Wouldn't the answer be Greedy Best First Search since the path cost $= 0$?
Which search strategy is simulated by the $A*$ search algorithm if we take cost function$ g(n) = 0$ if one assumes no further restrictions on the heuristic function $h$?
- Depth First Search
- Greedy Best First Search
- Uniform Cost Search
- None of the above.
Edit: I believe it is Greedy Best First Search because the GBFS always looks for the lowest next heuristic, while the A* search looks for the lowest sum of the pathcost + heuristic. If pathcost is 0, then there would be no difference between the two?
Greedy Best First Search (GBFS) is the correct answer. To demonstrate this, we need to address the two properties that differ between A* and GBFS:
Let $h(n)$ be the heuristic value of the search node $n$, and let $g(n)$ be the distance from the initial search node to $n$. Assuming similar tie-breaking for both algorithms, GBFS selects nodes from the OPEN LIST based on their heuristic value $h(n)$. A* selects nodes based on $f(n) = g(n) + h(n)$. Given the assumption that $g(n) = 0$ for every $n$, we have that $f(n) = h(n)$. Thus, the order of selecting nodes from the OPEN LIST is the same.
To demonstrate that the OPEN LIST and CLOSED LIST are similar for both GBFS and A* after each step of the algorithms, we need to show that the RE-OPEN subroutine is never activated for problems with a cost function of $g(n) = 0$.
The RE-OPEN subroutine is activated if the following conditions hold for the currently inspected node $n$: the search has already encountered and expanded the node $n$ ($n$ is in the CLOSED LIST), and the current path to $n$ is shorter than the path we previously expanded $n$ with. In this case, A* updates the path to $n$ and puts it back in the OPEN LIST. However, this is impossible in the case where $g(n) = 0$ for every $n$, as all paths to all nodes cost zero. Thus, the RE-OPEN subroutine is never activated.
Hence, A* behaves like GBFS for a zero-cost problem when using the same heuristic function and tie-breaking strategy.