Construct Matrix with given Permanent

[Albert Chan]

(01-03-2024 05:53 PM)Albert Chan Wrote:  To solve puzzle, we undo randomization, to make above "standard form".
Then, we just read off first row pattern for Permanent.

VA Square Matrix Permanent Puzzle
VA Solution posted yesterday, so this is not a spoiler anymore.

> m := [
[1,0,1,1,1,0,1,1,1,1,1,1],
[0,0,1,1,0,0,0,0,0,1,0,1],
[1,0,1,1,0,0,0,1,0,1,1,1],
[0,0,1,0,0,0,0,0,0,1,0,0],
[1,0,1,1,1,1,1,1,1,1,1,1],
[0,0,1,1,0,0,0,0,0,1,1,1],
[1,0,1,1,1,0,0,1,0,1,1,1],
[0,1,0,1,1,1,1,1,1,1,0,1],
[1,0,1,1,0,0,0,0,0,1,1,1],
[1,1,1,1,1,1,1,1,1,1,1,1],
[0,0,1,0,0,0,0,0,0,1,0,1],
[1,0,1,1,1,0,1,1,0,1,1,1]]

> s:=map(sum,m)      → [10,4,7,2,11,5,8,9,6,12,3,9]

If Permanent Puzzle setup as I predicted, "first row" has 9 1's
We make them slightly different for sorting purpose.

> s[0]:=9.5
> m:=map(sort(s), k->m[index(s,k)])

\(\left(\begin{array}{cccccccccccc}
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1
\end{array}\right)\)

> SWAPCOL(m,10,12)
> SWAPCOL(m,3,11)
> SWAPCOL(m,4,9)
> SWAPCOL(m,3,8)
> SWAPCOL(m,1,7)
> SWAPCOL(m,3,6)
> SWAPCOL(m,1,4)
> SWAPCOL(m,1,3)
> SWAPCOL(m,1,2)

\(\left(\begin{array}{cccccccccccc}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1
\end{array}\right) \)

"First row" bits (if we move 8th row to top) for Permanent

> 0b11111100110 + 1

2023