
TL;DR
This paper investigates the conditions under which almost complete packings of edge-colored complete graphs can be achieved using only a subset of smaller edge-colored complete graphs, revealing that most such graphs are avoidable.
Contribution
It introduces the concept of avoidable graphs and families in edge-coloring packings, providing sufficient conditions and showing most elements are avoidable, including all Eulerian graphs.
Findings
Most elements of (k) are avoidable
All Eulerian elements of (k) are avoidable
The set of all Eulerian elements is avoidable
Abstract
Erd\H{o}s and Hanani proved that for every fixed integer , the complete graph can be almost completely packed with copies of ; that is, contains pairwise edge-disjoint copies of that cover all but an fraction of its edges. Equivalently, elements of the set of all red-blue edge colorings of can be used to almost completely pack every red-blue edge coloring of . The following strengthening of the aforementioned Erd\H{o}s-Hanani result is considered. Suppose . Is it true that we can use elements only from and almost completely pack every red-blue edge coloring of ? An element is {\em avoidable} if has this property and a subset is avoidable if has this property. It seems difficult to determine all…
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Packing without some pieces
Raphael Yuster Department of Mathematics, University of Haifa, Haifa 31905, Israel. Email: [email protected]. This research was supported by the Israel Science Foundation (grant No. 1082/16).
Abstract
Erdős and Hanani proved that for every fixed integer , the complete graph can be almost completely packed with copies of ; that is, contains pairwise edge-disjoint copies of that cover all but an fraction of its edges. Equivalently, elements of the set of all red-blue edge colorings of can be used to almost completely pack every red-blue edge coloring of .
The following strengthening of the aforementioned Erdős-Hanani result is considered. Suppose . Is it true that we can use elements only from and almost completely pack every red-blue edge coloring of ? An element is avoidable if has this property and a subset is avoidable if has this property.
It seems difficult to determine all avoidable graphs as well as all avoidable families. We prove some nontrivial sufficient conditions for avoidability. Our proofs imply, in particular, that (i) almost all elements of are avoidable (ii) all Eulerian elements of are avoidable and, in fact, the set of all Eulerian elements of is avoidable.
MSC codes: 05C70, 05C35
1 Introduction
Throughout this paper a red-blue edge coloring of is synonymous with a graph on vertices where are the blue edges and are the red edges. We usually omit the word “edge” and just refer to red-blue colorings. Let be the set of all red-blue colorings of . Equivalently, we can view as the set of all graphs on vertices.
If are pairwise edge-disjoint cliques of size forming a packing of , then given any red-blue coloring of with color classes and , we can view the ’s as red-blue colorings of where the coloring of is given by and for . The main question of the paper is what possible -colorings and are forced to arise in asymptotic packings (packings that cover almost all of the edges of ).
More formally, for an -packing of a red-blue coloring of is a set of pairwise edge-disjoint subgraphs of this colored , where each subgraph is isomorphic to an element of . The size of the packing is . Obviously, .
We say that has the asymptotic packing property if every red-blue coloring of has an -packing of size at least . More formally, for every and all sufficiently large , there is an packing of every red-blue coloring of of size at least . The following was proved by Erdős and Hanani [2]:
Theorem 1
C*(k) has the asymptotic packing property.*
In other words, they proved that can be packed with edge-disjoint copies of so that only edges remain unpacked. This result has many applications and was generalized in several ways, most notably by Rödl for hypergraphs [10], by Wilson for exact graph decompositions [12] and by Keevash for exact hypergraph decompositions [8]. See also Glock et al. [5] for another, more general proof.
It is therefore interesting to determine to what extent can Theorem 1 be strengthened by requiring less than in its statement. Namely, which subsets of have the asymptotic packing property.
Problem 1
For every fixed , determine the subsets of that have the asymptotic packing property.
An element is avoidable if has the asymptotic packing property and a subset is avoidable if has the asymptotic packing property. Non-avoidable graphs or subsets are unavoidable. So Problem 1 can be reformulated as asking to determine all avoidable subsets and in particular all avoidable graphs.
For we trivially have that every nonempty subset of is unavoidable. It is also easy to verify that every nonempty subset of is unavoidable. In fact:
Proposition 1.1
For all , the graphs , and their complements are unavoidable. Also, , and and their complements are unavoidable.
Already for we do not know the complete solution for Problem 1.
Let denote the set of all unavoidable graphs on vertices. Our first main result is that almost all elements of are avoidable.
Theorem 2
.
Theorem 2 is a consequence of a result that gives a more general sufficient condition for avoidability in terms of the asymmetry of a graph (Lemma 3.6). It is natural to use random -vertex graphs as it is not difficult to prove that these are almost surely highly asymmetric (in a well-defined sense made later). The main technical issue is proving that this asymmetry property suffices for avoidability.
While Theorem 2 shows that graphs that are sufficiently asymmetric are avoidable, our second main result proves that a certain large class of graphs which contains some highly symmetric graphs is avoidable. This class of graphs, whose definition follows, includes all Eulerian elements of .
The degree set of a graph is the set . For a set of integers let be the set of all graphs on vertices whose degree set is contained in . So, is the set of all -regular graphs on vertices. Equivalently, is the set of all red-blue colorings of where the degree set of each blue graph is contained in . When is odd, a red-blue coloring of is Eulerian if the blue graph is Eulerian and the red graph is Eulerian. For example, a coloring of with a blue (and hence a red ) is Eulerian. Notice that all Eulerian red-blue colorings are contained in where , but the latter is more general already for . An immediate corollary of the following theorem is that the family of all Eulerian red-blue edge-colorings of is avoidable.
Theorem 3
For all odd positive integers , is avoidable.
Theorem 3 is a nontrivial consequence of a more general statement (Theorem 4) that gives a sufficient condition for the avoidability of in terms of the solvability of a certain parametric linear program. For relatively small we can determine if a solution exists and hence determine many additional such that is avoidable.
The tool of fractional packings will be useful in proving Theorem 2, Theorem 3, and their more generalized statements. We describe this tool in Section 2. Sections 3 and 4 prove Theorem 2 and Theorem 3 respectively. Section 5 contains the proof of Proposition 1.1. The final section contains some concluding remarks, most notably addressing the analogous problem where instead of an asymptotic packing we ask for an exact decomposition and consider the seemingly stronger property of decomposition avoidability. In particular, we prove there that is not decomposition avoidable.
2 Fractional packings
Let be a set of graphs of order . Let be a graph with . Let denote the set of all induced copies of in a graph (by induced copy we mean an induced subgraph of on vertices which is isomorphic to an element of ). Notice that in the special case that contains all induced -subgraphs of , then .
A function from to is a fractional -packing of if for each pair of distinct vertices we have
[TABLE]
For a fractional -packing , let
[TABLE]
The fractional -packing number, denoted by , is the maximum value of ranging over all fractional -packings . One observes that computing amounts to solving a linear programming maximization problem with constraints and variables. It can therefore be solved in polynomial time for fixed .
An -packing of is a fractional -packing whose image is . In other words, it is a set of induced copies of elements of in where any two copies do not share a pair of vertices (they are either disjoint or have a single vertex in common). Let denote the maximum size of an -packing of . As we restrict the values of in the definition of an -packing of , we have .
An important result of Haxell and Rödl [7] and later a slightly more general form (allowing for a “set of graphs” definition) by the author [13], both of which rely on Szemerédi’s regularity lemma [11], shows that the converse inequality is also asymptotically true, up to an additive error term which is negligible for dense graphs.
Lemma 2.1
For every and for every positive integer there exists such that the following holds. For any set of graphs of order and any graph with vertices, .
One can observe that Lemma 2.1 is extremely useful already by the following trivial use of it which implies the (nontrivial) result of Erdős and Hanani. Indeed, merely notice that if and , then clearly . Thus, .
3 Avoidable graphs
3.1 Decompositions and fractional decompositions
We say that has the decomposition property for if every red-blue coloring of has an -packing of size . Notice that having the decomposition property for is the same as having for every graph with vertices. Analogously, we say that has the fractional decomposition property for if . Trivially, has the fractional decomposition property for all , and a seminal result of Wilson [12] asserts that has the decomposition property for all sufficiently large that satisfy the necessary divisibility condition .
Let be a graph with vertices. For , let be the set of all induced subgraphs of on vertices. So, for example, if and , then .
Lemma 3.1
Let be a graph with vertices. Suppose that has the decomposition property for some . Then is avoidable.
Proof. Let be maximal such that has the decomposition property for some . Let be minimal subject to this, so only depends on .
Consider first the easy case where . In this case already has the decomposition property for . Then we can decompose every red-blue coloring of into pairwise edge-disjoint copies of where in each copy, the blue edges do not induce . By Theorem 1 (the Erdős-Hanani Theorem), has the asymptotic packing property. Thus, can be packed with edge-disjoint copies of so that only edges remain unpacked. This, in turn, implies that any red-blue coloring of can be packed with edge-disjoint copies of so that only edges remain unpacked, and in each copy, the blue edges do not induce . Thus, has the asymptotic packing property, which means that is avoidable.
Now consider the case where . By the result of Wilson mentioned earlier, there exists such that for all , if , then has a decomposition into pairwise edge-disjoint copies of , and as in the previous paragraph, each such can be decomposed into pairwise edge-disjoint copies of , such that in each copy of , the blue edges are isomorphic to an element of . Altogether, any red-blue coloring of has a decomposition into pairwise edge-disjoint copies of , such that in each copy of , the blue edges are isomorphic to an element of . Let denote the elements of this decomposition.
But recall that we want to prove that is avoidable (and not merely that is avoidable). To this end, let us design some fractional packing of . Consider some and recall that . There are vertices of that do not belong to . For any set of of these vertices (there are choices for ) consider the -subgraph of induced by the vertices of and the vertices of , call it . Notice that is a red-blue coloring of where the blue edges of do not induce a subgraph that is isomorphic to . Indeed, this is because is an induced -vertex subgraph of , so if were isomorphic to , then would have been a member of while by definition . We give the weight ( to be chosen later). We do this for every choice of and for every choice of , and they all get the same weight . So, altogether we obtain a fractional packing of consisting of
[TABLE]
elements, each one having weight , and each one being a red-blue coloring of with the blue edges not forming an . Since, by symmetry, the sum of the weights of each edge of is the same, we can choose the weight such that the total weight of this fractional packing is precisely (a fractional decomposition). In other words, where and is any graph on vertices.
There are still two small issues to take care of. First observe that the argument above assumed that (and recall that ). If is not of this form, that let be the largest integer such that . As , we can just ignore vertices (which touch edges) and thus where and is any graph on vertices. Finally, we can use Lemma 2.1 to obtain that . But this means that is avoidable, as required.
In the proof of Theorem 2 it would be very important to use Lemma 3.1 for which is very close to and for which is not too large. Quantitatively, this will be guaranteed by the following lemma.
Lemma 3.2
For every , there is such that has a decomposition into pairwise edge-disjoint copies of where .
Proof. Assume first that is a prime power. It is well-known that there is a finite projective plane of order , which means that decomposes into pairwise edge-disjoint copies of . So, in this case, the lemma holds for . In the case where is not a prime power, we can use the result of Baker, Harman, and Pintz [1] which states that there is always a prime strictly between and (this result is a significant extension of Chebyshev’s Theorem of Bertrand’s postulate). So, let be the largest integer such that is a prime power. Since , using the same argument of existence of projective plane of order , we have that decomposes into pairwise edge-disjoint copies of where .
It is important to note that if we wouldn’t have cared about the fact that is small (only a polynomial in ), then Lemma 3.2 would have worked already with since Wilson’s Theorem mentioned earlier guarantees that for some large , has a decomposition into . However, the bound in Wilson’s proof for such a does not suffice for our proof.
3.2 Graphs whose large subgraphs are asymmetric
Let be a graph on the vertex set . A permutation is an automorphism of if is an edge of if and only if is an edge of . The group of all automorphisms of is denoted by . We say that is asymmetric if consists only of the identity permutation. Otherwise, we say that is symmetric. The smallest graph (with more than one vertex) which is asymmetric is obtained from the path on vertices (in this order) by adding vertex and connecting it to vertices and . Erdős and Rényi [3] proved that almost all graphs are asymmetric.
For a graph , let be the smallest integer such that any two induced subgraphs of on at least vertices each, are non-isomorphic and further, any induced subgraph of on at least vertices is asymmetric. If is symmetric then define .
It is not difficult to prove that as it is well-known (Goodman [6]) that for any graph, there are two vertices that agree on at least other vertices, where and agree on if both are neighbors of or both are non-neighbors of . Asymmetric graphs are natural candidates for a graph with relatively small , but this is clearly not a sufficient condition, as it is easy to construct asymmetric graphs with . We will need graphs with relatively small as it would be possible to prove that such are avoidable.
Our next lemma proves that a randomly chosen graph on vertices has relatively small , with probability tending to one as increases. Recall that is the probability space of all graphs on vertices where each pair of vertices are connected with an edge with probability , and the choices are independent.
Lemma 3.3
Let be fixed and let . Then,
[TABLE]
Proof. Recall that . For a subset let be the subgraph of induced by . We will prove the following two claims.
- C1.
For every with , the probability that is symmetric is at most . 2. C2.
For any two distinct subsets with such that , the probability that and are isomorphic is at most .
There are less than subsets of size at least . For any such , the number of subsets with such that is less than . Also notice that so . Thus, if both claims hold we obtain by the union bound that
[TABLE]
where in the last inequality we have used the fact that which implies that .
We next prove Claim C1. Let with . Let be a permutation of which is not the identity. We would like to upper bound the probability that . As in the proof of Kim, Sudakov, and Vu [9], it would be useful to compute such a bound by considering the number of non-stationary points of . Let this number be . Notice that for any given , the number of possible with non-stationary points is less than .
Let be the set of non-stationary points of , so . Observe that we can always find pairs such that are distinct elements of and further for . Indeed, in each nontrivial orbit of of length we can obviously find such pairs. The worst case is when all nontrivial orbits are of length so we can only find one pair in each orbit, resulting in only pairs. Let .
For each and for each point consider the two pairs and . Since , in order for to be in we must have that and agree (both are edges or both are non-edges). Since agreement occurs with probability and since all the choices of and are independent with respect to the event of agreement (since they correspond to distinct pairs), we obtain that
[TABLE]
Now, since , we obtain that for all ,
[TABLE]
Notice that for sufficiently large, the left hand side is maximized when and already in this case the inequality holds since .
As there are less than permutations with non-stationary points we obtain by the union bound, the last inequality, and (2) that
[TABLE]
This completes the proof of Claim C1.
We next prove Claim C2 which is quite similar. Let with such that . Let be a bijection from to . We would like to upper bound the probability that is an isomorphism between and . Let be the number of non-stationary points of . Observe that and that the number of possible with non-stationary points is at most .
We claim that we can always find pairs such that all the vertices are distinct, , and . Indeed, the vertices of are all non-stationary, so we let them be and let their images be , respectively. Each may be either in or in . suppose of them are in . Then there are additional vertices in having images in . Denote the images by respectively. This still leaves non-stationary vertices of having images also in , so as in the proof of Claim C1, we can pick at least additional pairs such that for and such they are all distinct from the vertices in previously selected pairs. So, the least amount of selected pairs occurs when in which case we can still pick at least pairs. This proves the claim about the existence of . The reason we take to be the minimum between and is that we still want to leave sufficiently many vertices of that are not in these pairs. Let and observe that .
For each and for each point consider the two pairs and . Since , in order for to be an isomorphism we must have that and agree (both are edges or both are non-edges). Since agreement occurs with probability and since all the choices of and result in independent events (since they correspond to distinct pairs), we obtain that
[TABLE]
where in the last inequality we have used the fact that .
Consider first the case where . In this case we have for all sufficiently large that
[TABLE]
where we have used here the fact that and .
Consider next the remaining case where . Let so and . We now have that
[TABLE]
As there are less than bijections with non-stationary points we obtain by the union bound, the last two inequalities, and (3) that
[TABLE]
This completes the proof of Claim C2.
Let be a graph on vertices. The number of induced copies of in a graph is denoted by . Let denote the maximum of taken over all graphs with vertices. It is easy to observe that for we have . Indeed, let be a graph obtained from by selecting vertices of and duplicating them. Namely, selecting the vertices one by one, if is a selected vertex, then add another vertex and connect it precisely to all the neighbors of in the current graph. This creates a graph on vertices and a copy of in can be obtained by selecting each non-duplicated vertex, and one of the two copies of each duplicated vertex. The number of distinct copies of chosen in this way is . The following lemma shows that for graphs with where we in fact have at least when is not too large.
Lemma 3.4
Suppose where . Then for all we have .
Proof. Let be any graph on vertices. We prove that . We will assume that the vertices of are labeled by . The vertices of are labeled by . We may associate a copy of in with an injection . Let denote the set of all copies of in . Now, if , then and , thus . But since we have that if , then .
Let . So, . We may therefore assign a role to each , where the role of is if for some . By the above paragraph, roles are well-defined.
Hence may be partitioned into where all the vertices in have role . Now, every copy of in (i.e. every member of ) is formed by selecting one vertex from each for . So, the number of copies of in is at most .
Finally, we need the following simple lemma.
Lemma 3.5
Suppose that is a subgraph of on at least vertices. Then, .
Proof. Every subgraph of on at least vertices is also a subgraph of and hence is asymmetric. Any two subgraphs of on at least vertices are also two subgraphs of and hence are non-isomorphic.
Theorem 2 follows immediately from the following lemma and from Lemma 3.3.
Lemma 3.6
Let be a constant. For all sufficiently large, if , then is avoidable.
Proof. Applying Lemma 3.2, let be such that has a decomposition into pairwise edge-disjoint copies of , where .
Next, recall that is the set of all induced subgraphs of on vertices. We will prove that has the decomposition property for . Once we establish that, we are done since Lemma 3.1 implies that is avoidable.
Hence, it remains to prove that any red-blue coloring of can be decomposed into edge disjoint copies of (the fact that it can is already stated in the first paragraph of this proof) but with the additional requirement that in each copy of of this decomposition, the blue edges induce a subgraph which is not in .
Suppose now that . First observe that since , we have that is asymmetric. Furthermore, by Lemma 3.5,
[TABLE]
Using in Lemma 3.4 we obtain that for all we have .
Let be any graph on vertices (equivalently, a red-blue coloring of ). We next prove that the density of in , namely satisfies
[TABLE]
Assume the contrary. Then, for any such that , we would have a subgraph of on vertices such that the density of in is at least , namely
[TABLE]
We shall use . But then, . To arrive at the desired contradiction we only need to show that
[TABLE]
Now,
[TABLE]
and
[TABLE]
Thus indeed,
[TABLE]
Suppose the vertices of are . Let be some -decomposition of . Hence and any is an induced -vertex subgraph of . If each is an element of we are done, but the problem is that some might be isomorphic to some element of . For a permutation of , let be the -decomposition of corresponding to the permutation. That is, each now corresponds to where . We will prove that there exists such that each is an element of . As usual, it would be convenient to prove this counting argument using probabilistic language.
Suppose that is chosen uniformly among all permutations of . For a fixed , recall that the density of in is less than . As (and thus ) have elements, the probability that some element of is isomorphic to is less than . As there are at most elements in , we have that the expected number of elements of that are isomorphic to some element of is less than . Hence, there exists such that each is an element of . As was an arbitrary graph of vertices, we have proved that has the decomposition property for .
4 Proof of Theorem 3
4.1 A sufficient condition for the avoidability of
We now prove our main theorem of this section, from which Theorem 3 can be obtained as a (nontrivial) corollary. Recall that for a set of integers we let be the set of all graphs on vertices whose degree set is contained in . The next theorem gives a sufficient condition for to be avoidable.
Theorem 4
Suppose that for every real parameter , the following linear system of three equations in the variables has a nonnegative solution111A nonnegative solution is a solution where each coordinate is nonnegative.:
[TABLE]
Then is avoidable.
The proof of Theorem 4 is based on the following lemma.
Lemma 4.1
Let be the complement of , namely the set of all graphs on vertices whose degree set is not contained in . If the linear system of Theorem 4 has a nonnegative solution for every real parameter , then for any graph we have .
Notice that Lemma 4.1 together with Lemma 2.1 immediately implies Theorem 4, since we get that which means that is avoidable.
Proof of Lemma 4.1. Let and be fixed. As stated earlier, is the set of all graphs on vertices whose degree set is contained in and is the complement of .
Let be a graph with . Our goal is to design a fractional packing from to such that . This will prove that and yield a proof of Lemma 4.1.
We will construct as a sum of smaller fractional packings from to , one for each . So,
[TABLE]
We next define each and prove that satisfies the definition of a fractional packing, i.e. that (1) is satisfied for each pair of distinct vertices .
We first state a few properties that we require to have.
- P1.
only if and . 2. P2.
. 3. P3.
For any , the sum of the values of over all that contain the pair is . In other words,
[TABLE] 4. P4.
For any pair , the sum of the values of over all that contain the pair is . In other words,
[TABLE]
Let us see that if properties P2, P3, and P4 hold for each where , then indeed and is a valid fractional packing. First observe that by property P2, . Next, consider some pair . By P3, the sum of the values of over the elements that contain the pair is . Likewise, the sum of the values of over the elements that contain the pair is . By P4, for any , the sum of the values of over the elements that contain the pair is . So, the overall sum of values of over all elements that contain the pair is at most
[TABLE]
Hence, is a valid fractional packing with the claimed value.
We proceed to define . Let us first set for every with . This guarantees P1. For each let be a nonnegative real to be chosen later. Now, consider any subset of vertices of . Clearly induces a subgraph of on vertices which may or may not be in . Denote this subgraph by . If we must define . Recall that denotes the set of neighbors of in . Let and clearly . Set
[TABLE]
Notice that we do need to consider the case since it is possible that while .
We next define the values of the . These values will depend on Properties P2,P3,P4, on , and on , the degree of in . The number of elements that received the weight is the number of subsets of vertices of such that , which is
[TABLE]
So, to satisfy P2 we must have
[TABLE]
Consider some edge . How many elements that contain the edge received the weight ? For this to occur, must contain neighbors of , while is one of those neighbors. Hence, the number of such elements is
[TABLE]
To satisfy P3 we must therefore have that
[TABLE]
Similarly, consider some non-edge . How many elements that contain this non-edge received the weight ? For this to occur, must contain neighbors of , while is not one of those neighbors. Hence, the number of such elements is
[TABLE]
To satisfy P3 we must therefore have that
[TABLE]
Consider some pair such that both and . The number of elements that contain this pair and received the weight is
[TABLE]
To satisfy P4 we must therefore have that
[TABLE]
By similarly considering pairs such that both are non-neighbors of we get that in order to satisfy P4 we must have
[TABLE]
Finally, by considering pairs such that exactly one of is a neighbor of we get that in order to satisfy P4 we must have
[TABLE]
So, the question we remain with is whether we can find nonnegative reals such that equations (4-9) hold. To simplify notation, let us set and hence . Also let . Thus, in these terms, (4-9) become:
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
It is not difficult to see that the six equalities (e1-e6) are linearly dependent and have rank at most even without the allowed error term. Indeed,
[TABLE]
So, (e1), (e2), (e4) span the system of six equations. It will be slightly more convenient to work with (e4), (e5), (e6) as they all have the same right hand side. They also span the six equations since (12) shows that (e2) is spanned by (e4), (e6) and thus (11) shows that (e1) is also spanned by (e4), (e5), (e6) and thus (10) shows that (e3) is spanned by (e4), (e5), (e6) as well.
Finally, notice that the coefficients of the left hand side of each of (e4), (e5), (e6) are exactly the coefficients of the left hand sides of the equations stated in Theorem 4. Since equations (e4), (e5), (e6) have the same right hand side, solvability is maintained if we normalize to require that each right hand side is , as in the equations stated in Theorem 4. Finally, as we have no control over , and we require solvability for each (and different ’s may have different degrees, thus different ’s) we need to ensure solvability for each . This prove Lemma 4.1.
4.2 Sets that satisfy the conditions of Theorem 4
We start this section with an example showing that for some , the linear system of Theorem 4 can only be non-negatively solved for all , where has positive measure strictly less than . Hence, Theorem 4 cannot be applied to such sets.
Consider the case and . Observe that in this case, . The set of variables is thus just . The system in Theorem 4 therefore becomes:
[TABLE]
This system has a nonnegative solution only if .
Table 1 contains a list of all maximal sets222If Theorem 4 holds for a set , then it clearly holds for any subset of as one can set any additional variables to zero. for which Theorem 4 holds, for .
While the values in this table are verified by a computer program, one particular symmetric pattern that emerges is when is odd. Our goal is to prove that this holds for all odd , thereby proving Theorem 3.
Proof of Theorem 3. To prove Theorem 3 using Theorem 4, we need to prove that for all , the system
[TABLE]
has a nonnegative solution . Let us denote the matrix of coefficients by (so has rows and columns), and the vector of variables by . So we need to prove that has a nonnegative solution where is the column vector . By the classical Farkas’ Lemma [4] (or directly using linear programming duality), this holds if and only if for any vector such that is nonnegative, we must have .
So, suppose that is nonnegative. We must prove that . Consider first the product of with the first column of . The first column of corresponds to so it is the column vector . As we assume that is nonnegative, this implies that . Consider now the product of with the last column of . The last column of corresponds to so it is the column vector . As we assume that is nonnegative, this implies that .
We now consider the remaining inequalities of the form where is column of and . We sum all of these inequalities. This sum is an inequality of the form . Specifically,
[TABLE]
Observe that . So, we know that , that and that . This, in turn, implies that
[TABLE]
Thus,
[TABLE]
But observe that
[TABLE]
so indeed .
5 Some unavoidable graphs
We prove Proposition 1.1. The fact that is unavoidable for every is trivial. We prove next that is unavoidable for each . Let be a positive constant. Consider a partition of into sets with . Color by coloring all edges in blue, and all edges in red. Consider a -packing of this which leaves edges unpacked. There are at most elements in this packing that contain at least one edge of . Any such element contains at most edges of . So, altogether, all of these elements contain at most edges of . But these do not cover all edges of since
[TABLE]
Hence, there is an element of the packing which contains no edge of and does contain an edge of . This element is thus a red .
To see that is unavoidable, consider a partition of into sets with . Color by coloring red and coloring blue. Consider a -packing of this which leaves edges unpacked. Any element of the packing which contains an edge of is either a red or a red . They cannot all be red as otherwise, since any red occupies four red edges of and at least edges of are packed, there would have been elements in the -packing, but all together they would occupy edges of , while the latter only has less than edges. The same example shows that is unavoidable. Any element of the packing which contains an edge of is either a red or a red or a red . They cannot all be red or red as otherwise, since any red or red occupies at most red edges of and at least edges of are packed, there would have been at least elements in the -packing, but all together they would occupy at least edges of , while the latter only has less than edges.
The proof that is unavoidable is slightly more involved. Consider a partition of into sets with for . Color by coloring blue for and coloring blue as well. All other edges are red. Consider a -packing of this with leaves edges unpacked. We claim that must contain a red . Suppose it does not. As each element in the packing consists of edges, we have that . We partition the elements of into five types as follows. Type elements have two vertices in and two vertices in . Type elements have three vertices in and one in , or vice versa. Type elements have all their four vertices in or all their four vertices in . Type elements have two vertices in and no vertex in , or two vertices in and no vertex in . Type elements are all remaining elements. Let be the number of elements of type for .
Consider a packed edge where and and the element containing . We claim that is entirely in . Indeed, otherwise, has at least one vertex in . Suppose w.l.o.g. that it has a vertex in . Then, no matter where the fourth vertex resides, we obtain a red , a contradiction. Thus, we have that for any packed edge where and , the element of containing it is of type or of type . As there are packed edges in , we have that . Also, the number of edges of type , type and type elements in is .
There remain edges in that are not of type . Hence, . Now, any element of type has at least three vertices in . Also, each element of type contains a single edge of . As the number of edges in is less than we have that . It follows that
[TABLE]
contradicting the fact that .
6 Concluding remarks and open problems
In the proof of Theorem 2 we used Lemma 3.3 that shows that is highly asymmetric, namely it has for all , asymptotically almost surely. However, it is not difficult to modify the proof of Lemma 3.3 so that it holds for for any constant . This would cause the lower bound for to increase towards (but staying strictly less than ), changing some constants in the proof as the probability of the agreement event in the proof changes from to . Since in the proof of Lemma 3.6 we can choose to be any small positive constant, we obtain that for every fixed , the random graph is avoidable asymptotically almost surely.
Theorem 4 gives a sufficient condition for avoidability of the family of graphs , namely all -vertex graphs whose degrees are in . It seems interesting to determine all maximal sets for which is avoidable. While this is trivial for , the following proposition determines the case .
Proposition 6.1
* and are the only maximal sets for which for which is avoidable.*
Proof. The set is trivially avoidable because it is empty (no graph with an odd number of vertices can have all its degrees odd). The set is avoidable by Theorem 3. The set is unavoidable since it contains which is unavoidable by Proposition 1.1. Similarly, the complement of is unavoidable so is unavoidable. The sets and are unavoidable since and its complement are unavoidable. Hence, and are the only maximal sets for which for which is avoidable.
Similar to the way Problem 1 asks to generalize the result of Erdős and Hanani [2], it may be interesting to consider the analogous problem for exact decompositions, generalizing Wilson’s Theorem. Recall from Section 3 that has the decomposition property for if every red-blue coloring of has an -packing of size . Accordingly, we say that has the decomposition property if for all sufficiently large, has the decomposition property for whenever has the decomposition property for (namely, by Wilson’s Theorem, whenever ). Similarly, we can define decomposition avoidability for graphs and sets. The following problem analogous to Problem 1 emerges.
Problem 2
For every fixed , determine the subsets of that have the decomposition property.
It is straightforward to see that if is decomposition avoidable, then it is also avoidable. However, the following proposition might suggest that the converse is not true.
Proposition 6.2
* is decomposition unavoidable.*
Proof. Let be such that has a decomposition (in fact, this is known to hold for all ). Partition the vertices of into two parts and of sizes and . Color blue and all the other edges red. If our -decomposition avoids a blue , then any element of this decomposition occupies at most blue edges. As there are blue edges, the decomposition must contain a least elements. But this is impossible since it contains precisely elements.
Acknowledgment
I thank the referee for valuable comments.
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