Oriented discrepancy of Hamilton cycles

We propose the following extension of Dirac's theorem: if G $G$ is a graph with n≥3 $n\ge 3$ vertices and minimum degree δ(G)≥n∕2 $\delta (G)\ge n\unicode{x02215}2$ , then in every orientation of G $G$ there is a Hamilton cycle with at least δ(G) $\delta (G)$ edges oriented in the same direction. We prove an approximate version of this conjecture, showing that minimum degree n+8k2 $\frac{n+8k}{2}$ guarantees a Hamilton cycle with at least (n+k)∕2 $(n+k)\unicode{x02215}2$ edges oriented in the same direction. We also study the analogous problem for random graphs, showing that if the edge probability p=p(n) $p=p(n)$ is above the Hamiltonicity threshold, then, with high probability, in every orientation of G~G(n,p) $G\unicode{x0007E}G(n,p)$ there is a Hamilton cycle with (1−o(1))n $(1-o(1))n$ edges oriented in the same direction.


Introduction
Recently there has been much interest in discrepancy-type problems on graphs.In the general setting of combinatorial discrepancy theory, one is given a hypergraph H and should colour the vertices of H with two colours so that each edge is as balanced as possible.More precisely, given a colouring f : V (H) → {−1, 1}, define the imbalance of an edge e ∈ H to be D(e) = x∈e f (x) .The discrepancy of H is then defined as min f :V (H)→{−1,1} max e∈E(H) D(e).In other words, the discrepancy is the maximum guaranteed imbalance.We refer the reader to Chapter 4 in the book of Matoušek [19] for an overview of combinatorial discrepancy.
Recent works considered instances of the above general setting, in which the hypergraph H originates from a graph, that is, V (H) is the edge-set of some graph G, and E(H) is a family of certain subgraphs of G.The goal is to identify conditions on G which guarantee that there exists a high-discrepancy subgraph of the specific type.For example, Balogh, Csaba, Jing and Pluhár [1] proved that if G has n vertices and minimum degree (3/4 + ε)n, then in every 2-colouring of E(G), there is a Hamilton cycle with at least (1/2 + ε/64)n edges of the same colour (and hence discrepancy at least εn/32).This was generalised to an arbitrary number of colours (with a suitable definition of multicolour discrepancy) in [11] and independently [8].This result (for two colours) was extended to K r -factors and r-powers of Hamilton cycles in [2] and [4], respectively.In [10], the authors determined the multicolour discrepancy of Hamilton cycles in the complete graph, and showed that if p is above the Hamiltonicity threshold, then the discrepancy of Hamilton cycles in G(n, p) is essentially as large as in the complete graph.
In the present work, we consider an oriented version of graph discrepancy.Here, instead of colouring the edges of the graph with two colours, one considers orientations of the edges.Throughout this paper, we use {u, v} to denote an unordered edge (or pair), and (u, v) to denote the directed edge from u to v. Hamilton cycles are naturally suitable for the oriented setting since they have a notion of direction.The goal is thus to find a Hamilton cycle with as many edges as possible which are oriented in the direction of the cycle.More precisely, for an n-vertex graph G, an orientation D of G and a Hamilton cycle It is easy to see that δ(G) = δ.Now, orient all edges from A to B (and orient the edges inside B arbitrarily).Consider any Hamilton cycle in G, and fix a cyclic direction.Any edge pointing forward (with respect to that direction) ends at a distinct vertex of B; hence there are at most |B| = δ such edges.We note that Conjecture 1.1 holds when G is a complete graph due to the fact that every tournament has a directed Hamilton path.
Our first result is an approximate version of Conjecture 1.1.
Theorem 1.2.Let k ≥ 0 and let G be an n-vertex graph with n ≥ 30+4(k−1) and δ(G) ≥ n+8k 2 .Then in every orientation of G, there is a Hamilton cycle with at least n+k 2 edges in the same direction.
As mentioned above, Conjecture 1.1 can be viewed as an extension of Dirac's theorem, stating that Dirac's theorem holds in a "robust way".Namely, not only do graphs with large minimum degree (above n/2) have Hamilton cycles, but the collection of Hamilton cycles is robust enough to guarantee the existence of an unbalanced Hamilton cycle in every edgeorientation.It is worth mentioning that there is by now an extremely rich literature (certainly too rich to be comprehensively surveyed here) on extensions and generalisations of Dirac's theorem in various directions.We refer the reader to the surveys of Gould [12][13][14] and of Kühn and Osthus [17,18], as well as the survey of Sudakov [23] focusing on robustness-type results.
As our second result, we study oriented discrepancy of Hamilton cycles in random graphs.We show that if the edge probability p is above the Hamiltonicity threshold, then with high probability1 (whp for short), in every orientation of G(n, p) there is a Hamilton cycle with almost all edges oriented in the same direction.
Theorem 1.3.Let n be an integer and let p ≥ (log n + log log n + ω(1))/n.Then G ∼ G(n, p) is whp such that in any orientation of its edges there exists a Hamilton cycle with at least (1 − o(1))n edges in the same direction.
The proof of Theorem 1.3 combines machinery from our earlier paper [10] with a result on the oriented discrepancy of almost-spanning paths in sparse expanders (see Theorem 3.1).In the special case of random graphs, this result is as follows.
Theorem 1.4.For every δ > 0 there is C > 0 such that G ∼ G(n, C/n) whp satisfies the following: in any orientation of the edges of G there exists a path of length at least (1 − δ)n with at most δn edges going backwards.
Theorem 1.4 has the following interesting implication to unbalanced Hamilton cycles in tournaments.It is a well-known fact [22] that every tournament has a directed Hamilton path.Hence, in any tournament, there is a Hamilton cycle (of the underlying complete graph) in which all edges but at most one go in the same direction.It may be, however, that a tournament has only one directed Hamilton path (this is the case in the transitive tournament).It is then natural to relax the notion of "imbalance" by considering Hamilton cycles which have up to δn edges in the opposite direction, and ask how many such unbalanced Hamilton cycles are guaranteed to exist.The following corollary addresses this question.
Corollary 1.5.For every δ > 0 there is C > 0 such that every tournament T on n vertices contains at least C −n n n Hamilton cycles (of the underlying complete graph), each of which has at least (1 − δ)n edges in the same direction.
Corollary 1.5 is optimal up to the value of the constant C. Indeed, let T be the random tournament where each edge is oriented randomly (with probability 1/2 for each direction).For every Hamilton cycle C in the underlying complete graph of T , the probability that C has at least (1/2 + ε)n forward edges is at most e −Ω(ε 2 n) , by the Chernoff bound.Hence, the expected number of Hamilton cycles having at least 0.51n (say) forward edges, is at most e −Ω(n) n!.By a similar consideration, the expected number of Hamilton cycles having at least (1 − δ)n edges in the same direction is at most c −n n! where c = 2 1−O(δ log(1/δ)) .Determining the correct dependence of C on δ in Corollary 1.5 would be interesting.Theorem 1.2 is proven in the next section, Theorems 1.3 and 1.4 are proven in Section 3 and Corollary 1.5 is proven in Section 4.

Unbalanced Hamilton cycles in Dirac graphs
A diamond is the graph with vertices a, b, c, d and edges {a, b}, {a, c}, {a, d}, {b, c}, {b, d}.We call c, d the ends of the diamond, and a, b the centre of the diamond.Given an orientation of its edges, a diamond is called good (with respect to that orientation) if c and d relate in the same way to a, b.(The direction of the edge {a, b} will not be important, as the roles of a, b are the same.For convenience, we assume that a → b.)There are four good diamonds, depicted in Fig. 1.
We say that a path is positive (with respect to an orientation) if it consists of more forward edges than backward edges when traversing the path from its first vertex to its last.Similarly, it is negative if it consists of more backward edges than forward edges.The following is the key property of good diamonds that we will use.It can easily be verified by hand (see Fig. 1).Proof.Edwards [6] and independently Khadžiivanov and Nikiforov [15] proved that a graph with n vertices and n2 4 + 1 edges has an edge which is contained in at least n/6 triangles (earlier, Erdős [7] proved a bound of n/6 − O(1)).So let {a, b} ∈ E(G) be an edge contained in at least n/6 triangles, and let x 1 , . . ., x m , m ≥ n/6 ≥ 5 be such that a, b, x i is a triangle in G. Fix any orientation of G.By the pigeonhole principle, there are 1 ≤ i < j ≤ m such that x i , x j relate in the same way (in the orientation) to a, b.Then a, b, c := x i and d := x j form a good diamond in G, as required.
Proof.The proof is by induction on k.Fix any orientation of G.It is easy to check that e(G) ≥ n 2 /4+1 by the assumptions of the lemma.By Lemma 2.2, G contains a good diamond D (this also establishes the base case k = 1).Let G be the subgraph of By the induction hypothesis, G contains k−1 vertex-disjoint good diamonds, finishing the proof.
We remark that Lemma 2.3 is optimal in the following sense.Consider the complete nvertex tripartite graph G = (A ∪ B ∪ C, E) where |A| = k − 1 and |B|, |C| are nearly equal.It is not hard to check that G has n 2 /4 + Θ(kn) edges but at most k − 1 vertex-disjoint diamonds.
The last ingredient in the proof of Theorem 1.2 is the following lemma due to Pósa.

Lemma 2.4 ([20]
).Let t ≥ 0 and let G be a graph with n vertices and minimum degree at least . So E is the edge-set of a path forest, and |E| = 3k.By Lemma 2.4, there is a Hamilton cycle H of G which contains all edges in E; in other words, H contains P 1 , . . ., P k as subpaths.Fix a direction of H which matches the orientation of the majority of the edges in E(H) E. We now define a Hamilton cycle H as follows.
For i = 1, . . ., k, we say that D i matches the direction of H if, when traversing P i ⊆ H along the fixed direction of H, we go from c i to d i (and not from d i to c i ).Recall that P i is positive when traversing it from c i to d i (and hence negative when traversing it from d i to c i ).Let I be the set of i = 1, . . ., k such that P i does not match the direction of H. Now let H be the Hamilton cycle with edge-set In other words, for each i ∈ I, we replace P i with Q i .It is easy to see that H is indeed a Hamilton cycle, and that it has at least n+k 2 edges in the same direction.

Unbalanced Hamilton cycles in sparse expanders
To prove Theorem 1. Theorem 3.1.For every δ > 0 there exists β > 0 such that the following holds.Let G be a β-graph on n vertices.Then, in any orientation of the edges of G there exists a path of length at least (1 − δ)n with at most δn edges going backwards.
Theorem 1.4 follows from Theorem 3.1 by the well-known and simple fact that for every β > 0 there exists C > 0 such that G(n, C/n) is whp a β-graph.For a proof (of a stronger statement), see e.g.[10,Lemma 3.4].
To prove Theorem 3.1, we will need the following result from [3].Proof of Theorem 3.1.Fix any orientation of G. Let β = β(δ) be small enough (to be chosen later).Set : and m ≤ 2n +2 ≤ 3 √ βn.We claim that G contains a collection of m vertex-disjoint directed paths of length each.We find these paths one by one.Suppose we already found paths P 1 , . . ., P j , j < m.Let G be the graph obtained from G by deleting , where the last inequality uses that ( + 2) < ( + 1) 2 ≤ 1 2β .So χ(G ) ≥ + 1.By the Gallai-Hasse-Roy-Vitaver theorem (see e.g., [24, Theorem 5.1.21]),every orientation of a graph with chromatic number k contains a directed path of length k − 1.Hence, G contains a directed path of length .This proves our claim about the existence of P 1 , . . ., P m .
For each 1 ≤ i ≤ m, let x i (resp.y i ) be the first (resp.last) vertex of P i .Define an auxiliary directed graph F on [m] in which i → j if and only if {y i , x j } ∈ E(G).We claim that for every pair of disjoint A, B ⊆ [m] with |A|, |B| ≥ βn, there is an edge of F from A to B. Indeed, set X := {x i : i ∈ A} and Y := {y i : i ∈ B}.Since G is a β-graph, there is an edge in G between X and Y ; namely there are j ∈ A, i ∈ B with {y i , x j } ∈ E(G).This gives that i → j in F . and each Hamilton cycle is counted at most n times this way.Hence, the number of such Hamilton cycles is at least for a large enough constant C depending on C and δ. (P5) G is (β, p)-pseudorandom.
Lemma A.4.For every ζ > 0 there exists c > 0 such that the following holds.Let p = (log n + log log n + ω( 1))/n and let G ∼ G(n, p).Then, whp, G satisfies the following: for every Summing over all choices of W and H, we see that the probability that the assertion of the lemma does not hold is at most Setting g(t) := (2en log n/t) t , we note that g (t) = g(t) • (log(2en log n/t) − 1) > 0 for every t in the range of the sum in (1), assuming c < 1, say.Thus, this sum is not larger than Now, if c is small enough so that log(2e/c) • c < ζ 2 /32, we get that (1) tends to 0 as n tends to infinity.This completes the proof.
The following lemma ([10, Lemma 4.6]) states that a graph possessing certain simple properties is necessarily an expander.Such statements are common in the study of Hamiltonicity of random graphs.
Lemma A.5 ([10, Lemma 4.6]).Let m, d ≥ 1 be integers and let H be a graph on h ≥ 4m vertices satisfying the following properties: 4.There is an edge between every pair of disjoint sets Then H is an (h/4, 2)-expander.Proof.The proof follows that of [10,Lemma 4.7].We may and will assume ε is sufficiently small (it is enough to have ε ≤ min{1/10, 125ζ, c/2}, where c = c(ζ) is the constant from Lemma A.4).We will assume that the events defined in Lemmas A.3 and A.4 hold, where we use (P5) in Lemma A.3 with β = ε/500.These events hold whp.We will show that then, deterministically, the assertion of Lemma A.6 holds.
It will be convenient to set Now, let us bound (from above) the probability that |E H (U 1 , U 2 )| = 0 (where the randomness is with respect to the choice of H).Recall that H is defined by choosing, for each v ∈ W , a random set E(v) of min{d(v, W ), d 0 } edges of G[W ] incident to v, with all choices made uniformly and independently, and letting E(H) = v∈W E(v).Fix any u 1 ∈ U 1 with d(u 1 , U 2 ) ≥ 1, and let A u 1 be the event that there is no edge in E(u 1 ) with an endpoint in U 2 .Observe that if d(u 1 , W ) < d 0 then P(A u 1 ) = 0, and otherwise Here, in the last inequality we used Property (P1) in Lemma A.3.Note that the events (A u 1 : u 1 ∈ U 1 ) are independent, and that if where in the last inequality we used (2).By taking the union bound over all at most 2 2n choices of U 1 , U 2 , we see that with high probability, E H (U 1 , U 2 ) = ∅ for every pair of disjoint sets U 1 , U 2 ⊆ W of size m each.Finally, we apply Lemma A.5 to conclude that whp H is a (|W |/4, 2)-expander.From now on, we fix such a choice of H. Before establishing the assertion of the lemma, we first show that G[W ] is Hamiltonian.To find a Hamilton cycle in G[W ], we define a sequence of graphs H i , i ≥ 0, as follows.To begin, set H 0 = H.For each i ≥ 0, if H i is Hamiltonian then stop, and otherwise take a booster of H i contained in G[W ] and add it to H i to obtain H i+1 .That such a booster exists is guaranteed by Lemma A.4, as we will always have where c is the constant appearing in Lemma A.4.Note also that H i is a subgraph of G[W ] for each i ≥ 0. Evidently, this process has to stop (because as long as H i is not Hamiltonian, the maximum length of a path in H i is longer than in H Proof of Lemma 3.3.Set ε = δ/3.For convenience, we show the existence of a partition V (G) = V ∪ V with |V | ≤ 2εn instead of |V | ≤ εn (this clearly does not matter).We assume that G satisfies the properties detailed in Lemma A.3 for β < δ/40 and the assertion of Lemma A.6.These events happen whp.Let U 1 , U 2 be disjoint subsets of V = V (G) satisfying (P4).Set V = U 1 ∪ U 2 and V = V V , and let P ⊆ V be a path with |V (P )| ≤ (1 − δ)n and endpoints a 1 , a 2 .In particular, |V | ≤ 2εn.Our goal is to extend P to a Hamilton cycle of G. Write V = V V (P ), partition V = V 1 ∪V 2 as equally as possible, so |V i | ≥ (δ−2ε)n/2 ≥ δn/10.For i = 1, 2, let W i = V i ∪ U i and choose a neighbour w i of a i in W i ; this is possible since Note that we always fix one of the two directions of the Hamilton cycle C.Let e D (C) be the number of forward edges.The oriented discrepancy of Hamilton cycles in G is the maximum k such that for every orientation D of G, there is a Hamilton cycle C in G such that e D (C) ≥ n+k 2 .Generalising Dirac's theorem [5], we propose the conjecture that if δ(G) ≥ n+k 2 then the oriented discrepancy of Hamilton cycles in G is at least k.It is convenient to state this conjecture in the following form: Conjecture 1.1.Let n ≥ 3 and let G be an n-vertex graph with δ(G) ≥ n/2.Then, in any orientation of the edges of G, there exists a Hamilton cycle in which at least δ(G) of the edges are pointing forward.If true, Conjecture 1.1 would be best possible.To see this, fix n ≥ 3 and δ ≥ n/2, and consider the graph G with vertex-partition V (G) = A ∪ B, where |A| = n − δ and |B| = δ.The edge set is the set of all pairs which touch B.

Observation 2 . 1 .
Let D be a good diamond with ends c, d.Then D contains a positive path of length 3 from c to d, as well as a positive path of length 3 from d to c. Lemma 2.2.Let G be a graph with |V (G)| = n ≥ 30 vertices e(G) ≥ n 2 4 + 1.Then every orientation of G contains a good diamond.

Figure 1 :
Figure 1: The four good diamonds.

n+t 2 .• n+8k 2 = n 2 4
Let E ⊆ E(G) be the edge-set of a path forest 2 of size at most t.Then there exists a Hamilton cycle in G which uses all edges in E.Proof of Theorem 1.2.If k = 0 then the assertion follows immediately from Dirac's theorem.So suppose from now on that k ≥ 1. Fix any orientation of G.We have e(G) ≥ n 2 + 2kn ≥ n 2 4 + 2(k − 1)n − 4k 2 + 6k − 1, hence we may apply Lemma 2.3.Let D 1 , . . ., D k be vertex-disjoint good diamonds in G. Let us denote by c i , d i the ends of D i .For each 1 ≤ i ≤ k, let P i ⊆ D i be a positive path of length 3 from c i to d i , and let Q i ⊆ D i be a positive path of length 3 from d i to c i ; such paths exists by Observation 2.1.Let 3 we first prove Theorem 3.1 below, which extends Theorem 1.4 to the more general setting of graphs with certain expansion properties.More precisely, for β > 0, a graph G is called a β-graph if there is an edge in G between every pair of disjoint U, W ⊆ V (G) with |U |, |W | ≥ β|V (G)|.This notion was introduced in [9].

Lemma 3 . 2 (
Lemma 4.4  in[3]).Let m, k ≥ 1, let F be a directed graph on m vertices, and suppose that for every pair of disjoint A, B ⊆ V (F ) with |A|, |B| ≥ k, there is an edge of F from A to B. Then F has a directed path of length at least m − 2k + 1.

( P2 )
No vertex v ∈ V (G) with d(v) < log n/10 is contained in a 3-or a 4-cycle, and every two distinct vertices u, v ∈ V (G) with d(u), d(v) < log n/10 are at distance at least 5 apart.(P3) Every set U ⊆ V (G) of size at most εn/100 spans at most ε|U | log n/10 edges.(P4) There exist disjoint sets U 1 , U 2 ⊆ V (G) with |U 1 |, |U 2 | ≤ εn for which the following hold for every v ∈ V (G): (a) If d(v) ≥ log n/10 then d(v, U 1 ), d(v, U 2 ) ≥ ε log n/100; (b) If d(v) ≤ log n/10 then v and all of its neighbours are in U 1 .
or a 4-cycle, and every two distinct vertices u, v ∈ V (H) with d(u), d(v) < d are at distance at least 5 apart; 3. Every set U ⊆ V (H) of size at most 5m contains at most d|U |/10 edges;

Lemma A. 6 .
Let ε, ζ > 0. For p = (log n + log log n + ω(1))/n, the random graph G ∼ G(n, p) satisfies the following whp.Let W ⊆ V (G) be such that |W | ≥ ζn, and for every v ∈ W it holds that d(v, W ) ≥ min{d(v), ε log n}.Then for every w ∈ W there exists Y ⊆ W with |Y | ≥ ζn/4 such that for each y ∈ Y , there is a Hamilton path in G[W ] whose endpoints are w and y.
be as in the statement of Lemma A.6.We select a random spanning subgraph H of G[W ] as follows.For each v ∈ W , if d(v, W ) < d 0 then add to H all edges of G[W ] incident to v. Otherwise, namely if d(v, W ) ≥ d 0 , then randomly select a set of d 0 edges of G[W ] incident to v and add these to H. Note that |E(H)| ≤ |W | • d 0 ≤ εn log n.On the other hand, our assumption that d(v, W ) ≥ min{d(v), ε log n} for every v ∈ W implies that δ(H) ≥ min{δ(G), d 0 }.Hence, as d 0 ≥ 2 (for large enough n), we have δ(H) ≥ 2 by Property (P1) of Lemma A.3.We claim that with positive probability (in fact, whp), H is a (|W |/4, 2)-expander.In light of Lemma A.5, it is sufficient to show that with positive probability, H satisfies Conditions 1-4 in that lemma.Here, we will choose the parameters of Lemma A.5 as d := d 0 and m := εn/500.Note that |V (H)| = |W | = ζn ≥ εn/125 = 4m, as required by Lemma A.5.We already showed that δ(H) ≥ 2 (which is Condition 1 in Lemma A.5).Let us prove Condition 2. Observe that if v ∈ V (H) = W has degree less than d 0 in H, then also d(v, W ) < d 0 in G (by the definition of H).But then, since d(v, W ) ≥ min{d(v), d 0 } by the assumption of Lemma A.6, we have d(v) < d 0 = ε log n ≤ log n/10.Now, by (P2) in Lemma A.3, there is no 3-or 4-cycle in G containing a vertex v whose degree in H is less than d 0 , and every two such vertices are at distance at least 5 apart.Condition 3 in Lemma A.5 holds because H is a subgraph of G and due to Property (P3) in Lemma A.3 (note that 5m = εn/100).Let us now prove that Condition 4 holds.Let U 1 , U 2 ⊆ V (H) = W be disjoint sets satisfying |U 1 |, |U 2 | = m = εn/500.Since G is (β, p)pseudorandom with β = ε/500, we have i−1 ), thus showing that G[W ] must contain a Hamilton cycle, as claimed.Now let w ∈ W .As G[W ] is Hamiltonian, there exists a Hamilton path P of G[W ] such that w is one of the endpoints of P .Evidently, P is a longest path in G[W ].Furthermore, note that G[W ] is a (|W |/4, 2)-expander because H, a subgraph of G[W ], is such an expander.By Lemma A.1 (applied to G[W ]), there exists a set R ⊆ V (P ) = W such that for every y ∈ R there is a Hamilton path in G[W ] with endpoints w and y, and such that |N G[W ] (R)| ≤ 2|R|−1.Now, since G[W ] is a (|W |/4, 2)-expander, it must be the case that |R| > |W |/4 ≥ ζn/4.So we see that the assertion of the lemma holds with Y = R.This completes the proof.
[10,ize |W | ≥ ζn and for every (|W |/4, 2)-expander H on W which is a subgraph of G and has at most cn log n edges, if H is not Hamiltonian then G contains a booster with respect to H.Proof.The proof follows that of[10, Lemma 4.5].We use a first moment argument.Evidently, the number of choices for the set W is at most 2 n .Let us fix a choice of W .For each t, the number of choices of H for which |E(H)| = t is at most t .Now let H be a non-Hamiltonian (|W |/4, 2)-expander on W , and set t := |E(H)|.It is known and easy to show that if a graph H is a (|V (H)|/4, 2)-expander then H is connected.By Lemma A.2, H has at least (|W |/4) 2 /2 = |W | 2 /32 ≥ ζ 2 n 2 /32 boosters.Now, the probability that G contains H, but no booster thereof is at most