
Mathematician Disproves Hedetniemi’s Graph Theory Conjecture - headalgorithm
https://www.quantamagazine.org/mathematician-disproves-hedetniemis-graph-theory-conjecture-20190617/
======
FabHK
Beautiful article, though not sure why two examples are introduced (students x
instruments, and jobs x hobbies).

The abstract for Shitov's paper is also a gem of brevity, here reproduced in
its entirety:

"The chromatic number of G×H can be smaller than the minimum of the chromatic
numbers of finite simple graphs G and H."

(And, of course, the first reference in the paper is co-authored by Erdős.)

~~~
pmiller2
I’ll take 2 excellent examples over a definition and one opaque and
unmotivated example any day (which is what you get in a lot of papers).

~~~
edjrage
The text and the examples are good enough, but I'm sure just one example
accompanied by some nice illustrations would be much more immediately
illuminating.

------
MrXOR
My lesson for today:

"Sometimes, the reason that a conjecture is very hard to prove is simply that
it is false."

Worth to see:

[https://gilkalai.wordpress.com/2019/05/10/sansation-in-
the-m...](https://gilkalai.wordpress.com/2019/05/10/sansation-in-the-morning-
news-yaroslav-shitov-counterexamples-to-hedetniemis-conjecture/)

Congratulations to Yaroslav Shitov!

------
glitcher
> Network coloring problems, which were inspired by the question of how to
> color maps so that adjoining countries are different colors, ...

> Do four colors suffice to color any map? — took more than a century to
> answer (the answer is yes, in case you were wondering).

I know very little about graphs and found this bit surprising. After some
searching I found out it is called the Four Color Theorem.

Does anyone know of a resource that provides a more intuitive explanation for
this for non-graph-theorists?

~~~
hawkice
It required a notoriously hard-to-simplify proof, which required computer
assisted brute force analysis. But an intuition is that a complete graph with
four nodes is planar, and five is not. A (3,3) bipartite graph is also the
smallest non-planar bipartite graph. That's probably good for answering "why
would you suspect 4".

~~~
jessriedel
To save laymen a google: A complete graph is one where all the nodes are
connected. A planar graph is one that can be draw in the plane without two
lines intersecting.

~~~
rocqua
Notably one you can draw on either a flat piece of paper or a sphere. On a
torus, you need up to 7 colors.

~~~
pmiller2
Yep, and the proof that any graph that can be drawn without crossings on the
torus can be properly colored with no more than 7 colors is vastly easier than
the case for graphs drawn on the sphere/plane.

------
firstinstinct
Long time ago I failed to solve this conjecture. Nice to see it have been
solved. I shared some email with S. Hetdetniemi about some special cases. Good
old times.

~~~
pmiller2
What results and special cases did you work on?

~~~
firstinstinct
I cannot remember. I emailed him about some results, he said they were
interesting, but later I discovered those results were already published. I
told him : I will throw them to trash. Very impolite from my part!, that must
be 20 years ago.

------
pmiller2
This is absolutely huge. Steve Hetdetniemi is one of the giants of the field
(also a great guy, to boot). If he agrees the result is true, I believe it.
Now, off to read the paper.... :)

~~~
avaku
He commented (in the article) that he is delighted the resolution has been
found and that "it can be explained in two sentences".

~~~
C4stor
The construction can be explained in two sentences, the argument proving this
is a proper counter example not exactly I think !

~~~
_delirium
Although not that much more even for the proof... the paper is 2 1/2 pages,
and almost half of that is introduction and references.

------
urmish
Looks like this is the paper:

COUNTEREXAMPLES TO HEDETNIEMI’S CONJECTURE

[https://arxiv.org/pdf/1905.02167.pdf](https://arxiv.org/pdf/1905.02167.pdf)

~~~
svat
This is linked in the first sentence of the article, FYI. (Though I guess it's
easy to miss regardless…)

------
hammock
Tangential: I don't know anything about graph theory but I'm intrigued by the
student-instrument duet example.

Is there a way to include probability/fuzz into the tensor products? In other
words, how do you account for non-binary strength of connections between
nodes? Like I'm 80% good at piano but 50% good at oboe.

~~~
ky3
> how do you account for non-binary strength of connections between nodes

The adjacency matrix of a graph normally has entries only 0 or 1.

[https://en.wikipedia.org/wiki/Adjacency_matrix](https://en.wikipedia.org/wiki/Adjacency_matrix)

Generalize the adjacency matrices to include values _between_ 0 and 1.
Typically, you'd impose some normalization condition like rows and/or columns
sum to 1.

Then compute the Kronecker product of matrices to obtain the graph tensor
product.

[https://en.wikipedia.org/wiki/Kronecker_product](https://en.wikipedia.org/wiki/Kronecker_product)

------
Cogito
Is it true to say that a graph can be coloured by _k_ colours if and only if
the largest fully connected subgraph has _k_ or fewer members?

For example, is saying a planar graph can always be coloured by 4 colours the
same as saying a planar graph can never have more than 4 nodes that are all
connected to each other?

~~~
OscarCunningham
They're not quite the same. Certainly if there are five vertices all connected
to each other then the graph can't be four-coloured. But the converse isn't
true.

[https://standardwisdom.com/softwarejournal/2012/07/a-graph-t...](https://standardwisdom.com/softwarejournal/2012/07/a-graph-
that-requires-5-colors-but-does-not-contain-k5-complete-graph-on-5-vertices/)

~~~
Cogito
Thanks, that's a really great example. It definitely feels non trivial to come
up with an example.

------
2bitencryption
I've heard vaguely of the coloring problems before, but this one quote is
confusing me, can someone explain what I'm missing?

> Even the question that launched the field — Do four colors suffice to color
> any map? — took more than a century to answer (the answer is yes, in case
> you were wondering).

But what if my "map" includes a node that has more than four neighbors? If
there are only four colors, then one of the neighbors must be pidgeon-holed to
share a color with the node? Are there constraints on how many neighbors a
node can have?

~~~
meuk
The constraint is not on the number of neighbors directly. If there is just
one node (say, A) with 4 neighbors (say B1, B2, B3, B4), we can color A red
and B1, B2, B3, B4 blue. It only becomes a problem when all nodes are
interconnected (i.e. the graph is complete). So, the result does imply that
graphs which are not colorable with 4 colors are not planar. In particular,
the complete graph (where all nodes of the graph are connected) with 4 nodes
is not planar.

~~~
thaumasiotes
The complete graph with 4 nodes is planar, and obviously can be colored with
only 4 colors -- that gives every node a unique color.

The complete graph with 5 nodes is what isn't planar. (That, and the complete
bipartite graph with 3 nodes on each side.)

------
dylanfw
Could somebody clear up my confusion with these two statements that appear
contradictory?

"Do four colors suffice to color any map? — took more than a century to answer
(the answer is yes, in case you were wondering)."

and

"Returning our attention to colorings in which connected nodes are supposed to
be different colors, we have no guarantee that the five colors in our palette
will be sufficient to color the graph G"

How can it be that 4 colors is sufficient for any graph, but for our
hypothetical graph G we can't be sure that 5 colors are sufficient?

~~~
philipfweiss
Four colors is sufficient for a _planar_ graph, but insufficient for a general
graph. A planar graph is any graph where the edges can be drawn in such a way
that they only intersect at the endpoints (there is a more formal algebraic
definition but that helps with the intuition).

~~~
dylanfw
Thank you! That makes complete sense when I consider the 4 color example of a
world map. I didn't even consider that most graphs would have "overlapping
borders" on such a map.

An extra thank you for apparently creating a new account to assist me. (I
think that's what the green username indicates, at least).

------
thesz
From the article: "Now that mathematicians know Hedetniemi’s conjecture is
false, the natural next question is just how false it is..."

And from Stanislaw Lem works (Cyberiad): "Everyone knows that dragons don’t
exist. But while this simplistic formulation may satisfy the layman, it does
not suffice for the scientific mind."

Time to read some Lem again?.. ;)

