alternative process to prove no 16 and find all 17sFollowing the positive results on the 27 + X + Y using the facts that in a valid puzzle each band/stack must be solved against the corresponding gangster, I worked on a possible application to the search of new 17s if any or another way to prove that no 16 exists.
Here is a possible way.
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9..8......8....7......6..5.6.......8..4..9......3..6...3......9....2..4......7...
band 1 912875463586234791347961852 123456789457289631698317254 381 3
band 2 653742918824619537179358624 123456789457289163896731524 358 3
band 3 435186279798523146261497385 123456789457189623689732541 261 4
stack 1 312874965695132748478956213 123456789456789132789231546 23 3
stack 2 541298637736415829829763154 123456789456789123789132564 5 4
stack 3 267349581184527396953681472 123456789457189263689732145 160 4
123456789456789132789231546 i=23
nua=43 n1=12224
mladen 41
The first line is an existing 17 taken as start. A valid 17 is a kind of worst case in such a process, so the average situation will be much better than in that case.
The valid 17 has been solved to get the solution grid.
Next three items are the bands and stacks extracted from the solution grid.
For each item, we have
the values in the grid (27 cells in a band format)
the corresponding minlex values
the index of that minlex value in the 416 table I use
the minimum number of clues required to have a valid band in that "416"
In that situation, the best is likely to work on a morph of the puzzle with the final band order "stack1;stack2/stack3"
The target being to have a maximum of "minimum number of clues" if the pending bands
Doing so, we start with the "416" index 23 of my table (likely 24 in mladen's table)
In another program, I checked that "416" and got the following results:
I generated all UA's necessary to have all valid bands up to 9 clues. I have 43 UAs against 41 in mladen's table, but my list has to be cleaned for possible redundancy.
More interesting, I got only 12224 valid band 1.
Knowing that all band 1 with 9 clues can only be combined with 4 clues in band 2 and 4 clues in band 3 to stay within the 17 clues limit, This seems to be a very promising way to go.
The basic idea is to have the table of UA's prepared separately, although the cost to produce it is not that big for a batch of thousands solution grids (less than 4 seconds with my process for the 416 minlex starts)
then to morph a solution grid to the optimal form for the process and finally to extend bands and to check all the "band solutions against the stack".
Most of the solutions will be cancelled within the band expansion except, may be, as here, for solutions grids containing a valid 17.
- but we do know all the ways that any particular band of the 416 that can be covered in 3 clues ....