

The 17x17 challenge. "Worth $289.00. This is not a joke." - sp332
http://blog.computationalcomplexity.org/2009/11/17x17-challenge-worth-28900-this-is-not.html

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RiderOfGiraffes
To summarise the numbers given elsewhere ...

The number of possible 4-colorings of a 17x17 grid is 4^289 which is about
10^174, but that doesn't take into account the permutations of colors (4!) and
possible symmetries.

However, since almost all colorings have no symmetry that's largely
irrelevant, so the number of colorings of the grid (up to permutations of
colors) is about 4^289/24 which is about 4.122e172.

Within the 17x17 grid we must avoid X-Y-aligned rectangles with corners all
the same color. The number of such rectangles is obtained by choosing two rows
and two columns, and so is (17choose2)x(17choose2) = 23409.

~~~
Retric
This is also more brute forcible than 4^289 might suggest. Consider a 2 x 2
grid. The first square does not count, there are 2 options for the second
square, and 3 for the 3rd. But there is only 4 options for the fourth square
if the first 3 squares where not the same color. Resulting in (2 * 3 -1) * 4 +
3 = 23 options not 4^ 4 = 256.

~~~
RiderOfGiraffes
Although this succeeds in eliminating 90% of the search space for a 2x2 grid
(and even then I think your sums are slightly wrong) this doesn't even dent
the 17x17 case. The 2x2 square in the top left of that only has 23 (or fewer)
configurations, but now you can probably no longer permute the colors. The
savings are irrelevant in the larger cases.

Besides, to some extent you are preempting the problem. You've counted the
number of colorings of the 2x2 grid and found there to be only a small number.
The evidence suggests that the number of colorings of the 17x17 grid is either
1 or 0 (up to permutation of colors), but that doesn't mean the search is
trivial.

I may not have expressed that clearly, but I hope I've conveyed the point.

~~~
Retric
There is something like 12, valid 2x2 grids, and you can get rid of almost
another (17!)^2 options because rows and columns are interchangeable. Actually
thinking about it a little bit the only valid option for the top left square
is:

    
    
      0,0
      0,1
    

However, the vast majority of the grids are eliminated by the boxing
constriant before the grid get's all that large.

------
ars
Is the problem basically this game?
<http://www.kongregate.com/games/Mygame/blocky>

i.e. find a grid where you can not make any moves in the game?

~~~
dkasper
Yes

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PieSquared
I know the professor who writes this blog, Bill Gasarch. I was thinking of
interning with him (at our high school, we do an internship + report about it
in our senior year), but then decided against it because the math is a bit too
abstract and unapplicable to (as Gasarch would put it) make me "properly
enthused". He seems like a great guy. Anyway, while I haven't read the entire
thing, I would suggest reading the book he has online - it's quite
interesting, and it goes into more detail about colorings and such, so anyone
who likes this post will find the book interesting as well.

~~~
nandemo
> but then decided against it because the math is a bit too abstract and
> unapplicable

Combinatorics isn't really abstract as far as math goes. In fact some call it
_concrete mathematics_.

~~~
PieSquared
Combinatorics isn't, but... coloring grids of squares?

Not that I really have anything against abstract mathematics. It's still very
cool, but it's just not going to get me very excited - or at least this exact
branch of it, I guess.

~~~
drbaskin
I'm not one to say that coloring grids of squares sounds fascinating to me,
but nandemo's point still holds -- as far as mathematics goes, it's pretty
concrete.

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gort
I'm not the most mathy person in the world; can someone explain to me the
following statement:

"Exactly one of the following sets is in OBS4: 19x17, 18x17, 17x17."

If I've understood it correctly, OBS4 is the set of grid-sizes which are
impossible. But if 17x17 were impossible, surely the bigger ones would be too?

[Edit to answer my own question: the OBS4 set seems to be made up solely of
the smallest such grid-sizes that can be given without being redundant.]

~~~
jules
So if they say "Exactly one of a,b,c is in OBS4" then one of a,b,c is not
4-colorable?

~~~
jrp
OBS4 is supposed to be a list of all configurations which deny 4-colorability
for themselves, and any strictly larger (in both dimensions) configurations.

Ie, if (19,17) is in OBS4 then (19+x,17+y) is not 4-colorable either. So given
a shape (x,y) or even a more funky collection of grid points like a triangle,
to know if it's 4-colorable just check for all X in OBS4, if you can fit X
inside your shape.

~~~
jules
Got it, although the more interesting property is that smaller shapes than
those in OBS _are_ 4-colorable? (because it is trivially true that if nxk
cannot be colored then larger shapes can't either).

~~~
jrp
Right, he has apparently solved and proved a short list for OBS3 with the
property that X is 3-colorable if and only if X doesn't contain anything from
OBS3.

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rauljara
Took me a second to realize why 289. 17 * 17. One dollar per square. If you
could get him to offer a prize for a 1000 * 1000 grid, you have quite a nice
reward.

~~~
RiderOfGiraffes
You're not being serious, I know, but I'll answer seriously anyway. 1000x1000
is not an edge case - it can't be done - so offering a reward for a solution
is pretty pointless.

~~~
brent
A related and unsolved question (and thus legitimate) could be: what is the
smallest c such that 1000x1000 is c-colorable.

~~~
RiderOfGiraffes
The problem with that is knowing that you have the minimum. The good thing
about the question as asked is that it's asking for something specific and not
a proof of impossibility. Your question is legitimate and interesting, but
requires a proof of minimality.

That's difficult.

~~~
klipt
A different question would be "who can come up with the smallest c in 6
months", which probably wouldn't yield the true answer, but for a $1M prize it
might come quite close (and could lead to some interesting theoretical
advancements).

------
charlesju
Darn it. I thought this was a huge In & Out burger or something. I guess a
math game is more logical and fitting for this forum.

------
patrickgzill
Given the difficulty, should he not be offering more?

~~~
cellis
I think the cash is more symbolic than anything.

See: <http://en.wikipedia.org/wiki/Knuth_reward_check>

------
bshep
This was proven in 1976 - <http://en.wikipedia.org/wiki/Four_color_theorem>

~~~
Dove
The problem isn't what you think it is, otherwise the solution would be
trivial: alternate one pair of colors on even rows, and then alternate another
pair on odd rows. That satisfies the four color theorem.

This problem differs in two ways.

First, we're not looking for all of the vertices to be different colors. We're
looking for them to not all be the same color. This is easier to do.

Second, he's not trying to color a simple planar graph. He means _every_
rectangle on the grid -- for example, a 7 x 8 rectangle in the middle
somewhere. This is much, much harder to do.

(To be fair, I made the same mistake on a first reading of the problem.)

~~~
bshep
I see now! I completely missed this part: "Second, he's not trying to color a
simple planar graph. He means every rectangle on the grid -- for example, a 7
x 8 rectangle in the middle somewhere. This is much, much harder to do."

Otherwise the problem seemed trivial.

On the other hand... how many possible 17x17 4-colored grids are there? Can
this be brute forced?

~~~
cperciva
_how many possible 17x17 4-colored grids are there?_

There's 289 points on the grid, and 4 options for each. So... 2^578 options.

 _Can this be brute forced?_

That depends. Do you have a kilobit quantum computer available?

~~~
RiderOfGiraffes
Um, no. The rectangles are aligned with the X and Y axes, so you choose two
rows and two columns, giving 17C2x17C2 rectangles in total: 23409 possible
rectangles.

~~~
cperciva
He asked how many grids there were, not how many rectangles there were. :-)

~~~
RiderOfGiraffes
Ah - sorry, my turn to do a misreading. As you point out, I computed the
number of rectangles that aren't permitted to have all its vertices the same
color.

My bad - sorry.

