V = {A, B, C, D, E, F, G, H, I} E = {e1 = (A, B), e2 = (A, C), e3 = (A, D), e4 = (B, C), e5 = (B, F), e6 = (B, H), e7 = (C, D), e8 = (C, E), e9 = (C, F), e10 = (D, E), e11 = (D, I), e12 = (E, F), e13 = (E, G), e14 = (F, G), e15 = (F, H), e16 = (G, H), e17 = (G, I), e18 = (H, I)} W = {22, 9, 12, 35, 36, 34, 4, 33, 30, 18, 23, 24, 39, 25, 21, 19} ....................... is a Hamiltonian path in the above graph.
e1, e6, e15, e9, e7, e10, e13, e17
A Hamiltonian path visits every vertex of the graph exactly once, while an Eulerian path must use every edge exactly once. Here the graph has nine vertices and eighteen edges, so a Hamiltonian path has exactly eight edges and passes through all of A, B, C, D, E, F, G, H and I without repeating any vertex.
Step 1 — Trace e1, e6, e15, e9, e7, e10, e13, e17 using the edge list: e1 = (A, B), e6 = (B, H), e15 = (H, F), e9 = (F, C), e7 = (C, D), e10 = (D, E), e13 = (E, G), e17 = (G, I). The walk is A → B → H → F → C → D → E → G → I, which touches all nine vertices once each.
Step 2 — The sequence starting with e2 goes A → C → D → I → G → E → C, revisiting C. The sequence starting with e3 goes A → D → I → G → E → C → F → G, revisiting G. The sequence starting with e10 goes D → E → F → G → H → I → D, revisiting D. None of these is a Hamiltonian path.
Step 3 — No Eulerian path exists in this graph at all: A, C, F and I have odd degrees (3, 5, 5, 3), and an Euler path allows at most two odd-degree vertices. The question therefore tests the Hamiltonian property, which only the first sequence satisfies.
e1, e6, e15, e9, e7, e10, e13, e17 option is correct.
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