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From: Carl G. on 11 Jun 2010 12:57

"Leroy Quet" <qqquet(a)mindspring.com> wrote in message
news:9d86157d-cc25-4993-9b7b-6dc94c11f332(a)5g2000yqz.googlegroups.com...
> This for me is an unsolved puzzle.
>
> Start with a 9-by-9 grid.
> Place any number of black stones and whites stone on the squares of
> the grid so that each square has exactly one stone.
>
> For a given row of the grid, take the lengths of the runs of black
> stones and white stones and multiply these lengths.
> (By "run", it is meant a string of consecutive stones in the row (or
> column) all of the same color, bounded by stones of the opposite color
> or by the edge of the row (or column).)
>
> Do this for all rows and all columns to get 18 products.
>
> Is it possible to place the stones so that all 18 products are unique?
>
> I am almost certain that it is indeed possible, but I have not yet
> done so by hand.
>
> Try (by hand, ideally) to find a way to arrange the stones. Or maybe
> prove that there is no pattern of stones where the products are all
> unique.
>
> Thanks,
> Leroy Quet

My first impression is that obtaining 18 unique products is unlikely.
Looking at the factorizations, with factors less than or equal to 9, there
can only be 18 unique products (1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,24,
and 27). The products of 1 and 27 can only occur in two patterns each
(101010101 => 1, 111000111 => 27). If a solution exists, it will be
impressive. I do not have time to investigate this now, but my approach
would be to check all patterns with the 1 and 27 products.

-Carl G.



From: Leroy Quet on 11 Jun 2010 15:00
You all should know that Rob Pratt on the Sequence Fan email group has
found an 18-distinct-product solution.

I don't know how many 18 products solutions there are.
(Of course, there are at least 16, but besides Pratt's solution
rotated and reflected and color flipped, I don't know how many
solutions there are.)

Thanks,
Leroy Quet

From: Richard Heathfield on 11 Jun 2010 15:36
Leroy Quet wrote:
> You all should know that Rob Pratt on the Sequence Fan email group has
> found an 18-distinct-product solution.
>
> I don't know how many 18 products solutions there are.
> (Of course, there are at least 16, but besides Pratt's solution
> rotated and reflected and color flipped, I don't know how many
> solutions there are.)

I think I've got one. (I have spent *way* too much time on this!)

x o o o o o o o x 7
o o o x x x o o o 27
x o x o o o x o x 3
o o x x o o x x o 16
o o x o x o x x x 6
x o x o x o x x o 2
o o o o o x x o x 10
o o o o o x x o o 20
x o o o x x x x x 15
4 9 24 5 8 12 14 18 1

I think this counts as an existence proof. :-)

And now I can't resist looking for another one. Give me a tick...

Here!

x x x o x o x x x 9
o x o o x x x o x 6
o x x x x x o o x 10
o o o x x x o o o 27
x o x o x x o o x 4
o o o x o o o o o 15
o x x o x o x o x 2
o o o o o x x o o 20
o o x x x x x x x 14
12 18 1 8 5 16 24 7 3

This is not a reflection, rotation, or colour reversal of the first
solution, so there is no unique-modulo-RRCR solution.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
"Usenet is a strange place" - dmr 29 July 1999
Sig line vacant - apply within
From: Leroy Quet on 11 Jun 2010 16:27
Way to go! These two solutions are different than Rob Pratt's. So
there are at least 3 solutions modulo rotations, reflections, and
color reversals.

I bet there are a lot of such solutions.

Thanks,
Leroy Quet


Richard Heathfield wrote:
> Leroy Quet wrote:
> > You all should know that Rob Pratt on the Sequence Fan email group has
> > found an 18-distinct-product solution.
> >
> > I don't know how many 18 products solutions there are.
> > (Of course, there are at least 16, but besides Pratt's solution
> > rotated and reflected and color flipped, I don't know how many
> > solutions there are.)
>
> I think I've got one. (I have spent *way* too much time on this!)
>
> x o o o o o o o x 7
> o o o x x x o o o 27
> x o x o o o x o x 3
> o o x x o o x x o 16
> o o x o x o x x x 6
> x o x o x o x x o 2
> o o o o o x x o x 10
> o o o o o x x o o 20
> x o o o x x x x x 15
> 4 9 24 5 8 12 14 18 1
>
> I think this counts as an existence proof. :-)
>
> And now I can't resist looking for another one. Give me a tick...
>
> Here!
>
> x x x o x o x x x 9
> o x o o x x x o x 6
> o x x x x x o o x 10
> o o o x x x o o o 27
> x o x o x x o o x 4
> o o o x o o o o o 15
> o x x o x o x o x 2
> o o o o o x x o o 20
> o o x x x x x x x 14
> 12 18 1 8 5 16 24 7 3
>
> This is not a reflection, rotation, or colour reversal of the first
> solution, so there is no unique-modulo-RRCR solution.
>
> --
> Richard Heathfield <http://www.cpax.org.uk>
> Email: -http://www. +rjh@
> "Usenet is a strange place" - dmr 29 July 1999
> Sig line vacant - apply within
From: Carl G. on 11 Jun 2010 16:53

"Leroy Quet" <qqquet(a)mindspring.com> wrote in message
news:f6f2728d-cfb6-44bf-9a4a-b1401866a4d1(a)d37g2000yqm.googlegroups.com...
> Way to go! These two solutions are different than Rob Pratt's. So
> there are at least 3 solutions modulo rotations, reflections, and
> color reversals.
>
> I bet there are a lot of such solutions.
>
> Thanks,
> Leroy Quet
>

I pleased to know that my intuition was way off!

Next Challenge: Is there a solution with the products 1 to 9 all in the
row-products?

I noticed that both diagonal products of Richard Heathfield's solution are
3. What other matching diagonal products are possible?

Carl G.



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