Considere un gráfico G no dirigido donde no se permiten los bucles automáticos. El conjunto de vértices de G es {(i, j): 1 <= i <= 12, 1 <= j <= 12}. Hay una arista entre (a, b) y (c, d) si |a − c| <= 1 y |b − d| <= 1.
El número de aristas en este gráfico es __________.
(A) 500
(B) 502
(C) 506
(D) 510
Respuesta: (C)
Explicación:
Given: The vertex set of G is {(i, j): 1 <= i <= 12, 1 <= j <= 12}. There is an edge between (a, b) and (c, d) if |a − c| <= 1 and |b − d| <= 1. There can be total 12*12 possible vertices. The vertices are (1, 1), (1, 2) ....(1, 12) (2, 1), (2, 2), .... The number of edges in this graph? Number of edges is equal to number of pairs of vertices that satisfy above conditions. For example, vertex pair {(1, 1), (1, 2)} satisfy above condition. For (1, 1), there can be an edge to (1, 2), (2, 1), (2, 2). Note that there can be self-loop as mentioned in the question. Same is count for (12, 12), (1, 12) and (12, 1) For (1, 2), there can be an edge to (1, 1), (2, 1), (2, 2), (2, 3), (1, 3) Same is count for (1, 3), (1, 4)....(1, 11), (12, 2), ....(12, 11) For (2, 2), there can be an edge to (1, 1), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2), (3, 3) Same is count for remaining vertices. For all pairs (i, j) there can total 8 vertices connected to them if i and j are not in {1, 12} There are total 100 vertices without a 1 or 12. So total 800 edges. For vertices with 1, total edges = (Edges where 1 is first part) + (Edges where 1 is second part and not first part) = (3 + 5*10 + 3) + (5*10) edges Same is count for vertices with 12 Total number of edges: = 800 + [(3 + 5*10 + 3) + 5*10] + [(3 + 5*10 + 3) + 5*10] = 800 + 106 + 106 = 1012 Since graph is undirected, two edges from v1 to v2 and v2 to v1 should be counted as one. So total number of undirected edges = 1012/2 = 506.
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA