Iterate Point-Line Configurations Grow Doubly-Exponentially Joshua Cooper an Mark Walters July 9, 008 Abstract Begin with a set of four points in the real plane in general position. A to this collection the intersection of all lines through pairs of these points. Iterate. Ismailescu an Raoičić (003 showe that the limiting set is ense in the plane. We give oubly exponential upper an lower bouns on the number of points at each stage. The proof employs a variant of the Szemeréi-Trotter Theorem an an analysis of the minimum egree of the growing configuration. Consier the iterative process of constructing points an lines in the real plane given by the following: begin with a set of points P 1 = {p 1, p, p 3, p 4 } in the real plane in general position. For each pair of points, construct the line passing through the pair. This will create a set of lines L 1 = {l 1, l, l 3, l 4, l 5, l 6 }. Some of these constructe lines will intersect at points in the plane that o not belong to the set P 1. A any such point to the set P 1 to get a new set P. Now, note that there exist some pairs of points in P that o not lie on a line in L 1, namely some elements of P \ P 1. A these missing lines to the set L 1 to get a new set L. Iterate in this manner, aing points to P k followe by aing lines to L k. We assume that the original configuration is such that for every k N no two lines in L k are parallel. Now we introuce some notation for this iterative process. The k th stage is efine to consist of these two orere steps: 1. A each intersection of pairs of elements of L k to P k+1, an. A a line through each of pair of elements of P k to L k+1. 1
Uner this efinition, we say that stage 1 begins with the configuration of four points with six lines an stage k begins with n k points with m k lines. We will enote the set of points at the beginning of stage k by P k an likewise the set of lines at the beginning of stage k by L k. There are some trivial bouns on the number of points an lines at stage k that can be obtaine with this notation. Since a point in P k must lie at the intersection of at least two lines of L k 1 we know that at stage k, there are at most ( m k 1 points. Similarly, since a line in L k must contain at least two points from P k we know that at stage k there are at most ( n k lines. In other wors, n k ( mk 1 an m k ( nk. From this it follows that n k+1 ( mk ( < < ( n k = n k 4 8 an m k+1 ( nk+1 (( mk < ( mk ( < m k = m k 4 8. Note that a stage in this iterative process can be alternatively efine as follows: 1. Place a point at any intersection of a pair of lines for which a point oes not alreay exist.. Take the ual of the configuration of points an lines (points become lines an lines become points. 3. Return to step 1. Hence, points an lines play a very similar role in this process an we only nee to consier bouns on one of the two quantities. Henceforth we will only provie arguments concerning the bouns on n k. A trivial lower boun is given in the following: Proposition 1. For all k N, n k+1 n k + 1.
Proof. If this claim is false then we must have a stage at which the process stabilizes [1]. So, suppose that the process stabilizes at the beginning of stage k an let conv(p k enote the convex hull of P k, where conv(p k enotes the number of vertices of this convex hull. Suppose first that conv(p k 4. In this case, we can fin two nonajacent, nonparallel sies of the convex hull, which lie on lines that intersect outsie of the convex hull. This contraicts the stability supposition. So, conv(p k = 3. Let {a, b, c} be the set of vertices of the triangle forming the convex hull. Suppose that there exist points along at least two of the sies of the triangle efine by {a, b, c}, say x ab an y bc. In this case, the line forme by xy must intersect ac outsie the convex hull, again contraicting stability. So, there exist points along at most one of the sies of the triangle efine by {a, b, c}. Suppose that there exists some point x in the interior of {a, b, c} an efine y = ax bc, z = cx ab. In this case, we have y bc an z ab, a contraiction to the assumption that at most one sie of the triangle contains points. The only remaining possibility is that P k is comprise of n k 1 collinear points. But, the starting configuration of points an lines has the conition that for any line in L 1, there are at least two points of P 1 not passing through it. Since we never remove any points uring this process, then this must hol true for every stage, in particular stage k. This contraiction completes the proof. We efine the egree of a point p P k, enote k (p, to be the number of istinct lines incient upon p at the beginning of stage k. Similarly, the egree of a line l L k, enote k (l, is the number of istinct points through which it passes at the beginning of stage k. Also, let an δ k = min{ k (p p P k } an δ k = min{ k (l l L k } k = max{ k (p p P k } an k = max{ k (l l L k }. Define an n n gri to be any configuration of two collections of n parallel lines, where the one collection is not parallel to the other. Using these efinitions, we obtain the following observation: Proposition. For all k N, δ k 3. 3
Proof. Suppose to the contrary that there exists some k N with δ k < 3. Since there are no points of egree 1, we must have δ k =. So there exists p P k with k (p =, i.e., there exist two lines l, l L k with P k l l. Note that n = 7 an = 3 an so for all l, l L, P l l. Since we never remove points in this iterative process, we know that if there exists l, l L k with P k l l, then k <, i.e., k = 1. But we know that δ 1 = 3, a contraiction. We can obtain major improvements to the trivial lower boun using the following: Lemma 3. The minimum number of parallel lines require to pass through all of the intersections of an n n gri is n 1. Proof. Suppose that Q an R are sets of parallel lines that comprise an n n gri. Let S be a minimal witness set of s parallel lines passing through all intersections of the gri. We aim to show that s n 1. Without loss of generality, orient the gri so that the lines of S are vertical in the xy-plane an let X = {x 1, x,..., x s } be the x-intercepts of the lines of S. So X is the collection of projecte points, when we project the gri intersections onto the x-axis with this orientation. Let π(p enote the projection of a point p in the gri onto the x-axis. Arbitrarily choose lines l q, l r in the gri with l q Q an l r R. Let q 1, q,..., q n an r 1, r,..., r n be the points of intersection of l q with R an l r with Q, respectively, where an π(q 1 π(q π(q n π(r 1 π(r π(r n. Suppose also that q i = r j. Define A an B to be the sets of real numbers given by A = {π(q 1, π(q,..., π(q n } an B = {π(r 1 π(r j, π(r π(r j,..., π(r n π(r j }. Uner this setting we have that S = A + B an thus s = A + B. 4
It is well known that A + B n 1 for any pair A, B of sets of carinality n an that equality is achieve when A an B are arithmetic progressions []. It follows that s n 1, completing the proof. Using this lemma we can prove the following: Theorem 4. δ k+1 min{n k 1, δ k 3}. Proof. Let p P k. It suffices to show that k+1 (p min{n k 1, δ k 3}. First suppose each line in L k that passes through p has egree. In this case, it s easy to see that there are k (p + 1 points at the beginning of stage k an so k (p = n k 1. Since we never remove lines, we know that k+1 (p k (p = n k 1 min{n k 1, δ k 3}. Now suppose there exists a line l L k that passes through p with k (l 3. Let q, r P k be the other two points on l. Note that k (q δ k an k (r δ k an so there exist two sets of lines L q = {l q1, l q,..., l qn } L k \l an L r = {l r1, l r,..., l rm } L k \l, where n, m δ k 1 an the sets L q l an L r l consist of the lines incient upon q an r, respectively. Now, consier the real plane as a subset of the real projective plane in the stanar way an let l be the line at infinity. We restrict our attention to arbitrarily chosen subsets L q L q an L r L r, where L q = L r = δ k 1. These lines form a (δ k 1 (δ k 1 gri. Now in this gri we will place a point at each intersection for which one oes not alreay exist uring stage k. After oing so, we will construct a line through each pair of points for which one oes not alreay exist. In particular, we will o so for pairs of points of the form (p, x, where x lies at the intersection of lines from L q an L r. So, at the beginning of stage k + 1, there will be at least s lines incient upon p, where s enotes the 5
number of lines necessary to ajoin p with all of the intersections of the gri. In other wors, k+1 (p s. Note that any lines passing through p woul form a thir collection of parallel lines to a to the gri. Therefore, s is at least the minimum number of parallel lines require to pass through all of the intersections of a (δ k 1 (δ k 1 gri. Applying Lemma 1 yiels k+1 (p s (δ k 1 1 = δ k 3 min{n k 1, δ k 3}. Now by using techniques similar to the preceing proofs, we can obtain even faster growth of the minimum egree. We will then use the growth rate of δ k to provie arguments for a better lower boun on n k. First, let cr(g enote the crossing number of a graph, which is the minimum number of crossings in a planar rawing of the graph G. We will use the following lemma regaring crossing numbers (the proof can be foun in [3]: Lemma 5. If a graph G with n vertices an e eges has e > 7.5n, then we have e 3 cr(g 33.75n. We now use this crossing number inequality in the following theorem. The argument closely resembles Székely s proof ([4] of the Szemeréi-Trotter Theorem (first appearing in [5]. Theorem 6. Let F = {F 1, F,..., F N } be a collection of N 4 families, each of exactly k parallel lines, no two collections parallel to each other. Let P enote the collection of points that lie at the intersections of lines l i an l j, where l i F 1 an l j F j for some j k. Then where c is a positive real constant. P ck N 1/, Proof. Let A enote this configuration of P points an Nk lines. Let i be the number of point-line inciences in A. Note that there are N ifferent 6
families of parallel lines in A, each containing exactly k lines. For all families except F 1, each line contains exactly k points from P an thus contains exactly k 1 line segments which connect two points from A, call them eges. We know that k an so k 1 k/. Hence, each line contains at least k/ eges an if we a this up over all of the Nk lines, we see that the number of eges obtaine in this manner is at least half of the total number of inciences. In other wors, (total number of eges i. Now, we can count the exact number of eges in A. For the k lines of F 1, there are P k eges because all P points lie on the lines of F 1 an for each line we must subtract one to count the number of eges. For each of the remaining N 1 families there are exactly k lines, each containing exactly k 1 eges, yieling a total of (N 1k(k 1 eges. Aing these quantities together, we obtain a gran total of eges, which simplifies to P k + (N 1k(k 1 P + Nk(k 1 k. Now consier the graph G with V (G = P an E(G consisting of the aforementione eges. Since all of the eges lie on one of Nk lines, an any two lines intersect in at most one point, we have cr(g (Nk. Applying the crossing number inequality, we obtain that either P + Nk(k 1 k 7.5 P (1 or that (Nk ( P + Nk(k 1 k 3 33.75 P ( 7
In the case of (1 we get Nk(k 1 k 6.5 P which implies that Nk(k 1 k P. 6.5 Now, since we know that k an N 4, we have k 1 k/ an N N 1/. Combining this with the previous equation yiels P Nk(k 1 k 6.5 (N k 13 c 1 k N 1/ for some positive constant c 1. In the case of ( we have an so 33.75 P (Nk ( P + Nk(k 1 k 3 c P /3 (Nk /3 P + Nk(k 1 k for some positive constant c. Recall that the RHS of this inequality is E(G, which is at least i/. So we have i c P /3 (Nk /3. Now, since each of the Nk lines in A must pass through at least k points, then there are at least Nk inciences. From this it follows that for some positive constant c 3. Hence, an so Nk i c 3 P /3 (Nk /3 N 3 k 6 c 4 P (Nk P c 5 k N 1/ for some positive constants c 4 an c 5. So in both cases, we en up with our esire result. 8
Now, we can use the previous result to prove the following lemma regaring egree growth: Lemma 7. Given any point p P k with k (p =, there exists a positive real constant c such that 1/ k+1 (p cδ k Proof. Let p P k with k (p =. By the pigeonhole principle, there exists some line l through p with at least s = n k 1 points on it (excluing p. Since each of these s points has at least the minimum egree, we know that there are at least δ k 1 lines through each point (excluing l. Consier the real plane as a subset of the real projective plane in the stanar way an let l be the line at infinity. If we restrict our attention to only the points on l an the lines through them, then we obtain a gri of s + 1 families of parallel lines, one family for each of the points on l. Each family of parallel lines contains at least δ k 1 lines an no two families can be parallel (since they come from istinct points. We woul like to restrict our attention to families of exactly δ k 1 parallel lines. So for each family, except for the one generate by p, arbitrarily choose a subset of δ k 1 lines an isregar all other lines in that family. Let F be the family of lines through p an choose one family R to be a set of reference lines. Let P 0 enote the set of points that lie at the intersection of a reference line an one of the other s 1 families (excluing F. Now, uring stage k, a point must be ae to any intersection for which one oes not alreay exist, in particular all points of P 0. Also, a line must be ae to connect any pair of points for which one oes not alreay exist, in particular for the pairs in the set T = {(p, q q P 0 }. Let t enote the number of istinct lines generate by pairs in the set T. Note that any such line can pass through at most δ k 1 points of P 0 because all the points of P 0 lie in the family R, which contains exactly δ k 1 lines. It follows that k+1 (p t P 0 δ k 1 P 0 δ k, (3 with the first inequality holing because any line generate by the set T must pass through p, an hence contributes to its egree in the stage. Now, for the moment, exclue F from our collection of families an consier all other families of lines along with the points of P 0. Suppose s < 4. 9
Hence, n k 1 < 4 an so n k < 4 + 1, i.e., n k 4. It follows that 1/. Note that k+1 (p δ k for all p P k+1 an so k+1 (p cδ k hols true for c = 1. Hence, 1/ k+1(p cδ k for some positive real constant c, as esire. Now suppose that s 4. Since we also know that δ k 3, i.e., δ k 1 for all k N, we can apply Theorem to this configuration with F 1 = R, N = s, an k = δ k 1. It follows that P 0 c 1 (δ k 1 ( nk 1 1/ for some positive constant c 1. Now, note that δ k an n k 4, which implies that δ k 1 1δ k an n k 1 3n 4 k. Combining this with the previous equation, we obtain P 0 c 1 ( δk ( 1/ 3nk c δ k 4 1/ (4 for some positive constant c. If we combine (3 an (4 we get 1/ k+1 (p c δ k, (5 as esire. This completes the proof. Note that P 0 n k+1 an so n k+1 cδ k 1/ must hol for any point p P k with k (p = an some positive constant c. In particular, it must hol for p P k chosen with k (p = δ k. In this case ( 1/ nk n k+1 cδ k = cδ 3/ k n 1/ k. (6 δ k Now we are able to provie an improve lower boun on the minimum egree, which will be use to improve the lower boun on n k. 10
Lemma 8. Given any k N, ɛ 0, an any positive real constant c 1 such that δ k c 1 n k ɛ, there exists some positive real constant c such that δ k+1 c n k ( 1+ɛ 3. Proof. Suppose that δ k c 1 n k ɛ for some k N, ɛ 0, an positive real constant c 1. Define α R by α = 1 + ɛ. 3 Let p P k with k (p =. There are two cases: either < n k α or n k α. If < n k α, then by Lemma 7 we have ( 1/ nk 1 α k+1 (p c 0 δ k = c n α 0 δ k n k k for some positive real constant c 0. Since δ k c 1 n k ɛ an α = (1 + ɛ/3, we must have 1 α k+1 (p c 0 δ k n k c 0 c 1 n ɛ 1 ɛ k n k 3 = c n k ( 1+ɛ 3, where c = c 0 c 1. If instea n k α, then obviously we have k+1 (p n k α c n k ( 1+ɛ 3, where c 1. So, in both cases, we have the conclusion that k+1 (p c n k ( 1+ɛ 3 an this will hol true for any p P k. Since the choice of p P k was arbitrary, we have δ k+1 c n ( 1+ɛ k 3, (7 as esire. This completes the proof. 11
Now, we are able to obtain some numerical results from Lemma 8. Note first that 1/3 δ k c n k 1 (8 for some positive real constant c (letting ɛ = 0. Further recall that the trivial upper boun yiels Combining (8 an (9, we get that for all k N n k 1 (8n k 1/4. (9 δ k c n k 1 1/3 c [(8n k 1/4 ] 1/3 c 3 n k 1/1 for some positive real constant c 3. Now, we can apply Lemma 8 with ɛ = 1 1 for any k N. Since 1 + ( 1 1 = 7 3 18, we get 7/18 δ k c 4 n k 1 (10 for some positive real constant c 4. Now, we combine (10 with (9, to obtain that for all k N δ k c 4 n k 1 7/18 c 4 [(8n k 1/4 ] 7/18 c 5 n k 7/7 for some positive real constant c 5. This process can be iterate an the limiting value of ɛ > 0 is foun by setting which implies that ɛ = 1 + ɛ 1 ɛ = 0.1 + o(1. Now, using Lemma 8 (ɛ = 0.1 + o(1 with (6, we obtain n k+1 cδ k 3/ n k 1/ c (c n k 1 1+(0.1+o(1 3 c n k 1 1.1+o(1 3/ n 1/ k (11 for some positive constants c, c, an c. Using (11 along with the trivial upper boun, we obtain the following theorem: 1
Theorem 9. Given k N, there exists real positive constants c 1 an c such that c 1 4 1.0488k n k c 4 4k. (1 Proof. Note first that n 1 = 4 an n = 7. From repeate use of (11 we get that there exist real positive constants a 1, a, a 3, a 4 such that an a 1 4 (1.1+o(1k n k+1 a 4 4k+1 a 3 7 (1.1+o(1k 1 n k a 4 4 4k. Taking square roots, it follows that there exist real positive constants c 1 an c such that c 1 4 1.0488k n k c 4 4k, as esire. Theorem 9 shows that the growth of n k is inee oubly-exponential, as the easy upper boun suggests. However, a consierable gap still remains between the exponents. While we have no rigorous argument proviing improvements of either boun, computational results an heuristic reasoning suggest that the actual growth rate of n k is closer to the state upper boun. References [1] Dan Ismailescu an Raoš Raoičić. A ense planar point set from iterate line intersections. Computational Geometry, 7:57 67, 004. [] Melvyn B. Nathanson. Aitive Number Theory: Inverse Problems an the Geometry of Sumsets. Springer, 1996. [3] János Pach an Géza Tóth. Graphs rawn with few crossings per ege. Combinatorica, 17(3:47 439, 1997. [4] László A. Székely. Crossing numbers an har Erős problems in iscrete geometry. Combin. Probab. Comput., 6(3:353 358, 1997. [5] Enre Szemeréi an William T. Trotter, Jr. Extremal problems in iscrete geometry. Combinatorica, 3(3-4:381 39, 1983. 13