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Hedetniemi’s Conjecture in graph theory disproved

In Quanta Erica Klarreich recently wrote up Yaroslav Shitov’s new counter example which disproves Stephen Hedetniemi’s 50 year old conjecture, original dissertation, that the number of colors required to color the tensor product of two graphs is the lesser of the numbers used to color the original graphs. These colorings have applications in areas from scheduling to seating plans, and it is clear from Klarreich’s reporting that mathematicians are excited about this result. In fact, Hedetniemi responded very positively when asked by Klarreich about the counter example, saying it “has a certain elegance, simplicity and definitive quality to it.” The counter-example may show Hedetniemi’s conjecture is not true, but Klarreich points out that we do not yet know just how false it is. So, while Shitov has closed one door on this problem, there are still many which are open.

via Thomas Lin on Twitter.

LaTeX/TikZ to draw a star graph $K_{1,n}$

For a diagram for a class this week, I’ve written a LaTeX command to draw star graphs using TikZ. A star graph $K_{1,n}$ is a graph with a single central node, $n$ radial nodes, and $n$ edges connecting the central node to each radial node. I am sharing this here in case it is useful to anyone else.

The chromatic number of the plane is at least 5

A long-standing mathematical problem has had a recent breakthrough – scientist Aubrey de Grey has proved that the chromatic number of the plane is at least 5.

Timetabling choreography with maths

Earlier this week my sister-in-law (“SIL” from now on) sent me an email asking for help. She’s a dance teacher, and her class need to rehearse their group pieces before their exam. She’d been trying to work out how to timetable the groups’ rehearsals, and couldn’t make it all fit together. So of course, she asked her friendly neighbourhood mathmo for help.

My initial reply was cheery and optimistic. It’s always good to let people think you know what you’re doing: much like one of Evel Knievel’s stunts, it makes you look even better on the occasions you succeed.

I’d half-remembered Katie’s friend’s Dad’s golf tournament problem and made a guess about the root of the difficulty she was having, but on closer inspection it wasn’t quite the same. I’m going to try to recount the process of coming up with an answer as it happened, with wrong turns and half-baked ideas included.

This video about the proof of the Kelmans-Seymour conjecture is adorable

Theorem: every 5-connected non-planar graph contains a subdivision of $K_5$.

The above statement, conjectured independently by Alexander Kelmans and Paul Seymour in the 70s, is very easy to say. And the video below, starring Dawei He, Yan Wang, and Xingxing Yu, makes it look very easy to prove:

It’s like they got Wes Anderson to film an academic PR video. In the normally uninspiring world of maths press releases, it’s quite refreshing. And the written press release is pretty snappy, too. Let’s not make this a thing, though.

Steinberg’s conjecture is false

Conjecture   Every planar graph without 4-cycles and 5-cycles is 3-colourable.


In a paper just uploaded to the arXivVincent Cohen-Addad, Michael Hebdige, Daniel Kral, Zhentao Li and Esteban Salgado show the construction of a graph with no cycles of length 4 or 5 which isn’t 3-colourable: it isn’t possible to assign colours to its vertices so that no pair of adjacent vertices have the same colour, using only three different colours. This is a counterexample to a conjecture of Richard Steinberg from 1976.

The problem was listed in the Open Problem Garden as of “outstanding” importance.

Read the paper: Steinberg’s conjecture is false

via Parcly Taxel on Twitter