What is the total number of nodes generated by Iterative Deepening depth first search and Breadth First...
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What is the total number of nodes generated by Iterative Deepening depth first search and Breadth First search in terms of branching factor "b" and depth of shallowest goal "d"
algorithm breadth-first-search iterative-deepening
asked Feb 22 '15 at 0:52
AbhishekAbhishek
277
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What is the total number of nodes generated by Iterative Deepening depth first search and Breadth First search in terms of branching factor "b" and depth of shallowest goal "d"
algorithm breadth-first-search iterative-deepening
asked Feb 22 '15 at 0:52
AbhishekAbhishek
277
add a comment |
What is the total number of nodes generated by Iterative Deepening depth first search and Breadth First search in terms of branching factor "b" and depth of shallowest goal "d"
algorithm breadth-first-search iterative-deepening
asked Feb 22 '15 at 0:52
AbhishekAbhishek
277
What is the total number of nodes generated by Iterative Deepening depth first search and Breadth First search in terms of branching factor "b" and depth of shallowest goal "d"
algorithm breadth-first-search iterative-deepening
algorithm breadth-first-search iterative-deepening
asked Feb 22 '15 at 0:52
AbhishekAbhishek
277
asked Feb 22 '15 at 0:52
AbhishekAbhishek
277
asked Feb 22 '15 at 0:52
AbhishekAbhishek
277
asked Feb 22 '15 at 0:52
AbhishekAbhishek
277
asked Feb 22 '15 at 0:52
AbhishekAbhishek
277
277
add a comment |
add a comment |
2 Answers
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Here i found this on this website, it might help what you are looking for, the number really depends on the values for d and b :
https://mhesham.wordpress.com/tag/depth-first-search/
Iterative Deepening DFS worst case # of nodes:
N(IDS) = (b)d + (d – 1)b2 + (d – 2)b3 + …. + (2)bd-1 + (1)bd = O(bd)
BFS worst case # of nodes generated:
N(BFS) = b + b2 + b3 + b4 + …. bd + (bd + 1 – b) = O(bd + 1)
Example using real numbers:
branching factor = 10 and depth of shallowest goal = 5
N(IDS) = 50 + 400 + 3000 + 20000 + 100000 = 123450
N(BFS) = 10 + 100 + 1000 + 10000 + 100000 + 999990 = 1111100
answered Feb 22 '15 at 1:48
user3690249user3690249
538
add a comment |
As seen on Wikipedia:
"In an iterative deepening search, the nodes at depth d are expanded once, those at depth d-1 are expanded twice, and so on up to the root of the search tree, which is expanded d+1 times.[5] So the total number of expansions in an iterative deepening search is
IDS number of expansions
Example:
For b=10 and d=5
example
"
Reference:
https://en.wikipedia.org/wiki/Iterative_deepening_depth-first_search#Proof
answered Nov 14 '17 at 23:26
Antonio Alecrim JrAntonio Alecrim Jr
5616
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
Here i found this on this website, it might help what you are looking for, the number really depends on the values for d and b :
https://mhesham.wordpress.com/tag/depth-first-search/
Iterative Deepening DFS worst case # of nodes:
N(IDS) = (b)d + (d – 1)b2 + (d – 2)b3 + …. + (2)bd-1 + (1)bd = O(bd)
BFS worst case # of nodes generated:
N(BFS) = b + b2 + b3 + b4 + …. bd + (bd + 1 – b) = O(bd + 1)
Example using real numbers:
branching factor = 10 and depth of shallowest goal = 5
N(IDS) = 50 + 400 + 3000 + 20000 + 100000 = 123450
N(BFS) = 10 + 100 + 1000 + 10000 + 100000 + 999990 = 1111100
answered Feb 22 '15 at 1:48
user3690249user3690249
538
add a comment |
Here i found this on this website, it might help what you are looking for, the number really depends on the values for d and b :
https://mhesham.wordpress.com/tag/depth-first-search/
Iterative Deepening DFS worst case # of nodes:
N(IDS) = (b)d + (d – 1)b2 + (d – 2)b3 + …. + (2)bd-1 + (1)bd = O(bd)
BFS worst case # of nodes generated:
N(BFS) = b + b2 + b3 + b4 + …. bd + (bd + 1 – b) = O(bd + 1)
Example using real numbers:
branching factor = 10 and depth of shallowest goal = 5
N(IDS) = 50 + 400 + 3000 + 20000 + 100000 = 123450
N(BFS) = 10 + 100 + 1000 + 10000 + 100000 + 999990 = 1111100
answered Feb 22 '15 at 1:48
user3690249user3690249
538
add a comment |
Here i found this on this website, it might help what you are looking for, the number really depends on the values for d and b :
https://mhesham.wordpress.com/tag/depth-first-search/
Iterative Deepening DFS worst case # of nodes:
N(IDS) = (b)d + (d – 1)b2 + (d – 2)b3 + …. + (2)bd-1 + (1)bd = O(bd)
BFS worst case # of nodes generated:
N(BFS) = b + b2 + b3 + b4 + …. bd + (bd + 1 – b) = O(bd + 1)
Example using real numbers:
branching factor = 10 and depth of shallowest goal = 5
N(IDS) = 50 + 400 + 3000 + 20000 + 100000 = 123450
N(BFS) = 10 + 100 + 1000 + 10000 + 100000 + 999990 = 1111100
answered Feb 22 '15 at 1:48
user3690249user3690249
538
Here i found this on this website, it might help what you are looking for, the number really depends on the values for d and b :
https://mhesham.wordpress.com/tag/depth-first-search/
Iterative Deepening DFS worst case # of nodes:
N(IDS) = (b)d + (d – 1)b2 + (d – 2)b3 + …. + (2)bd-1 + (1)bd = O(bd)
BFS worst case # of nodes generated:
N(BFS) = b + b2 + b3 + b4 + …. bd + (bd + 1 – b) = O(bd + 1)
Example using real numbers:
branching factor = 10 and depth of shallowest goal = 5
N(IDS) = 50 + 400 + 3000 + 20000 + 100000 = 123450
N(BFS) = 10 + 100 + 1000 + 10000 + 100000 + 999990 = 1111100
answered Feb 22 '15 at 1:48
user3690249user3690249
538
answered Feb 22 '15 at 1:48
user3690249user3690249
538
answered Feb 22 '15 at 1:48
user3690249user3690249
538
answered Feb 22 '15 at 1:48
user3690249user3690249
538
538
add a comment |
add a comment |
As seen on Wikipedia:
"In an iterative deepening search, the nodes at depth d are expanded once, those at depth d-1 are expanded twice, and so on up to the root of the search tree, which is expanded d+1 times.[5] So the total number of expansions in an iterative deepening search is
IDS number of expansions
Example:
For b=10 and d=5
example
"
Reference:
https://en.wikipedia.org/wiki/Iterative_deepening_depth-first_search#Proof
answered Nov 14 '17 at 23:26
Antonio Alecrim JrAntonio Alecrim Jr
5616
add a comment |
As seen on Wikipedia:
"In an iterative deepening search, the nodes at depth d are expanded once, those at depth d-1 are expanded twice, and so on up to the root of the search tree, which is expanded d+1 times.[5] So the total number of expansions in an iterative deepening search is
IDS number of expansions
Example:
For b=10 and d=5
example
"
Reference:
https://en.wikipedia.org/wiki/Iterative_deepening_depth-first_search#Proof
answered Nov 14 '17 at 23:26
Antonio Alecrim JrAntonio Alecrim Jr
5616
add a comment |
As seen on Wikipedia:
"In an iterative deepening search, the nodes at depth d are expanded once, those at depth d-1 are expanded twice, and so on up to the root of the search tree, which is expanded d+1 times.[5] So the total number of expansions in an iterative deepening search is
IDS number of expansions
Example:
For b=10 and d=5
example
"
Reference:
https://en.wikipedia.org/wiki/Iterative_deepening_depth-first_search#Proof
answered Nov 14 '17 at 23:26
Antonio Alecrim JrAntonio Alecrim Jr
5616
As seen on Wikipedia:
"In an iterative deepening search, the nodes at depth d are expanded once, those at depth d-1 are expanded twice, and so on up to the root of the search tree, which is expanded d+1 times.[5] So the total number of expansions in an iterative deepening search is
IDS number of expansions
Example:
For b=10 and d=5
example
"
Reference:
https://en.wikipedia.org/wiki/Iterative_deepening_depth-first_search#Proof
answered Nov 14 '17 at 23:26
Antonio Alecrim JrAntonio Alecrim Jr
5616
answered Nov 14 '17 at 23:26
Antonio Alecrim JrAntonio Alecrim Jr
5616
answered Nov 14 '17 at 23:26
Antonio Alecrim JrAntonio Alecrim Jr
5616
answered Nov 14 '17 at 23:26
Antonio Alecrim JrAntonio Alecrim Jr
5616
5616
add a comment |
add a comment |
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