Assuming I have the values A B C D E and eight places 1 2 3 4 5 6 7 8. How can I get all possible combinations printed with two- and three-time repetitions?
Example:
1 2 3 4 5 6 7 8 _______________ A A B B C C D E A B B C C D D E (and so on...) A A A B B C D E A B B B C C D E (and so on...) Which combination is followed by which is not further important. There should not be any missing values like AABBCCDD (where E is missing)
A crude approach with zsh/ksh93/bash could be:
printf '%s\n' {A..E}{A..E}{A..E}{A..E}{A..E}{A..E}{A..E}{A..E} | sed '/A/!d; /B/!d; /C/!d; /D/!d; /E/!d; /\(.\)\(.*\1\)\{3\}/d' Where we use zsh-style {A..E} to generate all 58 combinations from AAAAAAAA to EEEEEEEE, and sed removes the ones that don't contain A, or don't contain B, C, D, E or more than 3 of the same character.
Add ;s/./ &/2g (GNU sed) or ;s/./ &/g;s/ // (any sed) to the sed code to insert spaces in-between the letters.
bash? If 1 then ask about the algorithm first. If 2 then show the algorithm (or a link to it) in the question. I think this question is not specifically related to UNIX or Linux and might better fit on stackoverflow.com.