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# How to solve a special linear equation Ax=b? A is a row vector.

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Benson Gou on 3 Jul 2019
Commented: Benson Gou on 19 Jul 2019
Dear All,
I am trying to solve the following linear equation Ax=b:
[a1 a2 ... an][x1 x2 ... xn]' = b. where b is a complex number, a_i is also complex, and x_i is 1 or 0. This equation is unsolvable because the number of variables is n while the equation is 1. But if we add another condition: x is the sparse solution (number of nonzero entries is known) , there exists a unique solution.
For example, [0.1+0.2i 0.2+0.3i 0.4+0.5i 0.6+7i 0.8+0.9i]*[x1 x2 x3 x4 x5 x6 x7 x8 x9]' = 0.5+0.7i. If we know there are 2 non-zeros in x, then we got a unique solution x1=x3=1, others are zero.
But would someone tell me how to solve this equation using Matlab code?
Thanks a lot.
Benson

#### 1 Comment

Steven Lord on 3 Jul 2019
This sounds somewhat like the knapsack problem or the change-making problem.
How large a value of n do you plan to use?
And no, the solution under the one condition that x is as sparse as possible does not guarantee a unique solution in general unless a must contain unique values. Let a = [1 1] and b = 1. Both x = [1 0] and x = [0 1] are solutions and both have only one nonzero element.
In fact, even uniqueness of values in a plus a sparsity restriction on x does not guarantee a unique solution. Let a = [1 2 3 4] and b = 5. Both [1 0 0 1] and [0 1 1 0] satisfy the problem and both have only two nonzero elements. None of the elements in a is greater than or equal to b, so any solution must have at least two nonzero elements.
So this problem, as described, seems to fall under the second of Cleve's Golden Rules of computation. Are there other requirements on a, b, and/or x you're not telling us that might make the solution unique?

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### Accepted Answer

Bruno Luong on 3 Jul 2019
Edited: Bruno Luong on 3 Jul 2019
If you have optimization toolbox
A=[0.1+0.2i 0.2+0.3i 0.4+0.5i 0.6+7i 0.8+0.9i]
b = 0.5+0.7i;
nz = 2;
AA = [real(A); imag(A)]*10;
bb = [real(b); imag(b)]*10;
AA(end+1,:) = 1;
bb(end+1) = nz;
n = size(AA,2);
fdummy = ones(n,1);
lb = zeros(n,1);
ub = ones(n,1);
x = intlinprog(fdummy,1:n,[],[],AA,bb,lb,ub)
Solution returned is:
x =
1
0
1
0
0
If there are more than 1 solution, they can be obtained by changing fdummy.

#### 8 Comments

Show 5 older comments
Benson Gou on 18 Jul 2019
Dear Bruno,
Thanks a lot for your prompt reply. But would you please show for the following problem how to implement intlinprog using "norm(A*x - b, p) <= noise_threshold"?
A = [0.103+0.201i 0.196+0.294i 0.401+0.499i 0.602+6.993i 0.803+0.892i]
b = 0.5+0.7i;
A is contaminated by noise.
Thanks a lot again.
Benson
Bruno Luong on 18 Jul 2019
I suggest you to try formuate with the case p=Inf, it's not difficult when you write down the equation.
Edit: p=1 more difficult than p=Inf, and not as I wrong stated previously
Benson Gou on 18 Jul 2019
I tried for p=1 case. I relaxed the absolute value of ||A*x-b||_1<=epthon by two inequalities: A*x-b<=epthon and -A*x+b<=epthon. Combined these two inequailties into one inequality and I got A1*x-b1<=epthon1. Then I applied intlinprog on this inequality, but I got nun x. I checked the error message and knew that intlinprog stopped running before reaching a solution. I do not know why. Thanks a lot again for your great help. -Benson

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### More Answers (1)

Bruno Luong on 18 Jul 2019
Edited: Bruno Luong on 18 Jul 2019
Code for p = Inf and noise_threshold = 0.01;
A = [0.103+0.201i 0.196+0.294i 0.401+0.499i 0.602+6.993i 0.803+0.892i]
b = 0.5+0.7i;
noise_threshold = 0.01;
AA = [+real(A);
+imag(A);
-real(A);
-imag(A)];
bb = [+real(b);
+imag(b);
-real(b);
-imag(b)] + noise_threshold;
n = size(AA,2);
fdummy = ones(n,1);
lb = zeros(n,1);
ub = ones(n,1);
x = intlinprog(fdummy,1:n,AA,bb,[],[],lb,ub)
if isempty(x)
fprintf('noise_threshold = %f too small', noise_threshold);
end
% Check the "fit" result
A*x
b

#### 1 Comment

Benson Gou on 19 Jul 2019
Dear Bruno,
Thanks a lot for your great help.
Best regards,
Benson

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