wrote:

You may order set A := {1 ... 8} any way you
want to but two different points (a a) and (b b)
of AxA will **never** be adjacent. They will be
always of the same (black) color.

Perhaps I expressed myself badly. I don't care
how you order the original set of A = [1..8], or
how you order AxA, but only about the method by which
you label the squares of the chessboard with the
labels (x y).

I quote: "identify chess squares with points (a b)
which have coordinates from the set 1 2 3 4 5 6 7 8."
That is, you leave it entirely to me to choose the
method: you only tell me what labels I may use.

So I choose to label the chess square A1 with (1 1),
and square A2 with (8 8). This is not against
the specification of your proof, yet it produces a
situation that causes the proof to fail. I win.

I quote again: "two different points (a a) and (b b)
of AxA will **never** be adjacent." yet without telling
me how the points are mapped to the board squares.
(I don't see you have improved the proof in any way
by specifying AxA -- the mapping function is still
missing. So I choose a mapping that is in my favour,
and again I win.

It's not my job as 'proof reader' to help you:
once I understand the proof mechanism, my job is to
find holes in the proof. If I find just one, the proof
is broken. Not irretrievably, though, only formally.
Specify the mapping function, and you seem to be on
dry ground.

But this is where my original question came in:
once the mapping function is specified, the similarities
between this proof and the original one are so large, that
I suspect the are the same, only differing as to
what the different label notations allow you to do.

--
Anders Thulin ath*algonet.se http://www.algonet.se/~ath

Michael wrote:
) David Richerby schrieb:
) [Ecellent proof snipped]
) Note that it is not guaranteed that a chessboard with two squares of each
) colour removed can be tiled with dominoes. Removing the two black squares
) closest to a white corner leaves a 1x1 region that cannot be covered.
)
) A chessboard with 2 squares of the same color removed can NEVER be
) covered.
)
) You have guaranteed by your proof that a chessboard with side length 2n
) can ALWAYS be covered when two tiles of opposite color are removed. In
) your example, you removed two tiles of the same color, hence your proof
) does not apply.

Read the first sentence you quoted again. Especially the bit between
'with' and 'removed'.


P.S. There's still the ObPuzzle: Can you find a chessboard with four
squares removed, two of each colour, but such that it is still in
one piece, that can not be tiled with dominoes ?


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

On 2004-12-20, Bill Smythe wrote:
"Prai Jei" wrote:
Now can you prove that if *any* two squares of opposite colour are removed
from the chessboard, the remaining squares can always be covered with 31
dominoes?

Find a way of walking through all the squares, so that you only take
single-square steps forwards, backwards, or sideways, and so that you
end up next to the square where you started. (That is: Find a
Hamiltonian cycle in the grid graph.)

Now you have a 64-square cycle where the colors alternate. If you remove
two squares of different color, you break the cycle into two bits.
The endpoints of a bit must be of a different color, since they are
adjacent to the squares that were removed, and the removed squares were
of different color. Therefore each bit has even length and can clearly
be covered by dominoes.

Harri H.

I believe that every proof necessarily relies on the parity proof in a
fundamental, unavoidable way, and the answer to the OP's original question
is a
resounding 'No can do'.

In other words you can write a proof that *seems* like it doesn't rely on
that
proof, but essentially it does.

I did wonder this when I posted originally. My very vague (mis)understanding
is that Wittgenstein said something along the lines that all mathematics is
essentially tautological (this occurred to me when I was doing my Maths
A-Level. I just couldn't see the external point of reference - Maths seemed
ultimately to be a language that talked about itself) and I wondered if the
equivalence of apparently different proofs of this problem was one
particular manifestation of that tautology. But then there are apparently
several hundred different proofs of Pythagoras. Who arbitrates on why one
proof is different from another I'm not sure.

cheers

dd

Harri Haanpaa wrote:
Now you have a 64-square cycle where the colors alternate. If you remove
two squares of different color, you break the cycle into two bits.
The endpoints of a bit must be of a different color, since they are
adjacent to the squares that were removed, and the removed squares were
of different color. Therefore each bit has even length and can clearly
be covered by dominoes.

That's a much better proof than mine. :-)


Dave.

--
David Richerby Erotic Watch (TM): it's like a
www.chiark.greenend.org.uk/~davidr/ precision chronometer but it's
genuinely erotic!

David Richerby about the proof posted
by Harri Haanpaa:

That's a much better proof than mine. :-)

To mathematicians such a (nice) proof,
based on a hamiltonian cycle is a reflex.
Also, it was published years ago in a book
on .. on ... on polyminos(sp??).

Now one may consider boards with even number
of squares, from which an even number (perhaps
smaller :-) of squares was removed without
disconnecting the board. To avoid too easy
impossible cases, due to a bottle neck, one
may assume that the removed set is sparse.
(One would consider boards of different
dimensions).

Best regards,

Wlod

Wlodzimierz Holsztynski (Wlod) wrote:
polyminos(sp??).

``Polyominoes''.


Dave.

--
David Richerby Slimy Goldfish (TM): it's like a fish
www.chiark.greenend.org.uk/~davidr/ but it's covered in goo!

David Richerby has rescued me:

polyminos(sp??).

``Polyominoes''.

Dave.

I see, the author has replaced "d"
in "dominoe" by "poly". Thank you,
Dave, see you,

-- Wlod

Anders Thulin:

I win.

If you are married I commisarate
your mother in law :-)

Regards,

Wlod