# Find random solutions of a system of inequalities

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Etienne Vaccaro-Grange
on 31 Mar 2021

Commented: Aditya Patil
on 6 Apr 2021

Hi everyone,

I am trying to find solutions of a system of inequalities, represented in matrix format by AX>0, where A is a known nxn matrix and X is an nx1 vector of unknown.

In a simple example where n=3, I would have:

a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}>0

a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}>0

a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}>0

Now, I do not want to find all solutions as there are probably an infinity. I would like to randomly pick a solution that works from the set of all possible solutions, without having to find them all. I wondered whether this could be feasible.

I have read about how to convert this linear program in standard form and the use the simplex algorithm. But as far as I understand, this would only give me a particular solution (especially if I use Matlab function "linprog" for that). Using the function solve did not seem to be satisfying either (although I may be wrong). Instead I would like at best to be able to write:

x(1)=random(truncate(makedist('Normal','mu',0,'sigma',1),ub,db),1,1)

where I can find ub and db from the system of equations, potentially also randomly from the set of possible ub and db.

Is there any way to do that mathematically and on Matlab?

Many thanks in advance for your help!

### Accepted Answer

Aditya Patil
on 5 Apr 2021

As the solutions are infinite, you can get solutions as a system of equations themselves. Then, you can either randomly put parameter values and verify the result, or you solve the system of equations by setting some of the parameters and solving for the other ones.

See the following code for example of randomly setting values.

syms x1 x2 x3;

A = [1 2 3; 4 5 6; 7 8 9];

eq1 = A(1, 1) * x1 + A(1, 2) * x2 + A(1, 3) * x3 > 0;

eq2 = A(1, 1) * x1 + A(1, 2) * x2 + A(1, 3) * x3 > 0;

eq3 = A(1, 1) * x1 + A(1, 2) * x2 + A(1, 3) * x3 > 0;

solutions = solve([eq1, eq2, eq3], [x1 x2 x3], 'ReturnConditions',true);

solutions.parameters

% select random parameters

inputs = randn(length(solutions.parameters), 1)';

% check if these inputs are valid

condWithValues = subs(solutions.conditions, solutions.parameters, inputs);

if isAlways(condWithValues)

subs(solutions.x1, solutions.parameters, inputs)

subs(solutions.x2, solutions.parameters, inputs)

subs(solutions.x3, solutions.parameters, inputs)

end

##### 2 Comments

Aditya Patil
on 6 Apr 2021

### More Answers (1)

Bruno Luong
on 5 Apr 2021

Edited: Bruno Luong
on 5 Apr 2021

For small dimensions, you might use existing tools in FEX to enumerate the vertexes of the polytopes.

If the domain is non bounded, you must bounded so it can give the vertexes that define the bounded domain.

clear

A = -magic(3);

% bounding box limits

lo = -1000;

up = 1000;

p = 1e5; % number of points

[m,n] = size(A);

AA = [-A; eye(n); -eye(n)];

b = [0;0;0];

BB = [b(:); up+zeros(n,1); -lo+zeros(n,1)];

% https://www.mathworks.com/matlabcentral/fileexchange/30892-analyze-n-dimensional-convex-polyhedra?s_tid=srchtitle

V=lcon2vert(AA,BB);

K = convhull(V);

% linear convex of the vertexes

W=-log(rand(p,size(V,1)));

W=W./sum(W,2);

X = W*V;

close all

plot3(X(:,1),X(:,2),X(:,3),'.');

hold on

for i=1:size(K,1)

xyz = V(K(i,:),:);

xyz = xyz([1 2 3; 2 3 1],:);

plot3(xyz(:,1),xyz(:,2),xyz(:,3),'-r');

end

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