Abstract 1 Introduction 2 Preliminaries 3 Proofs of hardness References

Complexity of Approximate Conflict-Free, Linearly-Ordered, and Nonmonochromatic Hypergraph Colourings

Tamio-Vesa Nakajima ORCID Department of Computer Science, University of Oxford, UK Zephyr Verwimp ORCID Department of Computer Science, University of Oxford, UK Marcin Wrochna ORCID Faculty of Mathematics, Informatics and Mechanics, University of Warsaw, Poland Stanislav Živný ORCID Department of Computer Science, University of Oxford, UK
Abstract

Using the algebraic approach to promise constraint satisfaction problems, we establish complexity classifications of three natural variants of hypergraph colourings: standard nonmonochromatic colourings, conflict-free colourings, and linearly-ordered colourings.

Firstly, we show that finding an -colouring of a k-colourable r-uniform hypergraph is 𝖭𝖯-hard for all constant 2k and r3. This provides a shorter proof of a celebrated result by Dinur et al. [FOCS’02/Combinatorica’05].

Secondly, we show that finding an -conflict-free colouring of an r-uniform hypergraph that admits a k-conflict-free colouring is 𝖭𝖯-hard for all constant 2k and r4, except for r=4 and k=2 (and any ); this case is solvable in polynomial time. The case of r=3 is the standard nonmonochromatic colouring, and the case of r=2 is the notoriously difficult open problem of approximate graph colouring.

Thirdly, we show that finding an -linearly-ordered colouring of an r-uniform hypergraph that admits a k-linearly-ordered colouring is 𝖭𝖯-hard for all constant 3k and r4, thus improving on the results of Nakajima and Živný [ICALP’22/ACM TocT’23].

Keywords and phrases:
hypergraph colourings, conflict-free colourings, unique-maximum colourings, linearly-ordered colourings
Category:
Track B: Automata, Logic, Semantics, and Theory of Programming
Funding:
Tamio-Vesa Nakajima: UKRI EP/X024431/1, Clarendon Fund Scholarship.
Zephyr Verwimp: UKRI EP/X024431/1.
Stanislav Živný: UKRI EP/X024431/1.
Copyright and License:
[Uncaptioned image] © Tamio-Vesa Nakajima, Zephyr Verwimp, Marcin Wrochna, and Stanislav Živný; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Theory of computation Design and analysis of algorithms
; Theory of computation Problems, reductions and completeness ; Theory of computation Constraint and logic programming
Acknowledgements:
We thank the anonymous reviewers for their feedback.
Editors:
Keren Censor-Hillel, Fabrizio Grandoni, Joël Ouaknine, and Gabriele Puppis

1 Introduction

Graph colouring

Graph colouring is one of the most studied computational problems: Given a graph G and an integer k, is there a k-colouring, i.e., an assignment of one of k colours to the vertices of the graph so that adjacent vertices are assigned different colours?

Deciding the existence of a 3-colouring is one of Karp’s 21 𝖭𝖯-complete problems [14]. Since finding a graph colouring with the smallest number of colours is 𝖭𝖯-hard, there has been much interest in the approximate graph colouring (AGC) problem: Given a graph G that admits a colouring with k colours, find a colouring with colours for some k. It is believed that for every constant 3k, this problem remains 𝖭𝖯-hard [12]. While some conditional results are known (i.e. AGC is 𝖭𝖯-hard if we assume the d-to-1 conjecture with perfect completeness [13]), proving unconditional results seems elusive. The strongest results known so far are for =2k1 [4] and =(kk/2)1 [15] (the first result is stronger for k=3,4, and equal to the second for k=5). As progress on proving the hardness of AGC seems to have hit a barrier, it is natural to try to attack variants of AGC, to see if any of the ideas and insights from those problems could apply to the AGC. In this paper, we will focus on hypergraph generalisations of graph colourings.

Recall that a hypergraph is a pair (V,E), where V is the vertex set and E2V the edge set. A hypergraph is called r-uniform if all the edges have size r. Thus, a 2-uniform hypergraph is a graph. A colouring of a graph (V,E) is an assignment of colours c(v) to the vertices vV such that for every edge {u,v}E we have c(u)c(v). Put differently in three different but equivalent ways, for every edge {u,v}E we have that (i) the set {c(u),c(v)} contains at least 2 elements, or (ii) some colour in the multiset {{c(u),c(v)}} appears exactly once, or (iii) the largest colour in the multiset {{c(u),c(v)}} appears exactly once.111This notion assumes that the colours are taken from a totally ordered set, e.g., the natural numbers. When these three definitions are applied to hypergraphs, we get three different notions, namely nonmonochromatic (NAE) colourings, conflict-free (CF) colourings, and linearly-ordered (LO) colourings. Note that any LO colouring is a CF colouring, and any CF colouring is an NAE colouring. The three notions of colourings are different already for 4-uniform222For 3-uniform hypergraphs, NAE and CF colourings coincide, but LO colourings are different. hypergraphs.333Indeed, the edge {a,b,c,d} could be assigned colours {{1,1,2,2}} in an NAE colouring, but not in the other two; whereas the edge could be assigned colours {{1,2,2,2}} in a CF colouring, but not in an LO colouring.

Promise CSPs

We will now review the most relevant literature on the three variants of hypergraph colourings. It will be convenient to present the existing results in the framework of so-called promise constraint satisfaction problems (PCSPs) [6], as we shall use the tools developed for understanding the computational complexity of PCSPs [4]. Constraint satisfaction problems (CSPs) are problems that can be cast as homomorphisms between relational structures. We will only need a special case of relational structures that contain only one relation. Formally, a relational structure 𝐀=(A,R𝐀) is a pair, where A is the universe of 𝐀 and R𝐀Ar is an r-ary relation. By abuse of language, we call r the arity of 𝐀.

An example of a relational structure is a graph or an r-uniform hypergraph, where the universe is the vertex set and the relation is the edge set – the arity of the relation is 2 for graphs and r for r-uniform hypergraphs. A homomorphism from one relational structure of arity r, say 𝐀=(A,R𝐀), to another of the same arity, say 𝐁=(B,R𝐁), is a map h:AB that preserves the relation: if (a1,,ar)R𝐀 then (h(a1),,h(ar))R𝐁. We denote the existence of a homomorphism from 𝐀 to 𝐁 by writing 𝐀𝐁.

Given two relational structures 𝐀 and 𝐁 with 𝐀𝐁, the promise constraint satisfaction problem with template (𝐀,𝐁), denoted by PCSP(𝐀,𝐁), is the following computational problem. Given a relational structure 𝐗 with 𝐗𝐀, find a homomorphism from 𝐗 to 𝐁. This is the search version of the problem. In the decision version, which reduces to the search version, one is given a relational structure 𝐗 with the same arity as 𝐀 and the task is to output Yes if 𝐗𝐀 and No if X↛𝐁.444Since the decision version reduces to the search version, solving the decision version is no harder than solving the search version. All of our results will hold for both versions – hardness results will hold even for the decision version, and tractability results will hold even for the search version.

In order to cast approximate hypergraph NAE/CF/LO-colourings as PCSPs, we will need to encode the NAE/CF/LO-colourability of a hypergraph by a homomorphism to a suitable relational structure. We will thus describe three families of relational structures capturing the three types of hypergraph colourings mentioned above (and therefore implicitly graph colouring). For any arity r2 and domain size k, we define:555For any integer k, we write [k] for the set {1,2,,k}.

𝐍𝐀𝐄kr =([k],{(x1,,xr)i,j[r]xixj}),
𝐂𝐅kr =([k],{(x1,,xr)i[r]j[r]i=jxixj}),
𝐋𝐎kr =([k],{(x1,,xr)i[r]j[r]i=jxi>xj}).

Observe that an r-uniform hypergraph 𝐗 has an NAE k-colouring if and only if 𝐗𝐍𝐀𝐄kr. The analogous statement holds for CF and LO colourings. Since NAE, LO and CF colourings are all identical to graph colouring on uniformity 2, we see that k vs.  AGC is the same as PCSP(𝐍𝐀𝐄k2,𝐍𝐀𝐄2) – or equivalently PCSP(𝐂𝐅k2,𝐂𝐅2) or PCSP(𝐋𝐎k2,𝐋𝐎2).

Nonmonochromatic colourings

The most studied hypergraph colourings are nonmonochromatic colourings, also known as weak hypergraph colourings. This is the weakest non-trivial restriction one can impose when colouring the vertices of a hypergraph, i.e., any type of hypergraph colouring (that excludes constant colourings) is also a nonmonochromatic colouring. As mentioned before, nonmonochromatic k-colourings of an r-uniform hypergraph correspond to homomorphisms from the hypergraph to 𝐍𝐀𝐄kr. Since nonmonochromatic colouring is 𝖭𝖯-hard for any uniformity r3 and number of colours k2, Dinur, Regev, and Smith investigated the approximate version, establishing the following result [9].

Theorem 1.

PCSP(𝐍𝐀𝐄kr,𝐍𝐀𝐄r) is 𝖭𝖯-hard for all constant 2k and r3.

In this paper we will show a simpler proof of this result. The proof in [9] relies on constructing a somewhat ad-hoc reduction and analysing its completeness and soundness. We recast this proof in the recent algebraic framework for PCSPs [4]. We also replace the use of Schrijver graphs with the simpler Kneser graphs, plus a (correct and very easy) case of Hedetniemi’s conjecture (cf. Lemma 9). We believe that our simplification is of interest since it replaces a more quantitative analysis of the polymorphisms with one that only deals with constants everywhere. In particular, this means that our proof does not require the full strength of the PCP theorem – only the “baby PCP” of Barto and Kozik [5].

Conflict-free colourings

A conflict-free hypergraph colouring is a colouring of the vertices in a hypergraph such that every hyperedge has at least one uniquely coloured vertex [10, 18]. As mentioned before, conflict-free k-colourings of an r-uniform hypergraph correspond to homomorphisms from the hypergraph to 𝐂𝐅kr. We shall determine the complexity of PCSP(𝐂𝐅kr,𝐂𝐅r) for all constants 2k and r3 (the case of r=3 corresponding to nonmonochromatic colourings, i.e., 𝐂𝐅k3=𝐍𝐀𝐄k3 for every k).

After the easy observation that PCSP(𝐂𝐅kr,𝐂𝐅r) reduces to PCSP(𝐂𝐅kr+t,𝐂𝐅r+t) for t2 (cf. Lemma 6 in Section 2), Theorem 1 directly implies 𝖭𝖯-hardness for promise conflict-free colouring for uniformity r5. The crux of the result is to deal with the case of uniformity r=4. Note that finding a conflict-free colouring of a 4-uniform hypergraph using 2 colours is identical to solving systems of equations of the form x+y+z+t1(mod2) over 2, and is hence in 𝖯 and consequently so is PCSP(𝐂𝐅24,𝐂𝐅4) for every 2. We resolve the only remaining case, showing in Section 3.3 that PCSP(𝐂𝐅k4,𝐂𝐅4) is 𝖭𝖯-hard for all 3k. Summarising, we have

Theorem 2.

PCSP(𝐂𝐅kr,𝐂𝐅r) is 𝖭𝖯-hard for all constant 2k and r3, except for k=2 and r=4, which is in 𝖯.

This also immediately implies the following (much weaker) corollary, which does not appear to have been known in the CF-colouring literature.

Corollary 3.

It is 𝖭𝖯-hard to approximate the conflict-free chromatic number666That is, the minimum number of colours needed to CF-colour a given hypergraph. of a hypergraph to within any constant factor, even if it is r-uniform for some constant r3.

Linearly-ordered colourings

A linearly-ordered [3] (or unique-maximum [8]) hypergraph colouring is a colouring of the vertices in a hypergraph with linearly-ordered colours such that the maximum colour in every hyperedge is unique. As mentioned before, linearly-ordered k-colourings of an r-uniform hypergraph correspond to homomorphisms from the hypergraph to 𝐋𝐎kr.

Barto at al. [3] conjectured that PCSP(𝐋𝐎k3,𝐋𝐎3) is 𝖭𝖯-hard for all constant 2k, but even the case PCSP(𝐋𝐎23,𝐋𝐎33) is still open. A recent result of Filakovský et al. [11] established 𝖭𝖯-hardness of PCSP(𝐋𝐎33,𝐋𝐎43) by generalising the topological methods of Krokhin et al. [15]. Nakajima and Živný [17] also showed 𝖭𝖯-hardness of PCSP(𝐋𝐎kr,𝐋𝐎r) for every 2k and rk+4. We strengthen this result, showing 𝖭𝖯-hardness of PCSP(𝐋𝐎kr,𝐋𝐎r) when 3k and r4.

Theorem 4.

PCSP(𝐋𝐎kr,𝐋𝐎r) is 𝖭𝖯-hard for all constant 3k and r4.

Observe that this theorem covers nearly all the cases from the result of [17]: the only case not covered is k=2 and r+2. In particular, Theorem 4 has no requirement on r in terms of , unlike the result in [17]. Indeed, Theorem 4 covers the full range of parameters except for the cases r=3 or k=2 (and thus the conjecture of Barto et al. remains open).

It is worth digressing somewhat to discuss the appearance of topological methods within these proofs. Our proof uses the chromatic number of the Kneser graph as an essential ingredient – this is a topological fact, and thus our proof is in some sense topological. (This is similar to the appearance of topology within hardness proofs for rainbow colourings [2].) On the other hand, the topological approach of [11], which proved that PCSP(𝐋𝐎33,𝐋𝐎43) is 𝖭𝖯-hard, is rather different. It assigns each relational structure an equivariant simplicial complex in a “nice enough” way so that the topological properties of these simplicial complexes imply the hardness of the original template. It would be interesting to see if these two approaches can be merged, or combined to strengthen both.

2 Preliminaries

Let (𝐀,𝐁) be a PCSP template with 𝐀 and 𝐁 of arity r. A polymorphism of arity n=ar(f) of (𝐀,𝐁) is a function f:AnB such that if f is applied component-wise to any n-tuple of elements of R𝐀 it gives an element of R𝐁. In more detail, whenever (aij) is an r×n matrix such that every column is in R𝐀, then f applied to the rows gives an r-tuple which is in R𝐁. We say that the rows of such a matrix are compatible. We denote by Pol(n)(𝐀,𝐁) the collection of n-ary polymorphisms of (𝐀,𝐁), and we let Pol(𝐀,𝐁)=nPol(n)(𝐀,𝐁).

For an n-ary function f:AnB and a map π:[n][m], we say that an m-ary function g:AmB is the minor of f given by π if g(x1,,xm)=f(xπ(1),,xπ(n)). We write f𝜋g if g is the minor of f given by π. Note that Pol(𝐀,𝐁) is closed under minors.

We use p to denote a polynomial-time many-one reduction.

Theorem 5 ([6]).

If Pol(𝐀,𝐁)Pol(𝐀,𝐁) then PCSP(𝐀,𝐁)pPCSP(𝐀,𝐁).

Lemma 6.

For any t2, PCSP(𝐂𝐅kr,𝐂𝐅r)pPCSP(𝐂𝐅kr+t,𝐂𝐅r+t).

Proof.

We will use Theorem 5 and show that Pol(𝐂𝐅kr+t,𝐂𝐅r+t)Pol(𝐂𝐅kr,𝐂𝐅r). Suppose fPol(n)(𝐂𝐅kr+t,𝐂𝐅r+t). Consider any r×n matrix A with rows 𝐚1,,𝐚r[k]n such that every column in the matrix has a unique entry. We can choose 𝐛[k]n such that each column in the (r+t)×n matrix A with rows 𝐚1,,𝐚r, and t copies of 𝐛 has a unique entry: choose element i of 𝐛 to be any value in [k] other than the unique entry in the i-th column of A. Since f is a polymorphism of (𝐂𝐅kr,𝐂𝐅r), we get that (f(𝐚1),,f(𝐚r),f(𝐛),,f(𝐛)) has a unique entry. Since t2, (f(𝐚1),,f(𝐚r)) must also have a unique entry. Thus fPol(n)(𝐂𝐅kr,𝐂𝐅r) as required.

An -chain of minors is a sequence of the form f0π0,1f1π1,2π1,f. We shall then write πi,j:[ar(fi)][ar(fj)] for the composition of πi,i+1,,πj1,j, for any 0i<j. Note that fiπi,jfj. We shall use the following 𝖭𝖯-hardness criterion for PCSPs.

Theorem 7 ([7]).

Suppose there are constants k, and an assignment sel which, for every fPol(𝐀,𝐁), outputs a set sel(f)[n] of size at most k, where n is the arity of f. Suppose furthermore that for every -chain of minors there are i,j such that πi,j(sel(fi))sel(fj). Then, PCSP(𝐀,𝐁) is 𝖭𝖯-hard.

For a graph G, the chromatic number of G, denoted by χ(G), is the smallest k such that GKk, where Kk is the clique on k vertices. We will rely on Lovász’s result for the chromatic number of Kneser graphs [16]. For 1h|A|, write A(h) for the family of subsets of A of size h. The Kneser graph is defined as KG(A,h)=(A(h),E), where {S,T}E if and only if ST=. For the special case A=[n], we use the notation KG(n,h)=KG([n],h).

Theorem 8 ([16]).

χ(KG(n,h))=n2h+2 for any n,h1.

Lemma 9.

Let χ(G)>n. Then χ(G×Kn+1)>n.

Proof.

We show the contrapositive: suppose there is a homomorphism G×Kn+1Kn. Equivalently, there is a homomorphism GKnKn+1, where KnKn+1 is the graph with vertex set {f:[n+1][n]}, and f and g adjacent if for every distinct i,j[n+1], f(i)g(j). This is equivalent since every homomorphism f:G×Kn+1Kn corresponds uniquely to the homomorphism f:GKnKn+1 given by f(u)=(vf(u,v)). There is also a homomorphism KnKn+1Kn: map any fV(KnKn+1) to an arbitrary repeating element in the range of f (at least one must exist by the pigeonhole principle). Thus GKn.

If 𝐀𝐀𝐁𝐁 then (𝐀,𝐁) is a homomorphic relaxation of (𝐀,𝐁). In this case it follows from the definitions that PCSP(𝐀,𝐁)pPCSP(𝐀,𝐁) [4].

3 Proofs of hardness

3.1 Avoiding sets imply hardness

Our hardness proofs will revolve around the notion of avoiding sets for polymorphisms [4], defined below. We denote by 𝟏X the indicator vector of X: (𝟏X)i=1 when iX and (𝟏X)i=0 otherwise. The overall length of the vector will be clear from context.

Definition 10.

Take A so that {0,1}A. Let f:AnB and TB. A T-avoiding set for f is a set S[n] such that for any RS, we have f(𝟏R)T. For t, we call a set S t-avoiding for f if it is T-avoiding for f for some subset TB of size t.

We will first collect some simple properties of avoiding sets.

Lemma 11.

Let f:AnB and =|B|.

  1. 1.

    There are no -avoiding sets for f.

  2. 2.

    [n] is an (1)-avoiding set for f.

  3. 3.

    If U is T-avoiding for f then so is every VU.

  4. 4.

    Take π:[n][m] and suppose f𝜋g for some g:AmB. Suppose S[n] is T-avoiding for f. Then π(S) is T-avoiding for g.

Proof.

For Item 1, observe that any -avoiding set would imply that f(𝟏[n])B, which is impossible. For Item 2, note that [n] is (B{f(𝟏[n])})-avoiding, and hence (1)-avoiding. Item 3 follows from the definitions. For Item 4, first observe that g(𝟏X)=f(𝟏π1(X)). For contradiction, suppose π(S) is not T-avoiding for g, i.e., there exists Rπ(S) with g(𝟏R)T. Then π1(R)S, and furthermore f(𝟏π1(R))=g(𝟏R)T. Hence S is not T-avoiding for f.

To apply Theorem 7, we want to build sel(f) out of (small) avoiding sets for f. This is a good idea because avoiding sets are preserved by minors, as shown in Item 4 of Lemma 11. The issue is that we might have too many avoiding sets. For the polymorphisms in this paper, many (small) t-avoiding sets which are pairwise disjoint imply the existence of a (small) (t+1)-avoiding set. Thus, since there can be no sets that avoid every output in the range, as shown in Item 1, there must be some maximal t for which a (small) avoiding set exists. By maximality, there cannot be too many disjoint t-avoiding sets. Thus, we can build sel(f) out of these disjoint “maximally avoiding” sets.

Theorem 12.

Let (𝐀,𝐁) be a PCSP template with {0,1}A and =|B|. Suppose that there exist constants N,{αt}t=1,{βt}t=1 such that every fPol(𝐀,𝐁) has the following properties:

  1. 1.

    f has a 1-avoiding set of size β1.

  2. 2.

    If f is of arity N and has a disjoint family of >αt many t-avoiding sets, all of size βt, then f has a (t+1)-avoiding set of size βt+1.

Then, PCSP(𝐀,𝐁) is 𝖭𝖯-hard.

Proof.

For each fPol(𝐀,𝐁), define t(f) to be the maximal t such that f has a t-avoiding set of size βt. By Assumption 1 and the lack of -avoiding sets (cf. Item 1 of Lemma 11), t(f) exists and 1t(f)<. For each fPol(𝐀,𝐁), let f be a maximal disjoint family of t(f)-avoiding sets of size βt. Define sel(f)=f. Then by Assumption 2, |sel(f)|max{N,max1t<αtβt}k.

Using Theorem 7, it remains to show that for every -chain of minors there are i,j such that πi,j(sel(fi))sel(fj). Let f0π0,1f1π1,2π1,f be such a chain. Since 1t(fi)<, there are distinct i,j such that t(fi)=t(fj)t. By Item 4 of Lemma 11, for every t-avoiding set S of fi, πi,j(S) is t-avoiding for fj. Hence by maximality of fj, πi,j(S) intersects sel(fj). Since sel(fi) is the union of such sets, certainly πi,j(sel(fi)) intersects sel(fj). The rest of this papers will show hardness of certain PCSP templates by showing that they have the two properties from Theorem 12.

3.2 Hardness of promise nonmonochromatic colouring

In this section we prove the advertised hardness results for NAE colouring. The most important case is PCSP(𝐍𝐀𝐄23,𝐍𝐀𝐄3) for 2; all other cases derive from this one by either gadget reductions or homomorphic relaxations.

Lemma 13.

Let 2 and n. Any fPol(n)(𝐍𝐀𝐄23,𝐍𝐀𝐄3) has a 1-avoiding set of size .

Proof.

By Item 2 of Lemma 11, [n] is an (1)-avoiding set and hence a 1-avoiding set. Thus if n, we are done. Otherwise assume n>. Let h=n2. Consider the Kneser graph KG(n,h), and colour each vertex S by f(𝟏S). By Theorem 8, χ(KG(n,h))=n2h+2>, so there are disjoint sets S,T[n](h) with the same colour f(𝟏S)=f(𝟏T)=b. Let X[n](ST).

We claim that X is a {b}-avoiding set and thus 1-avoiding. Since |X|=n2h this completes the proof. In order to prove the claim, consider any Y such that XY[n]; we want to show that f(𝟏Y)b. Construct a matrix in which each column corresponds to an element i[n] and whose rows are 𝟏S,𝟏T,𝟏Y. Observe that since STY=[n] (as S,T,X are a partition of [n]), no column contains only 0s. Similarly since STY= (as S and T are disjoint), no column contains only 1s. Hence all the columns of the matrix whose rows are 𝟏S,𝟏T,𝟏Y are tuples of 𝐍𝐀𝐄23. Since f is a polymorphism of (𝐍𝐀𝐄23,𝐍𝐀𝐄3), (f(𝟏S),f(𝟏T),f(𝟏Y)) must be a tuple of 𝐍𝐀𝐄3. But f(𝟏S)=f(𝟏T)=b, so f(𝟏Y)b as required. Thus X is 1-avoiding.

Lemma 14.

Let 1t< and n(+1)t++1. Suppose fPol(n)(𝐍𝐀𝐄23,𝐍𝐀𝐄3) has >(t) disjoint t-avoiding sets of size t. Then f has a (t+1)-avoiding set of size t+1.

Proof.

By assumption and the pigeonhole principle, f has +1 disjoint sets S1,,S+1[n] of size t that avoid the same T[] of size |T|=t. Let R[n](S1S+1). Let h=|R|2. We have h0 by the lower bound on n.

Consider subsets of [n] which are the union of exactly one of S1,,S+1, and a subset of R of size h; let 𝒮 be the collection of such subsets. We want to find two disjoint sets S,T𝒮 such that f(𝟏S)=f(𝟏T). Observe that there is a bijection between 𝒮 and the vertex set of K+1×KG(R,h), given by taking vertex (i,A) of K+1×KG(R,h) to SiA𝒮. Furthermore, this bijection extends to an isomorphism between the graph K+1×KG(R,h) and the graph whose vertex set is 𝒮 and which considers S,T𝒮 to be adjacent if and only if they are disjoint. By Lemma 9, χ(K+1×KG(R,h))>, so there must exist disjoint sets S,T𝒮 with f(𝟏S)=f(𝟏T)b.

Since SiS for some i, we have bT. Let X[n](ST). Identically to the reasoning in the proof of Lemma 13, observe that for any Y[n] with XY, f(𝟏Y)b. Moreover, XY implies SjY for (+1)21 different values of j (i.e. those for which Sj{S,T}), hence f(𝟏Y)T. Thus X is (T{b})-avoiding, so (t+1)-avoiding. Finally |X|((+1)2)t+|R|2h(1)t+t+1.

See 1

Proof.

The 𝖭𝖯-hardness of PCSP(𝐍𝐀𝐄23,𝐍𝐀𝐄3) for all 2 by Theorem 12, Lemma 13, and Lemma 14. For larger uniformity r>3, we have that PCSP(𝐍𝐀𝐄k3,𝐍𝐀𝐄3)pPCSP(𝐍𝐀𝐄kr,𝐍𝐀𝐄r) by a simple reduction: if I=(V,E) is a 3-uniform hypergraph, then let I=(V,E) be the r-uniform hypergraph with E={(x,y,z,,z)(x,y,z)E}. Then by homomorphic relaxation, as 𝐍𝐀𝐄2r𝐍𝐀𝐄kr𝐍𝐀𝐄r, we have PCSP(𝐍𝐀𝐄2r,𝐍𝐀𝐄r)pPCSP(𝐍𝐀𝐄kr,𝐍𝐀𝐄r).

3.3 Hardness of promise conflict-free and linearly-ordered colouring

In this section we prove the advertised hardness results for both LO and CF colourings. For the CF colourings, by Lemma 6 it suffices to establish hardness for r=4. However, our proof is the same for any r4 and thus we will present it that way. Since 𝐋𝐎kr𝐂𝐅kr, we can then do both LO and CF colourings “in one go” by proving the hardness of PCSP(𝐋𝐎3r,𝐂𝐅r) for all 3 and r4. In the following proofs, we let 𝟎 denote the vector whose elements are all 0, and we let 𝟐X denote a “scaled indicator vector”: 𝟐X=2𝟏X.

Lemma 15.

Let 3 and r4. Then any fPol(n)(𝐋𝐎3r,𝐂𝐅r) has a 1-avoiding set of size .

Proof.

By Item 2 of Lemma 11, [n] is an (1)-avoiding set and hence a 1-avoiding set. Thus if n, we are done. Otherwise assume n>, and set h=n2. Consider the Kneser graph KG(n,h), and colour each vertex S by f(𝟐S). By Theorem 8, χ(KG(n,h))=n2h+2>, so there exist disjoint sets S,T[n](h) such that f(𝟐S)=f(𝟐T). Let X[n](ST).

Now observe that for any Y[n] with XY, 𝟏Y is compatible with 𝟐S,𝟐T, and r3 many copies of 𝟎. To see why this is the case, consider the matrix whose rows are 𝟏Y,𝟐S,𝟐T and r3 copies of 𝟎. The columns of the matrix correspond to elements i[n]. We must show that each column contains a unique maximum element. Observe that for iS, the unique maximum is a 2 in the 𝟐S row; for iT the unique maximum is a 2 in the 𝟐T row; and for all other i the unique maximum is a 1 in the 𝟏Y row. Hence, since f is a polymorphism of (𝐋𝐎3r,𝐂𝐅r), we have that (f(𝟏Y),f(𝟐S),f(𝟐T),f(𝟎),,f(𝟎)) is a tuple of 𝐂𝐅r. Since f(𝟐S)=f(𝟐T) we deduce that f(𝟏Y)f(𝟎). Thus it follows that X is {f(𝟎)}-avoiding, and thus 1-avoiding. Noting that X has size n2h completes the proof.

Lemma 16.

Let 3 and r4. Let 1t<, and suppose that fPol(n)(𝐋𝐎3r,𝐂𝐅r) has a disjoint family of >(t) many t-avoiding sets, all of size t. Then f has a (t+1)-avoiding set of size (t+1).

Proof.

By assumption and the pigeonhole principle, there is a family of at least +1 disjoint T-avoiding sets of size t for some T[n] of size t. Let h,S,T,X be as in the proof of Lemma 15 – thus S,T[n](h),X[n](n2h),f(𝟐S)=f(𝟐T) and S,T,X form a partition of [n]. Recall that |X|.

Now, since the sets in are disjoint, at most of the sets in intersect X. Thus, since ||+1, there is some Z disjoint from X. Let C be any other set in and define C=CX. Note that |C|(t+1). Note that f(𝟏Z)T as Z is T-avoiding. We will show that C is (T{f(𝟏Z)})-avoiding, proving the result.

Since C is T-avoiding and CC, C is also T-avoiding (cf. Item 3 of Lemma 11). To see that C is {f(𝟏Z)}-avoiding, note that for every D[n] with CD, 𝟏D is compatible with 𝟐S,𝟐T, and r3 many copies of 𝟏Z. To see why this is the case, consider the matrix whose rows are 𝟏D,𝟐S,𝟐T and r3 copies of 𝟏Z. Each column corresponds to an element of i[n], and we want to show that each column has a unique maximum element. For iS (respectively iT), the unique maximum is given by a 2 in the 𝟐S row (respectively the 𝟐T row) – these two cases are disjoint since S,T are disjoint. Since S,T,X form a partition of [n], we now consider iX. Since S,T,Z are disjoint from X, all the 𝟐S,𝟐T,𝟏Z rows have a 0 in these columns. On the other hand, the (unique) 𝟏D row has a 1, since iXCD. Hence we see that every column in the matrix is an element of 𝐋𝐎3r. Applying f row-wise, we see that (f(𝟐S),f(𝟐T),f(𝟏D),f(𝟏Z),,f(𝟏Z)) must be a tuple of the relation of 𝐂𝐅3r.

Hence since f(𝟐S)=f(𝟐T) and since f is a polymorphism of (𝐋𝐎3r,𝐂𝐅r), we have that f(𝟏D)f(𝟏Z). Thus C is the required (T{f(𝟏Z)})-avoiding, so (t+1)-avoiding set.

Theorem 17.

PCSP(𝐋𝐎3r,𝐂𝐅r) is 𝖭𝖯-hard for all constant 3 and r4.

Proof.

By Theorem 12, Lemma 15 and Lemma 16. Theorem 2 and Theorem 4 follow immediately: See 2

Proof.

The case r=3 is given by Theorem 1 as 𝐍𝐀𝐄k3=𝐂𝐅k3 for every k2. The case r5 and 2k follows from Theorem 1 and Lemma 6. For r=4 and k=2, recall from Section 1 that 𝐂𝐅24 is identical to solving systems of mod-2 equations of the form x+y+z+t1(mod2), so CSP(𝐂𝐅24) is in 𝖯. By homomorphic relaxation, so is PCSP(𝐂𝐅24,𝐂𝐅4). For r=4 and 3k, the result follows by Theorem 17 and by homomorphic relaxation as 𝐋𝐎3r𝐋𝐎kr𝐂𝐅kr.

See 4

Proof.

By Theorem 17 and by homomorphic relaxation as 𝐋𝐎3r𝐋𝐎kr and 𝐋𝐎r𝐂𝐅r.

 Remark 18.

In the above proofs of Lemma 15 and Lemma 16, the only required property of 𝐂𝐅r is that if a tuple in the relation has two entries which are equal, then the remaining r2 entries in the tuple cannot all be equal. Thus, the same proof also shows a stronger result: Define 𝐁𝐍𝐀𝐄s,rs (Block-NAE) as the template on domain [], with a single relation which contains exactly the tuples for which if any s coordinates have the same value, then the remaining rs coordinates cannot all have the same value. (This relation with s=2 is strictly larger than the relation corresponding to 𝐂𝐅r when r6). Then we have shown that PCSP(𝐋𝐎3r,𝐁𝐍𝐀𝐄2,r2) is 𝖭𝖯-hard for all constant 3,r4. In fact, using the same proof technique, one can also show 𝖭𝖯-hardness of PCSP(𝐋𝐎3r,𝐁𝐍𝐀𝐄s,rs) for all constant 3,r4,2sr2 by considering the s-uniform Kneser hypergraph KG(s)(n,a), whose chromatic number is known to be ns(a1)s1 [1].

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