The Happy Ending Problem

Transcription

1 The Happy Ending Problem Neeldhara Misra STATUTORY WARNING This doument is a draft version 1 Introdution The Happy Ending problem first manifested itself on a typial wintery evening in 1933 These evenings would witness meetings of the Anonymous soiety (named so beause the meetings happened under the statue of Anonymous in the ity park of downtown Budapest, in Hungary), whih involved Paul Erdös, Esther Klein, Endre Makai, and György Szekeres The meetings served to feed their ommon passion for mathematis, and as the story goes a few things besides It was during suh a meeting that Klein posed the following question after some doodling: how many points do we need on a plane to guarantee that some four of them form a onvex quadrilateral? Clearly, four points are not enough: with some bad luk, three of them form a triangle and the fourth lies inside the triangle formed by the other three, and as a result, there are no four points that form a onvex quadrilateral On the other hand, and a little less learly, five points suffie Figure 1: Klein s proof for why any set of five points in general position must ontain four points that determine b a onvex quadrilateral Klien s reasoning was as follows: if the five points lie on a pentagon, then any four of them form a onvex d quadrilateral If four points form a onvex quadrilateral with the remaining point in the interior, then we are done again, for the four outer points take on the desired shape Finally, if three points form a triangle and both the remaining points lie in the interior of the triangle, then one of the sides of the triangle together with the line that joins the two interior points will form a onvex quadrilateral We remark that throughout this disussion, the points are always assumed to be in what is alled the general position, whih means that no three points are ollinear a Page 1 of 6

2 2 Some Notions and Notation Makai managed to show that one would need nine points to guarantee the existene of a onvex pentagon Interestingly, there is exatly one onfiguration of eight points for whih a onvex pentagon annot be formed when the first four points form a square and the last four points form a square surrounding the first The addition of a new point anywhere auses a onvex pentagon This, with some detail, amounts to a proof of Makai s laim As far as speifis go, the story ends here nobody knows the exat value of the smallest number of points one would need to guarantee a onvex hexagon What is known, however, is that the number is more than 17, and less than 37 Beyond the speifis, the natural generalization of this question that emerges is the following: given a natural number n, how many points do we need before we an guarantee that some subset of n verties will form a onvex polygon? The first thing to ask is whether the answer is even finite for every n, and if it is, then one might wonder how the answer behaves as a funtion of n Notationally, the tradition is to use g(n) to refer to the smallest number of points neessary to guarantee that a subset of n of them will form a onvex polygon Erdös and Szekeres answered the first question in the affirmative, showing upper bounds in a ouple of different ways The first method appealed to Ramsey theory, and shows that: g(n) R 4 (5, n) The other method relies on geometri onsiderations and leads to: g(n) ( ) 2n n 2 Figure 2: Makai s extremal example showing 8 points without a onvex pentagon We will desribe a ouple of proofs with a Ramsey flavor and then move to the geometri variant Inidentally, it was Erdös who dubbed this the Happy Ending Problem onsidering that shortly after these results were published, Szekeres and Klein got married 2 Some Notions and Notation In Eulidean spae, an objet is onvex if for every pair of points within the objet, every point on the straight line segment that joins them is also within the objet The onvex hull or onvex envelope of a set X of points in the Eulidean plane or Eulidean spae is the smallest onvex set that ontains X For instane, when X is a bounded subset of the plane, the onvex hull may be visualized as the shape formed by a rubber band strethed around X Given a set S of points on the plane, we say that the points determine a onvex objet if there are no points from S in the interior of the onvex hull of S We use R(s, t) to denote the Ramsey number on 2-olored omplete graphs That is to say, n = R(s, t) is the smallest integer suh that any oloring of K n with two olors (say red and blue) admits a red K s or a blue K t The notion an be generalized in several ways For instane, we use n := R k (s, t) to denote the Ramsey number on 2-olored omplete hypergraphs, where we are assured that in any 2-oloring of the k-sized subsets of K n, there is a subset of [n] of size s all of whose k-subsets are olored red, or there is a subset of [n] of size t all of whose k-subsets are olored blue On the other hand, we use n := R(s 1, s 2,, s t ) to denote the Ramsey number on Page 2 of 6

3 3 A few proofs of Erdös-Szekeres t-olored omplete graphs In this ase, in any oloring of the edges of K n with olors from [t], there exists i, 1 i t, for whih there is a subset of size s i olored with olor i We use [t] to denote the set {1, 2,, t} 3 A few proofs of Erdös-Szekeres We first show that g(n) R 4 (5, n) Suppose we are given some onfiguration of N := R 4 (5, n) points Label the points arbitrarily from 1, 2,, N We now desribe a oloring of the four-subsets of [N] Let S = {a, b,, d} N Consider the points labeled a, b, and d If these points form a onvex quadrilateral in the plane, then olor S blue, otherwise, olor S red By definition, we know that R 4 (5, n) either has a five-sized subset all of whose 4-sized subsets are olored red, or it has a n sized set all of whose 4-subsets are olored blue Reall, at this point, Klein s observation that any set of five points in the plane admits a subset of four points that reate a onvex quadrilateral Given this, our oloring ertainly annot admit a five-sized subset all of whose 4-sized subsets are olored red, for this would mean that we have identified five points in the plan all of whose four-sized subsets do not form onvex quadrilaterals, ontraditing our observation Thus, we must have a subset of size n all of whose four-sized subsets are olored blue, and the orresponding points by the definition of the oloring are suh that any four points form onvex regions in the plane From here, we laim that the entire set of points form a onvex n-gon Notie that it suffies to show that no point lies in the interior of the onvex hull of the remaining points However, this is easy to see, beause whenever a point lies in the interior of the onvex hull of the remaining points, one an form a triangle enlosing this point, whih would then imply a subset of four points that do not form a onvex quadrilateral a ontradition a b d Figure 3: If all triangles of a four-sized subset are monohromati and the points orresponding to the four-sized subset do not form a onvex objet, we obtain a ontradition based on the design of the triangles There is another elegant proof that uses the Ramsey numbers This argument is attributed to Tarsy, who was an undergraduate student when he proposed the proof while taking an exam in a ombinatoris ourse As before, let suppose we have N points on the plane, and this time, let N = R 3 (n, n) We now proeed to olor the three-element subsets of [N] Let S = {a, b, } be a 3-element subset where a < b < (WLOG) Color the set S blue if one enounters the points in the order (a, b, ) by passing lokwise around their onvex hull, and olor it red otherwise Sine N is R 3 (n, n), we are assured of a subset X of size n all of whose 3-sized subsets are olored with the same olor Now, we laim that the orresponding points form a onvex n-gon We will show that any four points among these n form a onvex region, and the proof follows as it did in the previous ase We abuse notation and ontinue to use X to refer to the subset of points on the plane that have labels orresponding to elements in X Consider any four points from X in the plane, and let S be the subset of N formed by the labels of these points Reall that, by the hoie of X, all 3-sized subsets of X Page 3 of 6

4 3 A few proofs of Erdös-Szekeres are olored with the same olor, and in partiular, all 3-sized subsets of S are olored with the same olor as well Without loss of generality, let the olor be blue Assume, for the sake of ontradition, that one of the points of S lies inside the triangle formed by the other three Let the endpoints of the triangle be a, b, with a < b < (WLOG) Let d be the point on the interior of b this triangle Sine d forms a triangle with a and that is olored blue, so we have a < d < Now, if d < b, then a < d < b, and the set orresponding to the triangle (a, d, b) should be olored red, and if d > b, then b < d <, and the set orresponding to the triangle (b, d, ) should be olored red In either situation, we have d arrived at a ontradition a We now turn to a proof with a geometri flavor Let us all a sequene of n line segments onseutive if the right endpoint of the i th line segment is also the left endpoint of the (i + 1) th, for the 1 i < n The length of suh a sequene is simply n, the number of points that determine the set of line segments We now propose the following definitions: A sequene of onseutive line segments is alled a ap if their slopes are monotonially dereasing A sequene of onseutive line segments is alled a up if their slopes are monotonially inreasing Figure 4: An example of a ap (left) and a up (right) Now, notie that if we have a ap of length n, then the n points involved in the ap atually determine a onvex n-gon The same an be said of a up So we are naturally interested in knowing if any suitably large set of points either has a ap of length n or a up of length n Let s name and onquer: denote by f(k, l) the smallest number of points suh that any olletion of f(k, l) points on the plane ontains a ap of length k or a up of length l Notie that g(n) f(n, n) We will now fous on showing a bound for f(k, l) First, observe that f(k, 3) = k This is quite easy to see: for example, given any k points x 1,, x k, if they form a ap of length k then we are done If not, then the slopes of (x 1, x 2 ),, (x k 1, x k ) are not monotonially inreasing, then there must be a onseutive pair of segments p, p + 1 for whih the slope of (p + 1) is less than the slope of p, and these two segments form a up of length 3, and again, we are done Along similar lines, one an argue that f(3, k) = k Thus we have: Now we propose the following reurrene: f(k, 3) = f(3, k) = k f(k, l) f(k 1, l) + f(k, l 1) + 1 To show this, onsider a olletion X of f(k 1, l) + f(k, l 1) 1 points in the plane If some k points form a k-ap, the we are done If not, let C be the olletion of all the (k 1)-aps formed by points in X We are now going to behead the aps Speifially, let L be the set of all the left-most endpoints of aps in C Consider X \ L Now, notie that in X \ L, even the longest aps have length stritly less than k 1 Therefore, if X\L f(k 1, l), then X\L must admit a up of length l, and we are done If, on the other hand, X\L < f(k 1, l), then notie that L > f(k, l 1) 1, or L f(k, l 1) Sine we assumed that X does not have aps of length k (reall that if this was the ase we would be done straightaway), the set L must admit a up of length at least (l 1) Now, the points of this up are atually also the left-most endpoints of (k 1)-length aps in X Page 4 of 6

5 a 4 Shur s Theorem The final observation that linhes this argument is the following: if a up of length (l 1) ends at the beginning of a ap of length (k 1), then either the ap or the up an be extended further, elongating it by one This is quite easy to see, beause if v is the point ommon to the ap and the up, and if u lies before v on the up, and w is next to v on the ap, then depending on the angle between uv and vw, the slope from uv to vw is either inreasing or dereasing, and either way, either the ap or the up an be extended by one more, whih brings us to the desired onlusion Figure 5: A ap and a up ombine to form a larger up (left) or a ap and a up ombine to form a larger ap (right) The reurrene that we established an be opened up so as to arrive at the following: ( ) k + l 2 f(k, l) + 1 k 2 The orretness of the expression above an be verified indutively: ( ) ( ) ( ) k + l 3 k + l 3 k + l 2 f(k, l) f(k 1, l) + f(k, l 1) = + 1 k 3 k 2 k 2 The last equality is obtained by applying the simple identity: ( ) n + 1 = k Notie that the bound on g(n) omes from setting k = l = n ( ) ( ) n n + k k 1 4 Shur s Theorem An interesting appliation of Ramsey numbers is in number theory Consider the following definition For a number r, let Sr the smallest number whih satisfies the following property: for any partition of {1, 2,, S r } into n parts, there exists a partition ontaining three numbers satisfying the equation: x + y = z With our training so far, we are ompelled to ask ourselves - does S r exist, and if it does, what does it look like? Consider the Ramsey number on r-olored graphs: let t be R(3, 3,, 3) In other words, t is suh that in any r-oloring of K t, we are assured of the existene of a monohromati triangle Ramsey s theorem tells us that suh a t exists, and is finite, for every r Page 5 of 6

6 4 Shur s Theorem We now laim that the number t is an exellent andidate for S n That is to say, if we have at least t numbers at hand, then any partition of [t] into r parts will lead to a part with the sum property Imagine that we want to assoiate the partitions of [t] with olors of K t, and we would like to engineer the assoiation in suh a way that monohromati triangles start orresponding to the equation x + y = z We urge the reader to ponder for a moment on what a suitable assoiation might be To begin with, suppose we are given a partition the numbers [t] into r parts Let P i be the numbers in the i th partition Imagine now K t to be the omplete graph on the vertex set [t], and onsider the following oloring of the edges of K t : an edge (p, q) gets olor i if (p q) P i Notie that we have used only r olors Sine we know that every r oloring of K t admits a monohromati triangle, our oloring in partiular must have one Let the monohromati triangle have olor j, and let its end points {u, v, w}, with u < v < w (WLOG) Clearly, (v u), (w v) and (w u) lie in P j Notie that we might set x to (v u), y to (w v) and z to (w u), to ahieve the desired effet This result was disovered by Issai Shur, a German mathematiian of Russian desent in 1916 Shur s motivation was the study of the the famous equation of Fermat, namely x n + y n = z n, restrited to the field of primes If there are integers x, y, z satisfying this equation, then for every prime p, they also solve the ongruene equation x n +y n = z n (modp) Shur used the result above to show that the ongruene equation has a non-trivial solution for all large primes p, showing that Fermat s Last Theorem is false in the field Z p for any suffiiently large prime p Page 6 of 6