Blah blah blah chromatic numbers of graphs.
This one isn’t too hard to prove. Given a -colouring of
, group the vertices by colour, then since the vertices with the same colour induce a null subgraph of
(correspondingly, a complete subgraph of
),
is at least the size of the largest group of vertices, which is necessarily at least
.
This follows from the above and the arithmetic-geometric mean inequality.
This isn’t too hard to prove with induction, but >induction. Proof later.
This follows from the above and the arithmetic-geometric mean inequality, again.
Back to trying to prove that. But then we’d need a digression first. This came up as an optional question for a graph theory assignment last year, and it was pretty much the killer question. I didn’t think of using induction then for whatever reason, and resorted to the power of the internet to get an answer. (Damn if I leave it blank.) I don’t actually recall much of the proof (neither did I understand much of it back then), but I recall finding it in some PDF file under a header ‘Some Slick Proofs’. I was like ‘screw it imma submit this’ (>me >actually doing homework), and so I did. And then Mr Lee Chan Lye (putting the full name here despite having done so for approximately one other teacher, because there’s a high chance I won’t remember anyone from school after a while) was like ‘you can also prove this using induction’. Story time over.
Lemma: Let be any graph with at least one vertex. Let
. Then
.
Proof: Clearly , since any
-colouring applied to
is still a valid colouring. By taking any
-colouring, and assigning
a new colour, a
-colouring for
is obtained, so
and
. ∎
Let be the proposition “
for all
such that
“.
is trivially true, since both
and
are 0. Suppose that
is true for some
. For the purposes of coming to a contradiction later, assume that
is false. There then exists a graph
with
such that
.
Let .
This is reminding me of /vg/ for obvious reasons. Then is a graph of k vertices, to which the proposition applies:
. Applying the power of subtraction,
. However, by applying the lemma, we also have
. From these:
and
.
is necessarily at least
since otherwise
, where choosing one of the colours not used for any vertex in
leads to a
-colouring. Similarly,
. Add these two inequalities to get
. However, since
, this leads to the absurd conclusion that
, hence
for any
, so by induction the proposition is true for all nonnegative integers. ∎
Trying to adapt this back for the first statement in the start of the blog post doesn’t seem to work too well, since it leads to the pair of statements and
, which are not by themselves false. Example: let
and
.
, and
for any choice of
.
Anyway. Time for a completely unrelated graph theory finding observation. Let be a graph with vertex set
(where
is an ordered pair) and two vertices being adjacent iff they have either the i-coordinate the same and j-coordinate differing by one (modulo 4, where 0 and 3 can be considered to differ by one), or j-coordinate the same and i-coordinate differing by one (again, mod 4). Appealing more towards a graphic visualisation, a 4-by-4 square lattice graph with edges wrapped. (See pics that I can’t upload here because >SVG.)
This graph happens to be isomorphic to the hypercube graph , and allows consideration of the 4-cube’s symmetries as automorphisms of the graph. The six squares around a point are fairly obvious when the graph is drawn normally, but when drawn as a grid, two of the six squares (the ones that wrap across the boundary) are less obvious, but still existent, of course.
It’s possible to label the vertices on a 4-cube such that each axis gets assigned a power of two from 1 to 8, resulting in a numbering of vertices from 0 to 15, where adjacent vertices differ by one bit in their binary representation. By sheer coincidence, by removing all the points divisible by 3, one obtains a graph with no cycles… which happens to be the only way of removing 6 vertices from to obtain a forest (up to symmetry; there are 8 ways total, as mentioned two years ago). Of course, I’ve already proven that that was the only way of removing 6 vertices to obtain a forest years back, but representing it as a 4-by-4 grid that wraps makes a lot of the symmetries more obvious (you get 16 from translation, multiply by 4 for rotation, and 2 again for reflection (which covers 128 of the 384 symmetries); contrast the typical drawing as a regular octagon, which only has 8 from rotation, multiplied by 2 for reflection).