hi again
what a stimulating R discussion! This is R-help at its very best!
I think I understand why you don't want pure inter-block permutations.
My solution would be to realize that weeding out "forbidden" permutations is quite difficult and time-consuming (also as several people have pointed out forbidden permutations are very rare, accounting for only a proportion of (3!)^4/12! ...about one in 400000).
The extra expense of this weeding process is likely to outweigh the slight loss of efficiency caused by generating a forbidden permutation. My solution would therefore be:
x <- c(1,2,3,4,5,6,7,8,9,10,11,12) x.new <- sample(x)
(note the not-inconsiderable advantage of code simplicity!)
I think my algorithm generated repeats because in it there are only (4!)^3=13284 distinct
permutations; see help(birthday). The system above has 12! ~= 4x10^8, much higher.
Hope this helps
Robin
} 7.- Robin Hankin a<-matrix(1,200,12) for (i in 1:200) { x <- c(1,2,3,4,5,6,7,8,9,10,11,12) dim(x) <- c(3,4) jj <- t(apply(x,1,sample)) a[i,]<-as.vector(jj) } ##In 200 permutations, there are 5 repetitions.
Thanks to all and sorryfor the confusion that have generated the "intra-block" permutation.
Jordi Altirriba PhD student Hospital Clinic - Barcelona - Spain
P.S. I think that I don't have forgot to anybody...(sorry if I have done it)
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-- Robin Hankin Uncertainty Analyst Southampton Oceanography Centre SO14 3ZH tel +44(0)23-8059-7743 [EMAIL PROTECTED] (edit in obvious way; spam precaution)
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