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} The weight of the shortest path from node A to I in the undirected graph is........... .
42
The figure shows an undirected weighted graph on nodes A to I with these edge weights: A–B 22, A–C 9, A–D 12, B–C 35, B–F 36, B–H 34, C–D 4, C–E 65, C–F 42, D–E 33, D–I 30, E–F 18, E–G 23, F–G 39, F–H 24, G–H 25, G–I 21 and H–I 19.
The weight of a path is the sum of its edge weights, so the shortest path is found with Dijkstra's algorithm, not by counting edges (the values 2 and 3 are only hop counts, not weights).
Dijkstra from A starts with distance 0 at A, then C = 9, D = min(12, 9 + 4) = 12, B = 22, E = min(9 + 65, 12 + 33) = 45, I = 12 + 30 = 42, F = min(22 + 36, 9 + 42, 45 + 18) = 51, H = 22 + 34 = 56, G = 45 + 23 = 68.
Node I is settled with distance 42 through the path A → D → I, since every other route to I (via H at 56 + 19 = 75 or via G at 68 + 21 = 89) is longer.
The route A → C → D → I costs 9 + 4 + 30 = 43, which is 1 more than the direct route through D, so 43 is not the minimum.
Check: the two edges on the chosen path are A–D = 12 and D–I = 30, and 12 + 30 = 42; no edge into I is lighter than 19, and reaching H already costs 56, so nothing beats 42.
Hence, 42.
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