David P Bird wrote:I was hoping that if a summary of this conclusion was produced I would at least have a chance of understanding it.

The subject is spread here and there and below I will attempt to describe the statistics produced by Blue in

this post.

The existence of low-clue puzzle which solution is a particular grid depends of grid's characteristics - the Unavoidable Sets.

Any valid puzzle hits all UA sets, else exchanging the values within the unhit UA set produces a secondary solution.

We divide the solution to a sub-structures - in this case 3 bands and 3 stacks.

We are forgetting about UA that are spread over multiple bands and investigate only those which are entirely located within a single band. Symmetrically, same for stacks.

Considering only a particular band, we can apply validity preserving transformations and represent it in its canonical form. It happens that there are only 416 different canonical bands. So we can investigate each of the 416 bands independently.

The opening post lists the minimal number of clues necessary to hit all UA sets within each canonical band. They are found by brute force. There, the absolute minimum is for band #29 (column 1) which can be completed by only 2 clues (column 4).

- Code: Select all
`Band Sol UA MinClues 54+2 ...`

29 516 81 2 9 ...

Irrelevant to the subject columns are the number of in-band UA sets (81, column 3) that, when unhit, lead to many solutions (516, column 2) but giving the required minimum of 2 clues in one of the 9 (column 5) ways resolves all of them.

Considering full grid and ONLY intra-band UA sets, a lower-band estimation of the minimal number of clues required to resolve the whole grid was done.

From one hand we need to resolve each of the 3 bands and it requires number of clues that is at least the sum of the minimally required number of clues for each of the 3 bands.

From other hand we need to resolve stacks in the same way, and this gives a second sum.

The worse (largest) sum from the bands and stacks gives the final lower-band estimation on the number of clues required to resolve the whole grid.

What Blue did is to scan all 5472730538 grids, cut-out them once to 3 bands and second time to 3 stacks, transform each of them to its canonical form, do a lookup for the minimal required clues for these 2 triples, find the larger sum, and display the count of grids having the respective sum.

blue wrote:- Code: Select all
`MCB | grids | 17CG | 17CG/grids | 17CGP`

----+------------+-------+------------+------

6 | 0 | | |

7 | 0 | | |

8 | 6 | | |

9 | 1239360 | 250 | 0.0201717% | 271

10 | 97729312 | 5750 | 0.0058836% | 6215

11 | 1082360670 | 20922 | 0.0019330% | 22235

12 | 2631838676 | 16691 | 0.0006342% | 17642

13 | 1364599022 | 2550 | 0.0001869% | 2654

14 | 263770800 | 126 | 0.0000478% | 129

15 | 28131065 | 10 | 0.0000355% | 10

16 | 2785923 | 1 | 0.0000359% | 1

17 | 228185 | | |

18 | 47519 | | |

----+------------+-------+------------+------

| 5472730538 | 46300 | | 49157

From the result we see that there is no valid solution grid composed entirely from band 29 (all 3 bands and stacks being morphs of it). Such grid would have MCB=2+2+2=6.

In the same way combination of band 29 + band 29 + one of the bands requiring 3 clues is impossible too. No grids for MCB=2+2+3=7.

But we see that band 29 can successfully be combined by itself and a band requiring 4 clues (MCB=2+2+4=8), or by 2 other bands each requiring 3 clues (MCB=2+3+3=8), and this produces 6 different grids with MCB=8 which are listed in the next post.

Is there strong correlation between the number of clues required to resolve the substructures of the grid (bands and stacks) and the number of 17-clue puzzles actually found in this grid? Obviously yes. Grids, requiring less clues for resolving all intra-band UA, have better chance to be entirely resolved by 17 clues (column 4 in Blue's table).

Chance is estimated against the 49157 known so far 17-clue puzzles that resolve 46300 solution grids, shown in the bottom row.

These estimations are in use when knowing the solution grid you can test whether it is resolvable by 17 clues. Present algorithms are doing it for for average grid in less than 30 seconds, and in few minutes for a grid capable to be resolved by 17 clues.