Re: [Algorithms] decompose onto non-orthogonal vectors

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

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.

.