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Let Pn be a path with n vertices and let Wm be a wheel of m + 1 vertices; namely, a graph consists of a cycle Cm with one additional vertex being adjacent to all vertices of Cm

# The Ramsey number of paths with respect to wheels

Discrete Mathematics, no. 3 (2005): 275-277

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Abstract

For graphs G and H, the Ramsey numberR(G,H) is the smallest positive integer n such that every graph F of order n contains G or the complement of F contains H. For the path P\"n and the wheel W\"m, it is proved that R(P\"n,W\"m)=2n-1 if m is even, m=4, and n=(m/2)(m-2), and R(P\"n,W\"m)=3n-2 if m is odd, m=5, and n=(m-1/2)(m-3).

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Introduction
• Let Pn be a path with n vertices and let Wm be a wheel of m + 1 vertices; namely, a graph consists of a cycle Cm with one additional vertex being adjacent to all vertices of Cm. The authors determine the following values for R(Pn, Wm).
• E.T. Baskoro, Surahmat / Discrete Mathematics 294 (2005) 275 – 277
• To prove R(Pn, Wm) 2n − 1 or 3n − 2 for m even or odd, respectively, the authors observe that no Pn is in the graph 2Kn−1 or 3Kn−1 and no Wm is in their complements, respectively.
Highlights
• For graphs G and H the Ramsey number R(G, H ) is defined as the smallest n such that every graph F of order |F | = n contains G or its complement, F, contains H .

Let Pn be a path with n vertices and let Wm be a wheel of m + 1 vertices; namely, a graph consists of a cycle Cm with one additional vertex being adjacent to all vertices of Cm
• R(Pn, Wm) 2n−1, if m is even: Let F be a graph of 2n−1 vertices and F contains no path
• For i = 1, 2, 3, . . . , (m − 4)/2 define couples Ai in path L1 are as follows: Ai =
• If L1 is the longest path in F and its endpoints are l11 and l1k, zl11, zl1k ∈/ E(F ) for any z ∈ V1 where V1 = V (F )\V (L1)
• F contains a wheel Wm formed by l11 and Wm−1 since l11 is not adjacent to any vertex in V1
Results
• R(Pn, Wm) 2n−1, if m is even: Let F be a graph of 2n−1 vertices and F contains no path
• L1k−1, l1k) be the longest path in F.
• L2t−1, l2t ) be the longest path in F [V1].
• Since |V (F )|=2n−1, there exists at least one vertex w ∈ V2 which is not adjacent to all endpoints l11, l1k, l21, l2t .
• M/2 let Li be the longest path in F [Vi−1], where Vi−1 =V (F )\
• Vertex w and all these endpoints form a wheel Wm in F .
• (m − 4)/2 define couples Ai in path L1 are as follows: Ai =
• Define couples Bi in path L2 are as follows: Bi =
• Since L1 is the longest path in F, there will exist one vertex of Ai for each i, say ai, which is not adjacent to w.
• Since L2 is the longest path in F \V (L1) there must be one vertex, say bi, in couple Bi which is not adjacent to w for each i.
• Since L1 is the longest path in F.
Conclusion
• Let F be a graph of 3n − 2 vertices.
• Assume F contains no path Pn. If L1 is the longest path in F and its endpoints are l11 and l1k, zl11, zl1k ∈/ E(F ) for any z ∈ V1 where V1 = V (F )\V (L1).
• Since |V1| 2n − 1, n (m − 1/2)((m − 1) − 2) and by the result for the case of m even, the complement of the subgraph F [V1] must contain a wheel Wm−1.
• F contains a wheel Wm formed by l11 and Wm−1 since l11 is not adjacent to any vertex in V1.
Study subjects and analysis
cases: 3
Since |V (F )|=2n−1, there exists at least one vertex w ∈ V2 which is not adjacent to all endpoints l11, l1k, l21, l2t. We distinguish three cases. Case 1: k < m − 2

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