Chapter 4, Geometric objects and transformations

Geometric ADT's

• Vectors
  • A vector is a position independant concept.
  • It consists of a direction and a magnitude
  • Given two points, P₁ and P₂ we can express the vector they determine by

    v = P2-P1
    	    

• Lines
  • A line is specified by a point P and a vector v.
  • And can be expressed by a point P₀ and a vector d.

    P(α) = P0+αd
          

  • This is a parametric form of the equation.
  • α varies from 0 to 1 (to draw the line segment)
• Dot product of two vectors.
  • if u and v are vectors. u·v is the dot product
    • u·v = u_xv_x + u_yv_y + u_zv_z
    • u·v = |u||v| cos(Θ), where Θ is the angle between the two vectors.
    • |u|² = u·u = u_x²+ u_y²+ u_z²
    • 
  • if u·v = 0, then cos(Θ)=0 and &Theta=nπ +π/2;
  • So the angle between two vectors can be calculated
  • cos(Θ) = u·v/|u||v|
• The cross product of two vectors.
  • if u and v are vectors, u×v is the cross product.
    • u×v = < u_yv_z - u_zv_y, u_zv_x - u_xv_z, u_xv_y - u_yv_x>
    • u×v = n|a||b|sin(Θ), where n is a unit vector orthognal to both u and v
  • We sometimes want to construct three mutually orthognal vectors in space, given two vectors in a plane
    • Given a, b, vectors in a planne.
    • u = a;
    • Compute w = a×b, a normal to both u and b.
    • Compute v = u×w, a normal to both u and w.
    • The three orthognal vectors are then u,v,w
  • Such a system can be either right handed or left handed
    • To tell the type of system, put your hand on the positive x axis, your thumb along the positive z axis, and curl your fingers towards the positive y axis.
    • The hand you use tells you the type of system.
    • 
• Convex Hulls
  • Given a set of points, Q, a convex hull is the smallest convex polygon which contains all of the points in Q either within the polygon or on the edges of the polygon.
  • Given a board, with nails at each point, a convex hull is created by snapping a rubber band around all of the nails.
  • This is an area known as computational geometry.
  • We are going to look at this because
    • It gives us an application for dot and cross products (in 2D)
    • It gives a graphical application to mess with
    • It is good fun!
  • There are a number of algorithms for finding a convex hull given a set of points.
  • We will look at one called the Graham's Scan

    	
     
    Graham-Scan(Q)
    
     Find p0, the point in Q with the minimum y coordinate, 
         in the case of a tie, select the point with the 
         minimal x coordinate.
    
          This is a linear search through the points
    
    
    
         We know that the plane is divided into a
         half-plane along the line x=p0.x
         
    
     Let p-1 be a point to the left of p0 with the same y coordinate.
     At each point pi, i>0, 
         find the angle described by  p-1, p0 and pi.
         If two points (or more) points have the same angle,
         discard the one(s) closest to p0
         
         This is accomplished by the dot product.  
         Values should be between π and 0. 
    
     
     Sort the points in descending order, based on the angle.
    
          Use any sort to do this.  
          The angle is the "key" in the sort
    
     
    
          We will now use a stack (S) to hold the
          verticies of the convex hull
      
     S.push(p0)
      
     S.push(p1)
      
     S.push(p2)
    
      
    
      
     for i <-- 0 to m, the number of points remaining do
      
         let pt be the point on the top of the stack
         let pt-1 be the the point next to top of the stack 
    
    
      
         while pt-1, pt and pi makes a non-left turn
      
                S.pop();
      
         S.push(pi)
      
    
    
    
      
    
    

  • All we need to do now is to determine if we are turning left or right.
    • This is accomplished via the cross product.
    • In this case, we will define the cross product to be
    • u×v = u_xv_y-u_yv_x
    • If the cross product is negatiive, we are turning left.
    • If the cross product is positive, we are turning right.
    • 
  • An example will help

    p0 = (1,2)
    p1 = (2,0)
    p2 = (3,1)
    
    u = p2p0 = <3-1,1-2> = <2,-1>
    v = p1p0 = <2-1,0-2> = <1,-2>
    u×v = (2)(-2) - (-1)(1) =  -4 +1 = -3  => left hand turn