 Hello everyone, this is Alice Gao. In this video, I'm going to solve the question regarding the optimality of A star search with multi-path pruning The question says, is there a situation in which A star search with multi-path pruning discards the optimal solution? The correct answer is yes. To prove this, we need to come up with an example to show that this happens Let's look at an example. A student came up with a similar example and I modified it for this lecture S is the start node and G is the goal node. There are two paths from S to G The top path is S, B, G with a total cost of 15 The bottom path is S, A, B, G with a total cost of 12 The bottom path is the optimal solution You can verify that the given heuristic is admissible Let's trace A star search on the search graph. You should expect that the algorithm finds and returns the top path Which is not optimal First, add the start state S to the frontier Next, remove S from the frontier and add S to the explore set S is not a goal node. Let's expand it S has two successors, A and B. Let's add S, A and S, B to the frontier S, A has the F value of 9 and S, B has the F value of 7 Next, S, B has the lower F value. Let's remove B from the frontier And add B to the explore set B is not a goal node. Let's expand it. B has one successor G Let's add S, B, G to the frontier with the F value of 15 Next, S, A has the smaller F value of 9. Let's remove A from the frontier and add A to the explore set A is not a goal. Let's expand it. A has one successor B Let's add S, A, B to the frontier with the F value of 4 Next, S, A, B has the smallest F value of 4. Let's remove B from the frontier B is already in the explore set. Multi-path pruning happens here And we will not add B's successors to the frontier Let's continue Next, S, B, G is the only path on the frontier Let's remove G from the frontier and add G to the explore set G is a goal. Let's stop and return the solution S, B, G with the cost of 15 This example shows you that a star search with multi-path pruning may return a suboptimal solution first Let's reflect on this example What went wrong? After expanding S, we chose to expand B before expanding A, since S, B has a smaller F value This causes us to find the top path to B first, but this is a longer path to B Later on, when we found a shorter path to B through A, we had to discard it due to multi-path pruning The real culprit is the heuristic value of A Suppose that we use the heuristic values to estimate the cost of the path from A to B This can be estimated as H of A minus H of B, which is equal to 8 minus 2, which is 6 Now, 6 is a gross overestimate of the actual cost, which is 1 Because of this overestimate, we assume that the bottom path is longer and decided to explore the top path first I invite you to take this example and think about which components of the example are necessary for this proof to work If you want to tweak this example, which component could you change and which component must stay the same That's everything on this question regarding A star search with multi-path pruning Thank you very much for watching. I will see you in the next video Bye for now