Counting models satisfying a boolean formula












1












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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







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    $begingroup$


    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









    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.







    $endgroup$















      1












      1








      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









      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






      share|cite|improve this question









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      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









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      share|cite|improve this question








      edited 3 hours ago









      David Richerby

      68.4k15103194




      68.4k15103194






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      asked 4 hours ago









      Rikard OlssonRikard Olsson

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      New contributor




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      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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      Check out our Code of Conduct.






















          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.






          share|cite|improve this answer









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











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          1 Answer
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          1 Answer
          1






          active

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          active

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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









          $endgroup$













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
















          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









          $endgroup$













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














          3












          3








          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









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          David RicherbyDavid Richerby

          68.4k15103194




          68.4k15103194












          • $begingroup$
            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




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










          Rikard Olsson is a new contributor. Be nice, and check out our Code of Conduct.










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