Question

My book here (Artificial intelligence A modern approach) says that the worst-case time and space complexity of a uniform-cost search algorithm would be O(b[C*/e]) , where b is the branching factor, C* is the cost of the optimal solution, and every action costs atleast e. But why is this so?

Answer 1

First, the complexity is O(B^(C/e)) [exponential in C/e].

To understand it, think of a simple example case first:

Let G=(V,E) be a graph, with branch factor B. The graph is unweighted (w(e) = 1 for each e).

Consider finding the shortest path from S to T.

In this case, the algorithm is actually a BFS, and it will discover all nodes in the path up to length SOL, where SOL is the length of the shortest path, which is O(B^|SOL|)

For the general case – the same idea holds, you need to discover all nodes up to cost C. So you discover nodes up to depth C/e, giving you O(B^(C/e)) total nodes needed to be explored.

The exponential factor is because: First level (root) has B^0=1 nodes, second level has B nodes. from each of these you discover B nodes, giving you B^2, ….

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EDIT:

Missed it so far, but the title asks for space complexity and not time complexity. However, the answer remains the same, since a uniform cost search holds a visited set, for already visited nodes. Since each node you discover is also added to it – the answer remains O(B^(C/e))