How does matlab get the combinations that meet certain conditions in the matrix

Suppose I have a matrix A = [1 2 3 4 5 6 7 8 9 10]] I want to find out the combination of the following conditions respectively how can I achieve 2A1 = A2 A1 + A2 = A3

[Jan]

What do you call "A1", "A2" and "A3"? Do you mean A(1), A(2), A(3)? Then you have written the solution already...

This sounds like a homework question. Then please post, what you have tried so far and ask a specific question. The forum will not solve your homework.

[peter huang]

Sorry, because I have a huge amount of data during the process, so I can only use this method. I want to achieve my expected effect through a simple example

Answer 1

[Walter Roberson]

A = [1 2 3 4 5 6 7 8 9 10];
[A1, A2, A3] = ndgrid(A);
dA1 = 2*A1;
mask = dA1 == A3 & dA1 == A2.*A1 + A2;
solutions = [A1(mask), A2(mask), A3(mask)]

solutions = 1×3

1 1 2

Yup, that works.

As the size() of A goes up, or as the number of variables involves increases, then the memory requirements can go up a lot for this approach, and you start to need other approaches.

[peter huang]

thanks~

Answer 2

[Bruno Luong]

You don't need MATLAB at all. From

2A1 = A2 A1 + A2,

We divide by A1, and factor A2 on rhs to get

2 = A2*(1+1/A1)

Meaning 2 is integer multiple of (1+1/A1).

But (1+1/A1) > 1 (since A1 > 0), or in other world

(1+1/A1) > 2/2 > 2/3 > 2/4 .... > 2/10

So A2 must be 1 (we just exclude A2 to be 2, 3, 4, ... 10)

and (1+1/A1) must be 2. Therefore A1 = 1, and A2=1, A3=2.

Answer 3

[John D'Errico]

Assuming you want to solve this problem:

2*A1 = A2

A1 + A2 = A3

then you could use the ndgrid solution, as shown by Walter. In fact, that is surely the solution for such a small problem, since there are only 1000 possible combinations.

Of course, on this specific problem, the answer is simple. Replace a2 in the second equation, and we see

A1 + 2*A1 = A3

So we now have two euations:

A2 = 2*A1

A3 = 3*A1

So we can choose any value for A1 that will not cause a problem with A3. Thus we trivially have

A1 = 1:3

A1 = 1×3

1 2 3

A2 = 2*A1

A2 = 1×3

2 4 6

A3 = 3*A1

A3 = 1×3

3 6 9

That is clearly the set of all possible solutions. Is this always a viable approach? If the system were far more complex, of course there could be problems. For example, suppose you have some more complex system? I chose a random set of numbers here.

syms x y z
A = [1     1     3
     4     4     2];
B = [27;36]

B = 2×1

27 36

Now suppose we want to find all solutions of this form

A*[x;y;z] == B

ans = 

where (x,y,z) all come from that set? Clearly, I have no clue as to whether such a solution even exists. We can find if a particular solution exists. For example, intlinprog will do it.

xyz = intlinprog(ones(3,1),[1 2 3],[],[],A,B,ones(3,1),ones(3,1)*10)

LP:                Optimal objective value is 12.600000.                                            


No feasible solution found.

Intlinprog stopped because no integer points satisfy the constraints.


xyz =

     []

Now we know that no solution exists at all, for that problem. If I change B, though,

B2 = [20;30]

B2 = 2×1

20 30

xyz = intlinprog(ones(3,1),[1 2 3],[],[],A,B2,ones(3,1),ones(3,1)*10)

LP:                Optimal objective value is 10.000000.                                            


Optimal solution found.

Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value, options.AbsoluteGapTolerance = 0 (the default value). The intcon variables are integer
within tolerance, options.IntegerTolerance = 1e-05 (the default value).

xyz = 3×1

4.0000 1.0000 5.0000

then we do find a solution. But not all possible solutions.

format rat
[R,basis] = rref([A,B2])

R = 
       1              1              0              5       
       0              0              1              5       

basis = 
       1              3       

which tells us the problem reduces to

z == 5, and

x + y == 5

And now we can trivially find all solutions from the admissable set.