Laurent_R
Veteran
/ Moderator
Nov 26, 2012, 2:09 PM
Post #7 of 13
(5513 views)

Re: [gevni] how to skip redundant permutations?
[In reply to]

Can't Post


Means {1,2}, {2,1} Here 1=A,B 2= C,D so resultant permutations are only {A,B,C,D} and {C,D,A,B} No, as far as I understand your explanations, I have to disagree. Read again what I wrote in post #2, you have six possible permutations, not two. If I summarize the letters with the group to which they belong, you have: {A,B,C,D} is 1122, and {C,D,A,B} is 2211. {A,C,B,D} (or 1212), {A,C,D,B} (1221), {C,A,B,D} (2112) and {C,A,D,B} (2121) are in no case equivalent to your first two permutations, nor is any of them equivalent to any of the others. So I maintain that you have six possible permutations. Or you have to explain an additional rule that you did not tell us about so far (or perhaps I missed some part of your explanation). This is central to your task, so we really have to clarify this before we can go any further in trying to achieve what you want.
