>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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