gdalgorithms-list

Re: [Algorithms] C++ inherited constructors

[Matt A. <de...@gm...>]

I need this for aesthetic reasons :)
I've got an abstract CThread class. If a want to create a new thread, I
build my process class and inherit CThread, which creates a new thread by
construction.
Now I want my main-threadloop-class to inherit CThread, too, to provide all
the thread functionality to my main thread. But that thread already exists
 created by compiler ), and shouldn't be created again by CThread.
Of course I can put CThread () and ~CThread () in functions like Create()
and Destroy() and leave the constructor / destructor empty. But it'd be
nicer if it could be done in another way...

----- Original Message -----
From: Favnir
To: gda...@li...
Sent: Saturday, July 15, 2000 1:32 PM
Subject: Re: [Algorithms] C++ inherited constructors

The only way you can do this is to define a (preferably) protected
do-nothing constructor for the base class, and inherit it in the subclass
constructor.

But, why in the world would you need to do this, in the first place?

Are,
F

----- Original Message -----
From: Matt Adams
To: gda...@li...
Sent: Saturday, July 15, 2000 12:22 PM
Subject: [Algorithms] C++ inherited constructors


Hi,

I got a question about the C++ class hierarchy.

class a
{
    a () {...}; // constr a
};

class b : a
{
    b () {...}; // constr b
};

b instance;

When creating an instance of b, constructors for both class a and class b
are called.
Is there a way to suppress the automatic calling of constructor a ?
Or something like 'overloading' the old constructor by a new one ?
I couldn't find anything like that in the compiler docs.

Any help appreciated,
Matt


_______________________________________________
GDAlgorithms-list mailing list
GDA...@li...
http://lists.sourceforge.net/mailman/listinfo/gdalgorithms-list

Re: [Algorithms] C++ inherited constructors

[Gabor F. <xl...@po...>]

i don't know if you can supress it, but why do you want to do it?

object "a" must be constructed before the construction of "b" begins ,
and the only way to construct-initialize "a" is thru an "a"-constructor.

if you have access to the sources of "a" you can create an empty constructor
for "a"....

but once again... why?

Gabor


Gabor Farkas - student / Comenius University ,Slovak Republic
----
If my mad scientist/wizard tells me he has almost perfected my Superweapon
but it still needs more testing, I will wait for him to complete the tests.
No one ever conquered the world using a beta version.
(www.eviloverlord.com)
----

----- Original Message -----
From: "Matt Adams" <de...@gm...>
To: <gda...@li...>
Sent: Saturday, July 15, 2000 12:22 PM
Subject: [Algorithms] C++ inherited constructors


> Hi,
>
> I got a question about the C++ class hierarchy.
>
> class a
> {
>     a () {...}; // constr a
> };
>
> class b : a
> {
>     b () {...}; // constr b
> };
>
> b instance;
>
> When creating an instance of b, constructors for both class a and class b
> are called.
> Is there a way to suppress the automatic calling of constructor a ?
> Or something like 'overloading' the old constructor by a new one ?
> I couldn't find anything like that in the compiler docs.
>
> Any help appreciated,
> Matt
>
>
> _______________________________________________
> GDAlgorithms-list mailing list
> GDA...@li...
> http://lists.sourceforge.net/mailman/listinfo/gdalgorithms-list
>

Re: [Algorithms] C++ inherited constructors

[Michael H. <he...@po...>]

Make class A's default constructor do nothing, and make a second one
that takes a useless argument.  You can do it like this:

class A {
protected:
    A()    {}    // B can call this, but users will get warnings if
they use it
public:
    A(bool x);
};

class B: public A {
public:
    B();        // calls A(), which does nothing
};

mike

----- Original Message -----
From: "Matt Adams" <de...@gm...>
To: <gda...@li...>
Sent: Saturday, July 15, 2000 3:22 AM
Subject: [Algorithms] C++ inherited constructors


> Hi,
>
> I got a question about the C++ class hierarchy.
>
> class a
> {
>     a () {...}; // constr a
> };
>
> class b : a
> {
>     b () {...}; // constr b
> };
>
> b instance;
>
> When creating an instance of b, constructors for both class a and
class b
> are called.
> Is there a way to suppress the automatic calling of constructor a ?
> Or something like 'overloading' the old constructor by a new one ?
> I couldn't find anything like that in the compiler docs.
>
> Any help appreciated,
> Matt
>
>
> _______________________________________________
> GDAlgorithms-list mailing list
> GDA...@li...
> http://lists.sourceforge.net/mailman/listinfo/gdalgorithms-list
>
>

Re: [Algorithms] C++ inherited constructors

[Favnir <fa...@wi...>]

The only way you can do this is to define a (preferably) protected
do-nothing constructor for the base class, and inherit it in the =
subclass
constructor.

But, why in the world would you need to do this, in the first place?

Are,
F

  ----- Original Message -----=20
  From: Matt Adams=20
  To: gda...@li...=20
  Sent: Saturday, July 15, 2000 12:22 PM
  Subject: [Algorithms] C++ inherited constructors


  Hi,

  I got a question about the C++ class hierarchy.

  class a
  {
      a () {...}; // constr a
  };

  class b : a
  {
      b () {...}; // constr b
  };

  b instance;

  When creating an instance of b, constructors for both class a and =
class b
  are called.
  Is there a way to suppress the automatic calling of constructor a ?
  Or something like 'overloading' the old constructor by a new one ?
  I couldn't find anything like that in the compiler docs.

  Any help appreciated,
  Matt


  _______________________________________________
  GDAlgorithms-list mailing list
  GDA...@li...
  http://lists.sourceforge.net/mailman/listinfo/gdalgorithms-list

[Algorithms] C++ inherited constructors

[Matt A. <de...@gm...>]

Hi,

I got a question about the C++ class hierarchy.

class a
{
    a () {...}; // constr a
};

class b : a
{
    b () {...}; // constr b
};

b instance;

When creating an instance of b, constructors for both class a and class b
are called.
Is there a way to suppress the automatic calling of constructor a ?
Or something like 'overloading' the old constructor by a new one ?
I couldn't find anything like that in the compiler docs.

Any help appreciated,
Matt

[Algorithms] Diatribe Re: decompose onto non-orthogonal vectors

[Jonathan B. <jo...@bo...>]

In advance I'll say that this diatribe is not directed straight at Ben;
it's to everyone on this list who isn't eager to learn math.


Ben Discoe wrote:

> I was asking if somebody knew the solution, not for how to do the math,
> which i'm willing to leave to people who enjoy re-deriving solutions to math
> problems.

The problem here is that modern graphics programming is *all about* math.
If you don't know the math, you can't play ball.

You shouldn't think of the question you asked as some isolated problem
that you ought to be able to look up in a book.  If this were the best
way to learn how to do things, then you'd need to look up hundreds of
little special-case problems in order to write a modern 3D program
that does interesting things.  And as you've seen, it's often hard
to find a lucid explanation of even one of these problems, even though
lots of people know how to do them.

To understand graphics publications and implement them, and
especially to do new work, you need to be fluent enough in certain
areas of math (linear algebra being a major one) that you just instinctively
know answers to problems like the one you posed.  Once you have this 
knowledge, you'll find that you look at problems in a different way.  
You'll skip over whole truckloads of questions that would have been really 
perplexing before, because now they're non-issues.  Months of programming
work that would have been misguided can now be eliminated.  The scope of things 
you can do will be vastly expanded.

Trying to do 3D graphics without linear algebra is like this: You want to
be a sculptor and create this cool statue called The Thinker, and you've
got the big block of stone, your hammer and chisel are sitting
there on the table... but unfortunately you cut your hands off last
month, so you're squeezing the chisel between two wrist-stumps and
holding the hammer in your mouth and trying to bang away.

It's not good for your dental work.

    -J.

Re: [Algorithms] decompose onto non-orthogonal vectors

[Scott J. S. <sjs...@um...>]

Okay, let me try and explain this as simply as possible.  You want to find
scalars u,v such that u*a + v*b = p.  Since these are 3-dimensional
vectors, you have 3 equations and 2 unknowns:

u*a1 + v*b1 = p1
u*a2 + v*b2 = p2
u*a3 + v*b3 = p3

where p1,p2,p3 are the coordinates of the point p, likewise for a1,a2,a3
and a, and b1,b2,b3 and b.

How do you solve a system of 3 linear equations with 2 unknowns?  If you
don't know Gaussian elimination, this can be done with some simple
algebraic rules and substituion.

If you know your constant values (a, b, p) now, it's very simple to just
do the necessary substitutions by hand.

On the other hand, if the values are determined at run-time, coding the
algebra isn't as easy, unless you use gaussian elimination, since gaussian
elimination is essentially an algorithm for solving linear equations.  The
problem with "coding" algebra is that there are a lot of special cases.

So I'll present you with one sample, "stupid" solution that will work,
assuming that a1 is non-zero.  Simply vary the approach if a1 is zero, and
choose the starting equation using any element of a that is non-zero (at
least one of them must be non zero, or else you reduce to a much simpler
equation, since the point is along b).

This is a dumb approach, and simply meant for a sample illustration of a
possible, yet ugly way, to solve a system like this.  It also assumes a
solution exists:


Start with equation 1.  Solve for u in terms of v.  This gives:

u = (p1 - v*b1)/a1

plug this value of u into equation 2.

This gives you:
a2* (p1 -v*b1)/a1  + v*b2 = p2

now factor out v:

v[ a2* (p1-v*b1)/a1 + b2 ] = p2

divide p2 by that big thing in brackets, so you have v by itself.

Now you have v expressed entirely in terms of known values, so you know
the value of v.  with this, it is trivial to find the value of u, just
plug back into the first equation.  By plugging the vales of u and v you
have determined into the third equation, you can see if in fact there is a
real solution. If you end up getting something like 0=1 for the third
equation upon substiting your discovered values of u and v, no solution
exists.

Note to Ron: obviously, this isn't rigorous at all, but the goal 
was merely give an impression of how the solution can be obtained.

You'd really want to use gaussian elimination for anything like this,
since all of the special cases (is a1 0? start with a different equation, 
is a1,a2,a3 all 0?  just reduce to one equation, etc.) are neatly
handled by it. In fact, I like to think of gaussian elimination as a
mathematical "encapsulation" for solving systems of linear equations, it
automatically takes care of the details for you.

Try doing a search on the net for guassian elimination, you will
definitely come across some hits.  I believe I learned it my sophomore
year of high school, algebra II class, so it's certainly not anything
remotely unusual.


Hope this helps.



--
Scott Shumaker
sjs...@um...

On Fri, 14 Jul 2000, Discoe, Ben wrote:

> 
> I thought it would be a simple problem, but all usual sources failed to
> answer, so perhaps it will be obvious to someone on this list.
> 
> Given a point p and two unit vectors a and b like this:
> 
>       b
>      /
>    v/----p
>    /    /
>   /    /
>  /    /
> ---------a
>     u
> 
> how to you get scalars (u,v) such that u*a + v*b = p?
> Ie. decompose p onto a and b.
> 
> I promise i consulted an academic, a linear algebra textbook and the WWW
> (all failed) before resorting the list :)
> 
> Thanks,
> Ben
> http://vterrain.org/
> 
> 
> _______________________________________________
> GDAlgorithms-list mailing list
> GDA...@li...
> http://lists.sourceforge.net/mailman/listinfo/gdalgorithms-list
> 

Re: [Algorithms] decompose onto non-orthogonal vectors

[Jim O. <j.o...@in...>]

> > I prefer this answer to mine from a theory standpoint, because it
> > expresses all of the gotchas that I had to list explicitly... code wise,
I
> > wouldn't want to actually make a matrix to solve it (i'm sure you
wouldn't
> > either).
>
> There was once a time when I would have thought this.  But once you get
> ...
> this kind of code in a bug-free manner.

I totally agree. I used to have all sorts of math functions, grabbed from
this tutorial here and that book there and the end result was that I didn't
even understand what was going on inside my own mathematics library - which
makes it *really* hard to find a bug.

Nowadays, I write down which matrices and matrix multiplications I need to
get the job done and then implement it using 'dumb' matrix multiplications.
Once I am sure it works, I feed the matrices and multiplications to this
little tool I have written, which filters out any unnecessary steps in the
multiplications (i.e. B.x.x = A.x.x * I.x.x + A.x.y * I.x.y would simply
become B.x.x = A.x.x as I.x.x = 1 and I.x.y = 0, where I is identity and A
and B are any given matrix).

Works like a charm :-)

Jim Offerman

Innovade
- designing the designer

RE: [Algorithms] decompose onto non-orthogonal vectors

[Discoe, B. <ben...@in...>]

> From: ro...@do... [mailto:ro...@do...]
>
> Chapter 1 in any linear algebra book.

I assure i looked in chapters 1 though 18 of my book ("calculus and analytic
geomtry", thomas/finney) and nothing resemling a formula for decomposing a
vector is anywhere in it.  In general, math textbooks tend to be concerned
with axioms and definitions, not with formulas that actually prove to be
useful.

> That is A SYSTEM OF TWO LINEAR EQUATIONS IN TWO UNKNOWNS u, v,
> which you ought to have learned to solve in intermediate algebra,

I was asking if somebody knew the solution, not for how to do the math,
which i'm willing to leave to people who enjoy re-deriving solutions to math
problems.

> 	u = (p1 b2 - p2 b1)/(a1 b2 - a2 b1)
> 	v = (a1 p2 - a2 p1)/(a1 b2 - a2 b1)

OK, i'll believe that after solving for v and substituting twice (a page or
so of algebra) you get that 2d solution.

> If the denominator (a1 b2 - a2 b1) is zero then it means that a and b
> are linearly dependent (i.e. collinear) and either it has no solution
> (if p is not also collinear with a and b) or infinitely many different
> solutions (if p is collinear with a and b)

The picture i drew showed a nondegenerate case.

> Now suppose that you are given this as a 3D problem, say
> 	a = (a1,  a2,  a3) and b = (b1, b2, b3)

Yes.  I'm sorry i forgot to state that i do have 3d vectors, and p is known
to lie in the plane formed by a and b.

> It may or may not have solutions..
> Cramer's rule does not apply here, but Gaussian elimination
> still does.

Is this your way of saying you don't know the answer?

It would be pretty sad if not a single person on the list knows the answer.
I was kinda hoping *you* would, Ron :)

> No, I will not belabor the list with a review of Gaussian elimination
> algorithms.  You will find one in Chapter 1 of your favorite linear
> algebra book.

The index in my textbook does not list "Gaussian elimination".

Thanks,
Ben
http://vterrain.org/

Re: [Algorithms] decompose onto non-orthogonal vectors

[Jonathan B. <jo...@bo...>]

Will Portnoy wrote:
 
> I prefer this answer to mine from a theory standpoint, because it
> expresses all of the gotchas that I had to list explicitly... code wise, I
> wouldn't want to actually make a matrix to solve it (i'm sure you wouldn't
> either).

There was once a time when I would have thought this.  But once you get
good at understanding transformations, it is *hugely* easier to just say
something like: "Hey, I'll make some matrix, invert it, and transform
something else by that to get my answer."  (I don't mean for this problem,
which requires no inversion, but.)  It is a lot nicer to be able to come
back to code like this later and understand what it's doing, as opposed
to some random math that's going on.  And it is also a lot easier to write
this kind of code in a bug-free manner.

Unless you really *really* care about speed, I'd recommend using a
transformation-style approach if there is one.  And if you do care
about speed, I'd still recommend using this kind of approach, until you
are sure the code works; *then* you optimize and keep the old code
in a comment as a backup.

> One thing I didn't mention, and I haven't proven it to myself
> mathematically yet:  if vectors a and b are linearly independent (which
> they don't seem to be from the original thread author's drawing), then
> there's only one solution for u and v. For example, in the "normal"
> orthonormal basis that we use ((1,0,0)(0,1,0)(0,0,1), there's only one
> linear combination of the basis that will give a point p.

They are linearly independent in the drawing.  Linearly independent does
not mean orthogonal, it just means that the two vectors are not fully
redundant (i.e. one is not a scalar multiple of the other).  However 
it is a basic theorem of linear algebra that there is only one 
solution for p regardless of the vectors so long as they are linearly 
independent.

    -J.

Re: [Algorithms] decompose onto non-orthogonal vectors

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

Will Portnoy  wrote:

>> how to you get scalars (u,v) such that u*a + v*b = p?
>> Ie. decompose p onto a and b.
>
>Regardless of what the answer is, you have one equation (ua + vb = p) and
>two unknowns (u and v).  

Nah, it's one vector equation, so two scalar equations.

>You need to come up with an equation involving u
>and/or v that is linearly independent of the first equation, and then you
>can solve the system of linear equations.
>
>You could make up something like:
>
>u + v = 1, which is the same as v = 1 - u
>
>and then substitute in to the first equation:
>

This part and the remainder is utter nonsense.

Re: [Algorithms] decompose onto non-orthogonal vectors

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

Jonathan Blow  wrote:

>"Discoe, Ben" wrote:
>> 
>> I thought it would be a simple problem, but all usual sources failed to
>> answer, so perhaps it will be obvious to someone on this list.
>> ...

>This is one of those "fundamental linear algebra things"
>that Ron and I were just talking about.
>
>Your original coordinates of 'p' are in the coordinate
>system that we are used to, that is, two basis vectors
>that are orthonormal.  We will call this space R.
>

Actually you don't need them expressed in an orthonormal basis, any
basis will do. (Although the solution is shorter when you have an
orthonormal basis and have learned the concept of "dot product", but
you might not get that until one of the later chapters.

>You want the coordinates of 'p' in a system in which
>the vectors are unit, but not orthogonal.  We will call
>this space S.
>
>Just build a transformation matrix from R to S, and
>call that matrix T.  Then p' = Tp and you are done.
>
>For instructions on how to make a matrix that goes from
>one space to another, given the basis vectors of each 
>space, consult any one of a trillion graphics books.

Again, you don't need matrices (maybe chapter 3) but just high school
algebra.

Re: [Algorithms] decompose onto non-orthogonal vectors

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

Discoe, Ben  wrote:

>
>I thought it would be a simple problem, but all usual sources failed to
>answer, so perhaps it will be obvious to someone on this list.
>
>Given a point p and two unit vectors a and b like this:
>
>      b
>     /
>   v/----p
>   /    /
>  /    /
> /    /
>---------a
>    u
>
>how to you get scalars (u,v) such that u*a + v*b = p?
>Ie. decompose p onto a and b.
>
>I promise i consulted an academic, a linear algebra textbook and the WWW
>(all failed) before resorting the list :)
>

Sheesh, this is a problem that you ought to be able to solve after
doing Chapter 1 in any linear algebra book.  In fact, under the most
natural interpretation of what you have stated, this is a problem in
intermediate (high school) algebra.

First, this is essentially a 2D problem because it has no solution
unless p lies in the plane spanned by a and b. So I suppose  that it
is posed as a 2D problem, then consider what to do if you are dealing
with it posed as a 3 or higher dimensional problem

So if you are "given"  a, b and p, then you must be given their
components with respect to SOME coordinate system (and it doesn't even
have to be an orthonormal coordinate system), So we suppose you are
given
	a = (a1, a2), b=(b1, b2) and p = (p1, p2), 
where a1, a2, b1, b2, p1, p2 are known scalars.  

(If you are not given this info, but just a diagram, then we are
reduced to a high school geometry compass and straight edge solution,
which also is fun, but I'll forbear)

Now when you write out your vector equation ua + vb = p in components
it is actually a system  of two linear equations

	u a1 + v b1 = p1
	u a2 + v b2 = p2

(Note that I refuse to use * for multiplication in algebra, even in
ASCII algebra)

That is A SYSTEM OF TWO LINEAR EQUATIONS IN TWO UNKNOWNS u, v,
which you ought to have learned to solve in intermediate algebra,
probably the simplest of all problems in the subject that we call
"linear algebra", no?

Gaussian elimination, the general means of attacking all linear
systems, works fine for this, but most people would have remembered
the Cramer's rule solution in terms of determinants.

	u = (p1 b2 - p2 b1)/(a1 b2 - a2 b1)
	v = (a1 p2 - a2 p1)/(a1 b2 - a2 b1)

Voilà la solution|  
(Where I use the French  in honor of July 14).

If the denominator (a1 b2 - a2 b1) is zero then it means that a and b
are linearly dependent (i.e. collinear) and either it has no solution
(if p is not also collinear with a and b) or infinitely many different
solutions (if p is collinear with a and b)

Sheesh.  What kind of "academic" did you consult, a cooking teacher?

Now suppose that you are given this as a 3D problem, say
	a = (a1,  a2,  a3) and b = (b1, b2, b3)

Then you have 

	u a1 + v b1 = p1
	u a2 + v b2 = p2
	u a3 + v b3 = p3

a system of THREE linear equations in TWO unknowns.  It may or may not
have solutions, in fact, as stated above, it will have a solution only
if p lies in the plane spanned by a and b, which requires that the
three equations be linearly dependent.   Cramer's rule does not apply
here, but Gaussian elimination still does.  As you apply your favorite
Gaussian elimination algorithm,  one of two things will happen:
either (1) you will get one of the equations into the form 0 = 0 , in
which case it is irrelevant and you are left with a system of two
equations in two unknowns, or (2) you will get one of the equations
into the form 0 = 1, in which case there is no solution, i.e. p does
NOT lie in the plane spanned by a and b

No, I will not belabor the list with a review of Gaussian elimination
algorithms.  You will find one in Chapter 1 of your favorite linear
algebra book. 

Sheesh.  You weren't trolling me were you?

RE: [Algorithms] decompose onto non-orthogonal vectors

[Discoe, B. <ben...@in...>]

Hi Jonathan!

> This is one of those "fundamental linear algebra things"
> that Ron and I were just talking about.

Yeah.  I wonder why it's still so hard to find and express an answer..

> Just build a transformation matrix from R to S

Could you give me a hint how this is done?

> For instructions on how to make a matrix that goes from
> one space to another, given the basis vectors of each 
> space, consult any one of a trillion graphics books.

The answer apparently isn't in the books i have..

> > I promise i consulted an academic
> 
> What kind of academic was that?

A guy from the Stanford graphics lab, just left the phd program to work on
distributed rendering architectures.. he started on about barycentric
coordinates but couldn't give an actual answer.

Thanks,
Ben

Re: [Algorithms] decompose onto non-orthogonal vectors

[Will P. <wi...@cs...>]

I prefer this answer to mine from a theory standpoint, because it
expresses all of the gotchas that I had to list explicitly... code wise, I
wouldn't want to actually make a matrix to solve it (i'm sure you wouldn't
either).

One thing I didn't mention, and I haven't proven it to myself
mathematically yet:  if vectors a and b are linearly independent (which
they don't seem to be from the original thread author's drawing), then
there's only one solution for u and v. For example, in the "normal"
orthonormal basis that we use ((1,0,0)(0,1,0)(0,0,1), there's only one
linear combination of the basis that will give a point p.

i'm assuming the basis vectors span the subspace for those mathematically
inclined people who want to jump on me. :)

Will

----
Will Portnoy
http://www.cs.washington.edu/homes/will

On Fri, 14 Jul 2000, Jonathan Blow wrote:

> "Discoe, Ben" wrote:
> > 
> > I thought it would be a simple problem, but all usual sources failed to
> > answer, so perhaps it will be obvious to someone on this list.
> > 
> > Given a point p and two unit vectors a and b like this:
> > 
> >       b
> >      /
> >    v/----p
> >    /    /
> >   /    /
> >  /    /
> > ---------a
> >     u
> > 
> > how to you get scalars (u,v) such that u*a + v*b = p?
> > Ie. decompose p onto a and b.
> 
> This is one of those "fundamental linear algebra things"
> that Ron and I were just talking about.
> 
> Your original coordinates of 'p' are in the coordinate
> system that we are used to, that is, two basis vectors
> that are orthonormal.  We will call this space R.
> 
> You want the coordinates of 'p' in a system in which
> the vectors are unit, but not orthogonal.  We will call
> this space S.
> 
> Just build a transformation matrix from R to S, and
> call that matrix T.  Then p' = Tp and you are done.
> 
> For instructions on how to make a matrix that goes from
> one space to another, given the basis vectors of each 
> space, consult any one of a trillion graphics books.
> 
>    -J.
> 
> 
> > I promise i consulted an academic
> 
> What kind of academic was that?
> 
> _______________________________________________
> GDAlgorithms-list mailing list
> GDA...@li...
> http://lists.sourceforge.net/mailman/listinfo/gdalgorithms-list
> 

RE: [GDAlgorithms-List] Re: [Algorithms] Message format

[Daniel R. <dan...@ho...>]

>
>Shorter is better for me... [GDA] would be sweet.
>
mee, too

because long subject get always lost:

RE: [GDAlgorithms-List] Re: Meshpr...
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Re: [Algorithms] decompose onto non-orthogonal vectors

[Will P. <wi...@cs...>]

> how to you get scalars (u,v) such that u*a + v*b = p?
> Ie. decompose p onto a and b.

Regardless of what the answer is, you have one equation (ua + vb = p) and
two unknowns (u and v).  You need to come up with an equation involving u
and/or v that is linearly independent of the first equation, and then you
can solve the system of linear equations.

You could make up something like:

u + v = 1, which is the same as v = 1 - u

and then substitute in to the first equation:

ua + vb = p

ua + (1 - u) b = p

ua - ub = p - b

u = (p - b) / (a - b)

and then v = 1 - u... assuming I didn't make any mistakes in this textual
algebra, which is hard for me to write (and I imagine hard to read).

the thing is, this made-up equation (u + v = 1) doesn't guarrantee a
positive u and positive v.  so I hope you weren't looking for that. :)

the other problem is you have to make sure vector a and vector b are
linearly independent, or there is likely to be no solution, no matter what
u, v, and the second equation is.

Will

Re: [Algorithms] decompose onto non-orthogonal vectors

[Jonathan B. <jo...@bo...>]

"Discoe, Ben" wrote:
> 
> I thought it would be a simple problem, but all usual sources failed to
> answer, so perhaps it will be obvious to someone on this list.
> 
> Given a point p and two unit vectors a and b like this:
> 
>       b
>      /
>    v/----p
>    /    /
>   /    /
>  /    /
> ---------a
>     u
> 
> how to you get scalars (u,v) such that u*a + v*b = p?
> Ie. decompose p onto a and b.

This is one of those "fundamental linear algebra things"
that Ron and I were just talking about.

Your original coordinates of 'p' are in the coordinate
system that we are used to, that is, two basis vectors
that are orthonormal.  We will call this space R.

You want the coordinates of 'p' in a system in which
the vectors are unit, but not orthogonal.  We will call
this space S.

Just build a transformation matrix from R to S, and
call that matrix T.  Then p' = Tp and you are done.

For instructions on how to make a matrix that goes from
one space to another, given the basis vectors of each 
space, consult any one of a trillion graphics books.

   -J.


> I promise i consulted an academic

What kind of academic was that?

RE: [Algorithms] decompose onto non-orthogonal vectors

[Ignacio C. <i6...@ho...>]

Discoe, Ben wrote:
> Given a point p and two unit vectors a and b like this:
> 
>       b
>      /
>    v/----p
>    /    /
>   /    /
>  /    /
> ---------a
>     u
> 
> how to you get scalars (u,v) such that u*a + v*b = p?
> Ie. decompose p onto a and b.

what you are trying to do is usually called 'change of basis'

look here for further explanations:
http://www.math.hmc.edu/calculus/tutorials/changebasis/


Ignacio Castano
ca...@cr...

RE: [GDAlgorithms-List] Re: [Algorithms] Message format

[Rob K. <rk...@bu...>]

Shorter is better for me... [GDA] would be sweet.

Rob

RE: [Algorithms] Message format

[Akbar A. <sye...@ea...>]

this might be overdoing this and i don't want to start ... <snip>
:D

peace.
akbar A.


-----Original Message-----
From: gda...@li...
[mailto:gda...@li...]On Behalf Of Tom
Hubina
Sent: Friday, July 14, 2000 6:33 PM
To: gda...@li...
Subject: Re: [Algorithms] Message format


At 08:01 PM 7/14/2000 -0500, you wrote:
>i second that if it's okay with tom.

How's this?

Tom



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[Algorithms] decompose onto non-orthogonal vectors

[Discoe, B. <ben...@in...>]

I thought it would be a simple problem, but all usual sources failed to
answer, so perhaps it will be obvious to someone on this list.

Given a point p and two unit vectors a and b like this:

      b
     /
   v/----p
   /    /
  /    /
 /    /
---------a
    u

how to you get scalars (u,v) such that u*a + v*b = p?
Ie. decompose p onto a and b.

I promise i consulted an academic, a linear algebra textbook and the WWW
(all failed) before resorting the list :)

Thanks,
Ben
http://vterrain.org/

Re: [Algorithms] Message format

[gl <gl...@nt...>]

perfect ;)
--
gl

----- Original Message ----- 
From: "Tom Hubina" <to...@3d...>
To: <gda...@li...>
Sent: Saturday, July 15, 2000 2:33 AM
Subject: Re: [Algorithms] Message format


> At 08:01 PM 7/14/2000 -0500, you wrote:
> >i second that if it's okay with tom.
> 
> How's this?
> 
> Tom
> 
> 
> 
> _______________________________________________
> GDAlgorithms-list mailing list
> GDA...@li...
> http://lists.sourceforge.net/mailman/listinfo/gdalgorithms-list
> 

Re: [Algorithms] Message format

[Tom H. <to...@3d...>]

At 08:01 PM 7/14/2000 -0500, you wrote:
>i second that if it's okay with tom.

How's this?

Tom

RE: [GDAlgorithms-List] Re: [Algorithms] Message format

[Akbar A. <sye...@ea...>]

i second that if it's okay with tom.

peace.
akbar A.

"We want technology for the sake of the story, not for its own sake. When
you look back, say 10 years from now, current technology will seem quaint"
Pixars' Edwin Catmull.


-----Original Message-----
From: gda...@li...
[mailto:gda...@li...]On Behalf Of gl
Sent: Friday, July 14, 2000 7:46 PM
To: gda...@li...
Subject: Re: [GDAlgorithms-List] Re: [Algorithms] Message format



I'm doing this publically Tom to get a consensus (or otherwise).  Can we
change the subject list title to be shorter?  Right now it eats too much
subject real-estate.

I suggest simply [GDAlgos-List], and if poss. drop the GD too.  It might not
seem like much, but it matters to me (opinions?).
--
gl

----- Original Message -----
From: "Tom Hubina" <to...@3d...>
To: "Algorithms List" <alg...@3d...>
Sent: Friday, July 14, 2000 11:22 PM
Subject: [GDAlgorithms-List] Re: [Algorithms] Message format


> At 09:43 AM 7/14/2000 -0700, you wrote:
> >There has been some kind of breakdown of message formatting since moving
the
> >list to the new host.
> >
> >I read the list on two different mail readers from two diffferent sites.
> >One is Outlook Express the other Forte Agent.  Since the changeover of
the
> >list host, the messages have been all but unreadable on Outlook Express,
and
> >they were perfectly OK before.  The header information, author, subject
etc.
> >are all blank where they used to appear, but the entire header
information ,
> >including all which used to be hidden,
> >   is in the body of the message.
> >
> >Also I get lots of  "=20"s at the end of lines and whole lines of  "=20"s
at
> >the end of
> >
> >This makes the whole list practically unusable from the site where I use
> >Outlook Express.
> >
> >Interestingly, there is no change in the way the messages are presented
in
> >Forte Agent.
>
>  From what I can tell, the temp server that was in place had garbled
> headers. The SourceForge stuff seems to be working correctly. If anyone
has
> a problem with a message with the new list name in the title, please let
me
> know at to...@3d...
>
> Tom
>
>
>
>
> _______________________________________________
> GDAlgorithms-list mailing list
> GDA...@li...
> http://lists.sourceforge.net/mailman/listinfo/gdalgorithms-list
>


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