OpenGL uses the same data type for several different things, not just geometric vectors. In particular, it uses them for colors. The componentwise multiplication is in fact very useful in this case. Code implementations of mathematical concepts are rarely completely equivalent to their theoretical version.In Opengl, the operation is component wise multiply, though it has no sence to be used in manner of geomteric need, it ields a vector space where column and row vectors are the same, implying certain inpropriatness of talking about transposing a matrix, since it transposes by its order with multiply operation (that is impossible in linear algebra becouse of NxM demand in mutipling matricies/vectors, implying also vectors to be nx1x1xm, but that algebra has its definition of operation, that opengl does differ in (it will multiply vectors without their dimension introduced), thus opengl has a vector space of different properties where multiplication is not associative but it is a vector space).
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apatriarca
Member Since 03 Jul 2006Online Last Active Today, 07:28 PM
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 Age 29 years old
 Birthday February 11, 1985

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Posts I've Made
In Topic: Is upper division Linear Algebra a good idea or necessary?
06 October 2014  01:49 AM
In Topic: OIT (Order Independent Transparency)
17 September 2014  10:10 AM
The Intel solution is based on a quite recent and, as far as I know, Intelonly extension. The Weighted Blended OIT is more general, but it is an approximation. There are also other methods (Abuffers for example). I don't think this is a solved problem and in many cases sorting is still probably the way to go. The best method depends on what you are trying to do and the scene you are trying to display.
In Topic: how to represent this number
17 September 2014  06:50 AM
While I agree that the probability of selecting a single number is zero, I think it is possible to make sense of something like "the probability of a natural number to be odd" by using a limit, i.e. P({n in {0, 1, 2, ..} : n is odd}) = lim_{M to infinity} P({n in {0, 1, 2, .., M1} : n is odd}). This is a well defined limit and its value is 1/2. But even if we use a definition like this one, the rest of the sentences in the problem still does not make any sense to me.
When I said the problem was equivalent to the one where we limit ourself to the first two bits I used the definition above. Indeed, if we take M = 2^m, we may represent the numbers from 0 to M1 as strings of m bits. The set we are sampling from is then equal to {0, 1}^m and the set of odd numbers is {1} x {0, 1}^{m1}. Since the remaining (m1) bits can be chosen at random without limitations, this is equivalent to the choice of just the first bit value (the rest does not matter for the probability). For this reason, we have that
P({n in {0, 1, 2, ..} : n is odd}) = lim_{M to infinity} P({n in {0, 1, 2, .., M1} : n is odd}) = lim_{m to infinity} P({n in {0,1, .. 2^(m1)} : n is odd} = lim_{m to infinity} P({n in {0, 1} : n == 1}) = P({n in {0,1} : n == 1}).
In Topic: Formulas, Math, and theories for RPG combat/leveling systems?
17 September 2014  04:09 AM
IMHO If you want low level players/monsters to have a chance against higher level ones, then you should rely less on stats and add more depth and strategy to your combat system. This is in fact the only way a weaker individual can beat a stronger one in real life too. If you have to fight with a boxer, you try to stay at the distance. If you are fighting with someone with a gun, you hide and try to disarm or use surprise or..
In Topic: how to represent this number
17 September 2014  03:32 AM
Each bit has a probability of 1/2 of being on and 1/2 off regardless of the state of the other bits and its position in the binary representation. You problem is thus (if I understand it correctly) equivalent to the following:
Choose a number between 0 and 3. You win if the number is 1 or 3, and you lose if the number is 3. It doesn't really make sense as a problem since the winning event and the losing events are not disjoint. Moreover, what happen in all other cases?