>Yes each ganster will a significant expansion element - if
>they have the same number of solutions then that is ok. I
>might have thought it was as much as 6 ber B2 and 6 for B3 ie
>36 - obviously not thankfully.
>
>Are you allowing 3 permutations for the gangster in B1B4B7 ?
yes
>That would then be 3*3 permutations for each B12347
once we fix B123 we have lots of permutations
>I think the whole enumeration is possible this way - but the
>usefulness perhaps is if we can get the B12347 with the least
>number of solutions.
...249183
...367459
...158726
213......
745......
869......
971......
582......
436......
with 1960 solutions. It also has only 199296 solutions when you solve the
complement 6*6.
>find the 4 box [a four box
>square or the four corners - they are the same] with the
>minimum number of solutions.
>
>[If my program would do it I would guess the minimum might be
>4 392s !]
Takes about 5s times 2802. All counts divisible by 144.
Maximum 3904*144, minimum 1370*144. 761 different counts,
average ~315464.02
>Put them together and we have one tight mama !!!!! [Pairs not
>present ???]
>
>.............but the number of solutions for a corner box is
>more than 130000 [sadman stalls here][I see 300000 is your
>figure]
>.............of course if we really went for the knarliest
>grid we would do the big calculation first.........leave it
>for now[plan B]
>
>
>
>............this is not quite the final answer
>............every grid has 9 different "middle box" "crunching
>exercises" which are all different
>............so we are not quite there yet................
>
>I am glad we are making progress with this now - and we have
>got something out of it.............. .
>
>Gfroyles 17er grid is starting to reveal itsself I fear
>We need to do the 9 crunching exercises for it - I reckon one
>of them will be a very low combination.
you mean, we do the B12347 calculation 9 times for that grid,
with each of the boxes as B1 ?
>The tight box combinations will coincide with the "important"
>6 boxes [which determine the grid"] in the 17er grid.
>
>The 6 -a formulation of these ones:
>XX-
>X-X
>-XX
>
>Do you follow ?
why are they important ?
-----------------
>dukuso wrote:
>there are essentially 2802 different B2B3B4B7
>sudokugrids
>with 1035 different nonzero solution-counts for the
>remaining
>6*6 square plus 3*3 square of blocks B5B6B8B9 and B1.
>Zero solutions only occur, when block1 can't be filled.
>Otherwise maximum is:6280 solutions,minimum is 1960.
>
>
>
>Do you think this is useful in searching for a 16?
>
>And why?
I don't know. Note that I posted to the math-thread,
not the minimum-clue-thread.
It could be useful - see Coloins ideas.
But if I'd just only been interested in the 16,
then I'd probably gone another way.
-----------------------------
>I hope it s useful
>
>We need to find the grid first.
>
>We need to know why Gfroyles 17ers have so many 17s in them -
>[I hope it isnt that he has just looked harder.]
>If they are special - we need to find out why - and then
>improve on it - to find a 16.
just speed up his calculation a bit by finding even more 17s ?!
>There are so many grids out there - I cant believe we have
>found the grid which has the best chance [but fails].
? just because there are so many, it's hard to find the best
>Anyhow - what we have done is to count the number of solutions
>for the following patterns
>
>X[X]X
>...X
>...X which is also
have we done
X.X
..X
..X
?
>...X
>X[X]X
>...X
>
>average 2693 solutions [middle box is forced if valid]
?
>and its opposite mate
>
>X...X
>......
>X...X
same as
XX.
XX.
...
>which has 320000 solutions approx [equiv to B5B6B7B8]
>By the way G - I cant open p files
what are p files ?
-------------------------------
>Tim wrote:
>
>I've now brute forced B1+B2+B4+B5 (took about 1 hour on
>a 4GHz pentium) and I get:
>9!*25609120*48*48 = 21411158320742400
why 48*48 ? Well, 44*3 for B1+B2 are sufficient, and then I get
about 3.3 million extensions
We have to group these solutions into the 2802 classes to recompute N
this is possible, but I didn't do it.
>different nonzero solution-counts - upper limit 625 [maybe]
>
>6,670,903,752,021,072,936,960 / 21411158320742400 = 311562
>average solutions [assuming no zero solutions]
>
>The B5689 [rather than the B12347] is going to be the more
>important variable I think
----------------------------------
>coloin wrote:
>
>Gfroyles 17er grid is starting to reveal itsself I fear
>We need to do the 9 crunching exercises for it - I
>reckon one of them will be a very low combination
>The tight box combinations will coincide with the
>"important" 6 boxes [which determine the grid"] in the
>17er grid.
>
>The "important" 6 -a formulation of these ones:
>XX-
>X-X
>-XX
>
>Do you follow ?
>
>
>
>Follow this - even better - a "Eureka" moment
>
>Boxes
>123
>456
>789
>
>lowest score in B1245 and a lowest score in B5678
you mean: B5689 ?
>put them together with the same box 5 and a valid box 3&7
>
>This is the tightest grid - this is the Gfroyle 17 secret
you can't just take 2802^2 combinations of B1245 and B5689 -
they interact. Well, we have at most 9! essentially different
B37s.
>In Gfroyles 17er grid
>one of the 9 is 245952 [average is 311000]
>It only needs one other low score [out of the four that
>intersect] to meld in and "autocomplete" the grid.
>The other boxes dont really matter - as long as they are valid
>- which they are.
>
>If Gfroyles 17 grid utilizes the two lowest B1245 / B5678
>combinations - which combine - we are stuffed. If there is
>scope for improvement we have a fighting chance.
>
>The other three intersecting 4 boxes are [solutions count]
>290744
>..........
>..........
>
>I await the computation !
----------------------------
>Follow this
>
>Boxes
>123
>456
>789
>
>Aim for a lowest score in B1245 and a lowest score in B5689
>
>put them together with the same box 5 and a valid box 3&7
>
>This is the tightest grid - this is the Gfroyle 17 secret
>
>In Gfroyles 17er grid
>one of the 9 is 245952 [average is 311000]
>It only needs one other low score [out of another four that
>intersect] to meld in and "autocomplete" the grid.
>
>
>If Gfroyles 17 grid utilizes the two lowest B1245 / B5678
>combinations - which combine - we are stuffed. If there is
>scope for improvement we have a fighting chance.
>
>Rearranging the Boxes
>123
>456
>789
>
>to
>
>213
>879
>546
>
>2187..........245952 solutions *
>2356..........291744 solutions
>7946..........246816 solutions *
>1346..........309744 solutions
>8956..........297072 solutions
>1379..........245232 solutions *
>2389..........285696 solutions
>8754..........337536 solutions
>2154..........314784 solutions
>
>
>We have 2 low solutions* intersecting - boxes 2187 & 7946
>
>Rearranging again
>
>132
>798
>465
>we also have another low solution *1379 which intersects with
>9865 [a below average count]
>
>Rearranging again
>
>123
>789
>456
>
>1278 also intersects with 8956
>
>
>I think this explains why the grid was so special -- Can we
>beat it ?
>
>Is the 16er with 2 solutions better ?
>
>I await the computation of the 6*6ers !
>
>We will have the answer soon
>
>Regards
>
>Colin
>
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