K J Ramsey Popular Books

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In combinatorics, Ramsey's theorem, in one of its graph-theoretic forms, states that one will find monochromatic cliques in any edge labelling (with colours) of a sufficiently large complete graph. To demonstrate the theorem for two colours (say, blue and red), let r and s be any two positive integers. Ramsey's theorem states that there exists a least positive integer R(r, s) for which every blue-red edge colouring of the complete graph on R(r, s) vertices contains a blue clique on r vertices or a red clique on s vertices. (Here R(r, s) signifies an integer that depends on both r and s.) Ramsey's theorem is a foundational result in combinatorics. The first version of this result was proved by Frank Ramsey. This initiated the combinatorial theory now called Ramsey theory, that seeks regularity amid disorder: general conditions for the existence of substructures with regular properties. In this application it is a question of the existence of monochromatic subsets, that is, subsets of connected edges of just one colour. An extension of this theorem applies to any finite number of colours, rather than just two. More precisely, the theorem states that for any given number of colours, c, and any given integers n1, …, nc, there is a number, R(n1, …, nc), such that if the edges of a complete graph of order R(n1, …, nc) are coloured with c different colours, then for some i between 1 and c, it must contain a complete subgraph of order ni whose edges are all colour i. The special case above has c = 2 (and n1 = r and n2 = s). Examples R(3, 3) = 6 Suppose the edges of a complete graph on 6 vertices are coloured red and blue. Pick a vertex, v. There are 5 edges incident to v and so (by the pigeonhole principle) at least 3 of them must be the same colour. Without loss of generality we can assume at least 3 of these edges, connecting the vertex, v, to vertices, r, s and t, are blue. (If not, exchange red and blue in what follows.) If any of the edges, (rs), (rt), (st), are also blue then we have an entirely blue triangle. If not, then those three edges are all red and we have an entirely red triangle. Since this argument works for any colouring, any K6 contains a monochromatic K3, and therefore R(3, 3) ≤ 6. The popular version of this is called the theorem on friends and strangers. An alternative proof works by double counting. It goes as follows: Count the number of ordered triples of vertices, x, y, z, such that the edge, (xy), is red and the edge, (yz), is blue. Firstly, any given vertex will be the middle of either 0 × 5 = 0 (all edges from the vertex are the same colour), 1 × 4 = 4 (four are the same colour, one is the other colour), or 2 × 3 = 6 (three are the same colour, two are the other colour) such triples. Therefore, there are at most 6 × 6 = 36 such triples. Secondly, for any non-monochromatic triangle (xyz), there exist precisely two such triples. Therefore, there are at most 18 non-monochromatic triangles. Therefore, at least 2 of the 20 triangles in the K6 are monochromatic. Conversely, it is possible to 2-colour a K5 without creating any monochromatic K3, showing that R(3, 3) > 5. The unique colouring is shown to the right. Thus R(3, 3) = 6. The task of proving that R(3, 3) ≤ 6 was one of the problems of William Lowell Putnam Mathematical Competition in 1953, as well as in the Hungarian Math Olympiad in 1947. A multicolour example: R(3, 3, 3) = 17 A multicolour Ramsey number is a Ramsey number using 3 or more colours. There are (up to symmetries) only two non-trivial multicolour Ramsey numbers for which the exact value is known, namely R(3, 3, 3) = 17 and R(3, 3, 4) = 30. Suppose that we have an edge colouring of a complete graph using 3 colours, red, green and blue. Suppose further that the edge colouring has no monochromatic triangles. Select a vertex v. Consider the set of vertices that have a red edge to the vertex v. This is called the red neighbourhood of v. The red neighbourhood of v cannot contain any red edges, since otherwise there would be a red triangle consisting of the two endpoints of that red edge and the vertex v. Thus, the induced edge colouring on the red neighbourhood of v has edges coloured with only two colours, namely green and blue. Since R(3, 3) = 6, the red neighbourhood of v can contain at most 5 vertices. Similarly, the green and blue neighbourhoods of v can contain at most 5 vertices each. Since every vertex, except for v itself, is in one of the red, green or blue neighbourhoods of v, the entire complete graph can have at most 1 + 5 + 5 + 5 = 16 vertices. Thus, we have R(3, 3, 3) ≥ 17. To see that R(3, 3, 3) = 17, it suffices to draw an edge colouring on the complete graph on 16 vertices with 3 colours that avoids monochromatic triangles. It turns out that there are exactly two such colourings on K16, the so-called untwisted and twisted colourings. Both colourings are shown in the figures to the right, with the untwisted colouring on the left, and the twisted colouring on the right. If we select any colour of either the untwisted or twisted colouring on K16, and consider the graph whose edges are precisely those edges that have the specified colour, we will get the Clebsch graph. It is known that there are exactly two edge colourings with 3 colours on K15 that avoid monochromatic triangles, which can be constructed by deleting any vertex from the untwisted and twisted colourings on K16, respectively. It is also known that there are exactly 115 edge colourings with 3 colours on K14 that avoid monochromatic triangles, provided that we consider edge colourings that differ by a permutation of the colours as being the same. Proof 2-colour case The theorem for the 2-colour case can be proved by induction on r + s. It is clear from the definition that for all n, R(n, 2) = R(2, n) = n. This starts the induction. We prove that R(r, s) exists by finding an explicit bound for it. By the inductive hypothesis R(r − 1, s) and R(r, s − 1) exist. Lemma 1. R ( r , s ) ≤ R ( r − 1 , s ) + R ( r , s − 1 ) . {\displaystyle R(r,s)\leq R(r-1,s)+R(r,s-1).} Proof. Consider a complete graph on R(r − 1, s) + R(r, s − 1) vertices whose edges are coloured with two colours. Pick a vertex v from the graph, and partition the remaining vertices into two sets M and N, such that for every vertex w, w is in M if edge (vw) is blue, and w is in N if (vw) is red. Because the graph has R ( r − 1 , s ) + R ( r , s − 1 ) = | M | + | N | .... Discover the K J Ramsey popular books. Find the top 100 most popular K J Ramsey books.