Happy New Year 2024 ... and 2023's last teaser !

[Ajaja]

There are many rows and columns with zeros. I think, the fastest way to calculate permanent of the matrix is to sort it first, then transpose, sort again and after that use J-F Garnier optimized subprogram. It turns this 12x12 permanent into sum of 8 9x9 permanents if I'm not mistaken.
And result should be the same:

Code:


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> with(LinearAlgebra):
> Permanent(<
> <0,0,0,0,0,1,1,1,1,1,1,1>|
> <0,0,0,0,1,0,0,0,0,0,0,1>|
> <0,0,0,0,1,0,0,0,0,0,1,1>|
> <0,0,0,0,1,0,0,0,0,1,1,1>|
> <0,0,0,0,1,0,0,0,1,1,1,1>|
> <0,0,0,0,1,0,0,1,1,1,1,1>|
> <0,0,0,0,1,0,1,1,1,1,1,1>|
> <0,0,0,1,0,1,1,1,1,1,1,1>|
> <0,0,1,1,1,1,1,1,1,1,1,1>|
> <0,1,1,1,1,1,1,1,1,1,1,1>|
> <1,1,1,1,0,1,1,1,1,1,1,1>|
> <1,1,1,1,1,1,1,1,1,1,1,1>>);
                                                         2023