Triangular Ramsey Numbers
aa r X i v : . [ m a t h . C O ] D ec Triangular Ramsey Numbers
Zachary Chaney Connor Mattes Jed MenardTimothy TrujilloSeptember 27, 2018
Abstract
The purpose of this paper is to introduce the idea of triangular Ram-sey numbers and provide values as well as upper and lower bounds forthem. To do this, the combinatorial game Mines is introduced; after somenecessary theorems about triangular sets are proved. This game is easyenough that young children are able to play. The most basic variationsof this game are analyzed and theorems about winning strategies and theexistence of draws are proved. The game of Mines is then used to definetriangular Ramsey numbers. Lower bounds are found for these triangularRamsey numbers using the probabilistic method and the theorems abouttriangular sets.
Combinatorial games make it possible to easily explain the underlying workingsof combinatorial problems, which in turn help build a deeper understanding ofcombinatorics. In fact, some combinatorial games are so simple that they whereinvented as tools to be used in grade school classrooms to introduce studentsto logical reasoning and mathematical concepts. For example, the game of Triwas introduced by Haggard and Schonberger in [1] with the goal of “developinglogical skills of evaluating alternatives and their consequences.” The game hadthe unintended learning outcome of developing the skill of visual disembedding, i.e. the skill of picking out simple figures from a more complex image. Haggardand Schonberger in [1] point out that this ability has been linked to success insolving mathematical problems. In this paper we introduce a new game in thespirit of Haggard and Schonberger. In theory, these games are simple enoughthat they can be used to help develop mathematical problem solving skills inprimary school students.This paper focuses on a new combinatorial game called Mines which we use tointroduce the notion of a triangular Ramsey number. The game is called Mines,because our original game boards were in the shape of a Reuleaux triangle,the logo of Colorado School of Mines (see Figure 1). Our main results concerntheorems about this game, such as the existence of a winner and the possibilityof a winning strategy. The existence of triangular Ramsey numbers follows1rom the work of Dobrinen and Todorcevic in [2]. The primary purpose ofintroducing the game of Mines is to provide a simplified presentation of thefinite-dimensional Ramsey theory of the infinite-dimensional topological Ramseyspace R introduced and studied by Dobrinen and Todorcevic in [2]. The gameprovides a simplified method for defining triangular Ramsey numbers which arethe direct analogue of the Ramsey numbers for the finite-dimensional Ramseytheory of R . Figure 1: Two different sized Mines game boardsIn Section 2, we introduce the games of Tri and Sim. Then we describe howthe games can be used to define Ramsey numbers. Near the end of the sectionwe provide a short survey of some known Ramsey numbers and some boundson unknown Ramsey numbers. The section concludes with a lemma about Trineeded later in the paper.In Section 3, in order to help precisely describe the gameboards, we introducethe concept of a triangular set. The remainder of Section 3 is devoted to provingcombinatorial results related to counting triangular sets.In Section 4, we introduce the game of Mines and prove some theorems aboutits game play. For example, certain variations of Mines have the property thatthey can never end in a draw.In Section 5, we use the game to define the notion of a triangular Ramseynumber. Our main results in this section involve finding exact values of sometriangular Ramsey numbers and bounds for other triangular Ramsey numbers.Section 5 ends by applying the combinatorial results from Section 3 and theprobabilistic method as pioneered by Erd˝os to find lower bounds for triangularRamsey numbers.In Section 6, we collect together the main results of the paper in Table 11.We then discuss the connection between the triangular Ramsey numbers andthe topological Ramsey space R introduced by Dobrinen and Todorcevic in[2]. We conclude with some open questions and problems related to Mines andtriangular Ramsey numbers. 2 Tri, Sim and Ramsey numbers
Two games that have attracted attention in the literature are Sim and Tri. Thegame Sim m was introduced by Simmons in 1969 in [3] and Tri m was introducedby Haggard and Schonberger in 1977 in [1]. Both Sim m and Tri m are two playergames played on a game board of m > (cid:0) m (cid:1) possible edges. Eachplayer chooses a color, players alternate turns coloring uncolored edges usingtheir color. Both games end when a monochromatic triangle is constructed(three vertices all of whose edges have the same color) or all edges have beencolored. In Tri m the winner is the player that constructs a monochromatictriangle. In Sim m a player wins if they can force the other player to constructa monochromatic triangle. In either game, if no monochromatic triangle isconstructed then we say the game ends in a draw. The finite Ramsey theoremfor pairs implies that there exists a natural number m such that neither Tri m nor Sim m ever ends in a draw. Problem 1.
Find the smallest natural number m > m nor Sim m ever ends in a draw.The solution to Problem 1 is m = 6. For n < m , the two games havenatural generalizations to Tri m ( n ) and Sim m ( n ). The only difference beingthat these versions end when a monochromatic complete graph with n verticesis constructed. In this notation Tri m and Sim m correspond to Tri m (3) andSim m (3). The Finite Ramsey Theorem for pairs implies that for all naturalnumbers n there exists a natural number m such that neither Tri m ( n ) norSim m ( n ) ever ends in a draw. Problem 2.
Let n be a natural number greater than 2. Find the smallestnatural number m ≥ n such that neither Tri m ( n ) nor Sim m ( n ) ever ends in adraw.The solution to Problem 2 for the natural number m is called the Ramseynumber for n and denoted by R ( n ). It is known that R (3) = 6 and R (4) = 18.However, R (5) still remains unknown. Figure 2 gives upper and lower boundsfor some small Ramsey numbers. For example, from the table we have 43 ≤ R (5) ≤
49. In other words, there is a game of Tri (5) that ends in a drawand no game of Tri (5) can end in a draw. The lower bound in the second tolast row of the table is the lower bound obtain by Erd˝os using the probabilisticmethod in [4]. The last row gives the best known upper and lower bounds.The next Lemma about Tri will be used later to obtain an upper bound fora small triangular Ramsey number. For natural numbers k the notation R k ( n )denotes the Ramsey number R ( R ( · · · ( R | {z } k − times ( n )) · · · )). Lemma 1.
Let k be a natural number. If k games of Tri R k (3) (3) are played onthe same game board then there exists a complete graph with three vertices thatis monochromatic for each of the k games. Lower Bound R ( n ) Upper Bound References3 - 6 - [5]4 - 18 - [5]5 43 ? 49 [6] [7]6 102 ? 165 [8] [9]7 205 ? 540 [10] [9] n n/ ? 4 n − [4][11] n n n/ [ √ /e + o (1)] ? n − C log n log log n n [12] [13]Figure 2: Table of upper and lower bounds of some Ramsey numbers and theirreferences. Proof.
By the definition of Ramsey number there is complete graph with R k − (3)vertices that is monochromatic for the first game. Restrict the second game tothis complete subgraph of the game board. Again by the definition of Ramseynumber there is a complete subgraph of this graph with R k − (3) vertices that ismonochromatic for the first and second games. Continuing this way for k stepswe obtain a complete graph with three vertices that is monochromatic for all n games. A triangular number is a number that can be represented by a triangular ar-rangement of equally spaced points. For example, the number 15 can be ar-ranged into a triangle with five levels (see Figure 3).Figure 3: Triangular arrangement of 15 points.For this reason, 15 is called a triangular number. The first four triangularnumbers are 1, 3, 6, and 10 whose arrangements are given in Figure 4. If T n n th triangular number then by construction T n +1 = T n + n + 1 with T = 1. It is well known that these numbers can be represented as follows, T n = n X i =1 i = n ( n + 1)2 = (cid:18) n + 12 (cid:19) . Figure 4: Triangular arrangement of 1, 3, 6, and 10 pointsLet N = { , , , . . . } denote the set of natural numbers. A subset X of N is triangular if | X | is a triangular number. Let { x , x , x , . . . , x T n } be anincreasing enumeration of X . The numbers in X can be naturally arranged intoa triangle with n levels as shown in Figure 5. x x x x x x ... x T n − +1 . . . x T n ← Level 1 ← Level 2 ← Level 3 ← Level n Figure 5: Triangular arrangement of { x , x , x , . . . , x T n } Let △ denote the collection of all triangular subsets of N . For k ∈ N , let △ k denote the triangular sets with k levels i.e. those subsets of N such that | X | = T k . The next partial order is an adaptation of the order on R consideredby Dobrinen and Todorcevic in [2] to our current setting. It can be seen as arestriction (to triangular sets) of the partial order used by Laflamme, whichinspired the work in [2], to study complete combinatorics in [14]. Definition 1.
For
X, Y ∈ △ , X Y means that X ⊆ Y and every level of X is contained in a single distinct level of Y .For example, if we let W = { , , , , , } , X = { , , } , Y = { , , } and Z = { } then W ∈ △ , X, Y ∈ △ and Z ∈ △ such that Z ⊆ X, Y and5 , Y ⊆ W . Each level of X and Y are contained in a distinct single level of W and Z , being only one element, is contained in X and Y . Figure 6 displays thisconfiguration and the associated Hasse diagram in the partial order ( △ , ≤ ).12 34 5 6 WX YZ ZX YW
Figure 6: Hasse diagram and possible configuration for
W, X, Y and Z Definition 2.
For k ∈ N and X ∈ △ , let △ k ( X ) = { Y ∈ △ k : Y ≤ X } .Suppose X ∈ △ and let { x , x , x , x , x , x } be an increasing enumerationof X . Then △ ( X ) contains the elements { x , x , x } , { x , x , x } , { x , x , x } , { x , x , x } , { x , x , x } , { x , x , x } , { x , x , x } , { x , x , x } , { x , x , x } , and { x , x , x } . Thus for any X ∈ △ , |△ ( X ) | = 10.Note that if Y ∈ △ then |△ ( Y ) | = 41. To see this, first note that thereare (cid:0) (cid:1) = 4 ways to choose three elements from the last row of Y . Each oneof these possibilities can be added onto any element of △ ( Y ′ ) where Y ′ is theelement of △ obtained by removing the last level of Y to obtain a distinctelement of △ ( Y ). In particular there are (cid:0) (cid:1) · |△ ( X ) | = 40 elements of △ ( Y )obtained this way. The only other element of △ ( Y ) is Y ′ . So |△ ( Y ) | = (cid:0) (cid:1) · |△ ( Y ′ ) | + 1 = 41. By a similar argument, one can show that for all k ∈ N and for all Y ∈ △ k , if Y ′ ∈ △ k − then |△ k − ( Y ) | = 1 + k |△ k − ( Y ′ ) | . To better express these types of combinatorial relationships we introduce avariant of the binomial coefficient (cid:0) nm (cid:1) . The next definition should be contrastedwith the recursive definition of the binomial coefficients using Pascal’s triangle. Definition 3. (cid:20) nk (cid:21) = (cid:20) n − k (cid:21) + (cid:18) nk (cid:19)(cid:20) n − k − (cid:21)(cid:20) n (cid:21) = 1 , (cid:20) nn (cid:21) = 1With this definition note that for all natural numbers k , (cid:20) kk − (cid:21) = (cid:20) k − k − (cid:21) + (cid:18) kk − (cid:19)(cid:20) k − k − (cid:21) = 1 + k (cid:20) k − k − (cid:21) . (cid:20) (cid:21) = 1 , the argument in the previous paragraph implies that for allnatural numbers k and for all Y ∈ △ k , |△ k − ( Y ) | = (cid:2) kk − (cid:3) as they both satisfythe same recursive formula. The next Theorem generalizes this result. Theorem 1. If k < n and X ∈ △ n then (cid:20) nk (cid:21) = |△ k ( X ) | . That is, (cid:20) nk (cid:21) countsthe number of △ k ’s in a given △ n .Proof. Let k < n and X ∈ △ n . We show that |△ k ( X ) | satisfies the samerecursive definition as (cid:2) nk (cid:3) . It is clear that |△ n ( X ) | = 1. If we consider ∅ to be the only element of △ then |△ ( X ) | = 1. Thus the base cases of therecursions are the same. We complete the proof by verifying that |△ k ( X ) | = |△ k ( X ′ ) | + (cid:0) nk (cid:1) |△ k − ( X ′ ) | where X ′ is the triangular set in △ n − ( X ) obtainedby removing the last level of X .Note that |△ k ( X ) | can be thought of as the number of △ k ’s in a △ n . Itshould be clear that |△ k ( X ) | = |△ k ( X ′ ) | + c where c is the amount of new △ ′ k s formed when the last level of X is added back to X ′ . Each △ k contributing to c must have its last level in the last level of X . There are (cid:0) nk (cid:1) possibilities forthose k points in the final level of X . For each collection of k points in the lastlevel of X there are |△ k − ( X ′ ) | possibilities for triangular sets in △ k ( X ) whoselast level is the given k points. Therefore c = (cid:0) nk (cid:1) |△ k − ( X ′ ) | and |△ k ( X ) | = |△ k ( X ′ ) | + (cid:0) nk (cid:1) |△ k − ( X ′ ) | .Later in the paper we use the next corollary to obtain estimates for upperand lower bounds on (cid:2) nk (cid:3) . These estimates are needed to apply the probabilis-tic method to our combinatorial game and obtain lower bounds on triangularRamsey numbers. Corollary 1.
For 0 < k < n , (cid:20) nk (cid:21) = X
We show that P
Let X denote the set of moves made by player one and Y theset of moves made by player two. Let x m denote a move done by player one onposition m and y m a move done by player two on position m .At any point in the game we have X ∩ Y = ∅ . If both players always choose tomark a position then given that player one goes first we also have | X | = | Y | − | X | = | Y | , and | X | + | Y | = T when all positions have been played.12 34 5 6 12 34 5 6Figure 7: Two alternative gameboards for Mines .Player one wins if △ ( X ) = ∅ . Player two wins if △ ( Y ) = ∅ . A draw occurs if all positions have been played and neither player has won the game.That is, △ ( X ) = △ ( Y ) = ∅ and | X | + | Y | = T .Figure 8 describes a possible game such that X = { x , x , x } and Y = { y , y , y } . Here △ ( Y ) = ∅ since { y , y , y } ∈ △ indicating that player twohas won the game. Theorem 2.
Both players cannot construct a △ . That is, it is impossible forboth △ ( X ) = ∅ and △ ( Y ) = ∅ . . Proof.
Assume this is not the case. In other words, there exists a situation suchthat △ ( X ) = ∅ and △ ( Y ) = ∅ . Without loss of generality, we may assumethat all positions have been played.The largest row of the game board has 3 positions and the largest level ofa △ has 2 positions. Therefore, both X and Y cannot construct their △ ’slargest level in the same row of the game board. By the pigeon hole principlethere exists a, b ∈ { , , } such that x a , x b ∈ X or y a , y b ∈ Y . If x a , x b ∈ X then y , y ∈ Y . In order for △ ( X ) = ∅ we must have x ∈ X and we find △ ( Y ) = ∅ , a contradiction. If instead we had y a , y b ∈ Y then x , x ∈ X . Onceagain, in order for △ ( X ) = ∅ we must have x ∈ X and we find △ ( Y ) = ∅ , acontradiction. Therefore it is impossible for both players to construct a △ . Theorem 3.
Mines never ends in a draw.Proof. Assume a full game of Mines has been played. By the pigeonhole prin-ciple, at least two of the bottom three elements must be of the same color. Ifany of the three elements above the bottom row are of the same color as the twoon the bottom, then a △ has been constructed in that color. For this not tohappen, the three elements in the top two rows must all be in the opposite color.If this is the case, then a △ has been constructed in the opposite color. There-fore, it is impossible for a game of Mines to be played in which neither playerconstructs a △ , and no game can be played in which both players construct a △ .An interesting variation of the game play exploits the fact that the gameboard has 120 ◦ rotational symmetry about its center. There are three directionsto it given by the perpendicular from any of the three edges to its adjacentvertex. We let D n denote the orientation of the game board in the n direction(see Figure 9). We use the notation △ k,n ( X ) to denote all △ k ’s on the gameboard in the D n direction. For example, { , , } ∈ △ , ( X ) but { , , } 6∈ , ( X ). Note that if △ k,n ( X ) = ∅ then there exists a △ k in the directionof D n contained in X . Likewise, if △ k,n ( Y ) = ∅ then there exists a △ k inthe direction of D n contained in Y . The player that wins in two of the threedirections wins this variation of the game.12 34 5 6Direction 1, D Direction 2, D Direction 3, D Figure 9: A game board for Mines with directions D , D and D labeled.By Theorem 2 and Theorem 3 we see that there can never be a draw in asingle direction. In addition, given that there are three directions one playermust win in at least two directions. Therefore this variation of Mines can neverend in a draw. The next Theorem is true for both variations of Mines . Wegive the proof for the omnidirectional case as it is more interesting. Theorem 4.
Player one has a winning strategy for Mines .Proof. In the first three moves of the game we have | X | = 2 and | Y | = 1.Therefore player one can guarantee there exists q, r ∈ { , , } such that x q , x r ∈ X . By rotating the game board 120 ◦ to the left or right, we can without lossof generality, assume x , x ∈ X . At this point in the game { y , y , y } 6⊆ Y and { x , x , x } 6⊆ X since | X | + | Y | = 3. On the next move player one playsposition 1, 2 or 3 whichever is available. Player one wins in direction D since △ , ( X ) = ∅ . Thus player one only needs to win in one other direction to winthe game. If x ∈ X then { x , x , x } = X and △ , ( X ) = ∅ . If x ∈ X then { x , x , x } = X and △ , ( X ) = ∅ . If x ∈ X then { x , x , x } = X and △ , ( X ) = ∅ . Thus, in any case, player one wins in at least two out of the threedirections. m ( p, q, k ) Let p, q, m and k be positive integers with k ≤ p, q ≤ m . The game Mines m ( p, q, k )is played on a game board of size △ m with two players each assigned oneof two markings and/or colors. Players have alternating turns in which theymay choose to mark one position, in this case positions are the △ k ’s on the10ame board, or not mark a position. If p = q = n then we use the notationMines m ( n, k ). If p = q = n and k = 1 then we use the notation Mines m ( n ).In this notation, Mines = Mines (2) = Mines (2 ,
1) = Mines (2 , ,
1) since thepositions played by both players in Mines are the △ ’s on the game board.We again let X ⊆ △ k denote the set of moves made by player one and Y ⊆ △ k the set of moves made by player two. At any point in the gamewe have X ∩ Y = ∅ . If both players always choose to mark a position thengiven that player one goes first we also have | X | = | Y | − | X | = | Y | , and | X | + | Y | = (cid:2) mk (cid:3) when all positions have been played. Player one wins if theyconstruct a Z ∈ △ p such that △ k ( Z ) ⊆ X before the second player is able toconstruct a Z ∈ △ q such that △ k ( Z ) ⊆ Y . Player two wins if they constructa Z ∈ △ q such that △ k ( Z ) ⊆ Y before the first player is able to construct a Z ∈ △ p such that △ k ( Z ) ⊆ X . A draw occurs if all positions have been playedand neither player has won the game. That is, | X | + | Y | = (cid:2) mk (cid:3) and for all Z ∈ △ n , △ k ( Z ) X and △ k ( Z ) Y . The game ends when all positions havebeen played or one of the players wins.For some small game boards it is unnecessary to keep track of which playerfirst constructs the winning triangular set. Instead the players can simply fillout the game board completely and then check to see who wins. For example,Mines (3) = Mines (3 , ,
1) has this property. Figure 10 gives two examples ofgame boards for Mines ( n ).12 34 5 67 8 9 1011 12 13 14 15 12 34 5 67 8 9 1011 12 13 14 15Figure 10: Two example game boards of size △ Theorem 5.
In the game of Mines (3) , both players cannot construct a △ .That is, it is impossible for both △ ( X ) = ∅ and △ ( Y ) = ∅ .Proof. Assume this is not the case. There exists a situation such that △ ( X ) = ∅ and △ ( Y ) = ∅ . The largest level of a △ has 5 positions and the largest levelof a △ has 3 positions. Therefore, both X and Y cannot construct their △ ’slargest level in the same row of the game board. By the pigeon hole principle11here exists a, b, c ∈ { , , , , } such that x a , x b , x c ∈ X or y a , y b , y c ∈ Y .Suppose that x a , x b , x c ∈ X . We have the following two cases: Case 1:
There exists d, e, f ∈ { , , , } such that y d , y e , y f ∈ Y . Since △ ( Y ) = ∅ , either there exists g, h ∈ { , , } such that y g , y h ∈ Y or y , y ∈ Y .If g, h ∈ { , , } then x , x ∈ X . If y , y ∈ Y then there exists i, j ∈ { , , } such that x i , x j ∈ X . In order for △ ( X ) = ∅ we must have x ∈ X whichcauses △ ( Y ) = ∅ , a contradiction. Case 2:
There exists d, e, f ∈ { , , } such that y d , y e , y f ∈ Y . Then eitherthere exists g, h ∈ { , , , } such that x g , x h ∈ X or x , x ∈ X . If x , x ∈ X then clearly △ ( Y ) = ∅ . If there exists g, h ∈ { , , , } such that x g , x h ∈ X then we must have y , y ∈ Y . In order for △ ( X ) = ∅ we must have x ∈ X which causes △ ( Y ) = ∅ , a contradiction.If instead y a , y b , y c ∈ Y then a similar argument gives a contradiction. So itis impossible for both △ ( X ) = ∅ and △ ( Y ) = ∅ .The previous Theorem can be extended, in the game of Mines n − ( n ) bothplayers cannot construct a △ n . That is, it is impossible for both △ n ( X ) = ∅ and △ n ( Y ) = ∅ . We leave the proof to the interested reader. The Finite Ramsey Theorem for R which follows from the work of Dobrinenand Todorcevic in [2] can be used to show that for all p, q, k ∈ N with k ≤ p, q there is a game board of size △ m with m ≥ p, q such that Mines m ( p, q, k ) neverends in a draw. Problem 3.
Let p, q and k be natural numbers such that k ≤ p, q . Find thesmallest natural number m ≥ p, q such that Mines m ( p, q, k ) never ends in adraw.The solution to the Problem when p, q = 2 and k = 1 is m = 3. The solutionto the Problem for the natural numbers p, q and k is called the triangular Ramseynumber for p, q and k and denoted by R ( p, q, k ). If p = q = n then we denote R ( p, q, k ) by R ( n, k ). Lemma 2.
For all natural numbers p and q , R ( p, q, > p + q − .Proof. Suppose a game of Mines p + q − ( p, q,
1) is to be played. It is possiblefor the bottom row, of size p + q −
2, to contain p − X and theremaining q − Y . Note that neither player has constructed thebottom row of a winning triangle in the bottom row. Since the rows decreasein size as players move up the triangle, it is possible for both players to failto construct the bottom row of their winning triangle anywhere on the board.Therefore R ( p, q, > p + q − Theorem 6.
For all natural numbers p and q , R ( p, q,
1) = p + q − . Inparticular, for all numbers n , R ( n,
1) = 2 n − . roof. By the previous Lemma R ( p, q, > p + q −
2. So the result follows byshowing via induction on p + q that R ( p, q, ≤ p + q −
1. The base case occurswhen p = q = 1, i.e. p + q = 2. The base case is trivial; the first person to coloran element creates a complete △ and wins. In other words, R (1 , ,
1) = 1.Now suppose p + q = n + 1 and the results holds when p + q = n . Notethat ( p −
1) + q = n and p + ( q −
1) = n . Thus by the inductive hypothesis, R ( p − , q ) = R ( p, q − ≤ p + q −
2. We can now prove by contradictionthat no game of Mines p + q − ( p, q,
1) ends in a draw. Toward a contradictionsuppose a full game of Mines p + q − ( p, q,
1) has been played on a board of height n and ends in a draw, i.e. △ p ( X ) = ∅ and △ q ( Y ) = ∅ . Since p + q − n and the bottom level of the game board contains n positions, the pigeon holeprinciple mandates that the bottom row of the game board must contain either p elements in X , considered Case 1 , or q elements in Y , considered Case 2 . Case 1:
The bottom row of the game board contains Z ′ , a set of p elements in X . By the equation above, R ( p − , q ) = p + q −
1, and the inductive hypothesis,we can see that there exists either Z ∈ △ p − ( X ) = ∅ or ζ ∈ △ q ( Y ) = ∅ . Ifthe first is the case, then Z ∪ Z ′ ∈ △ p ( X ) and the game is won by player 1. Ifthe second is the case, then the game is won by player 2. In either, we have acontradiction. Case 2:
The bottom row of the game board contains q elements in Y . Bya similar method to above, we can show that either Z ∈ △ q − ( Y ) = ∅ or ζ ∈ △ p ( X ) exists. In either situation, a fully colored triangle is made and wehave a contradiction.In either case, we obtain a contradiction. Therefore R ( p, q, ≤ p + q − p + q = n + 1.Next we define a sequence ( M n,k ) which we use to establish upper bounds fortriangular Ramsey numbers. We let R ( n, k ) denote the smallest size, number ofvertices, of a complete graph such that for any coloring of its complete subgraphswith k vertices with two colors there exists a complete subgraph with n verticeswhere the coloring is monochromatic. The existence of these Ramsey numbersalso follows from Ramsey’s Theorem and could also be introduced by general-izing the game of Tri. Here we let R l ( n, k ) denote R ( R ( · · · ( R | {z } l − times ( n, k ) , k ) · · · ) , k ). M n,k = n if k = 1 ,M n,k = n if n = k,M n +1 ,k = R [ R Mn,k,k − k − ]( n + 1 , k ) if n > k > . (1) Theorem 7.
Let ( M n,k ) be the sequence recursively defined by (1) . If n ≥ k then R ( n, k ) ≤ M n − ,k . Proof.
We begin by establishing a simpler result by induction on n . Claim 1.
Suppose that a game of Mines M n,k ( n, k ) is played to completion.There exists a Z ∈ △ n on the game board such that given any level i of Z , ither all △ k ’s contained in Z whose last level is contained in the i th level of Z are played by player 1 or all △ k ’s contained in Z whose last level is containedin the i th level of Z are played by player 2.Proof. First note that by the previous Theorem R ( n,
1) = 2 n − M n − , .Thus, the Claim holds when k = 1. Next fix k ≥
2. Consider the base case when n = k and M n,k = n . If a game of Mines n ( n, k ) has been played to completionthen, since there is only one playable position on the game board, the base caseholds trivially.Assume that the Claim holds for M n,k . Suppose that a game of Mines M n +1 ,k ( n +1 , k ) has been played to completion. As usual let X denote the moves madeby player one and Y denote those made by player two. For each element Z ∈ △ k − in the first R ( M n,k , k −
1) levels of the game board, we play agame of Tri M n +1 ,k ( n + 1 , k ) on the final level of the game board as follows:player 1 plays the k -element set { i , i , . . . i k } if Z ∪ { i , i , . . . , i k } ∈ X andplayer 2 plays the k -element set { i , i , . . . i k } if Z ∪ { i , i , . . . , i k } ∈ Y . Byargument similar to the proof of Lemma 1, there exists an ( n + 1)-elementset in the last level of the game board { z , z , . . . z n +1 } such that for all Z ∈△ k − in the first R ( M n,k , k −
1) levels of the game board either ( † ) for all i , i , . . . i k ∈ { z , z , . . . z n +1 } , Z ∪ { i , i , . . . i k } ∈ X or ( ‡ ) for all i , i , . . . i k ∈{ z , z , . . . z n +1 } , Z ∪ { i , i , . . . i k } ∈ Y .Next consider the following hypothetical game of Mines R ( M n,k ,k − ( M n,k , k −
1) played on the first R ( M n,k , k −
1) levels of the our original game board. Let¯ X denote the moves made by player one and ¯ Y denote those made by playertwo. In this game, player 1 plays position Z ∈ △ k − if ( † ) holds and player2 plays position Z ∈ △ k − if ( ‡ ) holds. By definition this game does not endin a draw. If player 1 wins this game then there exists W ∈ △ M n,k such thatall △ k ’s whose first k − W and whose last level is contained in { z , z , . . . z n +1 } are played by player 1. If player 2 wins this game then thereexists W ∈ △ M n,k such that all △ k ’s whose first k − W and whoselast level is contained in { z , z , . . . z n +1 } are played by player 2.By the induction hypothesis there exists a Z ′ ∈ △ n ( W ) such that given anylevel i of Z ′ , either all △ k ’s contained in Z ′ whose last level is contained in the i th level of Z ′ are played by player 1 or all △ k ’s contained in Z ′ whose last levelis contained in the i th level of Z ′ are played by player 2.Let Z ′′ = Z ′ ∪ { z , z , . . . z n +1 } ∈ △ n +1 . Then either all △ ’s contained in Z ′′ whose last level is contained in the i th level of Z ′′ are played by player 1 orall △ ’s contained in Z ′′ whose last level is contained in the i th level of Z ′′ areplayed by player 2. Therefore the Claim holds by induction.To prove the inequality, assume toward a contradiction that a game ofMines M n − ,k ends in a draw. By the previous Claim there exists a Z ∈ △ n − on the game board such that given any level i of Z , either all △ k ’s containedin Z whose last level is contained in the i th level of Z are played by player 1or all △ k ’s contained in Z whose last level is contained in the i th level of Z areplayed by player 2. By the pigeon hole principle there are either at least n level14here player 1 plays all △ k ’s or at least n levels where player 2 plays all △ k ’s.If there are at least n levels where player 1 wins then any W ∈ △ n whose levelscome from these n levels witnesses a win for player 1, a contradiction. Simi-larly, if there are at least n levels where player 2 wins then any W ∈ △ n whoselevels come from these n levels witnesses a win for player 2, a contradiction.Therefore, this game could not have ended in a draw.When n = k + 1 the previous proof can be simplified and we obtain smallerupper bounds. In fact, in this special case, induction on n is unnecessary. Theorem 8.
Suppose that k ≥ . Then R ( k + 1 , k ) ≤ R [ R k +1 ,k − k − ]( k + 1 , k ) . Proof.
Let M = R [ R k +1 ,k − k − ]( k + 1 , k ). Toward a contradiction suppose thata game of Mines M ( k + 1 , k ) ends in a draw. As usual, let X denote the movesmade by player one and Y denote those made by player two. For each element Z ∈ △ k − in the first R ( k + 1 , k −
1) levels of the game board, we play a gameof Tri M ( k + 1 , k ) on the final level of the game board as follows: player 1 playsthe k -element set { i , i , . . . i k } if Z ∪ { i , i , . . . , i k } ∈ X and player 2 plays the k -element set { i , i , . . . i k } if Z ∪ { i , i , . . . , i k } ∈ Y . By Lemma 1, there existsan ( k + 1)-element set in the last level of the game board { z , z , . . . z k +1 } suchthat for all Z ∈ △ k − in the first R ( k + 1 , k −
1) levels of the game board either( † ) for all i , i , . . . i k ∈ { z , z , . . . z k +1 } , Z ∪ { i , i , . . . i k } ∈ X or ( ‡ ) for all i , i , . . . i k ∈ { z , z , . . . z k +1 } , Z ∪ { i , i , . . . i k } ∈ Y .Next consider the following hypothetical game of Mines R ( k +1 ,k − ( k +1 , k − X denote the moves made by player one and ¯ Y denote those made byplayer two. In this game, player 1 plays position Z if ( † ) holds and player 2 playsposition Z if ( ‡ ) holds. In other words, ¯ X = { Z ∈ △ k − : Z ∪{ i , i , . . . i k } ∈ X } and ¯ Y = { Z ∈ △ k − : Z ∪ { i , i , . . . i k } ∈ Y } . By definition this game does notend in a draw.Suppose that player 1 wins and let W be some element of △ k +1 witnessinga win for player 1 (in our hypothetical game of Mines R ( k +1 ,k − ( k + 1 , k − △ ′ k s in W are played by player 2 (in our original game)because otherwise the game would not have ended in a draw. So, without lossof generality, we can assume that there is at least one W ′ ∈ △ k ( W ) playedby player 1 (in the original game). However, this is a contradiction because W ′ ∪ { z , z , . . . z k +1 } then witnesses a win for player 1 (in our original game).If instead player 2 wins, we can let ¯ Z be some element of △ ( ¯ Y ) and we obtaina similar contradiction. The work in this section follows closely from the probabilistic method describedby Erd¨os [4]. In our case, we apply it to a randomly played game of Mines.
Theorem 9.
Let p, q, k and m be natural numbers such that k ≤ p ≤ q ≤ m . If (cid:2) mp (cid:3) · − [ pk ] + (cid:2) mq (cid:3) · − [ qk ] < then R ( p, q, k ) > m. roof. Let p, q, k and m be given. To use the probabilistic method, we considertwo players playing a game of Mines m ( p, q, k ) game with a fair coin. The playerspick a position and flip the coin to see if they will play that triangular set or skiptheir turn. The game is then played to completion using these random moves.Consider the random variables X and Y where X counts the number ofwinning △ p ’s colored by player 1 and Y the number of winning △ q ’s colored byplayer 2. By Theorem 1, the probability that a randomly chosen △ p from thegame board witnesses a win for player one is 2 − [ pk ] and for a randomly chosen △ q the probability that it witnesses a win for player two is 2 − [ qk ]. By Theorem1, there are exactly (cid:2) mp (cid:3) possible △ p ’s on the game board and (cid:2) mq (cid:3) possible △ q ’s. Thus, (cid:2) mp (cid:3) · − [ pk ] = E [ X ] and (cid:2) mq (cid:3) · − [ qk ] = E [ Y ]. Thus E [ X + Y ] = E [ X ] + E [ Y ] = (cid:2) mp (cid:3) · − [ pk ] + (cid:2) mq (cid:3) · − [ qk ] <
1. Since the expected value of X + Y is less than one then there exists some game of Mines m ( p, q, k ) that ends in adraw. Therefore m < R ( p, q, k ).The previous Theorem can be used to find lower bounds for small values of p, q and k by searching for the largest value of m that satisfies the inequality.A summary of these values for small p, q and k can be found in Figure 11. Thenext Theorem provides asymptotic estimates for large values of p, q and k with p = q = n which don’t require the computation of (cid:2) nk (cid:3) nor (cid:2) mn (cid:3) for any m . Theorem 10.
For all natural numbers n and k with n ≤ k , m > (2 πT n ) Tn · r T n e · nk − kk kkTn − Proof.
Let m = R ( n, k ). By the previous Theorem we must have (cid:2) mn (cid:3) − [ nk ] +1 ≥ . Now we also know that (cid:20) mn (cid:21) ≤ (cid:18) T m T n (cid:19) This is because if we are counting △ n ’s (which has T n points) in a △ m (which has T m points). Therefore we are counting some T n sized subset of T m points withcertain properties. We know that (cid:0) mn (cid:1) ≤ m n n ! Substituting in these inequalitiesgives, T T n m T n ! · − [ nk ] +1 ≥ ⇒ ( m ( m + 1)) T n T n ! · − [ nk ] +1 − T n ≥ n ! > √ πn · (cid:0) ne (cid:1) n we have( m ( m + 1)) T n √ πT n · T T n n · e − T n · − [ nk ] +1 − T n ≥ ⇒ ( m + 1) T n √ πT n · T T n n · e − T n · − [ nk ] +1 − T n ≥ ⇒ ( m + 1) T n ≥ p πT n · (cid:18) T n e (cid:19) T n · nk ] + T n − From Corollary 1, (cid:2) nk (cid:3) = P
Use upper and lower bounds for triangular Ramsey numbers tocharacterize the Dedekind cuts that can be associated to ultrafilter mappingsfrom Ramsey for R ultrafilters.In the follow up paper [15], Dobrinen and Todorcevic introduce a hierarchyof spaces R α for α < ω that extend the space R . Problem 5.
Introduce a combinatorial game, similar to Mines, that provides asimplified presentation of the finite-dimensional Ramsey theory of the infinite-dimensional topological Ramsey spaces R α for α < ω defined and studied byDobrinen and Todorcevic in [15]. Then find upper and lower bounds for Ramseynumbers based on these games. 17 q k Lower Bound R ( p, q, k ) Upper Bound Result1 2 1 − − Thm. 62 2 1 − − Thm. 62 3 1 − − Thm. 63 3 1 − − Thm. 62 2 2 − − trivial2 3 2 − − trivial3 3 2 6 ? R (3) Thm. 83 4 2 6 ? M , Thm. 7&Thm. 94 4 2 25 ? M , Thm. 7&Thm. 93 3 3 − − trivial3 4 3 − − trivial4 4 3 20 ? R [ M , ](4 ,
3) Thm. 8&Thm. 94 5 3 20 ? M , Thm. 7&Thm. 95 5 3 9 . × ? M , Thm. 7&Thm. 94 4 4 − − trivial4 5 4 − − trivial5 5 4 3425 ? R [ M , ](5 ,
4) Thm. 8&Thm. 930 30 20 221 ? M , Thm. 7&Thm. 1035 35 20 4 . ? M , Thm. 7&Thm. 1040 40 20 7 . × ? M , Thm. 7&Thm. 10Figure 11: Table of some triangular Ramsey numbers.Upper and lower bounds for the Ramsey numbers associated to the spaces R α , α >
1, also have similar applications to characterizing the Dedekind cutsthat can be associated to ultrafilter mappings from Ramsey for R α ultrafilters.In addition to these open problems there are problems still open related toplaying the game of Mines.Theorem 4 provides an explicit description of a winning strategy for Mines .The proof of Zermelo’s theorem in [17] can be adapted to show that either playerone or player two must have a winning strategy for any game of Mines m ( n, k )where m ≥ R ( n, k ). A standard strategy stealing argument can then be usedto show that player one must have a winning strategy for the game. However,the proof of Zermelo’s theorem does not provide for an explicit description ofhow player one should play to win the game. In the game of Mines (3 , △ . In addition, some gameboards will have an odd number of positions giving player one an additional18osition over player two; perhaps this gives them an even bigger advantage.Regardless, giving an explicit description of the winning strategy for a gameboard of size △ m seems like a difficult problem. Problem 6.
Find an explicit description of the winning strategy for player onein a game of Mines m ( n, k ) where m ≥ R ( n, k ). In particular, describe thewinning strategy for player one in a game of Mines (3 , k players. In this variation, we have a different set of Ramseynumbers. It is clear that as the number of players increases the size of theassociated triangular Ramsey numbers also increase. Problem 7.
How does the game of Mines change when adding more players?In particular, find upper and lower bounds for triangular Ramsey numbers forvariations of Mines with more than two players.In the off-diagonal case of the game Mines m ( p, q, p < q , then player onehas an advantage as they are now constructing a smaller triangular set. Playertwo can be given an advantage in this game by allowing them to play morethan one △ on each turn. The question then becomes, how many △ s shouldplayer two be allowed to color per turn such that the game is fair? Note that,if m = p + q − q △ s per turn then they clearlyhave a winning strategy provided that p > Problem 8.
Suppose p and q are given with 1 < p < q . What is the smallestnumber of △ s player two can be allowed to color per turn such that player twohas the winning strategy?In addition to simplifying the finite Ramsey theory of the Ramsey space R ,a secondary goal was to introduce a game that was simple enough to playedby young children. In this way, the game can be played with the intention ofdeveloping logical skill. In the paper [1], the authors give evidence that thegame of Tri can be used to develop logical thinking skills in young children.Our final problem will be of interest mainly to researchers in math education. Problem 9.
Give concrete evidence that the game Mines can be played byyoung children and used to develop their visual disembedding skills.————
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