What you should be looking at are combinations instead of permutations, because what you are computing are all the different ways of choosing +/b 1s (or +/~b 0s) from i.#b. For example:
b=: 1 1 1 0 0 (i.#b) e."1 (+/b) comb #b 1 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 1 0 1 0 1 0 1 1 0 0 1 1 0 1 1 1 0 0 1 1 0 1 0 1 0 1 1 0 0 1 1 1 comb is from the for. page of the dictionary. http://www.jsoftware.com/help/dictionary/cfor.htm On Sun, May 29, 2016 at 2:12 PM, Louis de Forcrand <[email protected]> wrote: > Is there a faster (and relatively concise) way of generating > (~.@:A.~ i.@!@#) b > where b is a logical vector (of 0s and 1s)? > There are obviously (+/ ! #) b vectors in the result, but I can't > see any relation in their anagram indices: > > I.@(-:"1/~ ~.)@(A. ~i.@!@#) 1 1 1 0 0 > 0 1 6 7 24 25 30 31 48 49 54 55 > 2 4 8 10 26 28 32 34 50 52 56 58 > 3 5 9 11 27 29 33 35 51 53 57 59 > 12 14 18 20 36 38 42 44 60 62 66 68 > 13 15 19 21 37 39 43 45 61 63 67 69 > 16 17 22 23 40 41 46 47 64 65 70 71 > 72 74 78 80 84 86 96 98 102 104 108 110 > 73 75 79 81 85 87 97 99 103 105 109 111 > 76 77 82 83 88 89 100 101 106 107 112 113 > 90 91 92 93 94 95 114 115 116 117 118 119 > > The (< I , J )&{ of this table corresponds to the Jth occurence > of the Ith unique permutation of the vector in its table of all > possible permutations. > > Thanks, > Louis > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
