Tjyylli

Split a number into equal parts given the number of parts

Why are special aircraft used for the carriers in the United States Navy?

Levi-Civita symbol: 3D matrix

I've given my players a lot of magic items. Is it reasonable for me to give them harder encounters?

Correct physics behind the colors on CD (compact disc)?

Can the Shape Water Cantrip be used to manipulate blood?

Difference between 'stomach' and 'uterus'

Plagiarism of code by other PhD student

Being asked to review a paper in conference one has submitted to

Sometimes a banana is just a banana

Canadian citizen, on US no-fly list. What can I do in order to be allowed on flights which go through US airspace?

Why do phishing e-mails use faked e-mail addresses instead of the real one?

Misplaced tyre lever - alternatives?

How do I deal with being envious of my own players?

It doesn't matter the side you see it

How to get the first element while continue streaming?

Specific Chinese carabiner QA?

Should I use HTTPS on a domain that will only be used for redirection?

I encountered my boss during an on-site interview at another company. Should I bring it up when seeing him next time?

Where is this quote about overcoming the impossible said in "Interstellar"?

Is there a way to find out the age of climbing ropes?

Can an earth elemental drown/bury its opponent underground using earth glide?

3.5% Interest Student Loan or use all of my savings on Tuition?

How can I handle a player who pre-plans arguments about my rulings on RAW?

All possible A of Ax=b with constraints on A

FrobeniusSolve with solutions only being 0 or 1 being acceptableHow to augment the realm of functions Mathematica thinks it knows how to integrate symbolicallyEstimate error on slope of linear regression given data with associated uncertaintyGenerating a list of integers that roughly satisfy a distributionChoosing appropriate WorkingPrecision when solving numerical system of equations“Reduce” works fine for a non-linear system with 9 equations, but cannot solve it if 10 equations. Any ways to improve the code?Finding mean of data sets with unequal lengthIs it possible to speedup these simple linear algebra operationsAnalytic inversion of linear systemObtaining the determinant from a LinearSolveFunction objectPlotting confidence interval from P values (from DistributionFitTest) for 2 parameters

5

$begingroup$

I have a linear problem that I want to solve but the method is quite different from normal, the problem is still Ax=b. However,
in this instant I have A being unknown, apart from the fact that each entry in A can only be zero or 1. Further I have x being known and fixed, the same holds for b.

My question, is it a easy method for getting mathematica to spit out all possible A such that Ax=b given conditions on A.

I tried doing a number of for loops etc, however, I have completely given up after number of hours and the expectation that my effort is incorrect.

probability-or-statistics linear-algebra

share|improve this question

asked yesterday

ALEXANDERALEXANDER

634518

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should be able to use Tuples[] + Partition[] (after perhaps filtering out nonsingular candidates).
  $endgroup$
  – J. M. is computer-less♦
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please post a concrete example. Also note this is essentially the same as this recent MSE question
  $endgroup$
  – Daniel Lichtblau
  yesterday
  

  
  
  
  

add a comment | 

5

$begingroup$

I have a linear problem that I want to solve but the method is quite different from normal, the problem is still Ax=b. However,
in this instant I have A being unknown, apart from the fact that each entry in A can only be zero or 1. Further I have x being known and fixed, the same holds for b.

My question, is it a easy method for getting mathematica to spit out all possible A such that Ax=b given conditions on A.

I tried doing a number of for loops etc, however, I have completely given up after number of hours and the expectation that my effort is incorrect.

probability-or-statistics linear-algebra

share|improve this question

asked yesterday

ALEXANDERALEXANDER

634518

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should be able to use Tuples[] + Partition[] (after perhaps filtering out nonsingular candidates).
  $endgroup$
  – J. M. is computer-less♦
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please post a concrete example. Also note this is essentially the same as this recent MSE question
  $endgroup$
  – Daniel Lichtblau
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

I have a linear problem that I want to solve but the method is quite different from normal, the problem is still Ax=b. However,
in this instant I have A being unknown, apart from the fact that each entry in A can only be zero or 1. Further I have x being known and fixed, the same holds for b.

My question, is it a easy method for getting mathematica to spit out all possible A such that Ax=b given conditions on A.

I tried doing a number of for loops etc, however, I have completely given up after number of hours and the expectation that my effort is incorrect.

probability-or-statistics linear-algebra

share|improve this question

asked yesterday

ALEXANDERALEXANDER

634518

$endgroup$

share|improve this question

asked yesterday

ALEXANDERALEXANDER

634518

share|improve this question

asked yesterday

ALEXANDERALEXANDER

634518

asked yesterday

ALEXANDERALEXANDER

634518

asked yesterday

ALEXANDERALEXANDER

634518

2 Answers
2

active

oldest

votes

6

$begingroup$

Here's a way to code up your problem with Solve automatically:

x = {1, 2, 3};

b = Reverse[x];

sol[x_List, b_List] /; Length[x] === Length[b] := Module[{mat, vars},

   mat = Array[[FormalM], Length[x]*{1, 1}]; (* construct a matrix of variables*)

   vars = Flatten[mat];

   mat /. 

    Solve[And @@ Thread[mat.x == b] (* Construct the equations *) 

      && vars ∈ Integers && And @@ Thread[0 <= vars <= 1] (* constraints *),

     vars

    ]

   ];

matrixSolutions = sol[x, b]

{{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, {{1, 1, 0}, {0, 1, 0}, {1, 0, 0}}}

As you can see, 2 solutions were found. Check that the residuals of the solutions are zero vectors:

#.x - b & /@ matrixSolutions

{{0, 0, 0}, {0, 0, 0}}

It works, but I don't know how scalable this approach is. Instead of Solve, you can also use NSolve and/or FindInstance (you can just replace Solve with any of those two functions in the code above).

share|improve this answer

edited yesterday

MarcoB

36.9k556113

answered yesterday

Sjoerd SmitSjoerd Smit

3,850715

$endgroup$

add a comment | 

1

$begingroup$

Pretty much the same questions as

FrobeniusSolve with solutions only being 0 or 1 being acceptable

Using @Sjoerd's problem

x = {1, 2, 3};

b = Reverse[x];

Can write solution as

res = Tuples@(Select[FrobeniusSolve[x, #], Max@# <= 1 &] & /@ b)



(* {{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, 

   {{1, 1, 0}, {0, 1, 0}, {1, 0, 0}}}  *)

share|improve this answer

edited yesterday

answered yesterday

MikeYMikeY

3,258614

$endgroup$

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: "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/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
});


}
});

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%2f192643%2fall-possible-a-of-ax-b-with-constraints-on-a%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

Here's a way to code up your problem with Solve automatically:

x = {1, 2, 3};

b = Reverse[x];

sol[x_List, b_List] /; Length[x] === Length[b] := Module[{mat, vars},

   mat = Array[[FormalM], Length[x]*{1, 1}]; (* construct a matrix of variables*)

   vars = Flatten[mat];

   mat /. 

    Solve[And @@ Thread[mat.x == b] (* Construct the equations *) 

      && vars ∈ Integers && And @@ Thread[0 <= vars <= 1] (* constraints *),

     vars

    ]

   ];

matrixSolutions = sol[x, b]

{{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, {{1, 1, 0}, {0, 1, 0}, {1, 0, 0}}}

As you can see, 2 solutions were found. Check that the residuals of the solutions are zero vectors:

#.x - b & /@ matrixSolutions

{{0, 0, 0}, {0, 0, 0}}

It works, but I don't know how scalable this approach is. Instead of Solve, you can also use NSolve and/or FindInstance (you can just replace Solve with any of those two functions in the code above).

share|improve this answer

edited yesterday

MarcoB

36.9k556113

answered yesterday

Sjoerd SmitSjoerd Smit

3,850715

$endgroup$

add a comment | 

6

$begingroup$

Here's a way to code up your problem with Solve automatically:

x = {1, 2, 3};

b = Reverse[x];

sol[x_List, b_List] /; Length[x] === Length[b] := Module[{mat, vars},

   mat = Array[[FormalM], Length[x]*{1, 1}]; (* construct a matrix of variables*)

   vars = Flatten[mat];

   mat /. 

    Solve[And @@ Thread[mat.x == b] (* Construct the equations *) 

      && vars ∈ Integers && And @@ Thread[0 <= vars <= 1] (* constraints *),

     vars

    ]

   ];

matrixSolutions = sol[x, b]

{{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, {{1, 1, 0}, {0, 1, 0}, {1, 0, 0}}}

As you can see, 2 solutions were found. Check that the residuals of the solutions are zero vectors:

#.x - b & /@ matrixSolutions

{{0, 0, 0}, {0, 0, 0}}

It works, but I don't know how scalable this approach is. Instead of Solve, you can also use NSolve and/or FindInstance (you can just replace Solve with any of those two functions in the code above).

share|improve this answer

edited yesterday

MarcoB

36.9k556113

answered yesterday

Sjoerd SmitSjoerd Smit

3,850715

$endgroup$

add a comment | 

$begingroup$

Here's a way to code up your problem with Solve automatically:

x = {1, 2, 3};

b = Reverse[x];

sol[x_List, b_List] /; Length[x] === Length[b] := Module[{mat, vars},

   mat = Array[[FormalM], Length[x]*{1, 1}]; (* construct a matrix of variables*)

   vars = Flatten[mat];

   mat /. 

    Solve[And @@ Thread[mat.x == b] (* Construct the equations *) 

      && vars ∈ Integers && And @@ Thread[0 <= vars <= 1] (* constraints *),

     vars

    ]

   ];

matrixSolutions = sol[x, b]

{{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, {{1, 1, 0}, {0, 1, 0}, {1, 0, 0}}}

As you can see, 2 solutions were found. Check that the residuals of the solutions are zero vectors:

#.x - b & /@ matrixSolutions

{{0, 0, 0}, {0, 0, 0}}

It works, but I don't know how scalable this approach is. Instead of Solve, you can also use NSolve and/or FindInstance (you can just replace Solve with any of those two functions in the code above).

share|improve this answer

edited yesterday

MarcoB

36.9k556113

answered yesterday

Sjoerd SmitSjoerd Smit

3,850715

$endgroup$

x = {1, 2, 3};

b = Reverse[x];

sol[x_List, b_List] /; Length[x] === Length[b] := Module[{mat, vars},

   mat = Array[[FormalM], Length[x]*{1, 1}]; (* construct a matrix of variables*)

   vars = Flatten[mat];

   mat /. 

    Solve[And @@ Thread[mat.x == b] (* Construct the equations *) 

      && vars ∈ Integers && And @@ Thread[0 <= vars <= 1] (* constraints *),

     vars

    ]

   ];

matrixSolutions = sol[x, b]

#.x - b & /@ matrixSolutions

share|improve this answer

edited yesterday

MarcoB

36.9k556113

answered yesterday

Sjoerd SmitSjoerd Smit

3,850715

edited yesterday

MarcoB

36.9k556113

edited yesterday

MarcoB

36.9k556113

answered yesterday

Sjoerd SmitSjoerd Smit

3,850715

answered yesterday

Sjoerd SmitSjoerd Smit

3,850715

1

$begingroup$

Pretty much the same questions as

FrobeniusSolve with solutions only being 0 or 1 being acceptable

Using @Sjoerd's problem

x = {1, 2, 3};

b = Reverse[x];

Can write solution as

res = Tuples@(Select[FrobeniusSolve[x, #], Max@# <= 1 &] & /@ b)



(* {{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, 

   {{1, 1, 0}, {0, 1, 0}, {1, 0, 0}}}  *)

share|improve this answer

edited yesterday

answered yesterday

MikeYMikeY

3,258614

$endgroup$

add a comment | 

1

$begingroup$

Pretty much the same questions as

FrobeniusSolve with solutions only being 0 or 1 being acceptable

Using @Sjoerd's problem

x = {1, 2, 3};

b = Reverse[x];

Can write solution as

res = Tuples@(Select[FrobeniusSolve[x, #], Max@# <= 1 &] & /@ b)



(* {{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, 

   {{1, 1, 0}, {0, 1, 0}, {1, 0, 0}}}  *)

share|improve this answer

edited yesterday

answered yesterday

MikeYMikeY

3,258614

$endgroup$

add a comment | 

$begingroup$

Pretty much the same questions as

FrobeniusSolve with solutions only being 0 or 1 being acceptable

Using @Sjoerd's problem

x = {1, 2, 3};

b = Reverse[x];

Can write solution as

res = Tuples@(Select[FrobeniusSolve[x, #], Max@# <= 1 &] & /@ b)



(* {{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, 

   {{1, 1, 0}, {0, 1, 0}, {1, 0, 0}}}  *)

share|improve this answer

edited yesterday

answered yesterday

MikeYMikeY

3,258614

$endgroup$

x = {1, 2, 3};

b = Reverse[x];

res = Tuples@(Select[FrobeniusSolve[x, #], Max@# <= 1 &] & /@ b)



(* {{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}}, 

   {{1, 1, 0}, {0, 1, 0}, {1, 0, 0}}}  *)

share|improve this answer

edited yesterday

answered yesterday

MikeYMikeY

3,258614

answered yesterday

MikeYMikeY

3,258614

answered yesterday

MikeYMikeY

3,258614