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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
New contributor
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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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
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New contributor
edited 3 hours ago
David Richerby
68.4k15103194
68.4k15103194
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asked 4 hours ago
Rikard OlssonRikard Olsson
1082
1082
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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.
$endgroup$
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
3 hours ago
add a comment |
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1 Answer
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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.
$endgroup$
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
3 hours 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.
$endgroup$
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
3 hours 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.
$endgroup$
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 3 hours ago
David RicherbyDavid Richerby
68.4k15103194
68.4k15103194
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
3 hours ago
add a comment |
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
3 hours ago
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
3 hours ago
$begingroup$
Wow, thanks man!!
$endgroup$
– Rikard Olsson
3 hours 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.
Rikard Olsson is a new contributor. Be nice, and check out our Code of Conduct.
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