Red Ed wrote:I wonder if dobrichev, whose brief foray into subtree methods sparked this exchange, is feeling any better informed now?
Hi,
All the methods for estimation of all minimals are based on as larger as possible sample and incorporate weighting up to some predefined degree of details. Nothing new here, although claiming ownership on centuries old mathematical methods sounds a bit strange.
Playing with distribution of distributions is a little confusing to me. Determining the distribution of, say, 25-clue minimals over several samples is targeting the mean/median/average/whatever that finally will be included in the final distribution of the n-clue puzzles. We want the the individual n-clue probabilities to sum to 1 and I am not sure for the right way this should be done.
A typical usage of the general distribution - this over the whole space - is when you use it as a reference to underline something special = something non-average. Nevertheless it seems I posted in wrong topic, using the "real" distribution as reference. Sorry, Denis.
I hate dividing something close to zero by other thing close to zero. This limits the applicability of the general distribution. A possible workaround is doing local investigations targeting the proportion N(k) / N(k+1) and using the subtree as the only option. I have no idea if these local observations are reusable outside the same local context.
We know UA sets are relatively well defined abstraction which determines the grid and respectively its puzzles. We know that every solution grid has 6.67e21 (non-minimal) unavoidable sets. We know that for different grids the ratio between unaviodable rectangles and all UA is between 0 and 36 / 6.67e21. Aren't rectangles extremely rare? If so, why we don't just ignore them? Because of the significance of the weighting, and no subjections here. Denis?
I hope some day some will approximate this distribution by product of the yet unknown uniques and minimality functions.
Any low-clue puzzle kills the minimality of all its supersets which is huge number. Can we solve the opposite problem - estimate the low-clue (or high-clue) fertility of a given subgrid by processing the counts of the easily obtainable moderate-clue minimal puzzles?
For the hard puzzles hunters, below is my first observation related to the "real" distribution.
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And yes, I am feeling better informed now, even if the exchange only repeats or reformulates already discussed matter. Thank you guys.