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From my previous question, if I consider a list like this:

$\{$$\{$$\{$$1,2,3$$\}$,$\{$$4,5,6$$\}$$\}$, $\{$$\{$$1,2,4$$\}$,$\{$$3,5,6$$\}$$\}$, $\{$$\{$$1,2,5$$\}$,$\{$$3,4,6$$\}$$\}$, $\{$$\{$$1,2,6$$\}$,$\{$$3,4,5$$\}$$\}$, $\{$$\{$$1,3,4$$\}$,$\{$$2,5,6$$\}$$\}$, $\{$$\{$$1,3,5$$\}$,$\{$$2,4,6$$\}$$\}$, $\{$$\{$$1,3,6$$\}$,$\{$$2,4,5$$\}$$\}$, $\{$$\{$$1,4,5$$\}$,$\{$$2,3,6$$\}$$\}$, $\{$$\{$$1,4,6$$\}$,$\{$$2,3,5$$\}$$\}$, $\{$$\{$$1,5,6$$\}$,$\{$$2,3,4$$\}$$\}$$\}$

how can I delete all the permuted sublists containing in one of their subset $2$ different integers already present in one of the subsets of the previous permuted sublists? I hope the request is clear. In the showed case, the output would just be:

$\{$$\{$$1,2,3$$\}$,$\{$$4,5,6$$\}$$\}$

While considering the sublists of $6$ elements divided in subsets of length $2$, starting with

$\{$$\{$$1,2$$\}$,$\{$$3,4$$\}$,$\{$$5,6$$\}$$\}$

this one has to be deleted:

$\{$$\{$$1,3$$\}$,$\{$$2,4$$\}$,$\{$$5,6$$\}$$\}$

while this one (and others too) should be in the output:

$\{$$\{$$1,3$$\}$,$\{$$4,5$$\}$,$\{$$2,6$$\}$$\}$

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  • $\begingroup$ With all due respect, I personally think it not quite clear ;). $\endgroup$ Commented Feb 16, 2023 at 6:34
  • $\begingroup$ @ΑλέξανδροςΖεγγ Sorry for that! I try with an example: considering the list of 6 divided in subsets of length 2, starting with $\{$$\{$$1,2$$\}$,$\{$$3,4$$\}$,$\{$$5,6$$\}$$\}$ this one has to be deleted: $\{$$\{$$1,3$$\}$,$\{$$2,4$$\}$,$\{$$5,6$$\}$$\}$ while this one should be in the output: $\{$$\{$$1,3$$\}$,$\{$$4,5$$\}$,$\{$$2,6$$\}$$\}$ $\endgroup$ Commented Feb 16, 2023 at 6:49
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    $\begingroup$ @user967210 perhaps you should add the second example so people have two expected results to cross-check. this will help to get a correct answer $\endgroup$ Commented Feb 16, 2023 at 7:18

2 Answers 2

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This works for the example you gave:

d = {{{1, 2, 3}, {4, 5, 6}}, {{1, 2, 4}, {3, 5, 6}}, {{1, 2, 5}, {3, 4, 6}}, {{1, 2, 6}, {3, 4, 5}}, {{1, 3, 4}, {2, 5, 6}}, {{1, 3, 5}, {2, 4, 6}}, {{1, 3, 6}, {2, 4, 5}}, {{1, 4, 5}, {2, 3, 6}}, {{1, 4, 6}, {2, 3, 5}}, {{1, 5, 6}, {2, 3, 4}}};; i = 1; While[++i <= Length[d], If[Or @@ Flatten[Outer[Length[Intersection[#1, #2]] > 1 &, Flatten[d[[;; i - 1]], 1], d[[i]], 1]], d = Delete[d, i]; --i;] ]; d {{{1, 2, 3}, {4, 5, 6}}}

or:

d = {{{1, 2}, {3, 4}, {5, 6}}, {{1, 3}, {2, 5}, {4, 6}}, {{1, 4}, {2, 3}, {5, 6}}, {{1, 5}, {2, 4}, {3, 6}}, {{1, 6}, {2, 3}, {4, 5}}};

results in:

{{{1, 2}, {3, 4}, {5, 6}}, {{1, 3}, {2, 5}, {4, 6}}, {{1, 5}, {2, 4}, {3, 6}}, {{1, 6}, {2, 3}, {4, 5}}}
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  • $\begingroup$ Thank you again. I tried with other example using your function $ main[6, 3] $ and the output is {{{1,2},{3,4},{5,6}}, {{1,3},{2,5},{4,6}},{{1,4},{2,3},{5,6}}, {{1,5},{2,4},{3,6}}, {{1,6},{2,3},{4,5}}} so the first 2 and last 2 sublists are fine, but not the third one because the subset {5,6} was already present in the previous sublist and is missing {{1,4},{2,6},{3,5}}. I really appreciate your help, Thank you! $\endgroup$ Commented Feb 16, 2023 at 11:26
  • $\begingroup$ Sorry I misread: "previous permuted sublist" and not "previous permuted sublists". I fixed this. $\endgroup$ Commented Feb 16, 2023 at 15:14
  • $\begingroup$ Thank you, that's what I looked for! $\endgroup$ Commented Feb 16, 2023 at 16:23
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You can use two-argument form of DeleteDuplicates:

ClearAll[cleanUp] cleanUp = DeleteDuplicates[#, GreaterThan[1] @ Max @ Outer[Length @* Intersection, ##, 1] &] &;

Examples:

d1 = {{{1, 2, 3}, {4, 5, 6}}, {{1, 2, 4}, {3, 5, 6}}, {{1, 2, 5}, {3, 4, 6}}, {{1, 2, 6}, {3, 4, 5}}, {{1, 3, 4}, {2, 5, 6}}, {{1, 3, 5}, {2, 4, 6}}, {{1, 3, 6}, {2, 4, 5}}, {{1, 4, 5}, {2, 3, 6}}, {{1, 4, 6}, {2, 3, 5}}, {{1, 5, 6}, {2, 3, 4}}}; cleanUp @ d1 
{{{1, 2, 3}, {4, 5, 6}}} 
d2 = {{{1, 2}, {3, 4}, {5, 6}}, {{1, 3}, {2, 5}, {4, 6}}, {{1, 4}, {2, 3}, {5, 6}}, {{1, 5}, {2, 4}, {3, 6}}, {{1, 6}, {2, 3}, {4, 5}}}; cleanUp @ d2 
{{{1, 2}, {3, 4}, {5, 6}}, {{1, 3}, {2, 5}, {4, 6}}, {{1, 5}, {2, 4}, {3, 6}}, {{1, 6}, {2, 3}, {4, 5}}}
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