Regular dissections of an infinite strip

In the early 1970s, Bro. U. Alfred Brousseau asked for the number of regions formed in an infinite strip by the mn segments that join m equally spaced points on one edge to n equally spaced points on the other. Using projective duality, we express the number of points, segments, and regions formed by Brousseau's configuration in terms of the numbers Lk(m, n) of lines that meet an m x n lattice array in exactly k points. 1. A problem of Brousseau's During a conversation with G.L. Alexanderson and the author in the early 1970s, Bro. U. Alfred Brousseau posed the following question. Let ~ and ~/be parallel lines in the plane. Fix positive integers m and n; choose ra points X t , X2 . . . . . X , on ~ and n points Y~, Y2 . . . . . Yn on ~/; and let go be the line segment that joins X~ and ~. Into how many regions do the rnn line segments gij divide the infinite strip ~ between the lines ~ and ~/? Formulas for the number of edges and faces formed by this configuration in terms of the multiplicities of the points are readily found, by the following sweep-line argument, for example. A line parallel and close to ~ meets all mn segments and initially identifies mn edges and ran + 1 faces. As that line sweeps from ~ to ~/, new segments and faces appear only at points of intersection; and at such a point P through which pass 2(P) segments, the edge count increases by 2(P) and the face count by ;t(P) 1. Consequently, the numbers E of segments and F of regions formed inside the strip are given by the formulas: E = ran + ~ 2(P), F = 1 + ran + ~ () , (P)1), (1) P ~ PE~ where ~ is the set of points of intersection. Note that Euler's formula V E + F = 1 is an immediate consequence of these formulas. (An analogous situation in E 3 is investigated in [31.) 0166-218X(95)$09.50 © 1995 Elsevier Science B.V. All rights reserved SSDI O 0 1 2 3 6 5 X ( 9 4 ) 0 0 0 6 8 T 264 J.E. Wetzel /Discrete Mathematics 146 (1995) 263-269 Writing Vk for the number of points of multiplicity k that are formed inside the strip a by the intersecting segments, we can recast these formulas in the following useful form: V = ~, Vk, E = m n + ~, kVk, F = 1 + m n + ~, ( k -1 )Vk . (2) k~>2 k~>2 k>~2 Formulas analogous to (1) and (2) are established for an arbitrary dissected oval in [1]. It is obvious (and readily proved) that the maximum region count occurs in the generic case, when the points X1,X2 . . . . . Xm on ~ and Yl, Y2 . . . . . Y. on r/ are so arranged that no three of the segments ~ij are concurrent in a point between ~ and q. Then ~(m, n) = (~')(~) and ~(m, n) = 0 for k >/3; and it follows that V = (~')(~) points of intersection, E = m n + 2(~')(~) segments, and F = 1 + mn + (~')(~) regions are formed in a. Brousseau's principal interest was in the regular case, in which the m ~> 2 points X1, X2,. . . , X,. on ~ and the n ~> 2 points YI, )'2 . . . . . Y, on t /are equally spaced. This case is significantly more difficult to analyze, because the intersection numbers ~ are not easy to determine. In answer to a question raised by F. Hering and H. Harborth, Martini [4] recently gave a formula for the number V of points of intersection formed inside tr in the regular case. He showed that V = ~ l ( m i)(n j ) ~2(m 2i)(n -2j), where the first sum y~ is over all relatively prime pairs (i,j) with 1 ~< i ~< m 1 and 1 ~< j ~< n 1, and the second sum ~2 is over all relatively prime pairs (i,j) with 1 ~< i ~< 1⁄2(m 1) and 1 ~< j ~< 1⁄2(n 1). Unfortunately, the determination of the number of segments and regions that are formed by Brousseau's configuration in the regular case requires more specific information: one needs to know the number Vk of points of each multiplicity k >i 2. 2. Rectangular latt ice arrays Let Zm, = {(i,j): 1 ,<< i ~< m, 1 ~< j ~< n} be the m x n rectangular lattice array in the (Euclidean) plane; and for each k/> 2, let Lk(m, n) be the number of lines that meet Zm, in precisely k points. The numbers Lk(m, n) appear to be fundamental in various counting problems, but little seems to be known about them. Some elementary properties are readily established. We may as well suppose that m ~< n, because Lk(m,n)= Lk(n,m). When 2 ~< k < m, each line that meets 2~m, in precisely k points is paired with a different such line by reflection in the line 2y = n + 1, and it follows that the counter Lk(m,n) must be even for 2 ~< k < m. If r e < n , then L.(m,n)=m, and Lk(m,n)=O for m < k < n ; if r e = n , then L.(n, n) = 2n + 2. J.E. Wetzel / Discrete Mathematics 146 (1995) 263-269 265 No explicit general formula for Lk (m, n) in terms of more familiar counting functions seems to be known. The author's 1979 query [5] about Lk(m, n) in the A M S Notices elicited no responses. 3. The related projective arrangement and projective duality We begin our investigation of the regular case of Brousseau's problem by examining the projective arrangement formed by the lines that carry the segments ~ij of Brousseau's configuration. Regard the projective plane •2 as the Euclidean plane n :2 augmented by a line at infinity. It will be convenient to employ homogeneous coordinates in p2 in the form (x, y; z), where z = 0 is the line at infinity. Take ~ to be the line y = 0 and q the line y = z. For given positive integers m >~ 2 and n~>2, let X i=( i ,0 ;1 ) and Yi=(J , 1;1) for each i and j with l~ 2 formed by the lines 266 J.E. Wetzel / Discrete Mathematics 146 (1995) 263-269 of A,.. corresponds to a line in 6 2 that meets Zm. in exactly k points, and conversely. The following result is just a restatement of this assertion. Theorem 2. For each k >>, 2, the arrangement .4ran forms tk(A,..) = Lk(m, n) points of multiplicity k. As a consequence, we can express the number of vertices, edges, and faces formed by the arrangement Am. in terms of the numbers Lk(m, n). Corollary 3. The arrangement Amn forms exactly V(Amn) = 2 Lk(m,n) points of intersection, k>~2 E(A.,.) = ~, kLk(m,n) segments, and k>~2 F(Am.) = 1 + ~ ( k 1)Lk(m,n) regions. k>~2 Proof. The first of these is obvious from the theorem, and the second follows from the elementary fact that k points on a projective line divide that line into k segments. The last is a consequence of Euler's relation V E + F = I [2]. [] We note in passing the corresponding results for the arrangement ,~,.~ = A,.. w {~, ~/} formed by including the projective lines ~ (with equation y = 0) and q (with equation y = z) in A,.,. The counters for ,4,~ can be determined by inspection from those of Am.. Corollary 4. The arrangement ,4.,. = ,4,,. u { ~, rl} forms V(,4,.,) = V(Am,) + 1 = 1 + ~ Lk(m, n) points of intersection, k>~2 E(,'fm.)= E(Am.) + m + n + 2 = 2 + m + n + ~ kLk(m,n) k>~2 F(,4m~)= F(Am.) + m + n = l + m + n + ~ (k 1)Lk(m,n) k~>2 segments, and