Creating the most symmetrical layout based on a range of spans and quantities

[farshidkaviani]

Hi Again,
I went through some different scenarios and compared the results from your 2nd code block with the results from the Python script from @c.poupin and it seems that except 2 scenarios the Python scripts distributes the spans better than your second code block.

Thank you so much for the efforts you both made. I think you are so close to crack it and I do appreciate it if you could come up with a new solution.

Regards,
Farshid

[farshidkaviani]

@jacob.small

permutation

Hi Again,
Thank you so much for sharing this, though I need to learn how to work with it, as @jacob.small mentioned I could see there is some limitations in this node.

I do appreciate your time and help once again.

Regards,
Farshid

[farshidkaviani]

Hi,
Thank you for your input.

Regards,
Farshid

[c.poupin]

@jacob.small

Last I checked the only downfall is that version of the node will be limited to ~12 items to sort fully due to the limits of the number slider and the scope of the problem (a list of 12 items has 12! permutations which is 479,001,600 options). Is the last value in your number slider a reverse of the original list? If not you can add another function to join a list of inputs together for the values (two number sliders) which gets you to 21 or 22 deep lists (22! = 1.1240007e+21) but I recall Python starting to round off for me at after adding a 3rd input.

Hello @jacob.small
In fact in this case i don’t use a brute force permutation, I permute groups of sub-list (Most symmetrical number list group possible)
so I can have more than 12 items in the entry list

but if the number of sublist groups increases (number of items to swap) a solution it’s to use the node “Number Slider” instead of “Integer Slider”.

an example with 15 groups to be swapped (range 1 to 1e+13 for 1.307674368 E+12 max permutations )
before it seems to me that it was not possible we were limited to Int32