GEOMBINATORIAL MIXED HYPERGRAPH COLORING PROBLEMS Peter Johnson Department of Mathematics and Statistics Auburn University, AL 36849
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Vitaly Voloshin Department of Mathematics and Physics Troy University Troy, AL 36082
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Abstract Can you color Rn so that point sets of one sort are never monochromatic, while point sets of another sort always are? If so, how many colors did you use, and can you do the job with more or fewer colors?
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A hypergraph
A hypergraph is a pair (V, E) in which V is a set and E is a collection of subsets of V . It is conventional to require that e ∈ E implies |e| ≥ 2; that will be the case for all the hypergraphs in this paper, unless otherwise announced. More generally, the elements of E are multisubsets of V (i.e., elements of V are 1
allowed to “appear more than once” in, or on, elements of E), but such generality has no place in the problems to be discussed here. A typical hypergraph coloring problem would be to color (partition) V with as few colors (into as few sets) as possible so that no e ∈ E is monochromatic (contained in one the partition sets). There are also coloring problems in which it is required that all the elements of each e ∈ E have different colors; if |e| = 2 for every e ∈ E then the two kinds of problems are the same. This will be quite often the case in what follows, but when it is not, it is the non-monochromaticity of each e ∈ E that we are aiming for. In a mixed hypergraph coloring problem we have a triple (V, E, F ) (called a mixed hypergraph) such that both (V, E) and (V, F ) are hypergraphs and we aim to color V so that no e ∈ E is monochromatic and every f ∈ F is monochromatic. (Again, there are variations, but not here.) We will call such a coloring a proper coloring of (V, E, F ). The smallest number of colors necessary for such a coloring is of interest, but notice also that using more colors may make it more difficult to achieve the monochromaticity of each f ∈ F . Therefore it is useful to define the chromatic spectrum of (V, E, F ), denoted here by Chrom(E, F ), to be the set of cardinal numbers n (infinite cardinals are allowed) such that there is a proper mixed coloring of (V, E, F ) with a set of colors of cardinality n, with every color in the set appearing as a color. There has been a considerable amount of work already invested in the study of mixed hypergraph colorings—-the interested reader may consult monograph [8] and website [9].
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Geombinatorial Mixed Hypergraph Colorings on the Line
Where to start? How about with the simplest mixed hypergraph coloring problem of a geometric nature that we could think of: for which d > 0 can one color the real numbers, R, so that no two points a distance 1 apart are of the same color, yet any two points a distance d apart are of the same color. Here V = R, E = {{x, y} ⊆ R | |x − y| = 1}, and F = {{x, y} ⊆ R | |x − y| = d}, in the terms of the preceding section. And the discussion there refines the question: if you can color so as to satisfy the requirements, how many colors do you use, and what numbers of colors can you use? Even at this primordial level, the answers are intriguing (we think). If d = 1, clearly you cannot color. If d is a rational, pq in lowest terms, with p odd, then you cannot color with 2 colors, for, if you could, then 0 and q( pq ) = p would have the same color, and yet, because p = p · 1, and p is odd, and there are only 2 colors used, they would have different colors. The contradiction establishes that you cannot color in this case (p odd) with 2 colors. Let Z denote the set of integers, and let h1, di = {m + nd | m, n ∈ Z}, the additive subgroup of R generated by 1 and d. Let E0 , F0 denote the collections of doubletons from h1, di that are in E, F , respectively. [That is, E0 = {{m+nd, m+1+nd} | m, n ∈ Z} and F0 = {{m + nd, m + (n + 1)d} | m, n ∈ Z}.] The problem of determining Chrom (R, E, F ) will be completely solved if we solve the more interesting problem of determining Chrom (h1, di, E0, F0 ). For if there is a proper coloring of (h1, di, E0, F0 ) with k colors (actually appearing), then by copying that coloring on cosets of h1, di in R with either the same colors or with some different colors, we see that we can achieve a proper coloring of (R, E, F ) using any number of colors from k up to c, the cardinality of R (the continuum!).
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Lemma 1 Suppose that p and q are relatively prime positive integers, and p > 1. If p is even then for any k ∈ {2, . . . , p}, and for no other k, Z can be colored using k colors so that any two points a distance q apart are colored differently, and any two points a distance p apart are colored the same. If p is odd then the same is true except that such a coloring with 2 colors is impossible. Proof Clearly the coloring cannot be done with 1 color. If there is such a coloring then each congruence class mod p is monochromatic, and these cover Z, so there are at most p colors. Thus the set {k | there is such a coloring using k colors} is a subset of {2, . . . , p}. If Z is so colored with 2 colors then 0 and pq = p + p + · · · + p must have the same color; but, also, pq = q + · · · + q, and the color changes every time you add q. So p must be even. Thus if p is odd, the “feasible” numbers of colors are among {3, . . . , p}. Since p and q are relatively prime, if mp + nq = m0 p + n0 q, m, n, m0 , n0 ∈ Z, then q | m − m0 and p | n − n0 ; in particular, n ≡ n0 mod p. For k ∈ {2, . . . , p}, partition the congruence classes mod p into k (non-empty) sets A1 , . . . , Ak and for m, n ∈ Z color mp + nq with the color i if the congruence class of n mod p, n + pZ, is in Ai . By remarks above, this coloring is well defined—i.e., the color of an integer z does not depend on its representation in the form mp + nq. (That p and q are relatively prime implies that every integer has such a representation.) Clearly, for any such coloring adding p to an integer does not change the color. We can arrange that adding q to an integer will always change the color by being shrewd in forming the Ai . We need to choose A1 , . . . , Ak so that for each n ∈ Z, the congruence classes mod p of n and n+1 lie in different Ai . Letting Ai consist of these congruence classes t + pZ, t ∈ {0, . . . , p − 1}, such that t ≡ i mod k, will satisfy our need unless k divides p − 1; in that case, n + pZ and n + 1 + pZ are in different Ai unless n ≡ p − 1 mod p, in which case n + pZ, n + 1 + pZ ∈ Ak . When k = 2 and p is odd, there is no remedy for this prob4
lem, as we have seen, but if k ≥ 3 and k divides p − 1, define the Ai by putting t + pZ in that Ai such that t ≡ i mod k, if 0 ≤ t ≤ p − 2, and p − 1 + pZ in Ai for some i other than k (Ak contains p + pZ = pZ) and k − 1 (Ak−1 contains p − 2 + pZ). Theorem 1 Suppose d > 0 and h1, di, E0 , and F0 are as defined above. If d is irrational then Chrom (h1, di, E0, F0 ) = {2, 3, · · · } ∪ {ℵ0 }. If d is rational, d = p/q in lowest terms, then: if p = 1, Chrom (h1, di, E0, F0 ) = ∅; if p > 1 and p is odd, then Chrom (h1, di, E0, F0 ) = {3, . . . , p}; and if p is even, Chrom (h1, di, E0, F0 ) = {2, . . . , p}. Proof If d is irrational then 1 and d are linearly independent over the rationals, so the representation of elements of h1, di in the form m + nd, m, n ∈ Z, is unique. For k ∈ Z, k ≥ 2, color m + nd with the congruence class of m mod k, to see that k ∈ Chrom (h1, di, E0, F0 ). In the case of ℵ0 = |Z|, color m+nd with m. Since h1, di is countable, no larger number can be in Chrom (h1, di, E0, F0 ). If d = 1/q, q ∈ Z, then in any coloring of h1, di such that every member of F0 is monochromatic, 0 and 1 = q· 1q = 1q +· · ·+ 1q will have the same color. Thus Chrom (h1, 1/qi, E0, F0 ) = ∅. If d = p/q, p, q positive, relatively prime integers, with p > 1, then a proper coloring of (h1, di, E0, F0 ) is equivalent to a coloring of Z = qh1, di as postulated in the Lemma; as you go from one set to the other by multiplying or dividing by q, import any coloring of one to the other. The remaining conclusions of the Theorem now follow from the Lemma. Corollary 1 Suppose d > 0 and E and F are as defined above. If d is irrational, or a rational p/q in lowest terms, with p even, then Chrom (R, E, F ) is the set of all cardinal numbers from 2 through c = |R|. If d = 1/q, q ∈ Z, then Chrom (R, E, F ) = ∅. If d = p/q in lowest terms, p odd, p ≥ 3, then Chrom (R, E, F ) = {3, 4, . . . , c}.
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Corollary 1 follows from the Theorem and previous remarks. In Corollary 1, E is the set of all translates of the 2-point set {0, 1}, and F is the set of all translates of {0, d}. This observation points to two, or perhaps four, open questions that are natural sequels of the problem we have just solved. We can let one of E, F be the set of all translates of a 2-point subset of R, and the other the union of the sets of translates of two different 2-point subsets of R. Or we can let the other be the set of translates of a 3-point subset of R. The problem in all four situations is, of course, to determine Chrom (R, E, F ) for all the different choices of the sets being translated. These seem to be worthy and challenging problems. See [3] for a discussion of a related problem of the unmixed hypergraph coloring type.
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Geombinatorial Mixed Hypergraph Colorings in Rn, n > 1
On the line the only isometries are translation and reflection, while in Rn , n > 1, we also have rotation, and this makes one obvious analogue in Rn of the problem solved for R in the last section very easy to dismiss. Theorem 2 Suppose that n > 1 and S ⊆ Rn , |S| ≥ 2. Then Rn cannot be colored so that all subsets of Rn congruent to S are monochromatic, while any two points a distance 1 apart are colored differently. Proof Suppose Rn is colored so that all sets obtained from S by translation and rotation are monochromatic. Let d be the distance between two different points of S. Since any two points a distance d apart are in some copy of S, any two points d apart must have the same color. From here we could deduce that all of Rn is one color, but we needn’t go to all that trouble; clearly any two points md apart are the same color, if m is a positive integer, so every sphere of radius md is monochromatic, for all 6
such m, and clearly for m sufficiently large such a sphere will include points a distance 1 apart. However, here is a potentially very rich and interesting class of mixed hypergraph coloring problems in Rn , n > 1: For which S ⊆ Rn , |S| ≥ 2, can one color Rn so that all translates of S are monochromatic while any two points a distance one apart are different colors? And, when there is such a coloring, how many colors are used, and with which other numbers of colors can such a coloring be accomplished? Or, to return to the terminology of the preceding sections, what is Chrom (Rn , E, F ), where E is the set of 2-point subsets of Rn of points a distance 1 from each other, and F = F (S) is the set of all translates of S in Rn . A variant of this problem: fix m and ask for which S is m ∈ Chrom (Rn , E, F (S))? Before we go on to generalities, let us look at two fairly famous colorings of R2 in the new light hopefully cast by the concern for mixed hypergraph coloring. First, consider the HadwigerIsbell coloring of the plane (see [7]) with 7 colors so that points a distance 1 apart are different colors. In this coloring the plane is tiled with regular hexagons of diameter d, √27 ≤ d ≤ 1, and then 7 colors are assigned, one to each hexagon, in a regular way, so that no two points of the same color are a distance 1 apart—in each hexagon there are no points a distance 1 apart, unless d = 1, in which case care must be taken in coloring the boundaries of the hexagons, and for each hexagon the nearest hexagons of the same color are far enough away that no points of hexagons with the same color are a distance 1 apart. (If √ d = 2/ 7 then, again, care must be taken on the boundaries.) In this coloring a 7-hexagon tile T consisting of a hexagon and the 6 hexagons around it are colored with the 7 colors, and then this coloring is copied on translates of T which tile the plane. Clearly, if strict conventions are adopted with respect to the colors on the outer boundaries of the tiles T , then translation by certain vectors—in particular, the 6 vectors that T is translated by repeatedly to achieve the tiling of R2 by copies of T , will 7
preserve color. It is customary to say that such a coloring is periodic, and the non-zero vectors translation by which preserve color are the periods of the coloring. Here is another periodic coloring related to a “classic” Euclidean Ramsey problem (see [3] √ and [4]) in the plane. Color the √ 3 3 2 half-open strip {(x, y) ∈ R | n 2 ≤ x < (n + 1) 2 } red if n is even and blue if n is odd. No equilateral triangle with side length one has all of its vertices the same color. By the periodicity of the coloring we have a proper 2-coloring of (R2 , E, F ) where E is the collection of all 3-subsets of R2 in which the 3 points are the vertices of an equilateral triangle of side length 1, and F is the collection of all translates of the 2-element sets √ {(0, 0), (n 3, y)}, n ∈ Z, y ∈ R, n2 + y 2 > 0. To help us make sense of all this, let us consider an arbitrary abelian group (A, +, 0) and an arbitrary coloring f of A; that is, f is a function with domain A. We define P (f, A) = {u ∈ A | for all a ∈ A, f (a) = f (u + a)}. The letter P is the first letter of “period”, and, indeed, P (f, A)\{0} is the set of periods of the coloring f . Theorem 3 With A and f as above, P (f, A) is a subgroup of A, and U ⊆ A has the property that all translates of U in A are monochromatic (with respect to f ) if and only if U is a translate of a subset of P (f, A). Proof Let P = P (f, A). It is straightforward to verify that P is a subgroup of A. From the definition of P it follows easily that any translate of any subset of P is monochromatic (with respect to f ), and thus that any translate of any translate of a subset of P is monochromatic. Now suppose that U ⊆ A has the property that any translate of U is monochromatic. Suppose that x ∈ U and let U 0 = U − x = {u − x | u ∈ U}. To finish the proof, it suffices to show that U 0 ⊆ P . 8
Suppose that a ∈ A. Then U 0 + a = U + (a − x) is monochromatic and, since 0 ∈ U 0 , the color on U 0 + a is f (a). Thus f (u0 + a) = f (a) for all u0 ∈ U 0 , so, since a was arbitrary, U0 ⊆ P . In the class of geometric mixed hypergraph coloring problems in which V = A, an additive subgroup of some Rn , and F is to be the set of translates of some class of configurations, Theorem 3 says that all questions to be addressed are about the subgroups P (f, A) = P for colorings of f of A: for colorings f forbidding monochromatic hyperedges in E what can P look like? Or, for a given subgroup B of Rn , for which E can we find f forbidding monochromatic e ∈ E such that B ⊆ P (f, A)? Obviously there are a truckload of questions we could pose here, but, based on our limited experience with these matters, we confine ourselves to the following: Is there any coloring f of R2 with a finite number of colors such that no two points a distance 1 apart are the same color and P (f, R2) is dense in R2 ? Of course we could ask the same question about Rn for any n ≥ 2. The point is that it is hard to color so as to forbid the distance 1 and make P (f, Rn ) big; and in all the finite colorings of Rn , n ≥ 2, that forbid the distance 1 that we know of, it seems that P (f, Rn ) is finitely generated, and, therefore, discrete. But for n = 1 the situation is relatively easy. Theorem 4 For every cardinal k, 2 ≤ k ≤ ℵ0 , there is a kcoloring f of R which forbids the distance 1, such that P (f, R) is dense in R. Proof Extend {1} to a Hamel basis {1} ∪ B of R as a vector space over Q. [This involves the Axiom of Choice, Sasha!]. For X ⊆ R let sp(X) denote the linear span of X as a subset of a vector space over Q; i.e., sp(X) is the set of all real numbers expressible in the form r1 x1 +· · ·+rn xn , n ≥ 1, with r1 , . . . , rn ∈ Q and x1 , . . . , xn ∈ X. Color Q with k colors so that the distance one is forbidden and then color R by coloring r + x, r ∈ Q, x ∈ sp(B), with 9
whatever color r had in the coloring of Q. The resulting coloring f forbids translation by ±1, and sp(B) ⊆ P (f, R), so, since sp(B) contains non-zero multiples of Q, P (f, R) is dense in R. 3 It has long been known that Q can be colored with 2 colors so that the distance 1 is forbidden [2]. Theorem 5 Suppose that k ∈ {2, 3} and that f is a 2-coloring of Qk which forbids the distance 1. Then P (f, Qk ) is dense in Qk . Proof Let S k−1 = {u ∈ Rk | |u| = 1}, and X = Qk ∩ S k−1 ; it is well known that X is dense in S k−1 , in the (k −1)-dimensional topology on S k−1. If u ∈ X, v ∈ Qk , then f (u + v) 6= f (v). Therefore, because there are only two colors, X +X ⊆ P (f, Qk ). Therefore, because P (f, Qk ) is an additive subgroup of Qk , hX + Xi ⊆ P (f, Qk ), and, clearly, hX + Xi is dense in Qk . (In fact, X + X is dense in {u ∈ Rk | |u| ≤ 2}.) It is shown in [1] that for any d > 0, Q2 can be 2-colored to forbid the distance d; if d is a distance realized between points of Q2 then the group of periods of such a coloring will be dense in Q2 , by an argument similar to the proof of Theorem 5. Questions arising from Theorem 5 are obvious, but here they are. 1. Can Qk , k ∈ {2, 3}, be colored with more than 2 colors so as to forbid the distance 1 and make the group of periods of the coloring dense in Qk ? 2. Can a 4-coloring of Q4 forbidding the distance 1 ([2], [5]) be found with a dense group of periods? Can such a 4coloring be found with P = {0, 0, 0, 0}? 4 3 3. Same question as 2, with √ Q replaced by Q and distance 1 replaced by distance 2 (see [6]).
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References [1] Aaron Abrams and Peter Johnson, Yet another species of forbidden-distances chromatic number, Geombinatorics 10 (January, 2001), 89-95. [2] Miro Benda and Micha Perles, Colorings of metric spaces, Geombinatorics 9 (January, 2000), 113-126; the paper was widely circulated for over 20 years before its publication in 2000. [3] A. W. Bohannon, P. D. Johnson, Jr., and E. G. Thomas, An easier analogue of a difficult old Euclidean coloring problem, Geombinatorics 12 (January, 2003), 94-101. [4] P. Erd¨os, R. L. Graham, P. Montgomery, B. L. Rothschild, J. Spencer, and E. G. Straus, Euclidean Ramsey Theory I, J. Combinatorial Theory Ser. A 14 (1973), 341-363. [5] P. D. Johnson Jr., Two colorings of a dense subgroup of Qn that forbid many distances, Discrete Math. 70 (1989/1990), 191-195. [6] Peter Johnson, Andrew Schneider, and Michael Tiemeyer, B1 (Q3 ) = 4, Geombinatorics 16 (April, 2007), 356-361. [7] Alexander Soifer, Chromatic number of the plane and its relatives, part I: the problem and its history, Geombinatorics 12 (January, 2003), 131-148. [8] V. Voloshin, Coloring mixed hypergraphs: theory, algorithms and applications, AMS monograph, 2002. [9] http://spectrum.troy.edu/∼voloshin/mh.html
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