Bsrgvty

Is it safe to pay bills over satellite internet?

Uniform Roe algebra of virtually abelian group is type I C*-algebra?

Can you use a virtual credit card to withdraw money from an ATM in the UK?

Why is my paper "under review" if it contains no results?

Does Darwin owe a debt to Hegel?

Why do we need the coupling small when doing perturbative QFT calculation?

Does a restocking fee still qualify as a business expense?

How are Aircraft Noses Designed?

What could possibly power an Alcubierre drive?

How to increment the value of a (decimal) variable (with leading zero) by +1?

90's/00's anime about 3 girls who travel between the real world and a fantasy world

How to deal with people whose priority is to not get blamed?

What is the gold linker?

Car as a good investment

Does the US require a House vote to begin an impeachment inquiry?

Can a Pokemon that I tried to capture from field research run away?

How does a ball bearing door hinge work?

Does cover affect melee attacks?

Easy way of generating a 50-150W load @12V

When to use the gestalt principle of common region?

Can Chandra, Acolyte of Flame cast Heartfire without the additional cost?

Why is technology bad for children?

How to treat unhandled exceptions? (Terminate the application vs. Keep it alive)

First author doesn't want a co-author to read the whole paper

Create a list of all possible Boolean configurations of three constraints

Implementing a function which generalizes the merging step in merge sortList all possible elements of a setMatlab-style Find in Mathematica? Value list and Boolean List, how to select values according to the boolean list?How to combine all possible N boolean values?Create lists for all nested lists, all nested list lengthsList all possible license plate numbersFind positions of certain value in a big boolean listHow to test all possible input for a logic function?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;

3

$begingroup$

I have three nonnegativity constraints
beginequation
A_1>0, A_2>0, A_3>0.
endequation
I want to create a single List consisting of all possible/inequivalent Boolean configurations of none (trivial), one (also trivial), two, and three of them, using the AND, NOT and OR operations.

How long will this List be?

Well--in response to the initial comment/answer--I was really thinking of a (multi-element) List such as

Given such a List, I would make the substitutions

B1 -> A1 > 0, B2 -> A2 > 0, B3 -> A3 > 0

and then attempt Boolean integrations using Boole[each element of the resulting list].

list-manipulation boolean-computation

share|improve this question

edited May 11 at 15:24

Paul B. Slater

asked May 11 at 13:09

Paul B. SlaterPaul B. Slater

79344 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are there any equivalent pairs in this list? res = And @@ MapThread[Construct, #, a, b, c] & /@ Tuples[Not, Identity, 3]; res = Join[res, Not /@ res];
  $endgroup$
  – Szabolcs
  May 11 at 13:40
  

  
  
  
  

add a comment
 | 

3

$begingroup$

I have three nonnegativity constraints
beginequation
A_1>0, A_2>0, A_3>0.
endequation
I want to create a single List consisting of all possible/inequivalent Boolean configurations of none (trivial), one (also trivial), two, and three of them, using the AND, NOT and OR operations.

How long will this List be?

Well--in response to the initial comment/answer--I was really thinking of a (multi-element) List such as

Given such a List, I would make the substitutions

B1 -> A1 > 0, B2 -> A2 > 0, B3 -> A3 > 0

and then attempt Boolean integrations using Boole[each element of the resulting list].

list-manipulation boolean-computation

share|improve this question

edited May 11 at 15:24

Paul B. Slater

asked May 11 at 13:09

Paul B. SlaterPaul B. Slater

79344 silver badges1414 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are there any equivalent pairs in this list? res = And @@ MapThread[Construct, #, a, b, c] & /@ Tuples[Not, Identity, 3]; res = Join[res, Not /@ res];
  $endgroup$
  – Szabolcs
  May 11 at 13:40
  

  
  
  
  

add a comment
 | 

$begingroup$

I have three nonnegativity constraints
beginequation
A_1>0, A_2>0, A_3>0.
endequation
I want to create a single List consisting of all possible/inequivalent Boolean configurations of none (trivial), one (also trivial), two, and three of them, using the AND, NOT and OR operations.

How long will this List be?

Well--in response to the initial comment/answer--I was really thinking of a (multi-element) List such as

Given such a List, I would make the substitutions

B1 -> A1 > 0, B2 -> A2 > 0, B3 -> A3 > 0

and then attempt Boolean integrations using Boole[each element of the resulting list].

list-manipulation boolean-computation

share|improve this question

edited May 11 at 15:24

Paul B. Slater

asked May 11 at 13:09

Paul B. SlaterPaul B. Slater

79344 silver badges1414 bronze badges

$endgroup$

B1 -> A1 > 0, B2 -> A2 > 0, B3 -> A3 > 0

share|improve this question

edited May 11 at 15:24

Paul B. Slater

asked May 11 at 13:09

Paul B. SlaterPaul B. Slater

79344 silver badges1414 bronze badges

share|improve this question

edited May 11 at 15:24

Paul B. Slater

asked May 11 at 13:09

Paul B. SlaterPaul B. Slater

79344 silver badges1414 bronze badges

asked May 11 at 13:09

Paul B. SlaterPaul B. Slater

79344 silver badges1414 bronze badges

asked May 11 at 13:09

Paul B. SlaterPaul B. Slater

79344 silver badges1414 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

You can use BooleanCountingFunction:

exp = Array[Subscript[A, #] > 0 &, 3];
BooleanConvert[BooleanCountingFunction[1, 3] @@ exp]

TeXForm @ %

$smallleft(A_1leq 0land A_2leq 0land A_3>0right)lor left(A_1leq 0land A_2>0land A_3leq 0right)lor left(A_1>0land A_2leq 0land
A_3leq 0right)$

BooleanConvert[BooleanCountingFunction[2, 3] @@ exp] // TeXForm

$small left(A_1leq 0land A_2>0land A_3>0right)lor left(A_1>0land A_2leq 0land A_3>0right)lor left(A_1>0land A_2>0land A_3leq
0right)$

BooleanConvert[BooleanCountingFunction[3, 3] @@ exp] // TeXForm

$small A_1>0land A_2>0land A_3>0$

share|improve this answer

edited May 11 at 13:48

answered May 11 at 13:40

kglrkglr

220k1010 gold badges250250 silver badges507507 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

You can create a list of all 8 possible combinations:

tups = Tuples[And[B1, !B1, B2, !B2, B3, !B3]]

B1 && B2 && B3, B1 && B2 && ! B3, B1 && ! B2 && B3,
B1 && ! B2 && ! B3, ! B1 && B2 && B3, ! B1 && B2 && ! B3, ! B1 && ! B2 &&
B3, ! B1 && ! B2 && ! B3

Then, I think you want the power set:

sets = BooleanMinimize /@ Subsets[Or @@ tups];
sets[[;;20]]

(B1 && ! B2 && ! B3),
B2 && B3, (B1 && B2 && B3)

You examples are all included:

examples = ! (B2 && B3),
 ! B1 && B2 && B3
;
MemberQ[sets, #]& /@ BooleanMinimize /@ examples

True, True, True, True, True, True

share|improve this answer

answered May 11 at 18:54

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! I don't think I was considering more than 3 member combinations (hard enough to integrate with them as constraints)--but some of those (at most three) member combinations would have OR's in them--such as B1 || (B2 && B3). I don't see that standing by itself in your list (or any other two- or three-member group with an imbedded OR). Why is its presence tested as TRUE? Not sure as to the conversion rules of Boolean operations--can parentheses be removed? That's why I posted the question--to try to efficiently list all three-member distinct possibilities for my integration computations.
  $endgroup$
  – Paul B. Slater
  May 11 at 22:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PaulB.Slater I only showed the first 20 of the 256 elements of the power set. I don't understnad your criteria for what elements of the power set you want to consider.
  $endgroup$
  – Carl Woll
  May 11 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks again! I do now see the full 256 entries. How can I select those among them with n B's in them (n=2, 3, 4,...)? That seem like a resonable order in which to attempt my constrained integrations.
  $endgroup$
  – Paul B. Slater
  May 11 at 23:40
  
  

  
  
  
  

add a comment
 | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
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/4.0/"u003ecc by-sa 4.0 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
);



);

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%2fmathematica.stackexchange.com%2fquestions%2f198169%2fcreate-a-list-of-all-possible-boolean-configurations-of-three-constraints%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

You can use BooleanCountingFunction:

exp = Array[Subscript[A, #] > 0 &, 3];
BooleanConvert[BooleanCountingFunction[1, 3] @@ exp]

TeXForm @ %

$smallleft(A_1leq 0land A_2leq 0land A_3>0right)lor left(A_1leq 0land A_2>0land A_3leq 0right)lor left(A_1>0land A_2leq 0land
A_3leq 0right)$

BooleanConvert[BooleanCountingFunction[2, 3] @@ exp] // TeXForm

$small left(A_1leq 0land A_2>0land A_3>0right)lor left(A_1>0land A_2leq 0land A_3>0right)lor left(A_1>0land A_2>0land A_3leq
0right)$

BooleanConvert[BooleanCountingFunction[3, 3] @@ exp] // TeXForm

$small A_1>0land A_2>0land A_3>0$

share|improve this answer

edited May 11 at 13:48

answered May 11 at 13:40

kglrkglr

220k1010 gold badges250250 silver badges507507 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

You can use BooleanCountingFunction:

exp = Array[Subscript[A, #] > 0 &, 3];
BooleanConvert[BooleanCountingFunction[1, 3] @@ exp]

TeXForm @ %

$smallleft(A_1leq 0land A_2leq 0land A_3>0right)lor left(A_1leq 0land A_2>0land A_3leq 0right)lor left(A_1>0land A_2leq 0land
A_3leq 0right)$

BooleanConvert[BooleanCountingFunction[2, 3] @@ exp] // TeXForm

$small left(A_1leq 0land A_2>0land A_3>0right)lor left(A_1>0land A_2leq 0land A_3>0right)lor left(A_1>0land A_2>0land A_3leq
0right)$

BooleanConvert[BooleanCountingFunction[3, 3] @@ exp] // TeXForm

$small A_1>0land A_2>0land A_3>0$

share|improve this answer

edited May 11 at 13:48

answered May 11 at 13:40

kglrkglr

220k1010 gold badges250250 silver badges507507 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You can use BooleanCountingFunction:

exp = Array[Subscript[A, #] > 0 &, 3];
BooleanConvert[BooleanCountingFunction[1, 3] @@ exp]

TeXForm @ %

$smallleft(A_1leq 0land A_2leq 0land A_3>0right)lor left(A_1leq 0land A_2>0land A_3leq 0right)lor left(A_1>0land A_2leq 0land
A_3leq 0right)$

BooleanConvert[BooleanCountingFunction[2, 3] @@ exp] // TeXForm

$small left(A_1leq 0land A_2>0land A_3>0right)lor left(A_1>0land A_2leq 0land A_3>0right)lor left(A_1>0land A_2>0land A_3leq
0right)$

BooleanConvert[BooleanCountingFunction[3, 3] @@ exp] // TeXForm

$small A_1>0land A_2>0land A_3>0$

share|improve this answer

edited May 11 at 13:48

answered May 11 at 13:40

kglrkglr

220k1010 gold badges250250 silver badges507507 bronze badges

$endgroup$

exp = Array[Subscript[A, #] > 0 &, 3];
BooleanConvert[BooleanCountingFunction[1, 3] @@ exp]

TeXForm @ %

BooleanConvert[BooleanCountingFunction[2, 3] @@ exp] // TeXForm

BooleanConvert[BooleanCountingFunction[3, 3] @@ exp] // TeXForm

share|improve this answer

edited May 11 at 13:48

answered May 11 at 13:40

kglrkglr

220k1010 gold badges250250 silver badges507507 bronze badges

answered May 11 at 13:40

kglrkglr

220k1010 gold badges250250 silver badges507507 bronze badges

answered May 11 at 13:40

kglrkglr

220k1010 gold badges250250 silver badges507507 bronze badges

3

$begingroup$

You can create a list of all 8 possible combinations:

tups = Tuples[And[B1, !B1, B2, !B2, B3, !B3]]

B1 && B2 && B3, B1 && B2 && ! B3, B1 && ! B2 && B3,
B1 && ! B2 && ! B3, ! B1 && B2 && B3, ! B1 && B2 && ! B3, ! B1 && ! B2 &&
B3, ! B1 && ! B2 && ! B3

Then, I think you want the power set:

sets = BooleanMinimize /@ Subsets[Or @@ tups];
sets[[;;20]]

(B1 && ! B2 && ! B3),
B2 && B3, (B1 && B2 && B3)

You examples are all included:

examples = ! (B2 && B3),
 ! B1 && B2 && B3
;
MemberQ[sets, #]& /@ BooleanMinimize /@ examples

True, True, True, True, True, True

share|improve this answer

answered May 11 at 18:54

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! I don't think I was considering more than 3 member combinations (hard enough to integrate with them as constraints)--but some of those (at most three) member combinations would have OR's in them--such as B1 || (B2 && B3). I don't see that standing by itself in your list (or any other two- or three-member group with an imbedded OR). Why is its presence tested as TRUE? Not sure as to the conversion rules of Boolean operations--can parentheses be removed? That's why I posted the question--to try to efficiently list all three-member distinct possibilities for my integration computations.
  $endgroup$
  – Paul B. Slater
  May 11 at 22:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PaulB.Slater I only showed the first 20 of the 256 elements of the power set. I don't understnad your criteria for what elements of the power set you want to consider.
  $endgroup$
  – Carl Woll
  May 11 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks again! I do now see the full 256 entries. How can I select those among them with n B's in them (n=2, 3, 4,...)? That seem like a resonable order in which to attempt my constrained integrations.
  $endgroup$
  – Paul B. Slater
  May 11 at 23:40
  
  

  
  
  
  

add a comment
 | 

3

$begingroup$

You can create a list of all 8 possible combinations:

tups = Tuples[And[B1, !B1, B2, !B2, B3, !B3]]

B1 && B2 && B3, B1 && B2 && ! B3, B1 && ! B2 && B3,
B1 && ! B2 && ! B3, ! B1 && B2 && B3, ! B1 && B2 && ! B3, ! B1 && ! B2 &&
B3, ! B1 && ! B2 && ! B3

Then, I think you want the power set:

sets = BooleanMinimize /@ Subsets[Or @@ tups];
sets[[;;20]]

(B1 && ! B2 && ! B3),
B2 && B3, (B1 && B2 && B3)

You examples are all included:

examples = ! (B2 && B3),
 ! B1 && B2 && B3
;
MemberQ[sets, #]& /@ BooleanMinimize /@ examples

True, True, True, True, True, True

share|improve this answer

answered May 11 at 18:54

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! I don't think I was considering more than 3 member combinations (hard enough to integrate with them as constraints)--but some of those (at most three) member combinations would have OR's in them--such as B1 || (B2 && B3). I don't see that standing by itself in your list (or any other two- or three-member group with an imbedded OR). Why is its presence tested as TRUE? Not sure as to the conversion rules of Boolean operations--can parentheses be removed? That's why I posted the question--to try to efficiently list all three-member distinct possibilities for my integration computations.
  $endgroup$
  – Paul B. Slater
  May 11 at 22:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PaulB.Slater I only showed the first 20 of the 256 elements of the power set. I don't understnad your criteria for what elements of the power set you want to consider.
  $endgroup$
  – Carl Woll
  May 11 at 23:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks again! I do now see the full 256 entries. How can I select those among them with n B's in them (n=2, 3, 4,...)? That seem like a resonable order in which to attempt my constrained integrations.
  $endgroup$
  – Paul B. Slater
  May 11 at 23:40
  
  

  
  
  
  

add a comment
 | 

$begingroup$

You can create a list of all 8 possible combinations:

tups = Tuples[And[B1, !B1, B2, !B2, B3, !B3]]

B1 && B2 && B3, B1 && B2 && ! B3, B1 && ! B2 && B3,
B1 && ! B2 && ! B3, ! B1 && B2 && B3, ! B1 && B2 && ! B3, ! B1 && ! B2 &&
B3, ! B1 && ! B2 && ! B3

Then, I think you want the power set:

sets = BooleanMinimize /@ Subsets[Or @@ tups];
sets[[;;20]]

(B1 && ! B2 && ! B3),
B2 && B3, (B1 && B2 && B3)

You examples are all included:

examples = ! (B2 && B3),
 ! B1 && B2 && B3
;
MemberQ[sets, #]& /@ BooleanMinimize /@ examples

True, True, True, True, True, True

share|improve this answer

answered May 11 at 18:54

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

$endgroup$

tups = Tuples[And[B1, !B1, B2, !B2, B3, !B3]]

sets = BooleanMinimize /@ Subsets[Or @@ tups];
sets[[;;20]]

examples = ! (B2 && B3),
 ! B1 && B2 && B3
;
MemberQ[sets, #]& /@ BooleanMinimize /@ examples

share|improve this answer

answered May 11 at 18:54

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

answered May 11 at 18:54

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges

answered May 11 at 18:54

Carl WollCarl Woll

92.1k33 gold badges121121 silver badges233233 bronze badges