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No, it will not necessary be consistent or admissible. Consider this example, where $s$ is the start, $g$ is the goal, and the distance between them is 1. s --1-- g Assume that $h_0$ and $h_1$ are perfect heuristics. Then $h_0(s) = 1$ and $h_1(s) = 1$. In this case the heuristic is inadmissible because $h_0(s)+h_1(s) = 2 > d(s, g)$. Similarly, as an ...


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Yes, in both cases. Below I give two very simple proofs that directly follow from the definitions of admissible and consistent heuristics. However, in a nutshell, the idea of the proofs is that $h_{\max}(n)$ and $h_{\min}(n)$ are, by definition (of $h_{\max}$ and $h_{\min}$), equal to one of the given admissible (or consistent) heuristics, for all nodes $n$, ...


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Welcome to AI.SE @hpr16! Your understanding of when a heuristic is admissible is correct, but your heuristic is inadmissible. An admissible heuristic must always underestimate the cost to move from a given state to a goal state. Notice that states in the search are not the same as positions on the circle in your problem. A state needs to capture all the ...


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Consistency is a property of heuristics. You can think of consistency as the common sense idea that our guess at the time to go from $A \rightarrow B \rightarrow C$ cannot be more than the time to go from $A \rightarrow B$, plus our guess of the time to go from $B \rightarrow C$. Supposing we remove a given edge $c(n,m)$ from our graph, but that our ...


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It depends on what you mean by optimal. A* will always find the optimal solution (that is, the algorithm is admissible) as long as the heuristic is admissible. (Note that the definition of admissible is overloaded and means something slightly different for an algorithm and a heuristic.) If you talk about the set of nodes expanded by A*, then it expands the ...


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The issue is that you must include assumptions about hopping into your heuristic. In particular, if you are considering individual cars then you must assume that they might be able to hop all of the way to the goal. Thus, your heuristic for each car should be Manhattan distance divided by 2. That's guaranteed to be admissible when you take the max. If you ...


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If a heuristic is not admissible, can it be consistent? No. Consistency implies admissibility. In other words, if a heuristic is consistent, it is also admissible. However, admissibility does not imply consistency. In other words, an admissible heuristic is not necessarily consistent. Definitions Given a graph $G=(V, E)$ representing the search space, ...


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For a heuristic to be admissible, it must never overestimate the distance from a state to the nearest goal state. For a heuristic to be consistent, the heuristic's value must be less than or equal to the cost of moving from that state to the state nearest the goal that can be reached from it, plus the heurstic's estimate for that state. What this means is ...


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