Counting models satisfying a boolean formula
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I'm trying to implement the #2-SAT algorithm from the paper "Counting Satisfying Assignments in 2-SAT and 3-SAT" (Dahllöf, Jonsson and Wahlström, Theor. Comput. Sci. 332(1–3):265–291, 2005). A few lines into the algorithm description the authors denotes a sub algorithm and claims "The function $C_E$ computes #2-SAT by exhaustive search. It will be applied only to formulas of size ≤ 4 and can thus be safely assumed to run in O(1) time". The size of formulas is referred to the number of clauses.
I've been trying to find this exhaustive search algorithm that computes a #2-sat instance with number of clauses less than 4. But the results only returns algorithms for generally solving/counting models for #2 or #3-SAT and does not talk about a special case when size ≤ 4. First of all, is this claim true? Since the paper was published by a well known journal, I guess it is. But if so, does anyone know about this special case?
combinatorics satisfiability 2-sat
edited 2 days ago
David Richerby
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asked 2 days ago
Rikard OlssonRikard Olsson
1082
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I'm trying to implement the #2-SAT algorithm from the paper "Counting Satisfying Assignments in 2-SAT and 3-SAT" (Dahllöf, Jonsson and Wahlström, Theor. Comput. Sci. 332(1–3):265–291, 2005). A few lines into the algorithm description the authors denotes a sub algorithm and claims "The function $C_E$ computes #2-SAT by exhaustive search. It will be applied only to formulas of size ≤ 4 and can thus be safely assumed to run in O(1) time". The size of formulas is referred to the number of clauses.
I've been trying to find this exhaustive search algorithm that computes a #2-sat instance with number of clauses less than 4. But the results only returns algorithms for generally solving/counting models for #2 or #3-SAT and does not talk about a special case when size ≤ 4. First of all, is this claim true? Since the paper was published by a well known journal, I guess it is. But if so, does anyone know about this special case?
combinatorics satisfiability 2-sat
edited 2 days ago
David Richerby
68.6k15103194
asked 2 days ago
Rikard OlssonRikard Olsson
1082
New contributor
Rikard Olsson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
I'm trying to implement the #2-SAT algorithm from the paper "Counting Satisfying Assignments in 2-SAT and 3-SAT" (Dahllöf, Jonsson and Wahlström, Theor. Comput. Sci. 332(1–3):265–291, 2005). A few lines into the algorithm description the authors denotes a sub algorithm and claims "The function $C_E$ computes #2-SAT by exhaustive search. It will be applied only to formulas of size ≤ 4 and can thus be safely assumed to run in O(1) time". The size of formulas is referred to the number of clauses.
I've been trying to find this exhaustive search algorithm that computes a #2-sat instance with number of clauses less than 4. But the results only returns algorithms for generally solving/counting models for #2 or #3-SAT and does not talk about a special case when size ≤ 4. First of all, is this claim true? Since the paper was published by a well known journal, I guess it is. But if so, does anyone know about this special case?
combinatorics satisfiability 2-sat
edited 2 days ago
David Richerby
68.6k15103194
asked 2 days ago
Rikard OlssonRikard Olsson
1082
New contributor
Rikard Olsson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm trying to implement the #2-SAT algorithm from the paper "Counting Satisfying Assignments in 2-SAT and 3-SAT" (Dahllöf, Jonsson and Wahlström, Theor. Comput. Sci. 332(1–3):265–291, 2005). A few lines into the algorithm description the authors denotes a sub algorithm and claims "The function $C_E$ computes #2-SAT by exhaustive search. It will be applied only to formulas of size ≤ 4 and can thus be safely assumed to run in O(1) time". The size of formulas is referred to the number of clauses.
I've been trying to find this exhaustive search algorithm that computes a #2-sat instance with number of clauses less than 4. But the results only returns algorithms for generally solving/counting models for #2 or #3-SAT and does not talk about a special case when size ≤ 4. First of all, is this claim true? Since the paper was published by a well known journal, I guess it is. But if so, does anyone know about this special case?
combinatorics satisfiability 2-sat
combinatorics satisfiability 2-sat
edited 2 days ago
David Richerby
68.6k15103194
asked 2 days ago
Rikard OlssonRikard Olsson
1082
New contributor
Rikard Olsson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
David Richerby
68.6k15103194
asked 2 days ago
Rikard OlssonRikard Olsson
1082
New contributor
Rikard Olsson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
David Richerby
68.6k15103194
edited 2 days ago
David Richerby
68.6k15103194
edited 2 days ago
David Richerby
68.6k15103194
68.6k15103194
asked 2 days ago
Rikard OlssonRikard Olsson
1082
New contributor
Rikard Olsson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Rikard OlssonRikard Olsson
1082
asked 2 days ago
Rikard OlssonRikard Olsson
1082
1082
New contributor
Rikard Olsson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Rikard Olsson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Rikard Olsson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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1 Answer
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$begingroup$
For any fixed $k$, a $k$-CNF with at most four clauses has at most $4k$ variables. So you can count the satisfying assigments with
count = 0
j = number of variables
for v1 = 0 to 1 do
for v2 = 0 to 1 do
...
for vj = 0 to 1 do
if formula_value(phi, v1, ..., vj) == true
count = count + 1
This runs in time $Theta(2^j) = O(2^k) = Theta(1)$, since $k$ is fixed.
answered 2 days ago
David RicherbyDavid Richerby
68.6k15103194
$endgroup$
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
2 days ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For any fixed $k$, a $k$-CNF with at most four clauses has at most $4k$ variables. So you can count the satisfying assigments with
count = 0
j = number of variables
for v1 = 0 to 1 do
for v2 = 0 to 1 do
...
for vj = 0 to 1 do
if formula_value(phi, v1, ..., vj) == true
count = count + 1
This runs in time $Theta(2^j) = O(2^k) = Theta(1)$, since $k$ is fixed.
answered 2 days ago
David RicherbyDavid Richerby
68.6k15103194
$endgroup$
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
2 days ago
add a comment |
$begingroup$
For any fixed $k$, a $k$-CNF with at most four clauses has at most $4k$ variables. So you can count the satisfying assigments with
count = 0
j = number of variables
for v1 = 0 to 1 do
for v2 = 0 to 1 do
...
for vj = 0 to 1 do
if formula_value(phi, v1, ..., vj) == true
count = count + 1
This runs in time $Theta(2^j) = O(2^k) = Theta(1)$, since $k$ is fixed.
answered 2 days ago
David RicherbyDavid Richerby
68.6k15103194
$endgroup$
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
2 days ago
add a comment |
$begingroup$
For any fixed $k$, a $k$-CNF with at most four clauses has at most $4k$ variables. So you can count the satisfying assigments with
count = 0
j = number of variables
for v1 = 0 to 1 do
for v2 = 0 to 1 do
...
for vj = 0 to 1 do
if formula_value(phi, v1, ..., vj) == true
count = count + 1
This runs in time $Theta(2^j) = O(2^k) = Theta(1)$, since $k$ is fixed.
answered 2 days ago
David RicherbyDavid Richerby
68.6k15103194
$endgroup$
For any fixed $k$, a $k$-CNF with at most four clauses has at most $4k$ variables. So you can count the satisfying assigments with
count = 0
j = number of variables
for v1 = 0 to 1 do
for v2 = 0 to 1 do
...
for vj = 0 to 1 do
if formula_value(phi, v1, ..., vj) == true
count = count + 1
This runs in time $Theta(2^j) = O(2^k) = Theta(1)$, since $k$ is fixed.
answered 2 days ago
David RicherbyDavid Richerby
68.6k15103194
answered 2 days ago
David RicherbyDavid Richerby
68.6k15103194
answered 2 days ago
David RicherbyDavid Richerby
68.6k15103194
answered 2 days ago
David RicherbyDavid Richerby
68.6k15103194
68.6k15103194
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
2 days ago
add a comment |
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
2 days ago
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
2 days ago
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
2 days ago
add a comment |
Rikard Olsson is a new contributor. Be nice, and check out our Code of Conduct.
Rikard Olsson is a new contributor. Be nice, and check out our Code of Conduct.
Rikard Olsson is a new contributor. Be nice, and check out our Code of Conduct.
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