graph - Why is time complexity for BFS/DFS not simply O(E) instead of O(E+V)? -



graph - Why is time complexity for BFS/DFS not simply O(E) instead of O(E+V)? -

i know there's similar question in stack overflow, 1 person has asked, why time complexity of bfs/dfs not o(v).

the appropriate reply given e can big v^2 in case of finish graph, , hence valid include e in time complexity.

but, if v cannot greater e+1. so, in case not having v in time complexity, should work?

if given e = kv + c, real constants k , c then, o(e + v) = o(kv + c + v) = o(v) = o(e) , argument correct.

an illustration of trees.

in general, however, e = o(v^2), , cannot improve o(v^2).

why not write o(e) always? note there might cases when there no edges in graph @ (i.e. o(e) ~ o(1)). such cases, we'll have go each of vertex (o(v)), cannot finish in o(1) time.

thus, writing o(e) won't in general.

graph time-complexity depth-first-search breadth-first-search

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