 Hello everyone, this is Alice Gao. In this video, I'm going to trace lowest cost first search on a search graph. Let's recall the tie-breaking rule. If multiple paths have the same lowest cost, I will order the path by their best, their last nodes and choose the path that comes first in alphabetical order. During this process, I will keep track of a few things. The frontier, the cost of each path in the frontier and the search tree. I will also label the nodes in the order of expansion. Let's add the initial state S to the frontier and to the search tree. The path with the least cost is S. Let's remove S from the frontier. S is not a goal, let's expand it. S has two successors, B and C. Let's add SB with the cost of 1 and SC with the cost of 1 to the frontier. In the next few steps, we will explore all the paths with the cost of 1. SB and SC have the lowest cost of 1. Based on our tie-breaking rule, we'll remove SB from the frontier. B is not a goal, let's expand it. B has three successors, C, D and E. Let's add SBC, SBD and SBE to the frontier and keep track of their costs, 2, 10 and 2. Next, SC has the lowest cost of 1. Let's remove SC from the frontier. C is not a goal, let's expand it. C has one successor, H. Let's add SCH to the frontier with the cost of 2. In the next few steps, we will explore all the paths with the cost of 2. SBC, SBE and SCH have the lowest cost of 2. Based on our tie-breaking rule, let's remove SBC from the frontier. C is not a goal, let's expand it. C has one successor, H. Let's add SBCH to the frontier with the cost of 3. Next, SBE and SCH both have the lowest cost of 2. Let's remove SBE from the frontier. E is not a goal, let's expand it. E has one successor, F. Let's add SBEF to the frontier with the cost of 3. Next, SCH has the lowest cost of 2. Let's remove SCH from the frontier. H is not a goal, let's expand it. H has no successor, let's keep going. Next, we'll explore all the paths with the cost of 3. SBEF and SBCH both have the lowest cost of 3. Based on our tie-breaking rule, let's remove SBEF from the frontier. F is not a goal, let's expand it. F has no successor, let's keep going. Next, SBCH has the lowest cost of 3. Let's remove SBCH from the frontier. H is not a goal, let's expand it. H has no successor either, let's keep going. Next, we'll explore all the paths with the cost of 10. SBD has the lowest cost of 10. Let's remove SBD from the frontier. D is not a goal, let's expand it. D has 2 successors, F and G. Let's add SBDF and SBDG to the frontier with costs of 11 for both. Next, we'll explore all the paths with the cost of 11. SBDF and SBDG have the lowest cost of 11. Let's remove SBDF from the frontier. F is not a goal, let's expand it. F has no successor, let's keep going. SBDG has the lowest cost of 11. Let's remove SBDG from the frontier. G is a goal, let's return the solution SBDG with the total cost of 11. This completes the tracing process. As you can see, lowest cost first search explores the path in order of increasing total costs. That's everything for this example on tracing lowest cost first search. Thank you very much for watching. I will see you in the next video. Bye for now.