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From: Leroy Quet on 11 Jun 2010 08:12
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
From: Leroy Quet on 11 Jun 2010 11:40
Leroy Quet wrote:
> 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.

For those of you who are confused about what I am asking, here is a
sample 9-by-9 grid with just 16 distinct values that occur. (14
products don't occur elsewhere.)
There seem to be a number of 16-unique-products solutions, perhaps.

1 = black, 0 = white (although it could be the other way around just
as easily).

1 0 1 0 0 0 1 1 0
1 0 0 1 1 1 1 1 0
0 1 1 1 0 0 0 0 0
1 0 0 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 1 1 0 0 0 1
0 0 0 0 0 0 0 1 1
0 0 0 1 1 0 0 1 1
1 1 1 0 0 0 1 1 1

The row products are:
1*1*1*3*2*1 = 6, 1*2*5*1 = 10, 1*3*5 = 15, continuing:
16, 9, 18, 14, 24, 27.

And the column products:
2*1*1*4*1 = 8, 2*1*5*1 = 10, 1*1*1*5*1 = 5, continuing:
3, 1, 7, 12, 24, 20

So, the products 2 and 4 don't occur in my partial "solution".

Thanks,
Leroy Quet

[ ( [ ([( [ ( ([[o0Oo0Ooo0Oo(0)oO0ooO0oO0o]]) ) ] )]) ] ) ]


From: Richard Heathfield on 11 Jun 2010 11:44
Leroy Quet wrote:
> 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.

With a little fooling around, I got as high as 16:

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

Collisions on 4 and 12. I don't see a way to improve on this, but that
might just be because my eyes are beginning to cross!

--
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 11:57


Richard Heathfield wrote:
> Leroy Quet wrote:
> > 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.
>
> With a little fooling around, I got as high as 16:
>
> o x o o o o o o o 7
> x o x x o o o o o 10
> o o o x o x o o o 9
> o o x x x x x o o 20
> x x o x x x x x x 12
> x o x o x x x o o 6
> o x o x x x x x o 5
> o o o o x x x o o 24
> x x x x o x x x x 16
> 8 3 2 4 15 14 18 4 12
>
> Collisions on 4 and 12. I don't see a way to improve on this, but that
> might just be because my eyes are beginning to cross!
>


There seem to be a number of ways to get 16 distinct values/ 14
products occurring once each. The trick, if it is possible(!), is to
get all 18 products unique.

Maybe I should "allow" computers to be used here.

And, of course, there are at least 16 ways to do this if there is 1,
because of reflections, rotations, and flipping all of the stones'
colors.

Thanks,
Leroy Quet

From: Richard Heathfield on 11 Jun 2010 12:47
Leroy Quet wrote:
>
> Richard Heathfield wrote:
>> Leroy Quet wrote:
>>> 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.
>> With a little fooling around, I got as high as 16:
>>
>> o x o o o o o o o 7
>> x o x x o o o o o 10
>> o o o x o x o o o 9
>> o o x x x x x o o 20
>> x x o x x x x x x 12
>> x o x o x x x o o 6
>> o x o x x x x x o 5
>> o o o o x x x o o 24
>> x x x x o x x x x 16
>> 8 3 2 4 15 14 18 4 12
>>
>> Collisions on 4 and 12. I don't see a way to improve on this, but that
>> might just be because my eyes are beginning to cross!
>>
>
>
> There seem to be a number of ways to get 16 distinct values/ 14
> products occurring once each. The trick, if it is possible(!), is to
> get all 18 products unique.

This is getting silly. I've found a 17 - just one collision.

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

Tantalisingly close, but no cigar.

--
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
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