how to convert linear equation to cnf












1












$begingroup$


I'm working on reduction from binary puzzle problem into sat.
one of the game's rule is that in each row/column numbers of 1s equals to numbers of 0. I found a solution but it's exponential.

Therefore, I need to turn the equation:
$X_1+ X_2+..... +X_{2n} =n$ , $X_i$ is $0$ or $1$, to cnf.






logic satisfiability







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  • $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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    – José Carlos Santos
    Dec 6 '18 at 8:24
















1












$begingroup$


I'm working on reduction from binary puzzle problem into sat.
one of the game's rule is that in each row/column numbers of 1s equals to numbers of 0. I found a solution but it's exponential.

Therefore, I need to turn the equation:
$X_1+ X_2+..... +X_{2n} =n$ , $X_i$ is $0$ or $1$, to cnf.






logic satisfiability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 8:24














1












1








1





$begingroup$


I'm working on reduction from binary puzzle problem into sat.
one of the game's rule is that in each row/column numbers of 1s equals to numbers of 0. I found a solution but it's exponential.

Therefore, I need to turn the equation:
$X_1+ X_2+..... +X_{2n} =n$ , $X_i$ is $0$ or $1$, to cnf.






logic satisfiability







share|cite|improve this question











$endgroup$




I'm working on reduction from binary puzzle problem into sat.
one of the game's rule is that in each row/column numbers of 1s equals to numbers of 0. I found a solution but it's exponential.

Therefore, I need to turn the equation:
$X_1+ X_2+..... +X_{2n} =n$ , $X_i$ is $0$ or $1$, to cnf.






logic satisfiability




logic satisfiability






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share|cite|improve this question













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edited Dec 6 '18 at 9:01









Glorfindel

3,41981830




3,41981830










asked Dec 6 '18 at 8:13









Romy Adi BuchbutRomy Adi Buchbut

61




61












  • $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 8:24


















  • $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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    – José Carlos Santos
    Dec 6 '18 at 8:24
















$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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– José Carlos Santos
Dec 6 '18 at 8:24




$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
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Dec 6 '18 at 8:24










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$begingroup$

Another way to say, $X_1+ X_2+dots +X_{2n} =n$ where $X_i$ is $0$ or $1$ is to say, $exists_{=n}(X_1+ X_2+dots+X_{2n})$, meaning that exactly n of the literals are true.



The CNF for this can be split into two subproblems, "n or fewer of the literals are true" and "n or more of the literals are true". When these two groups of constraints are put together, the intersection is "exactly n of the literals are true".



1) To state the constraint: "n or more of the literals are true", generate all combinations of the literals taken $n+1$ at time.



For example, given the elements:



(x₁, x₂, x₃, x₄, x₅, x₆)


The CNF for "no fewer than 3 three are true" is generated by taking the literals four at a time:



(x₁, x₂, x₃, x₄)

(x₁, x₂, x₃, x₅)

(x₁, x₂, x₃, x₆)

(x₁, x₂, x₄, x₅)

(x₁, x₂, x₄, x₆)

(x₁, x₂, x₅, x₆)

(x₁, x₃, x₄, x₅)

(x₁, x₃, x₄, x₆)

(x₁, x₃, x₅, x₆)

(x₁, x₄, x₅, x₆)

(x₂, x₃, x₄, x₅)

(x₂, x₃, x₄, x₆)

(x₂, x₃, x₅, x₆)

(x₂, x₄, x₅, x₆)

(x₃, x₄, x₅, x₆)


2) To state the constraint: "n or fewer of the literals are true", generate all combinations of the negated literals taken $n+1$ at time.



For example, given the elements:



(x₁, x₂, x₃, x₄, x₅, x₆)


The CNF for "no more than 3 three are true" is generated by taking the negated literals four at a time:



(~x₁, ~x₂, ~x₃, ~x₄)

(~x₁, ~x₂, ~x₃, ~x₅)

(~x₁, ~x₂, ~x₃, ~x₆)

(~x₁, ~x₂, ~x₄, ~x₅)

(~x₁, ~x₂, ~x₄, ~x₆)

(~x₁, ~x₂, ~x₅, ~x₆)

(~x₁, ~x₃, ~x₄, ~x₅)

(~x₁, ~x₃, ~x₄, ~x₆)

(~x₁, ~x₃, ~x₅, ~x₆)

(~x₁, ~x₄, ~x₅, ~x₆)

(~x₂, ~x₃, ~x₄, ~x₅)

(~x₂, ~x₃, ~x₄, ~x₆)

(~x₂, ~x₃, ~x₅, ~x₆)

(~x₂, ~x₄, ~x₅, ~x₆)

(~x₃, ~x₄, ~x₅, ~x₆)


Last, concatenate the two CNFs to generate all solutions to the linear formula where the only half of the literals are true :-)






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    $begingroup$

    Another way to say, $X_1+ X_2+dots +X_{2n} =n$ where $X_i$ is $0$ or $1$ is to say, $exists_{=n}(X_1+ X_2+dots+X_{2n})$, meaning that exactly n of the literals are true.



    The CNF for this can be split into two subproblems, "n or fewer of the literals are true" and "n or more of the literals are true". When these two groups of constraints are put together, the intersection is "exactly n of the literals are true".



    1) To state the constraint: "n or more of the literals are true", generate all combinations of the literals taken $n+1$ at time.



    For example, given the elements:



    (x₁, x₂, x₃, x₄, x₅, x₆)


    The CNF for "no fewer than 3 three are true" is generated by taking the literals four at a time:



    (x₁, x₂, x₃, x₄)
    
(x₁, x₂, x₃, x₅)
    
(x₁, x₂, x₃, x₆)
    
(x₁, x₂, x₄, x₅)
    
(x₁, x₂, x₄, x₆)
    
(x₁, x₂, x₅, x₆)
    
(x₁, x₃, x₄, x₅)
    
(x₁, x₃, x₄, x₆)
    
(x₁, x₃, x₅, x₆)
    
(x₁, x₄, x₅, x₆)
    
(x₂, x₃, x₄, x₅)
    
(x₂, x₃, x₄, x₆)
    
(x₂, x₃, x₅, x₆)
    
(x₂, x₄, x₅, x₆)
    
(x₃, x₄, x₅, x₆)


    2) To state the constraint: "n or fewer of the literals are true", generate all combinations of the negated literals taken $n+1$ at time.



    For example, given the elements:



    (x₁, x₂, x₃, x₄, x₅, x₆)


    The CNF for "no more than 3 three are true" is generated by taking the negated literals four at a time:



    (~x₁, ~x₂, ~x₃, ~x₄)
    
(~x₁, ~x₂, ~x₃, ~x₅)
    
(~x₁, ~x₂, ~x₃, ~x₆)
    
(~x₁, ~x₂, ~x₄, ~x₅)
    
(~x₁, ~x₂, ~x₄, ~x₆)
    
(~x₁, ~x₂, ~x₅, ~x₆)
    
(~x₁, ~x₃, ~x₄, ~x₅)
    
(~x₁, ~x₃, ~x₄, ~x₆)
    
(~x₁, ~x₃, ~x₅, ~x₆)
    
(~x₁, ~x₄, ~x₅, ~x₆)
    
(~x₂, ~x₃, ~x₄, ~x₅)
    
(~x₂, ~x₃, ~x₄, ~x₆)
    
(~x₂, ~x₃, ~x₅, ~x₆)
    
(~x₂, ~x₄, ~x₅, ~x₆)
    
(~x₃, ~x₄, ~x₅, ~x₆)


    Last, concatenate the two CNFs to generate all solutions to the linear formula where the only half of the literals are true :-)






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Another way to say, $X_1+ X_2+dots +X_{2n} =n$ where $X_i$ is $0$ or $1$ is to say, $exists_{=n}(X_1+ X_2+dots+X_{2n})$, meaning that exactly n of the literals are true.



      The CNF for this can be split into two subproblems, "n or fewer of the literals are true" and "n or more of the literals are true". When these two groups of constraints are put together, the intersection is "exactly n of the literals are true".



      1) To state the constraint: "n or more of the literals are true", generate all combinations of the literals taken $n+1$ at time.



      For example, given the elements:



      (x₁, x₂, x₃, x₄, x₅, x₆)


      The CNF for "no fewer than 3 three are true" is generated by taking the literals four at a time:



      (x₁, x₂, x₃, x₄)
      
(x₁, x₂, x₃, x₅)
      
(x₁, x₂, x₃, x₆)
      
(x₁, x₂, x₄, x₅)
      
(x₁, x₂, x₄, x₆)
      
(x₁, x₂, x₅, x₆)
      
(x₁, x₃, x₄, x₅)
      
(x₁, x₃, x₄, x₆)
      
(x₁, x₃, x₅, x₆)
      
(x₁, x₄, x₅, x₆)
      
(x₂, x₃, x₄, x₅)
      
(x₂, x₃, x₄, x₆)
      
(x₂, x₃, x₅, x₆)
      
(x₂, x₄, x₅, x₆)
      
(x₃, x₄, x₅, x₆)


      2) To state the constraint: "n or fewer of the literals are true", generate all combinations of the negated literals taken $n+1$ at time.



      For example, given the elements:



      (x₁, x₂, x₃, x₄, x₅, x₆)


      The CNF for "no more than 3 three are true" is generated by taking the negated literals four at a time:



      (~x₁, ~x₂, ~x₃, ~x₄)
      
(~x₁, ~x₂, ~x₃, ~x₅)
      
(~x₁, ~x₂, ~x₃, ~x₆)
      
(~x₁, ~x₂, ~x₄, ~x₅)
      
(~x₁, ~x₂, ~x₄, ~x₆)
      
(~x₁, ~x₂, ~x₅, ~x₆)
      
(~x₁, ~x₃, ~x₄, ~x₅)
      
(~x₁, ~x₃, ~x₄, ~x₆)
      
(~x₁, ~x₃, ~x₅, ~x₆)
      
(~x₁, ~x₄, ~x₅, ~x₆)
      
(~x₂, ~x₃, ~x₄, ~x₅)
      
(~x₂, ~x₃, ~x₄, ~x₆)
      
(~x₂, ~x₃, ~x₅, ~x₆)
      
(~x₂, ~x₄, ~x₅, ~x₆)
      
(~x₃, ~x₄, ~x₅, ~x₆)


      Last, concatenate the two CNFs to generate all solutions to the linear formula where the only half of the literals are true :-)






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Another way to say, $X_1+ X_2+dots +X_{2n} =n$ where $X_i$ is $0$ or $1$ is to say, $exists_{=n}(X_1+ X_2+dots+X_{2n})$, meaning that exactly n of the literals are true.



        The CNF for this can be split into two subproblems, "n or fewer of the literals are true" and "n or more of the literals are true". When these two groups of constraints are put together, the intersection is "exactly n of the literals are true".



        1) To state the constraint: "n or more of the literals are true", generate all combinations of the literals taken $n+1$ at time.



        For example, given the elements:



        (x₁, x₂, x₃, x₄, x₅, x₆)


        The CNF for "no fewer than 3 three are true" is generated by taking the literals four at a time:



        (x₁, x₂, x₃, x₄)
        
(x₁, x₂, x₃, x₅)
        
(x₁, x₂, x₃, x₆)
        
(x₁, x₂, x₄, x₅)
        
(x₁, x₂, x₄, x₆)
        
(x₁, x₂, x₅, x₆)
        
(x₁, x₃, x₄, x₅)
        
(x₁, x₃, x₄, x₆)
        
(x₁, x₃, x₅, x₆)
        
(x₁, x₄, x₅, x₆)
        
(x₂, x₃, x₄, x₅)
        
(x₂, x₃, x₄, x₆)
        
(x₂, x₃, x₅, x₆)
        
(x₂, x₄, x₅, x₆)
        
(x₃, x₄, x₅, x₆)


        2) To state the constraint: "n or fewer of the literals are true", generate all combinations of the negated literals taken $n+1$ at time.



        For example, given the elements:



        (x₁, x₂, x₃, x₄, x₅, x₆)


        The CNF for "no more than 3 three are true" is generated by taking the negated literals four at a time:



        (~x₁, ~x₂, ~x₃, ~x₄)
        
(~x₁, ~x₂, ~x₃, ~x₅)
        
(~x₁, ~x₂, ~x₃, ~x₆)
        
(~x₁, ~x₂, ~x₄, ~x₅)
        
(~x₁, ~x₂, ~x₄, ~x₆)
        
(~x₁, ~x₂, ~x₅, ~x₆)
        
(~x₁, ~x₃, ~x₄, ~x₅)
        
(~x₁, ~x₃, ~x₄, ~x₆)
        
(~x₁, ~x₃, ~x₅, ~x₆)
        
(~x₁, ~x₄, ~x₅, ~x₆)
        
(~x₂, ~x₃, ~x₄, ~x₅)
        
(~x₂, ~x₃, ~x₄, ~x₆)
        
(~x₂, ~x₃, ~x₅, ~x₆)
        
(~x₂, ~x₄, ~x₅, ~x₆)
        
(~x₃, ~x₄, ~x₅, ~x₆)


        Last, concatenate the two CNFs to generate all solutions to the linear formula where the only half of the literals are true :-)






        share|cite|improve this answer











        $endgroup$



        Another way to say, $X_1+ X_2+dots +X_{2n} =n$ where $X_i$ is $0$ or $1$ is to say, $exists_{=n}(X_1+ X_2+dots+X_{2n})$, meaning that exactly n of the literals are true.



        The CNF for this can be split into two subproblems, "n or fewer of the literals are true" and "n or more of the literals are true". When these two groups of constraints are put together, the intersection is "exactly n of the literals are true".



        1) To state the constraint: "n or more of the literals are true", generate all combinations of the literals taken $n+1$ at time.



        For example, given the elements:



        (x₁, x₂, x₃, x₄, x₅, x₆)


        The CNF for "no fewer than 3 three are true" is generated by taking the literals four at a time:



        (x₁, x₂, x₃, x₄)
        
(x₁, x₂, x₃, x₅)
        
(x₁, x₂, x₃, x₆)
        
(x₁, x₂, x₄, x₅)
        
(x₁, x₂, x₄, x₆)
        
(x₁, x₂, x₅, x₆)
        
(x₁, x₃, x₄, x₅)
        
(x₁, x₃, x₄, x₆)
        
(x₁, x₃, x₅, x₆)
        
(x₁, x₄, x₅, x₆)
        
(x₂, x₃, x₄, x₅)
        
(x₂, x₃, x₄, x₆)
        
(x₂, x₃, x₅, x₆)
        
(x₂, x₄, x₅, x₆)
        
(x₃, x₄, x₅, x₆)


        2) To state the constraint: "n or fewer of the literals are true", generate all combinations of the negated literals taken $n+1$ at time.



        For example, given the elements:



        (x₁, x₂, x₃, x₄, x₅, x₆)


        The CNF for "no more than 3 three are true" is generated by taking the negated literals four at a time:



        (~x₁, ~x₂, ~x₃, ~x₄)
        
(~x₁, ~x₂, ~x₃, ~x₅)
        
(~x₁, ~x₂, ~x₃, ~x₆)
        
(~x₁, ~x₂, ~x₄, ~x₅)
        
(~x₁, ~x₂, ~x₄, ~x₆)
        
(~x₁, ~x₂, ~x₅, ~x₆)
        
(~x₁, ~x₃, ~x₄, ~x₅)
        
(~x₁, ~x₃, ~x₄, ~x₆)
        
(~x₁, ~x₃, ~x₅, ~x₆)
        
(~x₁, ~x₄, ~x₅, ~x₆)
        
(~x₂, ~x₃, ~x₄, ~x₅)
        
(~x₂, ~x₃, ~x₄, ~x₆)
        
(~x₂, ~x₃, ~x₅, ~x₆)
        
(~x₂, ~x₄, ~x₅, ~x₆)
        
(~x₃, ~x₄, ~x₅, ~x₆)


        Last, concatenate the two CNFs to generate all solutions to the linear formula where the only half of the literals are true :-)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 7:45

























        answered Dec 8 '18 at 18:12









        Raymond HettingerRaymond Hettinger

        46119




        46119






























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            Sphinx de Gizeh

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            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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            Dijon

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            Dijon.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} De haut en bas, de gauche à droite : Panorama du centre historique vu depuis la Cathédrale Saint-Bénigne, le tramway passant sous l'Auditorium , la place François Rude, les Tuiles vernissées de l'Hôtel Aubriot, le centre commercial Toison d'or et le jardin Darcy. Blason Logo Administration Pays   France Région Bourgogne-Franche-Comté (préfecture) Département Côte-d'Or (préfecture) Arrondissement Dijon (chef-lieu) Canton bureau centralisateur de 6 cantons : Dijon-1, 2, 3, 4, 5, 6 Intercommunalité Dijon Métropole Maire Mandat François Rebsamen 2015-2020 Code postal 21000 (comportait aussi le code 21100 quand la ville possédait deux bureaux distributeurs) Code commune 21231 Démographie Gentilé ...
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            Langue

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            Pour les articles homonymes, voir Langue (homonymie). Une langue est un système évolutif de signes linguistiques, vocaux, graphiques ou gestuels, qui permet la communication entre les individus. Langues du monde par familles :       Afro-asiatiques       Nigéro-congolaises       Nilo-sahariennes       Khoïsan       Indo-européennes       Caucasiennes       Altaïques       Ouraliennes       Dravidiennes       Sino-tibétaines       Austroasiatiques       Austronésiennes       Pama-nyungan       Papoues       Tai-Kadai       Amérindiennes       Na-dené       Eskimo-aléoutes       Isolat Sommaire 1 Définition : langue et langage 1.1 Langue et dialecte 1.2 Langue naturelle et langue construite 1.3 Langue vivante et langue morte 1.4 Autre 2 Histoire : origine des langues 3 Classification 3.1 Nombre de langues 3.2 Nombre de locuteurs par langue 3.3 Dénominations 4 Év...
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