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• For a admissable but not monotone heuristic function, A* and A* with properly designed cycling checker,both provide opTimal solution,Is it right or wrong ? we
Message 1 of 4 , Mar 9 10:37 AM
For a admissable but not monotone heuristic function, A* and A* with
properly designed cycling checker,both provide opTimal solution,Is it
right or wrong ? we have debat at class.no conclution.Pl some one help
us out.
• A* gives optimal result unless the heuristic function does not overestimates. see AI by kevin knight for a discription of this.... chinatigershanghai
Message 2 of 4 , Mar 10 9:16 PM
A* gives optimal result unless the heuristic function does not overestimates.
see AI by kevin knight for a discription of this....

chinatigershanghai <chinatigershanghai@...> wrote:
For a admissable but not monotone heuristic function, A* and A* with
properly designed cycling checker,both provide opTimal solution,Is it
right or wrong ? we have debat at class.no conclution.Pl some one help
us out.

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• ... overestimates. ... in business Artificial intelligence Artificial intelligence introduction ... Service. ... Consider: [s] 4/ 1 [b] --- [a]
Message 3 of 4 , Mar 18 4:54 PM
--- In aima-talk@yahoogroups.com, "\[3!|_/\\|_" <bilal_hayat_butt@...>
wrote:
>
> A* gives optimal result unless the heuristic function does not
overestimates.
> see AI by kevin knight for a discription of this....
>
> chinatigershanghai <chinatigershanghai@...> wrote:
> For a admissable but not monotone heuristic function, A* and A* with
> properly designed cycling checker,both provide opTimal solution,Is it
> right or wrong ? we have debat at class.no conclution.Pl some one help
> us out.
>
>
>
>
>
>
>
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Consider:
[s]
4/ \1
[b] --- [a]
5| 2
[c]
1|
[g]

h(a) = 2
h(b) = 1
h(c) = 1
h(g) = 0

I believe that the computer will output
{ S -> B, B->C , C -> G }
but the optimal solution will be
{S -> A, A->B, B -> C, C-> G }

The program produces different output because of cycle checking with a
bad heuristic. But as you can see, it is not optimal. So saying that
cycling checking always produces an optimal solutions is incorrect. Am
I right or am I wrong?
• ... Depends on the cycle check you use. But without it the A* algorithm will output the optimal solution. -- Ivan F. Villanueva B. artificialidea.com
Message 4 of 4 , Mar 21 4:06 AM
On Sun, Mar 19, 2006 12:54:19AM -0000, chinatigershanghai wrote:
> Consider:
> [s]
> 4/ \1
> [b] --- [a]
> 5| 2
> [c]
> 1|
> [g]
>
> h(a) = 2
> h(b) = 1
> h(c) = 1
> h(g) = 0
>
> I believe that the computer will output
> { S -> B, B->C , C -> G }
> but the optimal solution will be
> {S -> A, A->B, B -> C, C-> G }
>
> The program produces different output because of cycle checking with a
> bad heuristic. But as you can see, it is not optimal. So saying that
> cycling checking always produces an optimal solutions is incorrect. Am
> I right or am I wrong?

Depends on the cycle check you use. But without it the A* algorithm will output the
optimal solution.

--
Ivan F. Villanueva B.
artificialidea.com
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