All possible A of Ax=b with constraints on AFrobeniusSolve with solutions only being 0 or 1 being...
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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
$begingroup$
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
asked yesterday
ALEXANDERALEXANDER
634518
$endgroup$
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
asked yesterday
ALEXANDERALEXANDER
634518
$endgroup$
$begingroup$
You should be able to useTuples[]+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
asked yesterday
ALEXANDERALEXANDER
634518
$endgroup$
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
probability-or-statistics linear-algebra
asked yesterday
ALEXANDERALEXANDER
634518
asked yesterday
ALEXANDERALEXANDER
634518
asked yesterday
ALEXANDERALEXANDER
634518
asked yesterday
ALEXANDERALEXANDER
634518
asked yesterday
ALEXANDERALEXANDER
634518
634518
$begingroup$
You should be able to useTuples[]+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 |
$begingroup$
You should be able to useTuples[]+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
$begingroup$
You should be able to use
Tuples[] + Partition[] (after perhaps filtering out nonsingular candidates).$endgroup$
– J. M. is computer-less♦
yesterday
$begingroup$
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
$begingroup$
Please post a concrete example. Also note this is essentially the same as this recent MSE question
$endgroup$
– Daniel Lichtblau
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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).
edited yesterday
MarcoB
36.9k556113
answered yesterday
Sjoerd SmitSjoerd Smit
3,820715
$endgroup$
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}}} *)
answered yesterday
MikeYMikeY
3,248614
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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).
edited yesterday
MarcoB
36.9k556113
answered yesterday
Sjoerd SmitSjoerd Smit
3,820715
$endgroup$
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).
edited yesterday
MarcoB
36.9k556113
answered yesterday
Sjoerd SmitSjoerd Smit
3,820715
$endgroup$
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).
edited yesterday
MarcoB
36.9k556113
answered yesterday
Sjoerd SmitSjoerd Smit
3,820715
$endgroup$
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).
edited yesterday
MarcoB
36.9k556113
answered yesterday
Sjoerd SmitSjoerd Smit
3,820715
edited yesterday
MarcoB
36.9k556113
edited yesterday
MarcoB
36.9k556113
edited yesterday
MarcoB
36.9k556113
36.9k556113
answered yesterday
Sjoerd SmitSjoerd Smit
3,820715
answered yesterday
Sjoerd SmitSjoerd Smit
3,820715
answered yesterday
Sjoerd SmitSjoerd Smit
3,820715
3,820715
add a comment |
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}}} *)
answered yesterday
MikeYMikeY
3,248614
$endgroup$
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}}} *)
answered yesterday
MikeYMikeY
3,248614
$endgroup$
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}}} *)
answered yesterday
MikeYMikeY
3,248614
$endgroup$
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}}} *)
answered yesterday
MikeYMikeY
3,248614
edited yesterday
answered yesterday
MikeYMikeY
3,248614
answered yesterday
MikeYMikeY
3,248614
answered yesterday
MikeYMikeY
3,248614
3,248614
add a comment |
add a comment |
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$begingroup$
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