WIKIPEDIA Ramsey's theorem In combinatorial mathematics,Ramsey's theorem,in one of its graph-theoretic forms,states that one will find demonstrat edge nd sufficiently large complete two colours(say,blue red), let and s integers.1 louring of the complete grahon Rr)vertices contains a blue clique onverticesr an nsey's rem sta s that there exist least positive n every bl red edge vertices.(Here R(r,s)signifies an integer that depends on both r and s.) Ramsey's theorem isa foundational result in combinatorics.The first version of this resut was proved by F.P. I theory now called R msey theory,tha seeks regular mid diso ondition for the existence with regular properties.In this application it is a question An extension of this theorem applies to any finite number of colours,rather than just two.More precisely.the tha r any given numb urs,c,and any given integers n,.nthere R(n,...,n),such that if the edges of a complete graph of order R(n,..n)are coloured with c different colours,then for some ibetween 1 and c,it must contain a complete subgraph of order n;whose edges are all colour i.The special case above has c=2(and n=r and n2=s). Contents Example:R(3,3)=6 Proof of the theorem 2-colour case Case of more colours Ramsey numbers Asymptotics A multicolour example:R(3,3,3)=17 Extensions of the theorem Infinite Ramsey theorem Infinite version implies the finite Directed graph Ramsey numbers Ramsey computation and quantum computers See also Notes References External links Example:R(3,3)=6
Ramsey's theorem In combinatorial mathematics, 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. [1] 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 F. P. 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). Example: R(3, 3) = 6 Proof of the theorem 2-colour case Case of more colours Ramsey numbers Asymptotics A multicolour example: R(3, 3, 3) = 17 Extensions of the theorem Infinite Ramsey theorem Infinite version implies the finite Directed graph Ramsey numbers Ramsey computation and quantum computers See also Notes References External links Contents Example: 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,(r,s),(r,() (s,()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 Ke contains a monochromatic K3,and therefore R(3,3)s 6.The popular version of this is called the theorem on friends and strangers. A2-6 An altemative 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 0x 5=0(all edges from the vertex are the same colour),1 x 4=4(four are the same colour,one is the other colour),or 2 x 3=6(three are the same colour,two are the other colour)such triples.Therefore,there are at most 6x 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 Ks without creating any monochromatic K3.showing that R(3,3)>5. The uniquel21 colouring is shown to the right.Thus R(3,3)=6. The task of proving that R(3,3)s6 was one of the problems of William Lowell Putnam Mathematical Competition in 1953,as well as in the Hungarian Math Olympiad in 1947. Proof of the theorem 2-colour case The theorem for the 2-colour case,can be proved by induction on r+s.3]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. Lemma1.Rr,s）≤R(r-1,s+R(r,s-1) of(1+R(-1)vertice whose edges ed with N.suc w lours.Pick a ver R(revery ve .wi s in Mif ap"w is blu W) th )+R(r, and w rIM≥ R(r e grapl M≥R(, -1).In the former case,if M a red Ks then s does the original graph nd we are finished Otherwise M has a blue Kr-1 and so M U {v)has a blue Kr by the definition of M.The latter case is analogous.Thus the claim is true and we have completed the proof for 2 colours. In this 2-colocase if R(r,s)and R(r,s-1)are both even,the induction inequality can be strengthened to: R(r,s)s R(r-1,s)+R(r,s-1)-1
A 2-edge-labeling of K5 with no monochromatic K3 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, (r, s), (r, t), (s, t), 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[2] 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. The theorem for the 2-colour case, can be proved by induction on r + s. [3] 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): 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 (v, w) is blue, and w is in N if (v, w) is red. Because the graph has R(r − 1, s) + R(r, s − 1) = |M| + |N| + 1 vertices, it follows that either |M| ≥ R(r − 1, s) or |N| ≥ R(r, s − 1). In the former case, if M has a red Ks then so does the original graph and we are finished. Otherwise M has a blue Kr−1 and so M ∪ {v} has a blue Kr by the definition of M. The latter case is analogous. Thus the claim is true and we have completed the proof for 2 colours. In this 2-colour case, if R(r − 1, s) and R(r, s − 1) are both even, the induction inequality can be strengthened to:[4] R(r, s) ≤ R(r − 1, s) + R(r, s − 1) − 1. Proof of the theorem 2-colour case
Proof.Suppose p R(r-1,s)and g=R(r,s-1)are both even.Let t=p+q-1 and consider a two-coloured graph of t vertices.If di is degree of i-th vertex in the graph,then,according to the Handshaking lemma,is even.Given that tis odd,there must be an evenAssume dis even,M and Nare the vertices incident to vertex 1 in the blue and red subgraphs,respectively.Then both M=d and N=t-1-d are even.According to the Pigeonhole principle,either M>p-1,or N >g.Since M]is even,while p-1 is odd,the first inequality can be strengthened,so either M2p or N2g =R(r-1,).Then either the M subgraph has a red K and the p lete,or has a blue K-1 which along with vertex 1 makes a blue Kr.The case=R(r,s-1)is treated similarly. Case of more colours Lemma 2.If c>2,then R(m...n)s R(n..n.R(nc-1.nc)). Proof.Consider a complete graph of R(n n-2,R(n-,n)vertices and colour its edges with c colours and are th olou Thus the 1) chromatically coloured with colour ifor some 1sisc-2or aKg cooured in the blurred colourIn the former cas ecover our sight again and seefrom the definitionof either a(c 1)-mon chrome either case the proo is complete. Lemma 1 impli that any R(rs)is finite.The right hand side of the inequality in Lemma 2 expresses a Ramsey number for c colours in terms of Ramsey numbers for fewer colours.Therefore any R(n,...,n)is finite for any number of colours.This proves the theorem. Ramsey numbers The numbers R(r,s)in Ramsey's theorem (and their extensions to more than two colours)are known as Ramsey numbers.The ramsey number.R(m.n).gives the solution to the party problem.which asks the minimum number of guests.R(m,n).that must be invited so that at least m will know each other or at least n will not know each other.In the language of graph theory,the Ramsey number is the minimum number of vertices,v=R(m,n),such that all undirected simple graphs of order v,contain a clique of order m,or an independent set of order n.Ramsey's theorem states that such a number exists for all m and n. By symmetry,it is true that R(m,n)=R(n,m).An upper bound for R(r,s)can be extracted from the proof of the theorem,and other arguments give lower bounds.(The first exponential lower bound was obtained by Paul Erdos using the probabilistic method.)However,there is a vast gap between the tightest lower bounds and the tightest upper bounds.There are also very few numbers r and s for which we know the exact value of R(r.s). Computing a lower bound L forR(s) usually requires exhibiting a blue/red colouring of the graph KL-1 with no blue Kr subgraph and no red Ks subgraph.Such a counterexample is called a Ramsey graph. Brendan McKay maintains a list of known Ramsey graphs.[5]Upp oer bounds are often considerably more difficult to establish one gither has to check all nossible co olourings to confirm the absence of a counterexample,or to present a mathematical argument for its absence.A sophisticated c ter program does not need to look at all colourings individually in order to eliminate all of them:nevertheless it is a verv difficult computational task that existing software can only manage on small sizes.Each complete graph K
Proof. Suppose p = R(r − 1, s) and q = R(r, s − 1) are both even. Let t = p + q − 1 and consider a two-coloured graph of t vertices. If is degree of -th vertex in the graph, then, according to the Handshaking lemma, is even. Given that t is odd, there must be an even . Assume is even, M and N are the vertices incident to vertex 1 in the blue and red subgraphs, respectively. Then both and are even. According to the Pigeonhole principle, either , or . Since is even, while is odd, the first inequality can be strengthened, so either or . Suppose . Then either the M subgraph has a red and the proof is complete, or it has a blue which along with vertex 1 makes a blue . The case is treated similarly. Lemma 2. If c>2, then R(n1 , …, nc ) ≤ R(n1 , …, nc−2 , R(nc−1 , nc )). Proof. Consider a complete graph of R(n1 , …, nc−2 , R(nc−1 , nc )) vertices and colour its edges with c colours. Now 'go colour-blind' and pretend that c − 1 and c are the same colour. Thus the graph is now (c − 1)- coloured. Due to the definition of R(n1 , …, nc−2 , R(nc−1 , nc )), such a graph contains either a Kni monochromatically coloured with colour i for some 1 ≤ i ≤ c − 2 or a KR(nc − 1 , nc ) -coloured in the 'blurred colour'. In the former case we are finished. In the latter case, we recover our sight again and see from the definition of R(nc−1 , nc ) we must have either a (c − 1)-monochrome Knc−1 or a c-monochrome Knc . In either case the proof is complete. Lemma 1 implies that any R(r,s) is finite. The right hand side of the inequality in Lemma 2 expresses a Ramsey number for c colours in terms of Ramsey numbers for fewer colours. Therefore any R(n1 , …, nc ) is finite for any number of colours. This proves the theorem. The numbers R(r, s) in Ramsey's theorem (and their extensions to more than two colours) are known as Ramsey numbers. The Ramsey number, R(m, n), gives the solution to the party problem, which asks the minimum number of guests, R(m, n), that must be invited so that at least m will know each other or at least n will not know each other. In the language of graph theory, the Ramsey number is the minimum number of vertices, v = R(m, n), such that all undirected simple graphs of order v, contain a clique of order m, or an independent set of order n. Ramsey's theorem states that such a number exists for all m and n. By symmetry, it is true that R(m, n) = R(n, m). An upper bound for R(r, s) can be extracted from the proof of the theorem, and other arguments give lower bounds. (The first exponential lower bound was obtained by Paul Erdős using the probabilistic method.) However, there is a vast gap between the tightest lower bounds and the tightest upper bounds. There are also very few numbers r and s for which we know the exact value of R(r, s). Computing a lower bound L for R(r, s) usually requires exhibiting a blue/red colouring of the graph KL−1 with no blue Kr subgraph and no red Ks subgraph. Such a counterexample is called a Ramsey graph. Brendan McKay maintains a list of known Ramsey graphs. [5] Upper bounds are often considerably more difficult to establish: one either has to check all possible colourings to confirm the absence of a counterexample, or to present a mathematical argument for its absence. A sophisticated computer program does not need to look at all colourings individually in order to eliminate all of them; nevertheless it is a very difficult computational task that existing software can only manage on small sizes. Each complete graph Kn Case of more colours Ramsey numbers
has1)edges,so there would bea total ofgrapststhrough (for ccoou)f brte As described above,R(3,3)=6.It is easy to prove that R(4,2)=4,and,more generally,that R(s,2)=s for all s:a graph on s-1 nodes with all edges coloured red serves as a counterexample and proves that R(s,2)2S;among colourings of a graph on s nodes,the colouring with all edges coloured red contains a s node red subgraph,and all other colourings contain a 2-node blue subgraph(that is,a pair of nodes connected with a blue edge.) Using induction inequalities,it can be concluded that R(4,3)R(4,2)+R(3,3)-1=9,and therefore R(4,4)R(4,3)+R(3,4)<18.There are only two (4,4,16)graphs (that is,2-colourings of a complete graph on 16 nodes without 4-node red or blue complete subgraphs)among 6.4x 1022 different ouring of 16-node graphs,and only one (,4,17)graph (the Paley graph of order 17)am 2.46×1026 R(4,4)= (This was proven by Evans,Pulham and Sheehan in 1979.)It follows that 18 The fact that R(4,5)=25 was first established by Brendan McKay and Stanislaw The exact value of R(5.5)is unkno n,although it is known to lie between 43(Geoffrey Exoo (1989)8)) and 48(Angeltveit and McKay (2017)(inclusive). Erdos asks us to imagine ce,vastly more poweranding on Earth e ale. -Joel Spencer(10] In 1997,McKa m警场 Radzisski and Exoo employed computer-as conjecture that R(5,5)= R(r,s)with r,ss 10 are shown in the table below.Where the exact value is unknown,the table lists the best known bounds.R(r,s)with r,s<3 are given by R(1,s)=1 and R(2,s)=s for all values of s.The standard survey on the dev elopment of Ramsey number research is the Dynamic Survey 1 of the Electronic Journal of Combinatorics.121 It was written and is updated by Stanislaw Radziszowski.Its latest update was in March 2017.In general,there are a few y ears be een the updates.Wher not cited otherwise,entries in the table below are taken from this dynamic suvey.Note diagonal sice R(r,s)=R(s,r)
has 1 2 n(n − 1) edges, so there would be a total of c n(n − 1) 2 graphs to search through (for c colours) if brute force is used.[6] Therefore, the complexity for searching all possible graphs (via brute force) is O(c n 2 ) for c colourings and an upper bound of n nodes. As described above, R(3, 3) = 6. It is easy to prove that R(4, 2) = 4, and, more generally, that R(s, 2) = s for all s: a graph on s − 1 nodes with all edges coloured red serves as a counterexample and proves that R(s, 2) ≥ s; among colourings of a graph on s nodes, the colouring with all edges coloured red contains a snode red subgraph, and all other colourings contain a 2-node blue subgraph (that is, a pair of nodes connected with a blue edge.) Using induction inequalities, it can be concluded that R(4, 3) ≤ R(4, 2) + R(3, 3) − 1 = 9, and therefore R(4, 4) ≤ R(4, 3) + R(3, 4) ≤ 18. There are only two (4, 4, 16) graphs (that is, 2-colourings of a complete graph on 16 nodes without 4-node red or blue complete subgraphs) among 6.4 × 10 22 different 2-colourings of 16-node graphs, and only one (4, 4, 17) graph (the Paley graph of order 17) among 2.46 × 10 26 colourings. [5] (This was proven by Evans, Pulham and Sheehan in 1979.) It follows that R(4, 4) = 18. The fact that R(4, 5) = 25 was first established by Brendan McKay and Stanisław Radziszowski in 1995.[7] The exact value of R(5, 5) is unknown, although it is known to lie between 43 (Geoffrey Exoo (1989)[8] ) and 48 (Angeltveit and McKay (2017)[9] ) (inclusive). Erdős asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of R(5, 5) or they will destroy our planet. In that case, he claims, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they ask for R(6, 6). In that case, he believes, we should attempt to destroy the aliens. — Joel Spencer [10] In 1997, McKay, Radziszowski and Exoo employed computer-assisted graph generation methods to conjecture that R(5, 5) = 43. They were able to construct exactly 656 (5, 5, 42) graphs, arriving at the same set of graphs through different routes. None of the 656 graphs can be extended to a (5, 5, 43) graph.[11] For R(r, s) with r, s > 5, only weak bounds are available. Lower bounds for R(6, 6) and R(8, 8) have not been improved since 1965 and 1972, respectively. [12] R(r, s) with r, s ≤ 10 are shown in the table below. Where the exact value is unknown, the table lists the best known bounds. R(r, s) with r, s < 3 are given by R(1, s) = 1 and R(2, s) = s for all values of s. The standard survey on the development of Ramsey number research is the Dynamic Survey 1 of the Electronic Journal of Combinatorics. [12] It was written and is updated by Stanisław Radziszowski. Its latest update was in March 2017. In general, there are a few years between the updates. Where not cited otherwise, entries in the table below are taken from this dynamic survey. Note there is a trivial symmetry across the diagonal since R(r, s) = R(s, r)
Values/known bounding ranges for Ramsey numbers R(r,s)(sequence A212954 in the OEIS) s 1 4 5 7 8 9 10 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 8 9 10 3 6 9 14 18 23 28 36 40-42 4 18 250 36-41 49-61 5913194 73-115 92-149 43-48 5887 80-143 101216 133-316 14g[13 442 6 102-165 183-780 204-1171 205-540 217-1031 252 7 292-2826 1713 282-1870 3583 3436090 10 Asymptotics The inequality R(r,s)sR(r-1,s)+R(r,s-1)may be applied inductively to prove that s(中2） In particular,this result,due to Erdos and Szekeres,implies that whenr=s, B)s卫+oa √⑧ An exponential lower bound, R(8,8)≥[1+o(1)月 method.There is given by rds in1947 d wasnsnimen热of he pob如 01Rd0,16248620 thes ample, everth ial either bound improved to e and still There is y icit constc weo的aa通bos的R网s 1+oYy22i≤R6,≤geg/ser, e
Values / known bounding ranges for Ramsey numbers R(r, s) (sequence A212954 in the OEIS) s r 1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1 1 2 2 3 4 5 6 7 8 9 10 3 6 9 14 18 23 28 36 40–42 4 18 25 [7] 36–41 49–61 59 [13]–84 73–115 92–149 5 43–48 58–87 80–143 101–216 133–316 149 [13]– 442 6 102–165 115 [13]– 298 134 [13]– 495 183–780 204–1171 7 205–540 217–1031 252– 1713 292–2826 8 282–1870 329– 3583 343–6090 9 565– 6588 581– 12677 10 798– 23556 The inequality R(r, s) ≤ R(r − 1, s) + R(r, s − 1) may be applied inductively to prove that In particular, this result, due to Erdős and Szekeres, implies that when r = s, An exponential lower bound, was given by Erdős in 1947 and was instrumental in his introduction of the probabilistic method. There is obviously a huge gap between these two bounds: for example, for s = 10, this gives 101 ≤ R(10, 10) ≤ 48620. Nevertheless, exponential growth factors of either bound have not been improved to date and still stand at 4 and √2 respectively. There is no known explicit construction producing an exponential lower bound. The best known lower and upper bounds for diagonal Ramsey numbers currently stand at Asymptotics