lets see. optimization of division for mulitplication is for the vector macros.
instead of:
Normalize: vector/magnitde_of_vector
mag = sqrt(x*x+y*y+z*z);
x/=mag;
y/=mag;
z/=mag;
you could do
mag = 1.0/sqrt(x*x+y*y+z*z);
x*=mag;
y*=mag;
z*=mag;
which is generally faster.
division by a scalar MUST be a valid operation to be done to a vector, otherwise you could not normalize a vector.
also explain how:
vector*0.5;
differs from:
vector/2.0;
if you can prove they are different operation and result in a different answer, then you are correct and prove many mathmaticans wrong.
btw, a vector of ANY valid dimension can be used in any operation. there is no difference to the algorithm of what is done. a scalar is a single value, by defination. not an element of a matrix, not a subset of a vector. a single mathmatical number. 4.5 is a scalar. (2.3, 4.5, 7.8) is not.
you are confusing the STL vector class, with an actual vector. this is the MATH discussion. as such we are using mathmatical definations of things. dont confuse two different things.
also no matter what multiplication of a vector by a scalar value is defined as:
for(i=0; i element*=scalar;
division ironically is:
for(i=0; i element/=scalar;<br><br>LilBudyWizer, please go learn some math before continuing game programming. it will really help you.<br> </i>
math in quake
a person: Please watch it with the cutting remarks. Analyze why you do it: does it really help the person see their mistake? If not, then perhaps you''re doing it to gratify yourself, which doesn''t do anyone any good.
Peace,
ZE.
//email me.//zealouselixir software.//msdn.//n00biez.//
miscellaneous links
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Peace,
ZE.
//email me.//zealouselixir software.//msdn.//n00biez.//
miscellaneous links
[if you have a link proposal, email me.]
There is no need to be insulting here. The definition of a vector is an element of a vector space. Now admittedly I don''t know what restrictions there are on what you call a scalar. It could be the field axioms in which case you are arguably right. The problem is that all of those axioms don''t hold for vector spaces that I''m quite certain are valid. It could be that both of my linear algebra books were wrong to use the example of a vector space composed of vectors whose elements are n x n matrices. The commutative property for multiplication doesn''t hold and the zero matrix is not the only matrix that is singular. They are left commutative and right commutative, but not generally commutative. So perhaps you are right, but I don''t see much of a convincing arguement from you that you are right.
If you had of said no, a scalar must meet these conditions then I would be more convinced. The entire part of about a single number just baffles me. A scalar is a single number, but an element of a vector is what? The logic mystifies me. You are both arguing that it is not a single number and that it is a single number. Which is it? At least don''t contridict yourself. Is an element of a vector a scalar or not? Does a scalar have to be a real number or not? If a scalar must be a real number then what is a complex number? If it is not a scalar then does that mean you can''t have a vector composed of complex numbers? If you can does that mean you can only multiply that vector by real numbers? If you can also multiply be complex numbers and a complex number is not a scalar then what is that called? It can''t be scalar multiplication if it isn''t a scalar. If both a real number and complex number can be scalars then what determines what is a scalar? Does that definition exclude 2 x 2 matrices? Let''s at least be consistant here.
If you had of said no, a scalar must meet these conditions then I would be more convinced. The entire part of about a single number just baffles me. A scalar is a single number, but an element of a vector is what? The logic mystifies me. You are both arguing that it is not a single number and that it is a single number. Which is it? At least don''t contridict yourself. Is an element of a vector a scalar or not? Does a scalar have to be a real number or not? If a scalar must be a real number then what is a complex number? If it is not a scalar then does that mean you can''t have a vector composed of complex numbers? If you can does that mean you can only multiply that vector by real numbers? If you can also multiply be complex numbers and a complex number is not a scalar then what is that called? It can''t be scalar multiplication if it isn''t a scalar. If both a real number and complex number can be scalars then what determines what is a scalar? Does that definition exclude 2 x 2 matrices? Let''s at least be consistant here.
vector spaces exist over fields.
reference:
http://mathworld.wolfram.com/AbstractVectorSpace.html
Remember what got this started was someone tried to multiply by a multiplicative inverse. You harped on division as undefined against a vector. but its clearly defined.
First lets be blunt division is a defined operation: it must be translated to equation to have meaning. notice there is no reference to division in the field axioms. its not there because you need field axioms to define linear equations then having that you can define division. if you referenced division in field axioms you would have serious conceptual issues.
b/a is _defined_ as the unique solution of the equation aX = b. defined. not proven. just defined. where the equilvalence of b*1/a to b/a is easy to show. since we know that a*"inverse of a" = 1 and we say that 1/a must equal "inverse of a by" invoking uniqueness. they cant be different becuase they both solve the same equation.
Moving on again then is how does this relate to the original assertion you said:
" technically dividing a vector by a scalar is undefined"
citing field axioms that define a field. but what about the axioms that define a vector space ?
here:
http://mathworld.wolfram.com/VectorSpace.html
though i must say the wolfram.com definitions lack completeness. The concept is that to be a vector space over a field there must both exist a vector v element of V (of stuff you decide) AND AND!!!! there must exist "a" an element of a field with the following properties:
(STOP!!! intrinsic to understanding this is there has to exist a "a" an element of a field to have a vector! Also notice how the mathematica page defined it with a complete loss of generality skipping to an "a" of the field of reals where thats not required at all. it could be an "a" of field foofield. but moving on)
one requirement to be a vector space as per the above and every intro abstract linear algebra book in the world is that this be true and defined:
(ab)U = a(bU) where U is an element of the potential vector space and a,b are elements of a field space.
AHHH!!!
note that a,b have to abide by the field space axioms because they are in a field space. so "a" can take on the value 1. becuase 1 has to be defined in the field space. leaving 1*bU.
now note that b can take on the value you f^(-1) becuase f^(-1) has to be defined in that same field space where f is arbitrary.
so we have a vector U being multiplied by a mutliplicative inverse being not only DEFINED but REQUIRED for the vector space to exist as a vector space.
since division is defined as multiplication by a multiplicative inverse as above then we can safely say that division of a scalar (field) against a vector of scalars is not only defined but required to exist.
your original statement that it is technically undefined is quite wrong. i could be only slightly be more thorough in showing you why by writing a first chapter on abstract linear algebra.
reference:
http://mathworld.wolfram.com/AbstractVectorSpace.html
Remember what got this started was someone tried to multiply by a multiplicative inverse. You harped on division as undefined against a vector. but its clearly defined.
First lets be blunt division is a defined operation: it must be translated to equation to have meaning. notice there is no reference to division in the field axioms. its not there because you need field axioms to define linear equations then having that you can define division. if you referenced division in field axioms you would have serious conceptual issues.
b/a is _defined_ as the unique solution of the equation aX = b. defined. not proven. just defined. where the equilvalence of b*1/a to b/a is easy to show. since we know that a*"inverse of a" = 1 and we say that 1/a must equal "inverse of a by" invoking uniqueness. they cant be different becuase they both solve the same equation.
Moving on again then is how does this relate to the original assertion you said:
" technically dividing a vector by a scalar is undefined"
citing field axioms that define a field. but what about the axioms that define a vector space ?
here:
http://mathworld.wolfram.com/VectorSpace.html
though i must say the wolfram.com definitions lack completeness. The concept is that to be a vector space over a field there must both exist a vector v element of V (of stuff you decide) AND AND!!!! there must exist "a" an element of a field with the following properties:
(STOP!!! intrinsic to understanding this is there has to exist a "a" an element of a field to have a vector! Also notice how the mathematica page defined it with a complete loss of generality skipping to an "a" of the field of reals where thats not required at all. it could be an "a" of field foofield. but moving on)
one requirement to be a vector space as per the above and every intro abstract linear algebra book in the world is that this be true and defined:
(ab)U = a(bU) where U is an element of the potential vector space and a,b are elements of a field space.
AHHH!!!
note that a,b have to abide by the field space axioms because they are in a field space. so "a" can take on the value 1. becuase 1 has to be defined in the field space. leaving 1*bU.
now note that b can take on the value you f^(-1) becuase f^(-1) has to be defined in that same field space where f is arbitrary.
so we have a vector U being multiplied by a mutliplicative inverse being not only DEFINED but REQUIRED for the vector space to exist as a vector space.
since division is defined as multiplication by a multiplicative inverse as above then we can safely say that division of a scalar (field) against a vector of scalars is not only defined but required to exist.
your original statement that it is technically undefined is quite wrong. i could be only slightly be more thorough in showing you why by writing a first chapter on abstract linear algebra.
and now we come down onto the floor realising that our vector in a 3dgame is simply an array of 3 floats, and to get it to length 1 you have to divide each component through the length of the whole..
result:
float div = 1.f/sqrtf(x*x+y*y+z*z);
x *= div;
y *= div;
z *= div;
or even faster you get some fast rsq function and have no div left:
float div = rsq(x*x+y*y+z*z);
x *= div;
y *= div;
z *= div;
rsq is reverse square root..
rsq x = x^-.5
the optimized funcion for this can be found on flipcode if you search for a thread called "arcus cosinus"
but don''t use macros anymore..
carmack wrote/writes good engines. but his code is not the best designed..
but thats live.. it has to work no mather if ugly or not
"take a look around" - limp bizkit
www.google.com
result:
float div = 1.f/sqrtf(x*x+y*y+z*z);
x *= div;
y *= div;
z *= div;
or even faster you get some fast rsq function and have no div left:
float div = rsq(x*x+y*y+z*z);
x *= div;
y *= div;
z *= div;
rsq is reverse square root..
rsq x = x^-.5
the optimized funcion for this can be found on flipcode if you search for a thread called "arcus cosinus"
but don''t use macros anymore..
carmack wrote/writes good engines. but his code is not the best designed..
but thats live.. it has to work no mather if ugly or not "take a look around" - limp bizkit
www.google.com
Okay, I did not see anyone mention WHY Carmack would choose macros over overloaded inline operators. Of course, the answer is quite simple: He codes in C, not C++. I suppose he could hvae used inline functions (I am not sure if inline was part of ANSI C), but there is nothing really wrong with using macros in C, since you have to use them for constants anyway. But in C++, you should try to avoid using macros.
--TheMuuj
--TheMuuj
Well, that seems to prove one can refute a statement without being insulting and impart some knowledge in the process. I stand corrected. Whatever you use for a scalar must have a multiplicative inverse except for zero. My mistake was thinking that whatever your vector was made out of was your scalar. Looking in my linear algebra book they still used reals when the vector was composed of 2 x 2 matrices. I was correct that you have a choice in what you use as a scalar, but I was wrong that the requirements don''t include the existance of a multiplicative inverse.
I find it somewhat funny that the original supposition was just that he did it as a matter of style rather than performance. I was wrong to say that you may not be able to do anything equivalent to division, but it is a property of the scalar that a multiplicative inverse exists. Seperating that makes it clearer. You will catch more errors at compile time if you do not define the division operator for your vector than if you do.
I find it somewhat funny that the original supposition was just that he did it as a matter of style rather than performance. I was wrong to say that you may not be able to do anything equivalent to division, but it is a property of the scalar that a multiplicative inverse exists. Seperating that makes it clearer. You will catch more errors at compile time if you do not define the division operator for your vector than if you do.
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