Vwwxrf

Professor being mistaken for a grad student

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

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

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

Are ETF trackers fundamentally better than individual stocks?

What options are left, if Britain cannot decide?

What is a ^ b and (a & b) << 1?

ERC721: How to get the owned tokens of an address

Why does overlay work only on the first tcolorbox?

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

How to make healing in an exploration game interesting

Counting models satisfying a boolean formula

Why do tuner card drivers fail to build after kernel update to 4.4.0-143-generic?

Most cost effective thermostat setting: consistent temperature vs. lowest temperature possible

Is there a hypothetical scenario that would make Earth uninhabitable for humans, but not for (the majority of) other animals?

Meme-controlled people

This word with a lot of past tenses

How to pronounce "I ♥ Huckabees"?

The German vowel “a” changes to the English “i”

Problem with FindRoot

Instead of a Universal Basic Income program, why not implement a "Universal Basic Needs" program?

Why does a Star of David appear at a rally with Francisco Franco?

Do the common programs (for example: "ls", "cat") in Linux and BSD come from the same source code?

Is it insecure to send a password in a `curl` command?

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 3 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 3 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 3 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 3 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 3 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 3 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 3 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