In Edge decompositions of graphs with high minimum degree, Daniela Kühn, Allan Lo, Deryk Osthus and I proved that the edge sets of sufficiently dense graphs satisfying necessary divisibility conditions could be partitioned into copies of an arbitrary graph .  This result has since been generalised to other settings by various authors.  In this paper we present a simplified account of the latest version of the argument, specialised to the case where is a triangle.

I was explaining this problem to a colleague and they asked whether this graph was connected (it is) and whether that was still true if we restricted to rational coordinates.  It turns out this was addressed by Kiran B.Chilakamarri in 1988, and the answer is the rational unit distance graph is connected from dimension onwards.

To see that is not connected, consider a general unit vector where is coprime to .  Then .

Claim.  is divisible by at most once.

Proof. Squares mod are either , or .  If is divisible by then one of the is odd, hence squares to mod .  But then cannot be divisible by , which is a contradiction.

So the entries of in their reduced form do not contain any ‘s in their denominator, and so the same must hold for all sums of unit vectors.  Hence we can’t express, say, as a sum of unit vectors, and  is not connected to .

Connectedness in dimension (hence also later) uses Lagrange’s theorem on the sums of four squares.  We’ll show that can be expressed as a sum of unit vectors.  By Lagrange’s theorem, write .  Then

hence

is a sum of unit vectors.

Shagnik Das

A family of subsets of that shatters a -set has at least elements. How many -sets can we shatter with a family of size ? A block construction achieves . (Random is much worse.) Can in fact shatter constant fraction of all -sets. When , identify the ground set with , and colour by for .

Claim. A -set is shattered if and only if it is a basis for .

Proof. First suppose that is a basis. Then for any sequence , there is a unique vector such that . (We are just solving a system of full rank equations mod 2.)

Next suppose that are linearly dependent; that is, that they are contained in a subspace of . Choose orthogonal to . Then for any and any we have , so two of our colourings agree on .

We finish with the observation that random sets of vectors are fairly likely to span : the probability is

Blowing up this colouring gives a construction that works for larger .

At the other end of the scale, we can ask how large a family is required to shatter every -set from . The best known lower bound is , and the best known upper bound is , which comes from a random construction. Closing the gap between these bounds, or derandomising the upper bound, would both be of significant interest.

Andrew McDowell

At the start of his talk in Birmingham earlier this summer, Peter Hegarty played two clips from Terminator in which a creature first dissolved into liquid and dispersed, then later reassembled, stating that it had prompted him to wonder how independent agents can meet up without any communication. Andrew tackled the other half of this question: how can non-communicating agents spread out to occupy distinct vertices of a graph? He was able to analyse some strategies using Markov chains in a clever way.

Tássio Naia

A sufficient condition for embedding an oriented tree on vertices into every tournament on vertices that implies that almost all oriented trees are unavoidable in this sense.

A Latin square of order is an grid of cells, each of which contains one of distinct symbols, such that no symbol appears twice in any row or column.  There is a natural correspondence between Latin squares of order and partitions of into triangles.  We identify the three vertex classes of with the rows, columns and symbols of the Latin square.  A triangle corresponds to the symbol appearing in the intersection of row and column of the Latin square.  Since each cell contains exactly one symbol, and each symbol appears exactly once in each row and each column, the triangles corresponding to a Latin square do indeed partition .

What about partitions of into ‘s?  Identify the vertex classes of with rows, columns, red symbols and  blue symbols.  Then a corresponds to a red symbol and a blue symbol in the intersection of row and column .  If we look at just the red symbols or just the blue symbols then we see a Latin square.  But we also have the extra property that each pair of red and blue symbol appears in exactly one cell of the grid.  Two Latin squares with this property are called orthogonal.  So pairs of orthgonal Latin squares of order correspond to decompositions of into ‘s.  More generally, a sequence of  mutually orthogonal Latin squares of order  corresponds to a partition of the complete -partite graph on vertex classes of size into ‘s.

Suppose now that we have a partial Latin square of order , that is, a partially filled in grid of cells obeying the rules for a Latin square.  Can it be completed to a Latin square?  In the early 1980s several researchers proved that the answer is yes provided at most cells have been filled in total.  This is best possible, as if we place ‘s and a single on the main diagonal there is no legal cell in which to place the th .  The same example shows that it is not enough to ask only that each row and each column contains only a small number of non-empty cells.  But what if each row, column and symbol has been used at most times?  Can we then complete to a Latin square?  Daykin and Häggkvist conjectured that we can, provided .

What does this mean on the graph side?  Let be a subgraph of obtained by deleting a set of edge-disjoint triangles such that no vertex is in more than triangles.  Then should have a triangle-decomposition if .  In this paper we prove that has a triangle-decomposition provided .

In fact we prove something more general.  The ‘s obtained in this way have the properties that (i) each vertex has the same number of neighbours in each other vertex class and (ii) each vertex sees at least a proportion of the vertices in each other class (a partite minimum degree condition).  We prove that all such graphs have triangle-decompositions when .

The proof is based on that of the similar result for non-partite graphs in Edge decompositions of graphs with high minimum degree.  However, it is not simply a translation of that proof to the partite setting.  In the partite case we not only have to ensure that all of our gadgets can be embedded in partite graphs, we must also take care to ensure a stronger notion of divisibility is preserved throughout our decomposition process.  This makes the proof extremely technical.

We also prove the analogous result for -decompositions of complete -partite graphs, with the less impressive bound .  The connection to mutually orthogonal Latin squares is more complicated for , as the partially filled in cells only correspond to ‘s in the case where each non-empty cell contains one of each symbol, but we still show that there is some such that if each row, column and coloured symbol is used at most times then partial mutually orthogonal Latin squares can be complete.

As for the non-partite case, the bounds on are currently limited by available fractional or approximate decomposition results.  Improvements to these would lead automatically to improvements of the bounds in this paper.

Theorem.  Let be a spanning subgraph of .  If (i) or (ii) is -regular, then has a perfect matching.

Proof. Let be a maximal matching in with .

(i) Choose , .  We have and .  Since there is an such that is adjacent to and is adjacent to .  Then is an augmenting path.

(ii) Without loss of generality, is connected.  Form the directed graph on by taking the directed edge  () whenever is an edge of .  Add directed edges arbitrarily to to obtain a -regular digraph , which might contain multiple edges; since is connected we have to add at least one directed edge.  The edge set of decomposes into directed cycles.  Choose a cycle containing at least one new edge of , and let be a maximal sub-path of containing only edges of .  Let , be the start- and endpoints of respectively.  Then we can choose and , whence  is an augmenting path, where is the result of “pulling back” from to , replacing each visit to a in by use of the edge of .

(Philip) Hall’s theorem.  Let be a bipartite graph on vertex classes , .  Suppose that,  for every , .  Then there is a matching from to .

This is traditionally called Hall’s marriage theorem.  The picture is that the people in are all prepared to marry some subset of the people in .  If some people in are only prepared to marry into some set of people, then we have a problem; but this is the only problem we might have.  There is no room in this picture for the preferences of people in .

Proof. Suppose first that  for all .  Then we can match any element of arbitrarily to a neighbour in and obtain a smaller graph on which Hall’s condition holds, so are done by induction.

Otherwise  for some .  By induction there is a matching from to .  Let .  Then for any we have

hence and Hall’s condition holds on .  So by induction there is a matching from to , which together with the matching from to is a matching from to .

Corollary 1.  Every -regular bipartite graph has a perfect matching.

Proof. Counted with multiplicity, .  But each element of is hit at most times, so in the conventional sense.

Corollary 2. Let be a spanning subgraph of with minimum degree at least .  Then has a perfect matching.

This is very slightly more subtle.

Proof. If there is nothing to check.  Otherwise there is a .  Then and , so .  But for every .

Schrijver proved that a -regular bipartite graph with vertices in each class in fact has at least perfect matchings.  For minimum degree we have the following.

(Marshall) Hall’s theorem.  If each vertex of has degree at least and there is at least one perfect matching then there are at least .

This turns out to be easy if we were paying attention during the proof of (Philip) Hall’s theorem.

Proof. By (Philip) Hall’s theorem the existence of a perfect matching means that (Philip) Hall’s condition holds.  Choose a minimal on which it is tight.  Fix and match it arbitrarily to a neighbour .  (Philip) Hall’s condition still holds on and the minimum degree on this subgraph is at least , so by induction we can find at least perfect matchings.  Since there were at least choices for we have at least perfect matchings from to .  These extend to perfect matchings from to as in the proof of (Philip) Hall’s theorem.

Thanks to Gjergji Zaimi for pointing me to (Marshall) Hall’s paper via MathOverflow.

The first and most substantial concerns partition regularity.  It is a coherent presentation of all of the material from Partition regularity in the rationals, Partition regularity with congruence conditions and Partition regularity of a system of De and Hindman.

The remaining three chapters are expanded versions of Maximum hitting for sufficiently large, Random walks on quasirandom graphs and A note on balanced independent sets in the cube.¹

Erratum: the form of Talagrand’s inequality quoted as Theorem 6 in Chapter 4 has an incorrect proof in the book cited.  The proof can be modified to prove a slightly weaker result, with 60 replaced by 90 and an assumption that the median of is at least (a constant).  This is remarked in the published version of this chapter, Random walks on quasirandom graphs.

¹ “A note … ” was fairly described by one of my examiners as a “potboiler”.  It was also my first submitted paper, and completed in my second of three years.  Perhaps this will reassure anybody in the first year of a PhD who is worrying that they have yet to publish.

Seymour proved that every bridgeless graph has a nowhere zero -flow.  Thomas Bloom and I worked this out at the blackboard, and I want to record the ideas here.  First, a folklore observation that a minimal counterexample must be cubic and -connected.  We will temporarily allow graphs to have loops and multiple edges.

We first show that is -edge-connected.  It is certainly -edge-connected, else there is a bridge.  (If is not even connected then look at a component.)  If it were only -edge-connected then the graph would look like this.

Contract the top cross edge.

If the new graph has a nowhere zero -flow then so will the old one, as the flow in the bottom cross edge tells us how much flow is passing between the left and right blobs at the identified vertices and so the flow the we must put on the edge we contracted.  So is -edge connected.

Next we show that is -regular.  A vertex of degree forces a bridge; a vertex of degree forces the incident edges to have equal flows, so the two edges can be regarded as one.  So suppose there is a vertex of degree at least .

We want to replace two edges and by a single edge to obtain a smaller graph that is no easier to find a flow for.  The problem is that in doing so we might produce a bridge.

The -edge-connected components of   are connected by single edges in a forest-like fashion.  If any of the leaves of this forest contains only one neighbour of then there is a -edge-cut, so each leaf contains at least two neighbours of .

If there is a component of the forest with two leaves then choose and to be neighbours of from different leaves of that component.

Otherwise the -edge-connected components of are disconnected from each other.  Now any such component must contain at least neighbours of , else there is a -edge-cut.  If some contains neighbours of then we can choose and to be any two of them.  Otherwise all such contain exactly neighbours of , in which case there must be at least two of them and we can choose and to be neighbours of in different components.

So is -regular and -edge-connected.  If is only -connected then there is no flow between -connected components, so one of the components is a smaller graph with no nowhere zero -flow.  If is only -connected then because is so small we can also find a -edge-cut.

Finally, we want to get rid of any loops and multiple edges we might have introduced.  But loops make literally no interesting contribution to flows and double edges all look like

and the total flow on the pair just has to agree with the edges on either side.

We’ll also need one piece of magic (see postscript).

Theorem. (Tutte) Given any integer flow of there is a -flow of that agrees with the original flow mod .  (By definition, flows of in one direction and in the other direction agree mod .)

So we only need to worry about keeping things non-zero mod .

The engine of Seymour’s proof is the following observation.

Claim. Suppose that where each is a cycle and the number of new edges when we add to is at most .  Then has a -flow which is non-zero outside .  

Write for the set of edges added at the th stage.  We assign flows to in that order.  Assign a flow of in an arbitrary direction to ; now the edges in have non-zero flow and will never be touched again.  At the next stage, the edges in might already have some flow; but since there are only two possible values for these flows mod .  So there is some choice of flow we can put on to ensure that the flows on are non-zero.  Keep going to obtain the desired -flow, applying Tutte’s result as required to bring values back in range.

Finally, we claim that the we are considering have the above form with being a vertex disjoint union of cycles.  Then trivially has a -flow, and times this -flow plus the -flow constructed above is a nowhere-zero -flow on .

For , write for the largest subgraph of that can be obtained as above by adding cycles in turn, using at most two new edges at each stage.  Let be a maximal collection of vertex disjoint cycles in with connected, and let .  We claim that is empty.  If not, then the -connected blocks of are connected in a forest-like fashion; let be one of the leaves.

By -connectedness there are three vertex disjoint paths from to . At most one of these paths travels through ; let and be endpoints of two paths that do not.  These paths must in fact be single edges, as the only other way to get to would be to travel through .  Finally, since is -connected it contains a cycle through and , contradicting the choice of .

Postscript. It turns out that Tutte’s result is far from magical; in fact its proof is exactly what it should be.  Obtain a directed graph from by forgetting about the magnitude of flow in each edge (if an edge contains zero flow then delete it).  We claim that every edge is in a directed cycle.

Indeed, choose a directed edge .  Let be the set of vertices that can be reached by a directed path from and let be the set of vertices that can reach by following a directed path.  If  is not in any directed cycle then and are disjoint and there is no directed path from to .  But then there can be no flow in , contradicting the definition of .

So as long as there are edges with flow value at least , find a directed cycle containing one of those edges and push a flow of through it in the opposite direction.  The total flow in edges with flow at least strictly decreases, so we eventually obtain a -flow.

Together with Daniela Kühn, Allan Lo and Deryk Osthus I proved that for every graph there is a constant such that every “-divisible” graph on vertices with minimum degree at least has an -decomposition. In practice, the current obstacle to improving the bounds on is usually our knowledge of another quantity, the fractional decomposition threshold for cliques.

A graph has a fractional -decomposition if we can assign a non-negative weight to each copy of in such that the total weight of the copies of containing each fixed edge of is exactly 1. We prove that every graph with minimum degree at least has a fractional -decomposition. This greatly improves the previous bound of for large . We also prove a similar result for hypergraphs.

The proof begins with an approximate fractional -decomposition obtained by weighting every -clique in our graph equally. We then use small gadgets to make local adjustments to the total weight over each edge until we end up with a genuine fractional -decomposition.