Suboptimal solution given by A* search
I don't understand how the following graph gives a suboptimal solution with A* search.
The graph above was given as an example where A* search gives a suboptimal solution, i.e the heuristic is admissible but not consistent. Each node has a heuristic value corresponding to it and the weight of traversing a node is given. I don't understand how A* search will expand the nodes.
search artificial-intelligence heuristics
edited Nov 27 '18 at 11:58
nbro
5,68384996
asked Sep 13 '14 at 12:47
eejseejs
6913
add a comment |
I don't understand how the following graph gives a suboptimal solution with A* search.
The graph above was given as an example where A* search gives a suboptimal solution, i.e the heuristic is admissible but not consistent. Each node has a heuristic value corresponding to it and the weight of traversing a node is given. I don't understand how A* search will expand the nodes.
search artificial-intelligence heuristics
edited Nov 27 '18 at 11:58
nbro
5,68384996
asked Sep 13 '14 at 12:47
eejseejs
6913
Only A* with "graph search" will return a suboptimal solution. See stackoverflow.com/questions/10680180/….
– ziggystar
Sep 13 '14 at 14:39
Imho the pseudoimplementation of the "A* graph" is a really bad one but it can be the only reason for a subopt solution.
– Demplo
Sep 13 '14 at 14:56
add a comment |
I don't understand how the following graph gives a suboptimal solution with A* search.
The graph above was given as an example where A* search gives a suboptimal solution, i.e the heuristic is admissible but not consistent. Each node has a heuristic value corresponding to it and the weight of traversing a node is given. I don't understand how A* search will expand the nodes.
search artificial-intelligence heuristics
edited Nov 27 '18 at 11:58
nbro
5,68384996
asked Sep 13 '14 at 12:47
eejseejs
6913
I don't understand how the following graph gives a suboptimal solution with A* search.
The graph above was given as an example where A* search gives a suboptimal solution, i.e the heuristic is admissible but not consistent. Each node has a heuristic value corresponding to it and the weight of traversing a node is given. I don't understand how A* search will expand the nodes.
search artificial-intelligence heuristics
search artificial-intelligence heuristics
edited Nov 27 '18 at 11:58
nbro
5,68384996
asked Sep 13 '14 at 12:47
eejseejs
6913
edited Nov 27 '18 at 11:58
nbro
5,68384996
asked Sep 13 '14 at 12:47
eejseejs
6913
edited Nov 27 '18 at 11:58
nbro
5,68384996
edited Nov 27 '18 at 11:58
nbro
5,68384996
edited Nov 27 '18 at 11:58
nbro
5,68384996
5,68384996
asked Sep 13 '14 at 12:47
eejseejs
6913
asked Sep 13 '14 at 12:47
eejseejs
6913
asked Sep 13 '14 at 12:47
eejseejs
6913
6913
Only A* with "graph search" will return a suboptimal solution. See stackoverflow.com/questions/10680180/….
– ziggystar
Sep 13 '14 at 14:39
Imho the pseudoimplementation of the "A* graph" is a really bad one but it can be the only reason for a subopt solution.
– Demplo
Sep 13 '14 at 14:56
add a comment |
Only A* with "graph search" will return a suboptimal solution. See stackoverflow.com/questions/10680180/….
– ziggystar
Sep 13 '14 at 14:39
Imho the pseudoimplementation of the "A* graph" is a really bad one but it can be the only reason for a subopt solution.
– Demplo
Sep 13 '14 at 14:56
Only A* with "graph search" will return a suboptimal solution. See stackoverflow.com/questions/10680180/….
– ziggystar
Sep 13 '14 at 14:39
Only A* with "graph search" will return a suboptimal solution. See stackoverflow.com/questions/10680180/….
– ziggystar
Sep 13 '14 at 14:39
Imho the pseudoimplementation of the "A* graph" is a really bad one but it can be the only reason for a subopt solution.
– Demplo
Sep 13 '14 at 14:56
Imho the pseudoimplementation of the "A* graph" is a really bad one but it can be the only reason for a subopt solution.
– Demplo
Sep 13 '14 at 14:56
add a comment |
2 Answers
2
active
oldest
votes
The heuristic h(n) is not consistent.
Let me first define when a heuristic function is said to be consistent.
h(n) is consistent if
– for every node n
– for every successor n' due to legal action a
– h(n) <= c(n,a,n') + h(n')
Here clearly 'A' is a successor to node 'B'
but h(B) > h(A) + c(A,a,B)
Therefore the heuristic function is not consistent/monotone, and so A* need not give an optimal solution.
answered Aug 29 '16 at 8:11
Prakhar AgrawalPrakhar Agrawal
676717
add a comment |
Honestly I don't see how A* could return a sub-optimal solution with the given heuristic.
This is for a simple reason: the given heuristic is admissible (and even monotone/consistent).
h(s) <= h*(s) for each s in the graph
You can check this yourself comparing the h value in each node to the cost of shortest path to g.
Given the optimality property of A* I don't see how it could return a sub-optimal solution, which should be S -> A -> G of course.
The only way it could return a suboptimal solution is if it would stop once an action from a node in the frontier leading to the goal is found (so to have a path to the goal), but this would not be A*.
answered Sep 13 '14 at 14:45
DemploDemplo
421616
The heuristic is not monotone. It decreases from B to A.
– ziggystar
Sep 13 '14 at 15:14
True, not only there. Still is admissible and should guarantee optimality.
– Demplo
Sep 13 '14 at 15:33
Sorry, I'm a bit confused today. :)
– ziggystar
Sep 13 '14 at 16:02
Your comment was actually relevant. The heuristic is not consistent.
– Demplo
Sep 13 '14 at 16:26
1
@Crackej If the heuristic is not monotone, then you need to reopen nodes, as you can later encounter them on a cheaper path.
– ziggystar
Sep 13 '14 at 18:50
|
show 3 more comments
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
The heuristic h(n) is not consistent.
Let me first define when a heuristic function is said to be consistent.
h(n) is consistent if
– for every node n
– for every successor n' due to legal action a
– h(n) <= c(n,a,n') + h(n')
Here clearly 'A' is a successor to node 'B'
but h(B) > h(A) + c(A,a,B)
Therefore the heuristic function is not consistent/monotone, and so A* need not give an optimal solution.
answered Aug 29 '16 at 8:11
Prakhar AgrawalPrakhar Agrawal
676717
add a comment |
The heuristic h(n) is not consistent.
Let me first define when a heuristic function is said to be consistent.
h(n) is consistent if
– for every node n
– for every successor n' due to legal action a
– h(n) <= c(n,a,n') + h(n')
Here clearly 'A' is a successor to node 'B'
but h(B) > h(A) + c(A,a,B)
Therefore the heuristic function is not consistent/monotone, and so A* need not give an optimal solution.
answered Aug 29 '16 at 8:11
Prakhar AgrawalPrakhar Agrawal
676717
add a comment |
The heuristic h(n) is not consistent.
Let me first define when a heuristic function is said to be consistent.
h(n) is consistent if
– for every node n
– for every successor n' due to legal action a
– h(n) <= c(n,a,n') + h(n')
Here clearly 'A' is a successor to node 'B'
but h(B) > h(A) + c(A,a,B)
Therefore the heuristic function is not consistent/monotone, and so A* need not give an optimal solution.
answered Aug 29 '16 at 8:11
Prakhar AgrawalPrakhar Agrawal
676717
The heuristic h(n) is not consistent.
Let me first define when a heuristic function is said to be consistent.
h(n) is consistent if
– for every node n
– for every successor n' due to legal action a
– h(n) <= c(n,a,n') + h(n')
Here clearly 'A' is a successor to node 'B'
but h(B) > h(A) + c(A,a,B)
Therefore the heuristic function is not consistent/monotone, and so A* need not give an optimal solution.
answered Aug 29 '16 at 8:11
Prakhar AgrawalPrakhar Agrawal
676717
answered Aug 29 '16 at 8:11
Prakhar AgrawalPrakhar Agrawal
676717
answered Aug 29 '16 at 8:11
Prakhar AgrawalPrakhar Agrawal
676717
answered Aug 29 '16 at 8:11
Prakhar AgrawalPrakhar Agrawal
676717
676717
add a comment |
add a comment |
Honestly I don't see how A* could return a sub-optimal solution with the given heuristic.
This is for a simple reason: the given heuristic is admissible (and even monotone/consistent).
h(s) <= h*(s) for each s in the graph
You can check this yourself comparing the h value in each node to the cost of shortest path to g.
Given the optimality property of A* I don't see how it could return a sub-optimal solution, which should be S -> A -> G of course.
The only way it could return a suboptimal solution is if it would stop once an action from a node in the frontier leading to the goal is found (so to have a path to the goal), but this would not be A*.
answered Sep 13 '14 at 14:45
DemploDemplo
421616
The heuristic is not monotone. It decreases from B to A.
– ziggystar
Sep 13 '14 at 15:14
True, not only there. Still is admissible and should guarantee optimality.
– Demplo
Sep 13 '14 at 15:33
Sorry, I'm a bit confused today. :)
– ziggystar
Sep 13 '14 at 16:02
Your comment was actually relevant. The heuristic is not consistent.
– Demplo
Sep 13 '14 at 16:26
1
@Crackej If the heuristic is not monotone, then you need to reopen nodes, as you can later encounter them on a cheaper path.
– ziggystar
Sep 13 '14 at 18:50
|
show 3 more comments
Honestly I don't see how A* could return a sub-optimal solution with the given heuristic.
This is for a simple reason: the given heuristic is admissible (and even monotone/consistent).
h(s) <= h*(s) for each s in the graph
You can check this yourself comparing the h value in each node to the cost of shortest path to g.
Given the optimality property of A* I don't see how it could return a sub-optimal solution, which should be S -> A -> G of course.
The only way it could return a suboptimal solution is if it would stop once an action from a node in the frontier leading to the goal is found (so to have a path to the goal), but this would not be A*.
answered Sep 13 '14 at 14:45
DemploDemplo
421616
The heuristic is not monotone. It decreases from B to A.
– ziggystar
Sep 13 '14 at 15:14
True, not only there. Still is admissible and should guarantee optimality.
– Demplo
Sep 13 '14 at 15:33
Sorry, I'm a bit confused today. :)
– ziggystar
Sep 13 '14 at 16:02
Your comment was actually relevant. The heuristic is not consistent.
– Demplo
Sep 13 '14 at 16:26
1
@Crackej If the heuristic is not monotone, then you need to reopen nodes, as you can later encounter them on a cheaper path.
– ziggystar
Sep 13 '14 at 18:50
|
show 3 more comments
Honestly I don't see how A* could return a sub-optimal solution with the given heuristic.
This is for a simple reason: the given heuristic is admissible (and even monotone/consistent).
h(s) <= h*(s) for each s in the graph
You can check this yourself comparing the h value in each node to the cost of shortest path to g.
Given the optimality property of A* I don't see how it could return a sub-optimal solution, which should be S -> A -> G of course.
The only way it could return a suboptimal solution is if it would stop once an action from a node in the frontier leading to the goal is found (so to have a path to the goal), but this would not be A*.
answered Sep 13 '14 at 14:45
DemploDemplo
421616
Honestly I don't see how A* could return a sub-optimal solution with the given heuristic.
This is for a simple reason: the given heuristic is admissible (and even monotone/consistent).
h(s) <= h*(s) for each s in the graph
You can check this yourself comparing the h value in each node to the cost of shortest path to g.
Given the optimality property of A* I don't see how it could return a sub-optimal solution, which should be S -> A -> G of course.
The only way it could return a suboptimal solution is if it would stop once an action from a node in the frontier leading to the goal is found (so to have a path to the goal), but this would not be A*.
answered Sep 13 '14 at 14:45
DemploDemplo
421616
answered Sep 13 '14 at 14:45
DemploDemplo
421616
answered Sep 13 '14 at 14:45
DemploDemplo
421616
answered Sep 13 '14 at 14:45
DemploDemplo
421616
421616
The heuristic is not monotone. It decreases from B to A.
– ziggystar
Sep 13 '14 at 15:14
True, not only there. Still is admissible and should guarantee optimality.
– Demplo
Sep 13 '14 at 15:33
Sorry, I'm a bit confused today. :)
– ziggystar
Sep 13 '14 at 16:02
Your comment was actually relevant. The heuristic is not consistent.
– Demplo
Sep 13 '14 at 16:26
1
@Crackej If the heuristic is not monotone, then you need to reopen nodes, as you can later encounter them on a cheaper path.
– ziggystar
Sep 13 '14 at 18:50
|
show 3 more comments
The heuristic is not monotone. It decreases from B to A.
– ziggystar
Sep 13 '14 at 15:14
True, not only there. Still is admissible and should guarantee optimality.
– Demplo
Sep 13 '14 at 15:33
Sorry, I'm a bit confused today. :)
– ziggystar
Sep 13 '14 at 16:02
Your comment was actually relevant. The heuristic is not consistent.
– Demplo
Sep 13 '14 at 16:26
1
@Crackej If the heuristic is not monotone, then you need to reopen nodes, as you can later encounter them on a cheaper path.
– ziggystar
Sep 13 '14 at 18:50
The heuristic is not monotone. It decreases from B to A.
– ziggystar
Sep 13 '14 at 15:14
The heuristic is not monotone. It decreases from B to A.
– ziggystar
Sep 13 '14 at 15:14
True, not only there. Still is admissible and should guarantee optimality.
– Demplo
Sep 13 '14 at 15:33
True, not only there. Still is admissible and should guarantee optimality.
– Demplo
Sep 13 '14 at 15:33
Sorry, I'm a bit confused today. :)
– ziggystar
Sep 13 '14 at 16:02
Sorry, I'm a bit confused today. :)
– ziggystar
Sep 13 '14 at 16:02
Your comment was actually relevant. The heuristic is not consistent.
– Demplo
Sep 13 '14 at 16:26
Your comment was actually relevant. The heuristic is not consistent.
– Demplo
Sep 13 '14 at 16:26
1
1
@Crackej If the heuristic is not monotone, then you need to reopen nodes, as you can later encounter them on a cheaper path.
– ziggystar
Sep 13 '14 at 18:50
@Crackej If the heuristic is not monotone, then you need to reopen nodes, as you can later encounter them on a cheaper path.
– ziggystar
Sep 13 '14 at 18:50
|
show 3 more comments
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Only A* with "graph search" will return a suboptimal solution. See stackoverflow.com/questions/10680180/….
– ziggystar
Sep 13 '14 at 14:39
Imho the pseudoimplementation of the "A* graph" is a really bad one but it can be the only reason for a subopt solution.
– Demplo
Sep 13 '14 at 14:56