Question
(a) The complement of the Petersen graph is 2-connected. Prove or disprove
(b) Consider a graph G. Let x, y ∈ V (G) be such that x ↔ y. Show that for all z ∈ V (G), |d(x, z) − d(y, z)| ≤ 1
(c) Check whether the following graphs G and H are isomorphic or not
Answer :
Word Count : 1137
Alright, let's go through this step-by-step. --- ## (a) The complement of the Petersen graph is 2-connected. Prove or disprove Petersen graph \(P\): \(|V|=10\), each vertex degree \(3\), known to be 3-regular, diameter 2, girth 5, highly symmetric. Complement \(\overline{P}\): Each vertex in \(P\) has degree \(3\), so in \(\overline{P}\) each vertex has degree \(10-1-3 = 6\). So \(\overline{P}\) is \(6\)-regular on \(10\) vertices. 2-connected means no cut-vertex, i.e., removing any one vertex leaves a connected graph. --- Possible approach: Since Petersen graph \(P\) is strongly regular with parameters \((10,3,0,1)\), its complement is also strongly regular with parameters \((10,6,3,4)\). Strongly regular graphs with \(k=6\) here: \(v=10\), \(\lambda=3\), \(\mu=4\). We can check small vertex cuts: But one key property — strongly regular graphs with \(\mu>0\) are connected, but what about vertex connectivity? It's known that for a \(k\)-regular graph, vertex connectivity \(\kappa\) satisfies \(\kappa \ge k - \lambda\) sometimes? Not exactly; but here \(k=6\), \(\lambda=3\), \(\mu=4\). Actually known fact: For strongly regular graphs, vertex connectivity equals degree unless it's a conference graph or has special structure? Let's test: In \(\overline{P}\), removing one vertex \(v\) leaves 9 vertices, originally \(v\) was connected to 6 others (now gone), but those 6 are interconnected among themselves and to the remaining 3. Better: Draw structure mentally: Remove vertex \(v\): in \(\overline{P}\), \(v\)'s neighbors: all vertices except itself and its 3 non-neighbors in \(P\). Wait: In \(P\): vertex \(v\) has 3 neighbors, 6 non-neighbors (including itself? No, 10 vertices total: v, 3 neighbors, 6 others). So in \(\overline{P}\): \(v\) is connected to the 6 vertices that are not its neighbors in \(P\) (since in \(P\) they are non-adjacent, in complement they are adjacent). Also connected to the 3 neighbors in _____ ___ ____ ________ _________ __________ _______ _____.
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Alright, let's go through this step-by-step. --- ## (a) The complement of the Petersen graph is 2-connected. Prove or disprove Petersen graph \(P\): \(|V|=10\), each vertex degree \(3\), known to be 3-regular, diameter 2, girth 5, highly symmetric. Complement \(\overline{P}\): Each vertex in \(P\) has degree \(3\), so in \(\overline{P}\) each vertex has degree \(10-1-3 = 6\). So \(\overline{P}\) is \(6\)-regular on \(10\) vertices. 2-connected means no cut-vertex, i.e., removing any one vertex leaves a connected graph. --- Possible approach: Since Petersen graph \(P\) is strongly regular with parameters \((10,3,0,1)\), its complement is also strongly regular with parameters \((10,6,3,4)\). Strongly regular graphs with \(k=6\) here: \(v=10\), \(\lambda=3\), \(\mu=4\). We can check small vertex cuts: But one key property — strongly regular graphs with \(\mu>0\) are connected, but what about vertex connectivity? It's known that for a \(k\)-regular graph, vertex connectivity \(\kappa\) satisfies \(\kappa \ge k - \lambda\) sometimes? Not exactly; but here \(k=6\), \(\lambda=3\), \(\mu=4\). Actually known fact: For strongly regular graphs, vertex connectivity equals degree unless it's a conference graph or has special structure? Let's test: In \(\overline{P}\), removing one vertex \(v\) leaves 9 vertices, originally \(v\) was connected to 6 others (now gone), but those 6 are interconnected among themselves and to the remaining 3. Better: Draw structure mentally: Remove vertex \(v\): in \(\overline{P}\), \(v\)'s neighbors: all vertices except itself and its 3 non-neighbors in \(P\). Wait: In \(P\): vertex \(v\) has 3 neighbors, 6 non-neighbors (including itself? No, 10 vertices total: v, 3 neighbors, 6 others). So in \(\overline{P}\): \(v\) is connected to the 6 vertices that are not its neighbors in \(P\) (since in \(P\) they are non-adjacent, in complement they are adjacent). Also connected to the 3 neighbors in _____ ___ ____ ________ _________ __________ _______ _____.
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