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RavNaz

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About RavNaz

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  1. fastest builds I've ever seen have been with unity builds. http://buffered.io/2007/12/10/the-magic-of-unity-builds/ wickedly fast.
  2. any others .. ? c++ would be better rather than java... also +1 for Big O.
  3. Which part? The projection to 2D, the creation of a perpendicular vector, or the cross product? Regardless Wikipedia has the answer... [/quote] the projection maths that luca uses to create the tangent
  4. out of interest, can anyone point me at a math tut or something where this is discussed?
  5. luca-deltodesco, thank you very much, that was the magic bit I couldn't quite recall and couldn't find reference to anywhere!! (I should have been more clear in the original question). thanks again.
  6. its the radical plane formed by the circle made when two spheres intersect.
  7. :-) of course it doesn't. I have a point on the plane. Basically, I'm after getting the points of a circle from this equation. so imagine a circle of known radius and centre is drawn onto that plane, so to speak.
  8. Am having a mental block with some maths I'm hoping someone can help me out with. I have an equation of a plane in the form Ax + By + Cz + D = 0 I know A,B,C and D. I have the centre point of the plane and the distance from the centre to the edge. Is it possible to figure out from this two vectors at right angles to each other going towards the edge? e.g. from centre point to top edge and centre point to right edge?
  9. RavNaz

    two O(n) in a row?

    thanks guys. I think I'm just having trouble visualising how some of this fits together. for example: my understanding is that if you were to have say : for ( int i = 0 ; i < 5 ; i++ ) { for ( int j = 0 ; j < 5 ; i++ ) { for( int k = 0 ; k < 5 ; i++ ) { } } } this is O(n-cubed).... but.. how would it change if you had say: for ( int i = 0 ; i < 5 ; i++ ) { for ( int j = 0 ; j < 5 ; i++ ) { } for( int k = 0 ; k < 5 ; i++ ) { } } Is that then O(n-squared)?
  10. RavNaz

    two O(n) in a row?

    OK. void SomeFunc1( int iNum ) { for ( int i = 0 ; i < (iNum/2) ; i++ ) { } } void main( void ) { int iLength = 5; SomeFunc1( iLength ); // executes in O(n); int iCtr = 0; while( bDone == false ) { if ( iCtr < iLength ) { SomeFunc1(iCtr); } else { SomeFunc1(iCtr); bDone = true; } iCtr++; } } Not sure if this helps... Big O looks at worst case, so in the while..for nest I have above, given then iLength would be fixed, that is O(n-squared) right?
  11. RavNaz

    two O(n) in a row?

    Updated original code.. hopefully that clears it up?
  12. Hi everyone, if I've got something like the following: void SomeFunc1( int iNum ) { for ( int i = 0 ; i < (iNum/2) ; i++ ) { } } void main( void ) { int iLength = 5; SomeFunc1( iLength ); // executes in O(n); int iCtr = 0; while( bDone == false ) { if ( iCtr < iLength ) { SomeFunc1(iCtr); } else if ( something else ) { SomeFunc1(iCtr); bDone = true; } iCtr++; } } using Big O, what would this be? I have what appears to be O(n) with the somefunc1 function, followed by the while loop that I guess effectively means I have a nested loop with the SomeFunc1 being called inside... so overall I would say this algorithm was O(n-squared).... is that right?
  13. I need to describe this in a simple statement.. ish... I can give an example of Polymorphism.. as in class Base { virtual something() {..} }; class Derived : public Base { virtual something() {..} }; Base* a = new Derived; a->something(); // where 'something' called will go through Derived::something() but how do I describe in 1-2lines, so its' easy to understand?
  14. Alright guys, anyone know how to do this?? I'm exporting using max9 and am using the DirectX shader material. Opening the DAE file I can see all the information is in there, I just can't seem to navigate the FCollada classes to find it. any ideas?
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