Re: [Algorithms] decompose onto non-orthogonal vectors

[<ro...@do...>]

I  wrote:

>Peter Dimov  wrote:
>
>>> >This works for any number of dimensions. Whenever a solution exists, the
>>(u,
>>> >v) pair determined by this approach will have the property u * a + v * b
>>=
>>> >p. (Easy to prove.)
>>>
>>> Agree that it works for any number of dimensions provided that p does
>>> indeed lie in the plane of a and b.  My point was that it will not
>>> detect the case that p does not lie in that plane, that is, will not
>>> detect the case that no solution exisits.
>>
>>Hmm, it must be my English. :)
>>
>>1. Find the (u, v) pair as above.
>>2. Check whether u * a + v * b = p.
>>3. If yes, this (u, v) pair is the solution.
>>4. If no, no solution exists.
>>
>
>OK, you've stated a more complete algorithm now, and I agree that it
>works.  
>
>In any case, as I stated previously, I think your solution is the best
>and have long used it myself.  The Gaussian elimination approach,
>however, is based more directly on the most basic and fundamental
>concepts of linear algebra, which is why I stressed it..  The dot
>product approach is based on somewhat more advanced ideas.
>

And further, it appears to me that if you don't already know that p is
a linear combination of a and b, then the Gaussian elimination
solution is somewhat shorter than your algorithm above.