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Counting models satisfying a boolean formula

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1

$begingroup$

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

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edited 4 hours ago

David Richerby

68.4k15103194

asked 5 hours 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.
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$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this question

edited 4 hours ago

David Richerby

68.4k15103194

asked 5 hours 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.

$endgroup$

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

share|cite|improve this question

edited 4 hours ago

David Richerby

68.4k15103194

asked 5 hours 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.

$endgroup$

share|cite|improve this question

edited 4 hours ago

David Richerby

68.4k15103194

asked 5 hours 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.

share|cite|improve this question

edited 4 hours ago

David Richerby

68.4k15103194

asked 5 hours 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 4 hours ago

David Richerby

68.4k15103194

edited 4 hours ago

David Richerby

68.4k15103194

asked 5 hours 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 5 hours ago

Rikard OlssonRikard Olsson

1082

1 Answer
1

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oldest

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3

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

David RicherbyDavid Richerby

68.4k15103194

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, thanks man!!
  $endgroup$
  – Rikard Olsson
  4 hours ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

David RicherbyDavid Richerby

68.4k15103194

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, thanks man!!
  $endgroup$
  – Rikard Olsson
  4 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

David RicherbyDavid Richerby

68.4k15103194

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, thanks man!!
  $endgroup$
  – Rikard Olsson
  4 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.

share|cite|improve this answer

answered 4 hours ago

David RicherbyDavid Richerby

68.4k15103194

$endgroup$

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

share|cite|improve this answer

answered 4 hours ago

David RicherbyDavid Richerby

68.4k15103194

answered 4 hours ago

David RicherbyDavid Richerby

68.4k15103194

answered 4 hours ago

David RicherbyDavid Richerby

68.4k15103194