aa r X i v : . [ m a t h . C O ] M a y A note on short cycles in a hypercube
Maria Axenovich ∗ Ryan Martin † Abstract
How many edges can a quadrilateral-free subgraph of a hypercube have? Thisquestion was raised by Paul Erd˝os about 27 years ago. His conjecture that such asubgraph asymptotically has at most half the edges of a hypercube is still unresolved.Let f ( n, C l ) be the largest number of edges in a subgraph of a hypercube Q n containingno cycle of length l . It is known that f ( n, C l ) = o ( | E ( Q n ) | ), when l = 4 k , k ≥ f ( n, C ) ≥ | E ( Q n ) | . It is an open question to determine f ( n, C l ) for l = 4 k + 2, k ≥
2. Here, we give a general upper bound for f ( n, C l ) when l = 4 k + 2 and providea coloring of E ( Q n ) by 4 colors containing no induced monochromatic C . Let Q n be a hypercube of dimension n . We treat its vertices as binary sequences of length n or as subsets of a set [ n ] = { , , . . . , n } , whichever is more convenient. The edges of Q n correspond to pairs of sets with symmetric difference of size 1 or, equivalently, to pairs ofsequences with Hamming distance 1. We denote the set of subsets of [ n ] of size k by (cid:0) [ n ] k (cid:1) .We say that a subset of edges forms an i th edge layer of a hypercube if these edges joinvertices in (cid:0) [ n ] i (cid:1) and (cid:0) [ n ] i +1 (cid:1) . It is a classical question to find a dense subgraph of a hypercubewithout cycles of a certain fixed length. Moreover, it is of interest to consider the Ramseyproperties of cycles in Q n . We say that a cycle C l has a Ramsey property in Q n , if for a every k , there is an N such that if n > N and the edges of Q n are colored in k colors then thereis always a monochromatic C l in a such coloring.It is easy to see that there is a coloring of E ( Q n ) in two colors with no monochromatic C .Indeed, color all edges in each edge-layer in the same color, using two alternating colors onsuccessive layers. Each color class is a quadrilateral-free subgraph of Q n , with the larger onehaving at least | E ( Q n ) | edges. There is a coloring by Conder [5] using three colors on E ( Q n )and containing no monochromatic C , in particular providing a hexagon-free subgraph of Q n with at least | E ( Q n ) | / C and C do not have Ramsey propertyin a hypercube.In general, if f ( n, C l ) is the largest number of edges in a subgraph of Q n with no cycleof length l then the following facts are known: Fan Chung [4] proved that f ( n, C ) ≤ ∗ Department of Mathematics, Iowa State University, Ames, IA 50011, [email protected] † Department of Mathematics, Iowa State University, Ames, IA 50011, [email protected] | E ( Q n ) | . The conjecture of Erd˝os that f ( n, C ) = (1 / o (1)) | E ( Q n ) | is still open.Chung, [4] also gave the following upper bound for cycles of length 0 modulo 4: f ( n, C k ) ≤ cn (1 / k ) − (1 / | E ( Q n ) | , k ≥ C k has the Ramsey property for k ≥
2. In [4], Chung also raised the question aboutRamsey properties of cycles C k +2 , k ≥
2, in particular about C . Recently, Alon, Radoi˘ci´c,Sudakov and Vondr´ak, [1], completely settled the problem about Ramsey properties of cyclesin the hypercube by proving that for any l ≥ C l has a Ramsey property. In this note, weinvestigate the Ramsey properties of induced cycles in the hypercube and show that C , asan induced subgraph of Q n , does not have the Ramsey property. Theorem 1.1
There is a coloring of E ( Q n ) using colors such that there is no inducedmonochromatic C , C , or C . We prove this theorem and describe some properties of the corresponding coloring whichare of independent interest in Section 2. For completeness, we provide a general upper boundon the maximum number of edges in a subgraph of a hypercube containing no cycle of length C k +2 in Section 3. For the general graph-theoretic definitions we refer the reader to [8]. C x , let w ( x ) be the weight (or number of 1’s) in x .If the vertices corresponding to the binary sequences x and y form an edge in Q n (i.e., theydiffer in exactly one position) let the common substring that precedes the change be calledthe common prefix and be denoted p = p ( xy ). Description of a coloring c For an edge xy , with x ∈ (cid:0) [ n ] k (cid:1) , y ∈ (cid:0) [ n ] k +1 (cid:1) and having common prefix p = p ( xy ),let c ( xy ) = ( c , c ), where c ≡ k (mod 2) , c ≡ w ( p ) (mod 2) . We shall prove that the coloring c does not produce induced monochromatic cycles of length4, 6, or 10. Since the sets of colors used on consecutive edge-layers of Q n are disjoint, themonochromatic cycles can occur only within edge-layers. There is no C within an edge-layer,so there is no monochromatic C .Let C be a monochromatic cycle in Q n under coloring c . Then C = x , y , . . . , x m , y m , x ,where x i ∈ (cid:0) [ n ] k (cid:1) , y i ∈ (cid:0) [ n ] k +1 (cid:1) , for i = 1 , . . . , m and for some k ∈ { , . . . , n − } . Note also that y i = x i ∪ x i +1 , for i = 1 , . . . , m where addition is taken modulo m . In particular, we havethat dist( x i , x i +1 ) = 2 and dist( y i , y i +1 ) = 2, for i = 1 , . . . , m .In the following lemma, we settle the case with C mostly to illustrate the properties ofthe coloring. 2 emma 2.1 There is no monochromatic C in Q n under coloring c .Proof. Let C = x y x y x y x be a monochromatic C . Then x i ’s can be written as a a a a , a a a a , a a a a , for some binary words a , . . . , a . The correspond-ing y i ’s are a a a a , a a a a , a a a a . Therefore, the set of common prefixes onthe edges of C contains a a and a a . These two prefixes have different weights modulo2, thus C is not monochromatic under coloring c . (cid:4) For a word u = u u · · · u n , let the reverse of u be u = u n u n − · · · u . Observation 1.
Let x, x ′ ∈ (cid:0) [ n ] k (cid:1) and y, y ′ ∈ (cid:0) [ n ] k +1 (cid:1) . Let xy , x ′ y ′ be edges of Q n such that x and y have a common prefix p and a common suffix s and x ′ and y ′ have a common prefix p ′ and a common suffix s ′ . Then w ( p ) ≡ w ( p ′ ) (mod 2) iff w ( s ) ≡ w ( s ′ ) (mod 2). This is alsoequivalent to saying that if q is the common prefix of x and y and q ′ is the common prefixof x ′ and y ′ then w ( p ) ≡ w ( p ′ ) (mod 2) iff w ( q ) ≡ w ( q ′ ) (mod 2).For a binary sequence x of length n , we denote by x [ I ] its restriction to positions from I .For example, if x = 0010010110 and I = { , , , , } then x [ I ] = 01010. Main idea of the proof.
Let C = x , y , . . . , x , y , x be an induced C in Q n with x i ∈ (cid:0) [ n ] k (cid:1) and y i ∈ (cid:0) [ n ] k +1 (cid:1) , i = 1 , . . . ,
5. Let I ( C ) = { i , . . . , i m } , i < i < · · · < i m , be theset of positions where x i and x i +1 differ for some i ∈ [5].We define a 5 × m matrix A = A ( C ) where the i th row of A is x i [ I ].First, we shall show that m = 5. Second, we shall show that none of the 5 × A ( C ) for an induced cycle C monochromatic under coloring c . Note.
For the rest of the proof we assume that C is monochromatic under coloring c andthat it is an induced cycle in Q n . We take the addition modulo 5 unless otherwise specified. Lemma 2.2 | I ( C ) | = 5 . Moreover, ’s occur in consecutive positions (modulo ) in eachcolumn of A ( C ) .Proof. Let | I ( C ) | = { i , . . . , i m } . Observe first that all five rows of A ( C ) form distinctbinary words of the same weight. Case 1.
Let m ≤
3. Then there are at most 3 distinct binary words of the same weight withlength m , a contradiction. Case 2.
Let m = 4. If the number of 1’s in each row is i , we have that the total possiblenumber of distinct rows is (cid:0) i (cid:1) . This number is at least 5 only if i = 2. Thus each row of A has two 1’s. Therefore, y i [ I ] has weight three for each i = 1 , . . . ,
5. But there are only fourbinary words of length four and weight three, a contradiction to the fact that we have fivedistinct y i ’s. Case 3.
Assume that m ≥
5. We shall introduce a (multi-)graph H on vertices v i , . . . , v i m .Let v i s v i t ∈ E ( H ) if some pair of consecutive x i ’s differ exactly in positions i s and i t . Note3hat since C is a cycle, a graph H must have even nonzero degrees, thus 5 = | E ( H ) | ≥| V ( H ) | = m . Hence, m = 5 and each vertex has degree 2. Since a vertex of H correspondsto a column of A ( C ), and the degree of a vertex in H corresponds to a number of times 1changes to 0 and 0 changes to 1 on consecutive elements in a corresponding column of A ( C )(in cyclic order), we must have 1’s occur consecutively in each column of A ( C ). (cid:4) We say that a matrix A has bad prefixes if there are two pairs of consecutive rows r i , r i ′ and r j , r j ′ in A , i ′ = i ± j ′ = j ± r i starts with a b r i ′ starts with a b r j starts with a b r j ′ starts with a b r i starts with a b r i ′ starts with a b r j starts with a b r j ′ starts with a b a , b .For a matrix A with rows v , . . . , v , let A be a matrix with rows v , . . . , v in the sameorder. Lemma 2.3
Let A = A ( C ) , then neither A nor A have bad prefixes.Proof. Assume that the matrix A has bad prefixes. Then x i = a a a , x i ′ = a a a ′ ,and y = x i ∪ x i ′ = a a a , for some binary words a , a , a , a ′ . Here, y = y i or y = y i ′ .The common prefix of x i and y is a a .In part (1) we also have that x j = a a b , x j ′ = a a b , y ′ = x j ∪ x j ′ = a a b , forsome binary word b . Here y ′ = y j or y ′ = y j ′ . The common prefix of x j ′ and y ′ is a a .Thus, c ( x j ′ y ′ ) = c ( x i y ) since the weights of corresponding prefixes are different modulo 2.In part (2) we also have that x j = a a b , x j ′ = a a b ′ , y ′ = x j ∪ x j ′ = a a b ′ , forsome binary words b , b ′ . Here y ′ = y j or y ′ = y j ′ . The common prefix of x j ′ and y ′ is a a .Thus, c ( x j ′ y ′ ) = c ( x i y ).In both cases we have a contradiction to the assumption that the cycle C is monochro-matic.When A has bad prefixes, we use Observation 1 to arrive at the same conclusion. (cid:4) Lemma 2.4
The matrix A = A ( C ) has exactly two ’s in each row.Proof. Since all x i ’s have the same weight, each row of A contains the same number of 1’s.Assume that each row of A contains exactly one 1. There are three consecutive rows of A which form binary words in increasing lexicographic order. It is a routine observation tosee that either A or A has bad prefixes because of these rows.Assume that there are three 1’s and two 0’s in each row of A . Then y i [ I ] has weight4, for i = 1 , . . . ,
5. There are exactly 5 binary words of length 5 and weight 4, namely11110 , , . . . , A has consecutive 1’s, it has consecutive 0’s.If the number of 0’s in some column, i , is more than two, then there are at least two words,4 t [ I ] and y q [ I ] which have 0 in the i th position, a contradiction. Thus each column of A must have exactly two consecutive 0’s and three consecutive 1’s (in cyclic order). Then each x i = a ∗ a ∗ a ∗ a ∗ a ∗ a , where a j ’s are some binary words and ∗ ∈ { , } are locatedin positions from I .If there is a row with two consecutive 0’s in the first and second column then we imme-diately get bad prefixes since the first two columns must be as follows, up to cyclic rotationof the rows: Thus we can assume that the first two columns of A or A are as follows, up to cyclicrotation of the rows: Now, we see that a (3 ,
3) = 0, otherwise a (3 ,
4) = a (3 ,
5) = 0 and we arrive at bad prefixes in A , similarly to the previous argument. Then we have two possible cases for the first threecolumns of A : and . Note that in the first case we have prefixes a a a and a a a , which have differentweights modulo 2, and in the second case we have prefixes a a a and a a a , which havedifferent weight modulo 2, each case is a contradiction to the fact that C is monochromatic.Finally, the rows of A cannot contain more than three 1’s each since in this case y i = y j for i, j ∈ [5]. (cid:4) Note that Lemma 2.4 implies that the total number of ones in A is 10. Lemma 2.5
If there are indices i , i , i and j , j such that A ( C ) has entries a ( i , j ) = a ( i , j ) = a ( i , j ) = a ( i , j ) = a ( i , j ) = a ( i , j ) = 0 , then C has a chord in Q n .Proof. Consider x i , x i , x i . Note that x i ∪ x i = x i ∪ x i = x i ∪ x i . At least two of thesethree vertices must be successive vertices of C in a layer (cid:0) [ n ] k (cid:1) , say without loss of generality5hat x i = x and x i = x . But then x i = x , x i = x , otherwise we shall have y = y or y = y . Thus x i = x and y x forms a chord of C in Q n . (cid:4) Now, we are ready to complete the proof of the Theorem 1.1. Recall first that Lemma2.2 implies that 1’s occur consecutively in each column of A (up to cyclic rotation of therows). Second, observe that there are no two consecutive rows of A such that one has two0’s in columns j, j ′ and another has two 1’s in the same columns j, j ′ (otherwise x i and x i +1 will have a distance at least four between them). We see from Lemma 2.5 that A has atmost one column with at most one 1. Since the total number of 1’s in A is 10, we havethat each column of A has at at most three 1’s. Moreover there is at most one column withthree 1’s. Let A have exactly one column with exactly one 1. Then, there must be a columnwith exactly three 1’s. Consider possible submatrices formed by these two columns (up tocyclically rotating the rows): B = , B = . Since there are exactly two 1’s in each row of A and Hamming distance between consec-utive rows is 2, we have, in case of B , that there must be a column i , where i / ∈ { , } , suchthat a (2 , i ) = a (3 , i ) = 1, which implies that a (1 , i ) = 1. Thus there are two columns withexactly three 1’s, a contradiction. B is possible only if all five columns up to permutation are as follows. (1)(This is true because each row has two 1’s and because the Hamming distance betweenconsecutive rows is 2.)If we apply Lemma 2.3 to all pairs of columns in (1), we see that the only possible pairsfor the first two and the last two columns are { , } , { , } , { , } . Thus it is impossible tochoose two acceptable last columns and two acceptable first columns at the same time. Thisconcludes the argument that A cannot have columns with exactly one 1.Therefore A has exactly two consecutive 1’s in each column. Consider the first twocolumns. There are only two possibilities: when there is a row i such that a ( i,
1) = a ( i,
2) = 1and when there is no such row. In both situations, it is again a routine observation to seethat matrix A has bad prefixes.Thus no 5 × A ( C ) for some monochromatic in-duced 10-cycle C . This concludes the proof of Theorem 1.1. (cid:4) emark. The above proof basically reduces the analysis of the coloring c on E ( Q n ) to theanalysis of the coloring c on E ( Q ). We believe that a similar result should hold for chordless C k +2 in Q n , k > The following result of Fan Chung [4] states that dense subgraphs of a hypercube containall “not too long” cycles of lengths divisible by 4.
Theorem 3.1
For each t , there is a constant c = c ( t ) such that if G is a subgraph of Q n with at least cn − / | E ( Q n ) | edges then G contains all cycles of lengths k , ≤ k ≤ t . On the other hand, we can conclude that a dense subgraph of a hypercube contains acycle of length 4 k + 2 for some k which follows from the following strengthening of a classicaltheorem of Bondy and Simonovits [3] by Verstra¨ete [7]. Theorem 3.2 [7] Let q ≥ be a natural number and G a bipartite graph of average degreeat least q and girth g . Then there exist cycles of ( g/ − q consecutive even lengths in G . Unfortunately, these results still do not guarantee the existence of a cycle of length 4 k + 2for some small fixed k in dense subgraphs of a hypercube. For completeness, we include ageneral bound on the maximum number of edges in a C k +2 -free subgraph of a hypercube. Theorem 3.3 f ( n, C k +2 ) ≤ (1 + o (1)) √ n n − , k ≥ .Proof. Let k ≥ G be a subgraph of Q n containing no cycles of length4 k + 2.Let d v be the degree of a vertex v in G . For each v ∈ V ( Q n ), we introduce a graph H v = ( V, E ) such that V = { u ∈ V ( Q n ) : uv ∈ E ( Q n ) } and E = {{ u, w } : there is a 2 − path from u to w in G different from uvw } . If uw, u ′ w ′ are two distinct edges in H v ( u might coincide with u ′ ), then there is a unique( u − w )-path uxw , x = v and a unique ( u ′ − w ′ )-path u ′ x ′ w ′ , x ′ = v in G , moreover x = x ′ .We note first that H v does not have C k +1 . To that end, consider u , . . . , u k +1 , u , a cycle in H v . The previous observation implies that there is a cycle u , x , u , x , . . . , u k +1 , x k +1 , u in G , a contradiction. If n is sufficiently large, then any graph on n vertices with no copy of C k +1 has at most n / H v has at most n / n large enough.Then we have X v ∈ V ( Q n ) | E ( H v ) | ≤ n n / . (2)Next, we count p ( G ), the number of paths of length two in G in two ways. Trivially, p ( G ) = X v ∈ V ( Q n ) (cid:18) d v (cid:19) .
7n the other hand, p ( G ) = X v ∈ V ( Q n ) | E ( H v ) | , because if uv ′ w is a path of length two in G then { u, w } ∈ E ( H v ) for the unique v , v = v ′ ,which is adjacent to both u and w in Q n . Thus, we have X v ∈ V ( Q n ) (cid:18) d v (cid:19) = X v ∈ V ( Q n ) | E ( H v ) | . (3)Using the Cauchy-Schwarz inequality, we have X v ∈ V ( Q n ) (cid:18) d v (cid:19) = X v ∈ V ( Q n ) ( d v / − d v / X v ∈ V ( Q n ) d v / − | E ( G ) |≥ − n − X v ∈ V ( Q n ) d v − | E ( G ) | = 2 − n − (2 | E ( G ) | ) − | E ( G ) | . (4)Combining (2), (3), and (4), we have2 − n − (2 | E ( G ) | ) − | E ( G ) | ≤ n n / . If | E ( G ) | = an n , then a ≤ n + p /n ≤ / √ ǫ , where ǫ → n → ∞ . (cid:4) Acknowledgments.
The authors thank the referees for careful reading and helpful sug-gestions which improved the presentation of the results.
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