Am Dienstag, 8. Mai 2012 10:44:04 UTC+2 schrieb smichr:
>
> I don't know how best to ask this question. If I have a equation with more
> unknowns that knowns and want to know the possible solutions and get
> something like this,
>
> >>> solve(a*x+a+b*y-c-d, a, b)
> [(0, (c + d)/y), (c/(x + 1), d/y), (d/(x + 1), c/y), ((c + d)/(x + 1), 0)]
>
In this case, you have an underdetermined linear system, so you can solve
this problem using linear algebra.
Suppose you want to solve
ax + a + by = z = c + d
for a and b. Written in matrix notation:
>>> A = Matrix([[x+1, y]])
>>> b = Matrix([z])
We want to find x, such that A*x == b.
Using Gaußian elimination, we obtain the reduced row echelon form of (A|b):
>>> A.row_join(b).rref()[0]
⎡ y z ⎤
⎢1 ───── ─────⎥
⎣ x + 1 x + 1⎦
(Which is trivial in this case.)
Now we make the A part quadratic by inserting rows of the form (0...0, -1,
0...0) such that only 1 and -1 are on the diagonal:
>>> _.col_join(Matrix([[0, -1, 0]]))
⎡ y z ⎤
⎢1 ───── ─────⎥
⎢ x + 1 x + 1⎥
⎢ ⎥
⎣0 -1 0 ⎦
This is all we need to obtain the solution space: The last column is a
special solution to the inhomogeneous case (A*x == b). The linear
combinations of columns which contain a -1 we added are all possible
solutions to the homogeneous case (A*x == 0). So our solution is just the
sum of the special inhomogeneous solution and the general homogeneous
solution:
x = alpha * Matrix([y/(x + 1), -1]) + Matrix([z/(x +1), 0])
where alpha is any real number. (In this case our solution is a
1-dimensional vector space, in general it can have a higher dimension of
course.)
Please note that you can obtain the homogeneous solution directly using
A.nullspace(). I don't know whether there is already some function for the
inhomogeneous solution.
> Is there some significance of these 4 solutions? Are these solutions
> analogous to the case of `solve(a + b, a)` giving a = -b? The problem that
> I see is that if you write
>
I get something else:
>>> solve(a*x + a + b*y - c - d, a, b)
⎡⎧ -b⋅y + c + d⎫⎤
⎢⎨a: ────────────⎬⎥
⎣⎩ x + 1 ⎭⎦
It does not really make sense to me to return exactly 4 solutions. However,
you can use two special solutions to get the general one (constructing a
line):
>>> x1 = Matrix([0, z/y])
>>> x2 = Matrix([z/(x + 1), 0])
>>> x2 - x1
⎡ z ⎤
⎢─────⎥
⎢x + 1⎥
⎢ ⎥
⎢ -z ⎥
⎢ ── ⎥
⎣ y ⎦
>>> _ / (z/y)
⎡ y ⎤
⎢─────⎥
⎢x + 1⎥
⎢ ⎥
⎣ -1 ⎦
As you can see, alpha * (x2 - x1) + x2 is exactly the same result as above.
> >>> solve(a*x+a+b*y-4, a, b)
> [(0, 4/y), (4/(x + 1), 0)]
>
> Those are only 2 of an infinite number of solutions since there are an
> infinite number of ways to partition up the 4. If we partition 4 as c + d
> (which then gives a solution that we obtained above) then substituting in
> different values of c and d gives a variety of solutions:
>
> >>> ans=solve(a*x+a+b*y-c-d,a, b)
> >>> [Tuple(*ai).subs({c:1,d:3}) for ai in ans]
> [(0, 4/y), (1/(x + 1), 3/y), (3/(x + 1), 1/y), (4/(x + 1), 0)]
> >>> [Tuple(*ai).subs({c:2,d:2}) for ai in ans]
> [(0, 4/y), (2/(x + 1), 2/y), (2/(x + 1), 2/y), (4/(x + 1), 0)]
> >>> [Tuple(*ai).subs({c:4,d:0}) for ai in ans]
> [(0, 4/y), (4/(x + 1), 0), (0, 4/y), (4/(x + 1), 0)]
>
> (Solving this type of equation arose in doing the
> multinomial_coefficients-without-expansion problem.)
>
If you always get linear systems, it is doable using the algorithm I
described above.
Vinzent
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