Week 8 Homework -- Dec 6, 2011
					 see images: 08week_HW_Q*.jpg

HW8Q1 State Space Question
 what is the dimension of the state space of a dynamic system
 using a soccer ball which is rotationally invarient
  -- he really doesn't care about the absolute orientation of the ball
   but it might still have spin.

  kinematic state of ball
		(in a two dimensional ground plane)     on the ground:  2
		  -- just like car on ground but without heading --

		(might, or not, add another dimension)     in mid air:  3
		  -- still just positon, without heading --

  dynamic state of ball (ignoring spin, just velocities,
        as a person far away would observe them)
											    on the ground:  4
		  -- just like car moving on ground,
		      now has velocity, heading, and yaw = 5
 *****>>>     OR could just have x,y velocities??? = 4 --

											       in mid air:  6
		  -- 3 positions and a velocity in each direction --

  dynamic state of ball including spin             in mid air: 9 !!!NOT=12!!!
   (tricky question...not surprised if we fail)
   		  -- now the same as helicopter,
		      3 positons, 3 axis rotations,
			  and velocity for each position and rotation --
	( but not the same as heli because we have no absolute orientation
	  for each spin direction, just velocities
	   so 3 xyz, 3 dxdydz, and 3 d-spin... I guess you could get this
	   from the same logic as reducing 2d kinematic from 3 to 2 because
	   there is no heading...)


HW8Q2 Dynamic Programming Question 1
  a robot with three cooridinates x,y,theta
  target location is in the top right corner, facing East
  initial location is in the bottom left corner, facing North
  goal is worth 100
  going straight costs -1
  can turn on the spot at cost of -5

  what is the value of the start state?:  88

  7 moves @ -1  = -7
  1 turn  @ -5  = -5
         total  = -12  + 100 = 88


HW8Q3 Dynamic Programming Question 2
  in same system as Q2 except
   turn clockwise costs -10
   and counter-clockwise costs 0

  what is the value of the start state?: 93

 go along the bottom by turning ccw 6 times
 or along the top and turn CCW 3 times...
  7 moves @ -1  = -7
  3 turn  @  0  =  0
         total  = -7  + 100 = 93


HW8Q4 Particle Question 1
  particle filter question

  a single particle in checkerboard space as shown, facing East
  with x, y, and theta (heading) cooridinates

  draw 5 new particles from this particle for the motion of going right
   as shown with red dots

  each new particle will obtain an importance weight
   as now the robot measured a black square -- Measurement = Black
   where the prob of seeing Black on a black square -- P(B | blackSq) = 0.8
   and the prob of seeing Black on a white square -- P(B | whiteSq) = 0.1

   what are the total importance weights for particles in each square
    (add the importance weight of particles in same square)

	a2 = .0526
			P(B|w) = .1

	b2 = .8421
			P(B|b) = .8
			P(B|b) = .8
	
	c2 = .0526
			P(B|w) = .1

	b3 = .0526
			P(B|w) = .1

		    total  = 1.9


HW8Q5 Particle Question 2
  using the same layout as before (but somehow 4x4 now instead of 3x3...)
  in resampling on the next motion step lets assume the three (black)
   particles are used but ignoring the others
   the used ones are: two particles in b2 and one in c2

   they move right and get 5 new (red) particles

   using the same probabilities as before
   we measure a White square

   what are the cumulative importance weights for the new squares:

   	b3 = .5806
		(two particles in white square)
									P(W|w) =  .9
									P(W|w) =  .9

	b4 = .1290
		(two particles in black square)
									P(W|b) =  .2
									P(W|b) =  .2

	c4 = .2903
		(one particle in white square)
									P(W|w) =  .9
								total      = 3.1


HW8Q6 Stanley Question
 our robot Stanley performed as follows in the DARPA Challenge in 2005...

 first -- YAY!!!
 second
 third
 forth or below


HW8Q7 Motion Model Question 
 approximate motion model for this robot
 use the equations given even though you might be suspicious of it's accuracy
 where h = theta, w = omega

 	x' = x + v*dt*cos(h)
 	y' = y + v*dt*sin(h)
	h' = h + w*dt

 at time t=0 the corridinates are x=0, y=0, h=0
  dt = 4,  v = 10,  w = pi/8 (or 22.5 degrees)

 run the update every 4 timesteps,
  what will the postion be after 4 updates, or 16 total time steps
  (...schip notes that the angle changes on each update
  (    so can't just do all 16 in one swoop...)

   x = 0
   y = 0
   h = 0