Research Article | Open Access
Siyuan Cai, Gillian Grindstaff, András Gyárfás, Warren Shull, "Noncrossing Monochromatic Subtrees and Staircases in 0-1 Matrices", Journal of Discrete Mathematics, vol. 2014, Article ID 731519, 5 pages, 2014. https://doi.org/10.1155/2014/731519
Noncrossing Monochromatic Subtrees and Staircases in 0-1 Matrices
The following question is asked by the senior author (Gyárfás (2011)). What is the order of the largest monochromatic noncrossing subtree (caterpillar) that exists in every 2-coloring of the edges of a simple geometric ? We solve one particular problem asked by Gyárfás (2011): separate the Ramsey number of noncrossing trees from the Ramsey number of noncrossing double stars. We also reformulate the question as a Ramsey-type problem for 0-1 matrices and pose the following conjecture. Every 0-1 matrix contains zeros or ones, forming a staircase: a sequence which goes right in rows and down in columns, possibly skipping elements, but not at turning points. We prove this conjecture in some special cases and put forward some related problems as well.
A geometric graph (see ) is a graph whose vertices are in the plane in general position and whose edges are straight-line segments joining the vertices. A subgraph of a geometric graph is noncrossing if no two edges have a common interior point. A geometric bipartite graph is a geometric graph, whose vertices are in two disjoint -element sets and , and its edges are some segments with and . The following representation, apparently studied first in , seems to be a more natural subclass of balanced geometric bipartite graphs (in fact a standard way of drawing bipartite graphs). The partite sets of in are and and the edge is the line segment joining and . This representation is called a simple .
Problem 1 (see ). Find , the order of the largest monochromatic noncrossing subtree that exists in every -coloring of the edges of a simple geometric . The lower bound is and the upper bound is for even and for odd .
Noncrossing subgraphs of simple -s can be easily characterized.
Proposition 2 (see ). Every connected component of a noncrossing subgraph of simple is a caterpillar (a tree in which the vertices of degree larger than one form a path).
In fact, the lower bound is proved in a stronger form.
Theorem 3 (see ). In every -coloring of the simple there is a noncrossing monochromatic double star with at least vertices. This bound is asymptotically best possible.
Theorem 4. Consider , where .
We give a new construction showing that the upper bound in Problem 1 can be decreased by one when is odd.
Proposition 5. Consider for .
Theorem 3 and Proposition 5 show that the noncrossing condition is important because it is easy to see that for odd there is a monochromatic tree of order in every 2-coloring of the edges of . We suspect that the bound of Proposition 5 is best possible and risk the following conjecture.
Conjecture 6. For every , .
We also reformulate Conjecture 6 (and Problem 1) into a more attractive (and perhaps more inspiring) form. Let be an 0-1 matrix. A 0-staircase is a sequence of zeroes in which goes right in rows and down in columns, possibly skipping elements, but zero at each turning point. A 1-staircase is defined similarly on ones of . A homogeneous staircase in is either a - or a -staircase. The length of a homogeneous staircase is the number of elements in it. Let be the maximum among the lengths of the homogeneous staircases of . Finally, set An example with is the matrix below where a -staircase of length 6 is shown:
One can easily compute for any matrix with the following procedure. We define an matrix , the staircase matrix of , where is the length of the longest homogeneous staircase of ending in position . The matrix can be easily computed recursively from , following a linear order of the elements of , such that comes after all elements of and also after all elements of . Note that for all and . A natural linear order with this property can be defined by taking the first row left to right then the first column downwards starting at and repeating this for the remaining matrix. Under this linear ordering, we define inductively where the maximum of the empty set is defined to be zero; for example, With the example matrix , we get the following staircase matrix , showing that and providing an easy way to trace back the -staircase of length 8 in ending at . The 0-values are shown in bold:
The following easy but important observation shows that the Ramsey-type problems for noncrossing subtrees in simple geometric and for staircases in 0-1 matrices are the same problems in different formulations.
Theorem 7. Consider .
Proof. A -coloring of the edges of a simple geometric can be considered as an 0-1 matrix where the element in row and column is zero or one according to the color of the edge . By Proposition 2, every noncrossing monochromatic subtree is a caterpillar. The edges of the base path of the caterpillar correspond to turning points of a homogeneous staircase and the pending edges of the caterpillar correspond to elements in the same row or column between consecutive turning points (except at the beginning and the end). Thus, a monochromatic noncrossing subtree on vertices ( edges) corresponds to a homogeneous staircase of length . The correspondence works backward as well, proving the theorem.
Conjecture 8. Every 0-1 matrix has a homogeneous staircase of size .
Another advantage of the staircase formulation is that it gives “a proof without words” to Proposition 5. Indeed, let be the matrix with if or and zero otherwise (see below for ): One can immediately see that the longest homogeneous staircase of has elements, proving Proposition 5.
We have the following propositions supporting Conjecture 8 for various special matrices.
Proposition 9. Assume for the matrix and there is no -staircase from to . Then there is a -staircase of size at least .
The next proposition settles Conjecture 8 for four of the 16 corner configurations of the lower left submatrix of .
Proposition 10. Suppose that the matrix satisfies and . Then .
The next special case is when there are two consecutive columns whose binary sum is the all 1 vector (switching columns).
Proposition 11. Suppose that is an 0-1 matrix with switching columns; that is, there exists such that for every , one has . Then .
Finally, we prove Conjecture 8 for and note that its proof method still works for but would be much longer. On the other hand, the cases where are obvious and left to the reader.
Proposition 12. Every 0-1 matrix has a monochromatic staircase of length at least 7.
2. Proofs of Special Cases of Conjecture 6
Proof of Proposition 9. First, observe that , because otherwise there exists a -staircase connecting to . Suppose there are , , , zeroes in the first row, first column, last row, and last column, respectively. A zero from the first row and a zero from the last row cannot share the same column, because otherwise a -staircase connecting and would be present. Therefore, ; similarly . Since , the ones in the first row and in the last column form a -staircase of length . In the same way, there is a -staircase consisting of all ones in the first column and in the last row, of length . The two lengths add up to , so at least one of them must be at least .
Proof of Proposition 10. Without loss of generality, assume , . Consider two cases as follows.
Case 1. . Define to be the number of zeroes in column 1, to be the number of zeroes in column 2, to be the number of zeroes in row , and to be the number of zeroes in row . The following homogeneous staircases therefore exist:(i)the -staircase in column 1 and row , turning at with length ;(ii)the -staircase in column 2 and row , turning at with length ;(iii)the -staircase in column 1 and row , turning at with length ;(iv)the -staircase in column 2 and row , turning at with length . The sum of lengths of these four staircases is so their average length is ; thus at least one of them must have length or more.
Case 2. . Define to be the number of zeroes in column 1 and to be the number of zeroes in row . The following homogeneous staircases therefore exist:(i)the -staircase in column 1 and row , turning at with length ;(ii)The -staircase in column 1 and row , , turning at with length .The sum of lengths of these two staircases is so at least one of them must have length or more.
Proof of Proposition 11. Assume columns , are switching columns. Let and be the smallest and largest values of for which , and likewise let and are the smallest and largest values of for which . We may assume and all exist and are different, otherwise, column contains at most one 1 or at most one 0 and the proof is finished.
Define to be the number of zeroes to the left of in row , so there are ones there. Define to be the number of zeroes to the right of in row , so there are ones there.
Define to be the number of ones to the left of in row , so there are zeroes there. Define to be the number of ones to the right of in row , so there are zeroes there.
Define to be the number of zeroes in column , so there are ones in column , and there are zeroes and ones in column .
The -staircase in row , column , and row , which turns at and , contains zeroes.
The -staircase in row , column , and row , which turns at and , contains ones.
The -staircase in row , column , and row , which turns at and , contains ones.
The -staircase in row , column , and row , which turns at and , contains zeroes.
The length sum of these homogeneous staircases is = , so the average length is . Therefore, one of them has length at least .
Proof of Proposition 12. Let be an 0-1 matrix; without loss of generality, suppose . By contradiction, assume all homogeneous staircases have length 6 or less. In the first row and last column combined, there can be at most 6 zeroes, so there must be at least 9 ones. Without loss of generality, assume the first row has at least as many ones as the last column, by assumption at most 6.
Case 1. First row contains 6 ones. The column of the rightmost 1 must contain either 7 zeroes or a 1, both giving a homogeneous staircase of length 7, a contradiction.
Case 2. First row contains 5 ones; consequently, the last column contains at least 4 ones. Let be the set of indices with and let be the set of indices for which . We have , . Then , if is the smallest element of and is one of the last 4 elements of (otherwise we get a -staircase of length at least 7). Similarly , if is one of the first 4 elements of and is the largest element of . This gives seven elements of with , forming a -staircase, a contradiction finishing the proof.
3. Proof of Theorem 4
Proof of Theorem 4. Consider an arbitrary red-blue coloring of the edges of a balanced geometric bipartite graph . Let and denote the red and blue subgraphs of . Set
We will show that there is a noncrossing subtree in or in with vertices, where is computed during the proof.
Assume first that ; without loss of generality, the maximum is attained at in the red color. Let denote the smallest index for which is red. If has at least red neighbors in , we have a red noncrossing double star on spanning at least vertices. Otherwise has at least blue neighbors in giving a blue star that is as large as required.
From now on we consider the case ; assume (w.l.o.g.) that edge is red. Now—from the definition of —both and are strictly greater than ; therefore, we have a noncrossing red double star on with more than vertices in both and . We may assume that has less than vertices on both and ; otherwise has at least vertices. Therefore, is adjacent in blue to a set such that . Similarly, is adjacent in blue to a set such that . Let be the smallest index for which and is the largest index for which .
Observe that edges from to must be red apart from an initial segment such that because otherwise we have a noncrossing double star in blue of the required size. Similarly, edges from to must be red apart from an end segment of such that . Thus we have a red double star on the red edge containing vertices of . Thus has at least vertices in both and (and also at most ; otherwise we have the required large noncrossing red subtree). Note also that we may assume that ; otherwise if , then
We note next that there are less than blue edges from to ; otherwise the blue star can be extended to a noncrossing double star on vertices. Thus more than red edges go from to .
Set , and observe that there are less than red edges from to ; otherwise we can extend to a noncrossing red double star with vertices.
The two previous statements and imply that at least red edges go from to . The same argument shows that at least red edges go from to the set . In particular, .
Since sends at most red edges to , it sends at least blue edges to set . Suppose that the first vertex of ( with smallest ) sends blue edges to . Then we have a blue noncrossing tree with at least Vertices; thus we may assume . We conclude that some vertex of sends at least red edges to .
Now, we define a noncrossing blue tree as follows. Let be the first element of . Take the blue star from to ; it has at least edges. Continue with the sequence of blue edges from to ending with the blue edge , . This sequence must contain at least vertices because at most red edges go from to ; otherwise would be extended. Finally, we add blue edges from to a subset of . To estimate how many, observe that there is a red star with center with at least leaves in and at least leaves in , thus altogether at least leaves. If there is a red edge from to the th vertex of (from left), then we have a noncrossing red tree with at least vertices. This gives that and we conclude . Altogether, has at least vertices when .
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
The authors are grateful for remarks of a referee, especially for pointing out that the following conjecture (stronger than Conjecture 6) is not true: every 0-1 matrix has a -staircase and a -staircase with total length at least .
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