Noncrossing Monochromatic Subtrees and Staircases in 0-1 Matrices

1 Department of Mathematics, Northwestern University, 2033 Sheridan Road, Evanston, IL 60208-2730, USA 2Department of Mathematics, Pomona College, 640 North College Avenue, Claremont, CA 91711, USA 3Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences, Budapest, P.O. Box 127, Budapest 1364, Hungary 4Department of Mathematics and Computer Science, Emory University, Atlanta, GA 30322, USA


Introduction
A geometric graph (see [1]) is a graph whose vertices are in the plane in general position and whose edges are straightline 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 2 vertices are in two disjoint -element sets  and , and its edges are some segments  with  ∈  and  ∈ .The following representation, apparently studied first in [2], 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  2 are  = { 1 = (1, 0),  2 = (2, 0), . . .,   = (, 0)} and  = { 1 = (1, 1),  2 = (2, 1), . . .,   = (, 1)} and the edge     is the line segment joining   ∈  and   ∈ .This representation is called a simple (, ).
Problem 1 (see [5]).Find (), the order of the largest monochromatic noncrossing subtree that exists in every 2coloring of the edges of a simple geometric  , .The lower bound is 4/5 and the upper bound is  for even and  + 1 for odd .
Noncrossing subgraphs of simple (, )-s can be easily characterized.
Proposition 2 (see [10]).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 4/5 ≤ () is proved in a stronger form.
Theorem 3 (see [5]).In every 2-coloring of the simple  , there is a noncrossing monochromatic double star with at least 4/5 vertices.This bound is asymptotically best possible.
Since Theorem 3 is asymptotically sharp, it was asked in [5] whether one can separate () and 4/5.Here, we answer this question positively, although with a small margin.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  + 1 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  ≥ 2, () = .
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 0or a 1-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  = 7 is the matrix  below where a 1staircase 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 (, ) = { , : 1 ≤  < ,  , =  , } and also after all elements of (, ) = { , : 1 ≤  < ,  , =  , }.Note that (, 1) = (1, ) = 0 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  2,1 and repeating this for the remaining ( − 1) × ( − 1) 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 () = 8 and providing an easy way to trace back the 0-staircase of length 8 in  ending at  6,7 .
The 0-values are shown in bold: (5) 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.Proof.A 2-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 ( − 1 edges) corresponds to a homogeneous staircase of length −1.The correspondence works backward as well, proving the theorem.Another advantage of the staircase formulation is that it gives "a proof without words" to Proposition 5. Indeed, let  be the  ×  matrix with  , = 1 if  +  ≤  or  =  =  and zero otherwise (see below for  = 7): 1 1 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 0 0 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 (6) One can immediately see that the longest homogeneous staircase of  has  − 1 elements, proving Proposition 5.
We have the following propositions supporting Conjecture 8 for various special matrices.Proposition 9. Assume  1,1 =  , = 0 for the  ×  matrix  = [ , ] and there is no 0-staircase from  1,1 to  , .Then there is a 1-staircase of size at least  − 1.The next special case is when there are two consecutive columns whose binary sum is the all 1 vector (switching columns).
Finally, we prove Conjecture 8 for  = 8 and note that its proof method still works for  = 9 but would be much longer.On the other hand, the cases where  < 8 are obvious and left to the reader.Proposition 12. Every 8 × 8 0-1 matrix has a monochromatic staircase of length at least 7.These propositions will be proved in Section 2, and Theorem 4 is proved in Section 3.

Proofs of Special Cases of Conjecture 6
Proof of Proposition 9. First, observe that  1, =  ,1 = 1, because otherwise there exists a 0-staircase connecting  1,1 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 0-staircase connecting  1,1 and  , would be present.Therefore,  +  ≤ ; similarly  +  ≤ .Since  1, = 1, the ones in the first row and in the last column form a 1-staircase of length ( − ) + ( − ) − 1 = 2 −  −  − 1.In the same way, there is a 1staircase consisting of all ones in the first column and in the last row, of length 2 −  −  − 1.The two lengths add up to (2−−−1)+ (2−−−1) = 4−(+)−(+)−2 ≥ 2−2, so at least one of them must be at least  − 1.
The sum of lengths of these four staircases is so their average length is  − 1; thus at least one of them must have length  − 1 or more.The sum of lengths of these two staircases is 2 − 2 so at least one of them must have length  − 1 or more.
Proof of Proposition 11.Assume columns , +1 are switching columns.Let  and  be the smallest and largest values of  for which  , = 0, and likewise let  and  are the smallest and largest values of  for which  , = 1.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  − 1 −  ones there.Define  to be the number of zeroes to the right of  ,+1 in row , so there are  −  − 1 −  ones there.
Define  to be the number of ones to the left of  , in row , so there are −1− zeroes there.Define  to be the number of ones to the right of ,  + 1 in row , so there are  −  − 1 −  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  + 1.

Theorem 7 Conjecture 8 .
allows to put forward Conjecture 6 in a more challenging form.Every  ×  0-1 matrix has a homogeneous staircase of size  − 1.