function[newpop]=crossover（oldpoppoplengthpc）;%交叉概率
popsize=size（oldpop1）;
for?i=1:popsize/2
????%前半個個體和后半個個體進行交叉
????[newpop（i:）newpop（popsize/2+i:）]=crossover1（oldpop（i:）oldpop（popsize/2+i:）pcpoplength）;
end

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????目錄???????????0??2014-05-26?08:16??bizhang\
?????文件?????????244??2013-05-03?16:51??bizhang\crossover.asv
?????文件?????????244??2013-05-03?18:39??bizhang\crossover.m
?????文件?????????923??2013-05-03?18:45??bizhang\crossover1.m
?????文件????????2340??2013-05-03?21:37??bizhang\initialize.asv
?????文件????????2283??2013-05-10?20:05??bizhang\initialize.m
?????文件????????2156??2013-05-12?14:23??bizhang\muate.asv
?????文件????????2128??2013-05-12?14:26??bizhang\muate.m
?????文件?????????656??2013-05-03?16:47??bizhang\roulette.m
?????文件????????2094??2013-05-12?14:13??bizhang\shiyan.asv
?????文件????????1352??2013-05-10?20:17??bizhang\Untitled.asv
?????文件????????1364??2013-05-10?21:23??bizhang\Untitled.m