solving equation system

[TG]

Hi there.

Anyone knows how to use numpy / scipy in order to solve this ?

* A is an array of shape (n,)
* X is a positive float number
* B is an array of shape (n,)
* O is an array of shape (n,) containing only zeros.

A.X - B = O
min(X)

thanks.

 

[Ben C]

Hi there.

Anyone knows how to use numpy / scipy in order to solve this ?

* A is an array of shape (n,)
* X is a positive float number
* B is an array of shape (n,)
* O is an array of shape (n,) containing only zeros.

A.X - B = O
min(X)

Are we solving for A, B or X? And what do you mean by min(X)?

If we're solving for X there will be many combinations of A and B for
which there is no solution.

 

[Cameron Laird]

Hi there.

Anyone knows how to use numpy / scipy in order to solve this ?

* A is an array of shape (n,)
* X is a positive float number
* B is an array of shape (n,)
* O is an array of shape (n,) containing only zeros.

A.X - B = O
min(X)

thanks.

In general, no X solves the equality, so it's easy to minimize X.

If A is a non-zero positive multiple of B, exactly one X solves
the equality, so it's again easy to minimize.

If A and B vanish, choose X as the smallest positive float.

I suspect you intended a slightly different problem. It's hard
to guess what it is.

 

[TG]

Are we solving for A, B or X? And what do you mean by min(X)?

If we're solving for X there will be many combinations of A and B for
which there is no solution.

Sorry for the poor explanation. I'm trying to put it clear now.

i've got A and B. I'm looking for X. I made a mistake in my equation
:-/

It's more like :

A.X - B >= O

Well, maybe it will be much more simple if I explain the underlying
problem :

I have an array of N dimensions (generally 2).
- A first calculation gives me a set of integer coordinates inside this
array, which I will call the point W.
- After several other calculations, I've got a set of coordinates in
this N dimensional space that are floating values, and not bound to the
limits of my original N-array. This is the point L.

What I want to do is to translate the point L along the vector LW in
order to get a point L' which coordinates are inside the original
N-dimensional array. Then it will be easy to get the closest integer
coordinates from L'.


I'm not sure this is clear ... pretty hard to talk about maths in
english.

 

[faulal]

Hi there.

Anyone knows how to use numpy / scipy in order to solve this ?

* A is an array of shape (n,)
* X is a positive float number
* B is an array of shape (n,)
* O is an array of shape (n,) containing only zeros.

A.X - B = O
min(X)

thanks.

 

[faulal]

Hi there.

Anyone knows how to use numpy / scipy in order to solve this ?

* A is an array of shape (n,)
* X is a positive float number
* B is an array of shape (n,)
* O is an array of shape (n,) containing only zeros.

A.X - B = O
min(X)

thanks.

 

[faulal]

Hi there.

Anyone knows how to use numpy / scipy in order to solve this ?

* A is an array of shape (n,)
* X is a positive float number
* B is an array of shape (n,)
* O is an array of shape (n,) containing only zeros.

A.X - B = O
min(X)

thanks.

 

[Ben C]

Sorry for the poor explanation. I'm trying to put it clear now.

i've got A and B. I'm looking for X. I made a mistake in my equation
:-/

It's more like :

A.X - B >= O

How about this:

from random import *

def solve(A, B):
return reduce(max, (float(b) / a for a, b in zip(A, B)))

def test():
A = [random() for i in range(4)]
B = [random() for i in range(4)]

x = solve(A, B)

for a, b in zip(A, B):
print a, b, a * x - b

test()

This only works if all elements of both A and B are positive.

Well, maybe it will be much more simple if I explain the underlying
problem :

I have an array of N dimensions (generally 2).
- A first calculation gives me a set of integer coordinates inside this
array, which I will call the point W.

Is this an array of points, or an array of values, that contains only
one point?

- After several other calculations, I've got a set of coordinates in
this N dimensional space that are floating values, and not bound to the
limits of my original N-array. This is the point L.

What I want to do is to translate the point L along the vector LW

Do you mean the vector L - W? (LW is a scalar, assuming dot product).

in order to get a point L' which coordinates are inside the original
N-dimensional array. Then it will be easy to get the closest integer
coordinates from L'.

I'm not sure this is clear ... pretty hard to talk about maths in
english.

Not very clear to me I'm afraid!

 

[Carl Banks]

Hi there.

Anyone knows how to use numpy / scipy in order to solve this ?

* A is an array of shape (n,)
* X is a positive float number
* B is an array of shape (n,)
* O is an array of shape (n,) containing only zeros.

A.X - B = O
min(X)

thanks.

Looks like an incorrectly specified degenerate linear least squares
problem. The function numpy.linalg.linear_least_squares might be able
to do what you want.


Carl Banks

 

[Carl Banks]

Sorry for the poor explanation. I'm trying to put it clear now.

i've got A and B. I'm looking for X. I made a mistake in my equation
:-/

It's more like :

A.X - B >= O

Well, maybe it will be much more simple if I explain the underlying
problem :

I have an array of N dimensions (generally 2).
- A first calculation gives me a set of integer coordinates inside this
array, which I will call the point W.
- After several other calculations, I've got a set of coordinates in
this N dimensional space that are floating values, and not bound to the
limits of my original N-array. This is the point L.

What I want to do is to translate the point L along the vector LW in
order to get a point L' which coordinates are inside the original
N-dimensional array. Then it will be easy to get the closest integer
coordinates from L'.

I see. You have a simple linear programming problem. These can be
tricky in general. Because you only have one variable, it's probably
ok to use brute force. Try this:

given X, A, and B:

E = A*X-B
C = numpy.where(E<0,B/A,X)
X = min(C)

This assumes that you've designed the problem such that B/A would be
less than X where A*X-B<0 (if the opposite were true then of course
you'd need max(C)).


Carl Banks