Generate 2000 points from two equally weighted sphericalGaussiansN(0, I), N((3,0, . . . ,0), I) inRp, p= 1,11,21, . . .101 (note, youhave to first flip a coin to decide from which Gaussian to sample), whereIis the identity matrix and the centers of Gaussians are distance 3 apart. Implement 1-NN and 3-NN classifiers. Test the resulting classifier on aseparately generated dataset with 1000 pts. Plot the error rate as a functionofp. Observations?

struct SG

{

   float3 Amplitude;

   float3 Axis;

   float Sharpness;

};

float3 EvaluateSG(in SG sg, in float3 dir)

{

   float cosAngle = dot(dir, sg.Axis);

   return sg.Amplitude * exp(sg.Sharpness * (cosAngle - 1.0f));

}

SG SGProduct(in SG x, in SG y)

{

   float3 um = (x.Sharpness * x.Axis + y.Sharpness * y.Axis) /

               (x.Sharpness + y.Sharpness);

   float umLength = length(um);

   float lm = x.Sharpness + y.Sharpness;

   SG res;

   res.Axis = um * (1.0f / umLength);

   res.Sharpness = lm * umLength;

   res.Amplitude = x.Amplitude * y.Amplitude *

                   exp(lm * (umLength - 1.0f));

   return res;

}

its a straight line parallel to the time axis

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."