# How do I show that uniform-cost search is a special case of A*?

How do I show that uniform-cost search is a special case of A*? How do I prove this?

UCS uses the evaluation function $$f(n) = g(n)$$, where $$g(n)$$ is the length of the path from the starting node to $$n$$, whereas A* uses the evaluation function $$f(n) = g(n) + h(n)$$, where $$g(n)$$ means the same thing as in UCS and $$h(n)$$, called the "heuristic" function, is an estimate of the distance from $$n$$ to the goal node. In the A* algorithm, $$h(n)$$ must be admissible.
UCS is a special case of A* which corresponds to having $$h(n) = 0, \forall n$$. A heuristic function $$h$$ which has $$h(n) = 0$$, $$\forall n$$, is clearly admissible, because it always "underestimates" the distance to the goal, which cannot be smaller than $$0$$, unless you have negative edges (but I assume that all edges are non-negative). So, indeed, UCS is a special case of A*, and its heuristic function is even admissible!
To see this with an example, just draw a simple graph, and apply the A* algorithm using $$h(n) = 0$$, for all $$n$$, and then apply UCS to the same graph. You will obtain the same results.