e), {1;2}(line), {1;3}(line)... and on the other side to have {0;0}(all lines except {0}(0)), {0;1} (all lines except {0}(1)), {0;2}(all lines except {0}(2)), {0;3}(all lines except {0}(3)), {1;0} (all lines except {1}(0)), {1;1} (all lines except {1}(1)), {1;2} (all lines except {1}(2)) ,{1;3} (all lines except {1}(3))...The first tree is easy to achieve, simply grafting a branch for each element, and the other, what I've done is to copy all lines of each tree ({0},{1},{2},{3}), to have them in all branches of each tree ({0;0}(elements of {0}), {0;1}(elements of {0}),,{1;0}(elements of {1}), {1;1}(elements of {1})..., and then remove in the first branch({0;1} the first element(0), in the second branch the second element, the third branch the third element...And so correctly you compare each line with all the other within each branched tree.Aaaaapufff XD…
;0;1;1;0}
{0;0;1;2;0} ...
{0;0;2;0;0}
{0;0;2;1;0}
{0;0;2;2;0} ...
{0;0;3;0;0}
{0;0;3;1;0}
{0;0;3;2;0} ...
...
and I would like to have this in two lists separated:
{0;0;0;0;0}
{0;0;0;1;0}
{0;0;0;2;0} ...
{0;0;2;0;0}
{0;0;2;1;0}
{0;0;2;2;0} ...
...
{0;0;1;0;0}
{0;0;1;1;0}
{0;0;1;2;0} ...
{0;0;3;0;0}
{0;0;3;1;0}
{0;0;3;2;0} ...
...
How can I do that?…