
The 17x17 problem solved - DanielRibeiro
http://blog.computationalcomplexity.org/2012/02/17x17-problem-solved-also-18x18.html
======
jameskilton
For those of us who are confused what "monochromatic rectangles" actually
means, this article shows it perfectly:

[http://mathlesstraveled.com/2012/02/09/17x17-4-coloring-
with...](http://mathlesstraveled.com/2012/02/09/17x17-4-coloring-with-no-
monochromatic-rectangles/)

~~~
guard-of-terra
Do I understand correctly that you can shuffle the image's rows (or columns)
without losing that effect?

That would dramatically reduce the amount of computations we need to perform
in order to find a match.

Basically we are looking for a system of two sets of sets of 17-bit numbers.

First set of sets is comprised of all sets of 17-bit numbers which when ANDed
to each other yield a number with no more than one bit set.

Second set of sets is comprised of all sets of 17-bit numbers which do not
have any common bits set.

Then from those sets we should draw a 17 4-tuples of 17-bit numbers where:

every tuple is a subset of one of the sets in the second set

every set of i-elements of tuples is a subset of one of the sets on the first
set.

Sounds trivially computable on modern hardware to me! I'm sure that I'm wrong,
but where?

~~~
philh
> Do I understand correctly that you can shuffle the image's rows (or columns)
> without losing that effect?

Yes, but a dramatic reduction in complexity still leaves a lot of complexity.

Your conditions seem to be a lot simpler than they originally appear. If I
interpret you correctly, we're drawing seventeen 4-tuples of seventeen-bit
numbers such that

> every tuple is a subset of one of the sets in the second set

"the elements of each tuple have no bits in common", and

> every drawn number is contained in one of sets in the first set

"each number has at most one bit set".

But there are only eighteen seventeen-bit numbers with at most one bit set, or
seventeen if you exclude 0 (which seems reasonable). So isomorphically, we're
trying to find seventeen 4-tuples of distinct numbers less than seventeen.

This is indeed easy (there are (17 choose 4) = 2380 of them), but I don't
understand what you hope to do with these. At any rate, I don't think they
particularly help solve the problem at hand.

~~~
guard-of-terra
Nope, the second condition would be "each i-number is from a set where no two
numbers share more than one common bit"

You can pre-compute that set of sets (of course throwing away numbers with
just 0, 1 or 2 bits set because we suspect they would slow us down without
delivering) fit them in memory and then crunch the problem like a bug.

I've thrown together two simple scripts (in perl) which show that the known
solution fits both of my rules. Hopefully I can write a program that would
find solutions for any given NxM (or tell that no solution exists) tomorrow or
something like that.

Still I suspect the solution would hit the wall because e.g. sets would be too
big therefore the program would take actual ages to run, but one can surely
hope?

~~~
philh
Oh, I read AND but was doing OR. My mistake.

> Still I suspect the solution would hit the wall because e.g. sets would be
> too big therefore the program would take actual ages to run, but one can
> surely hope?

I suspect _something_ will go wrong, and this seems to me the obvious place
for it to fail. But I expect it to fail mostly based on "if it was that easy,
someone would probably have noticed", not because I have any particular
insight into the problem.

~~~
guard-of-terra
I have a same gut feeling, but one of my professors told us how the
discoveries are being made:

Everybody know that some thing is impossible to do. But one person who didn't
know that accidentally does it.

Right now I have one type-1 set having 36 numbers in it. The first numbers are
7, 25 and 42. I like it.

------
ColinDabritz
"I asked them if they did it for the money (which I doubted). No, but the
money made them more aware of the problem."

"The only grid that we do not know if it is 4-colorable is 12x21. This is
still open and you are URGED to work on it. Sorry, no cash on this one. Do it
for the glory!"

This strikes me as an excellent opportunity to use KickStarter to amplify
interest. If the people interested in 12x21 kicked in even $10 each, you could
have a noticeable prize. Perhaps this would be a good way to focus attention
on certain problems?

~~~
cgag
I was thinking the same thing. I love the idea of using something like kick
starter to publicly fund prizes for math / science, not necessarily related to
4-color rectangles.

~~~
ColinDabritz
You'd have to be careful about it, and have clear rules about who holds it and
when the reward is given out. You could conceivably fund research that way as
well, such as things that big business or the government shy away from, but
has popular demand or value. It sounds approachable.

------
experiment0
Best way of finding if there are rectangles:

> We encode the rows of the grid in a set of bit vectors–four vectors for each
> row, representing the four possible colors. For example, the red vector for
> a row has a 1 at each position where the corresponding node is red, and
> zeroes elsewhere. The blue vector has 1s for blue nodes, and so forth. Now
> we can detect a rectangle merely by taking the logical AND of two rows (an
> operation that could be a single machine instruction). A rectangle exists if
> and only if at least two bits are set in the resulting vector.

\-- <http://bit-player.org/2009/the-17x17-challenge>

Thats some lovely hacking.

------
endtime
>The only grid that we do not know if it is 4-colorable is 12x21. This is
still open and you are URGED to work on it.

Is this a typo for 21x21? If not, then presumably 21x21 is known to be
4-colorable...and it seem to me that removing a column from the edge* of a
4-colored grid results in another 4-colored grid (since the resulting subgrid
couldn't have contained any MRs if the original grid didn't)...so a 12x21
4-colored grid should result from removing nine columns from a 21x21 such
grid.

I don't actually think the above solves the problem - rather, I must be
misunderstanding it. Can someone explain how?

*The column doesn't have to be from the edge, but it's a sufficient claim, and is easier to see...I couldn't be bothered to write the small proof that removing a column from the middle works.

~~~
panic
Your argument only works if the 21x21 grid is known to be 4-colorable. But in
fact, the 21x21 grid is known _not_ to be 4-colorable.

------
ntkachov
Aside from being just interesting, Does this have and applications? It seems
like it might have some applications in data structuring but is this just a
math problem or are there actually people who are waiting on the solution?

~~~
kylemaxwell
I don't know if anyone has been waiting on this specific result for an
application, but in general graph coloring has lots of applications[1]. Also,
mathematical advances often predate their applications (sometimes by a very
long time). Not to mention that we know something know we didn't know before,
and that's _always_ worthwhile.

[1]: <http://en.wikipedia.org/wiki/Graph_coloring#Applications>

------
sp332
Lots of discussion from when the challenge was issued a couple years back:
<https://news.ycombinator.com/item?id=968577>

------
pavel_lishin
I'd really like to hear more about why this was not a trivial challenge.

~~~
cheald
Short answer: 4^(17*17) combinations to try, and each combination requires an
expensive test before you can pass/fail it. Plug that number into Wolfram
Alpha alongside things like "age of the universe" for some perspective on just
how massive the search space is.

~~~
rorrr
The search space a lot smaller, since you can cut many incorrect branches very
early.

~~~
cheald
Yes, but as noted in the post mvzink linked, you can't cut it to the point
that an exhaustive search is feasible.

------
akawry
Tried to apply a genetic algorithm to arrive at a solution:
<https://gist.github.com/1820794>.

Doesn't work too well just yet; maybe someone can take a look/play around with
it/give me some pointers :)

------
kcl
Wrote this up in response to the renewed attention to the problem:

17x17 is NP-Complete: <http://hackerne.ws/item?id=3584732>

------
sebastianavina
can someone explain me why this problem is important?

or it just happens to be another problem to be solved?

