Zrjydtj

Is it true that good novels will automatically sell themselves on Amazon (and so on) and there is no need for one to waste time promoting?

Why does energy conservation give me the wrong answer in this inelastic collision problem?

Fastest way to pop N items from a large dict

Shortcut for setting origin to vertex

Is "upgrade" the right word to use in this context?

Python if-else code style for reduced code for rounding floats

As a new Ubuntu desktop 18.04 LTS user, do I need to use ufw for a firewall or is iptables sufficient?

et qui - how do you really understand that kind of phraseology?

Brexit - No Deal Rejection

Why does overlay work only on the first tcolorbox?

Welcoming 2019 Pi day: How to draw the letter π?

Is it good practice to use Linear Least-Squares with SMA?

My adviser wants to be the first author

How to make healing in an exploration game interesting

combinatorics floor summation

Why is the President allowed to veto a cancellation of emergency powers?

A single argument pattern definition applies to multiple-argument patterns?

Why do newer 737s use two different styles of split winglets?

What did “the good wine” (τὸν καλὸν οἶνον) mean in John 2:10?

How to explain that I do not want to visit a country due to personal safety concern?

Can I use USB data pins as power source

What options are left, if Britain cannot decide?

How to pronounce "I ♥ Huckabees"?

How do you talk to someone whose loved one is dying?

Counting models satisfying a boolean formula

Prove NP-completeness of deciding satisfiability of monotone boolean formulaHow to represent a 0-valid boolean formula?What is wrong with this seeming contradiction with a paper about AND-compression of SAT?Why do we care about random Boolean SAT formula?What does a square mean in a Boolean formulaUnrolling closures into SAT boolean formulaEfficient alternatives to inclusion-exclusionCounting (enumerating) minimal solutions of a dual horn formulan-DNF boolean formula k satisfiabilityCalculating the number of assignments satisfying a general propositional formula

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 3 hours ago

David Richerby

68.4k15103194

asked 4 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 | 

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 3 hours ago

David Richerby

68.4k15103194

asked 4 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 3 hours ago

David Richerby

68.4k15103194

asked 4 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 3 hours ago

David Richerby

68.4k15103194

asked 4 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 3 hours ago

David Richerby

68.4k15103194

asked 4 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 3 hours ago

David Richerby

68.4k15103194

edited 3 hours ago

David Richerby

68.4k15103194

asked 4 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 4 hours ago

Rikard OlssonRikard Olsson

1082

1 Answer
1

active

oldest

votes

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 3 hours ago

David RicherbyDavid Richerby

68.4k15103194

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "419"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

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

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f105674%2fcounting-models-satisfying-a-boolean-formula%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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 3 hours ago

David RicherbyDavid Richerby

68.4k15103194

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

David RicherbyDavid Richerby

68.4k15103194

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.

share|cite|improve this answer

answered 3 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 3 hours ago

David RicherbyDavid Richerby

68.4k15103194

answered 3 hours ago

David RicherbyDavid Richerby

68.4k15103194

answered 3 hours ago

David RicherbyDavid Richerby

68.4k15103194