimal solution from a mathematical point of view for a Smith-Waterman alignment, because Ref ....c**aaaag..... R1 ....ccaaaaag..... R2a ....c*a R2b ....c*a R2c ....c*a or even Ref ....c**aaaag..... R1 ....ccaaaaag..... R2a ....c**a R2b ....c**a R2c ....c**a are mathematically less optimal for each individual read, although it does make more sense in the complete alignment that way. To be honest, I don't have a really good solution for that (and I tried quite a lot). Fortunately, 36mers are a thing of the past, 75 and 100mers take over and there the problem is much less pronounced. B. -- You have received this mail because you are subscribed to the mira_talk mailing list. For information on how to subscribe or unsubscribe, please visit http://www.chevreux.org/mira_mailinglists.html