
Revisiting the Mutilated Chessboard - ColinWright
http://www.solipsys.co.uk/new/TheMutilatedChessboardRevisited.html?HN_20171126
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tzs
> Proof: Consider a tour that visits every square exactly once ...

This would be clearer if the word "closed" were inserted before "tour".

~~~
ColinWright
That's a useful comment - thanks - that will go up in the next 20 minutes or
so.

Cheers.

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Mz
_because now it would now take 31.5 dominos._

Really minor nit: You could drop one of your _nows._

Tangentially, my sons told me that when they start card tricks with an
unopened pack of cards and strongly highlight that fact like it proves the
cards have not been tampered with, the trick there is that card decks all
start in the same order. So an unopened deck will come with a particular card
order and they can know the order of the cards and can use that fact to help
them with their card trick. It is part of how they fool the audience.

~~~
ColinWright
>> _... because now it would now take 31.5 dominos._

> _Really minor nit: You could drop one of your nows._

Good catch - change in the process of being uploaded.

WRT cards in an unopened deck, there are two different orderings that are
used. Sometimes there are "kissing kings" and sometimes not. If you always use
the same make of cards, probably the order is always the same.

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barrkel
For dealing out the cards, it seems that the only thing that would stop you
from drawing a straight via one card from each pile would be if any one pile
had one card in all 4 suits, along with one of its adjacent cards in the
straight; e.g. 4 of hearts, clubs, diamonds and spades, and any 5 or 3. (If
you pluck off all the same-suited cards first, then you'll end up with two
adjacent cards for the final draw in the same pile). Since the piles are
constrained to have no more than 4 cards, this situation can't happen.

~~~
ColinWright
That's very much an argument along the lines of: "I can't imagine anything
going wrong, so it must be OK."

What you say is true, but on its own it's not enough to be convincing.

~~~
mikekchar
It would be helpful to explain _why_ the answer is not convincing. After all,
you challenged in your article to find a proof. Replying with "Nope, that's
not it" is not really helpful.

In this case, if we consider the situation where we take a card from a pile,
this leaves 3 cards left in the pile. We can't take another card from this
pile this round, but there are only 3 cards left which guarantees that there
is still one card available in another pile. This much is correct, but as you
say it is not enough to prove the issue.

Once we take a card from one pile, three other cards are not available to draw
this round. When we take another card, three more cards are not available. How
do we know that there were not 2 of one card in the first pile and 2 of
another card in the second pile (or some other annoying combination)? Indeed,
this situation _will_ exist. In order to prove the original assertion, you
have to prove that there will _always_ be another solution that will allow us
to avoid doing that. That's much harder.

~~~
ColinWright
This isn't the right place to have this discussion because it's going to end
up long, threaded, context ambiguous, and generally frustrating for all
concerned. However, this is the _only_ place we have, so I'll try to address
your points here.

But let me begin by saying that it's true that a straight can always be drawn,
and showing that someone's false argument of a true statement is indeed false
is hard, because you can't give counter-examples. Even so, let's look at
what's been said:

> barrkel: _For dealing out the cards, it seems that the only thing that would
> stop you from drawing a straight via one card from each pile would be if any
> one pile had one card in all 4 suits, along with one of its adjacent cards
> in the straight; e.g. 4 of hearts, clubs, diamonds and spades, and any 5 or
> 3. (If you pluck off all the same-suited cards first, then you 'll end up
> with two adjacent cards for the final draw in the same pile). Since the
> piles are constrained to have no more than 4 cards, this situation can't
> happen._

> ColinWright: _That 's very much an argument along the lines of: "I can't
> imagine anything going wrong, so it must be OK." What you say is true, but
> on its own it's not enough to be convincing._

So here barrkel is describing the only way it can go wrong. _Why_ is this the
only way it can go wrong? What if I choose a card from each pile bar one, and
then find that the remaining pile doesn't have any of the final card I'm
looking for? Why can't that happen? Well, it _can_ happen, and I need to fix
my previous choices. How do I know I can fix my previous choices? Why can't it
be the case that my first 12 choices _always_ leave me in this predicament?
[X]

In short, the situation described is only one of the gazillion things that
might go wrong, and there is no discussion of why they can't happen. _That 's_
why it's not convincing.

Then, to go on with your comment:

> _if we consider the situation where we take a card from a pile, this leaves
> 3 cards left in the pile._

Yes.

> _We can 't take another card from this pile this round, but there are only 3
> cards left which guarantees that there is still one card available in
> another pile._

I don't understand at all what you're trying to say here. What is available in
the other piles? (Based on text later in your comment I suspect this doesn't
matter.)

 _Added in edit:_ Ah - I think you're saying that for any card we want to pick
next it must be available somewhere, because there are four of them, and only
three can be in the now-forbidden pile. OK.

> _This much is correct, but as you say it is not enough to prove the issue._

> _Once we take a card from one pile, three other cards are not available to
> draw this round._

Yes.

> _When we take another card, three more cards are not available._

Yes.

> _How do we know that there were not 2 of one card in the first pile and 2 of
> another card in the second pile (or some other annoying combination)?
> Indeed, this situation will exist. In order to prove the original assertion,
> you have to prove that there will always be another solution that will allow
> us to avoid doing that. That 's much harder._

Yes, exactly.

So you have outlined one way that it can all go wrong that isn't covered by
the original argument put forth by barrkel. I hope you agree that my comment
[X] above also provides enough explanation of why barrkel's argument is
unconvincing.

~~~
mikekchar
I felt bad about posting the above about 7 hours after I posted it, but
couldn't edit it :-P. I should have stayed out of the discussion because I
don't think I helped. But basically what I was trying to say was that I don't
think your response was clear enough -- in fact I had to think about it quite
a bit before I understood what you were trying to get at. At the time I felt
that it probably would have been better not to say anything at all than to
leave a reply that was difficult to understand. I think the OP left feeling
bad about the situation when they could have left thinking, "I learned
something today". I hope you can appreciate the irony in typing that, given
that I did the same thing!

Anyway, I enjoyed your blog post and it gave me lots to think about. Thanks
for doing it.

~~~
ColinWright
I'm glad you did post it - your comment made me think about it more carefully.
In math circles my originally reply would have been considered overly verbose,
and when I'm talking about math I tend to drop into that style of
communication. I needed to kick to get me out of it and explain in more
detail.

But this is a standard problem when teaching proof in math. All too often a
proposed "proof" simply doesn't prove the theorem, and if the theorem is
actually true it can be extremely difficult to explain why the "proof" is
wrong, inadequate, or incomplete. The most common form is "Well, I can't see
how anything could go wrong." and that's notoriously hard to get around.

But I'm pleased you posted, and I'm pleased you enjoyed the post. Thank you.

