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Computation of pi in CUDAPowerPoint Presentation

Computation of pi in CUDA

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Computation of pi in CUDA

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- Measure circumference of the circle by counting pixels on the edge of the circle.
- Compute value of pi using this circumference

Say I have a digital camera and magnification system which gives me pixel size of 1 unit on object plane

The image obtained is binary i.e. ‘1’ inside the circle and ‘0’ elsewhere.

What are my limitations in measurement of local radius of curvature ?

My hypothesis is that if I can measure circumference or compute ‘pi’. I can measure that radius of curvature.

Pixel counting performed in region

0≤ x ≤ R/sqrt(2). In this region,

Not Possible

state

Possible states

‘d’ is the counter which keep tracks

of contour length

pi = 4 x d/R

Change ‘R’ and see the error in pi

The region from 0 ≤ x ≤ R/sqrt(2) is further divided into segments.

The contour length of each segment is computed independently by separate threads.

The sum of these contour lengths gives us 1/8th of the circumference of the circle.

‘pi’ value and error value are computed and outputted at the end of the program.

dist_h = (float*)malloc(fsize);

status_h = (int*)malloc(isize);

start_h = (int*)malloc(isize);

end_h = (int*)malloc(isize);

float R = 10.0*sqrt(2.0);

int Nt = 8;

Variables for inputting Radius

And # of threads required for computation

Variables for tracking contour length,

Errors, start and end ‘x’ value for each thread

/* Kernel */

__global__ void distance(int *start, int *end, float *dist, int *status, float R)

{ int yold = floor(0.5+(sqrt(R*R-(start[threadIdx.x]-1.0)*(start[threadIdx.x]-1.0))));

int d = 0;

int flag = 0;

for (int k=start[threadIdx.x]; k <= end[threadIdx.x]; k++)

{ int ynew = floor(0.5+(sqrt(R*R-k*k)));

if (ynew == yold) { d = d + 1.0; }

else

{ if (ynew < yold) { d = d + 1.41421356; }

else { flag = 1; } }

yold = ynew; }

dist[threadIdx.x] = d;

status[threadIdx.x] = flag;}

Kernel function

The function computes contour length

of a circular segment with radius ‘R’

float sum = 0;

for (int i=0; i<Nt; i++)

{

sum = sum + dist_h[i];

if (status_h[i] == 1) printf("Error has occured",'\n'); }

sum = sum*4.0/R;

float error;

error = sum - 3.141592654;

printf(" value of pi = %2.15f\n ",sum);

printf(" error = %2.15f\n ",error);

Computation of ‘pi’ and the error

Printing of the values obtained

Result (for R = 10xsqrt(2))

value of pi =3.11126995

error = -0.0303227

Curvature change is not picked up because of poorer pixel resolution.

Causes the circumference to be underestimated

Large ‘R’ compared to pixel resolution

Computed pi = 2.8567 for R=100xsqrt(2)

Highly curved boundaries are not captured because of poorer pixel resolution.

Causes the circumference to be overestimated

Small ‘R’ compared to pixel resolution

Computed pi = 3.7712for R=3xsqrt(2)

- Camera pixels record light intensity. If we can predict the intensity distribution close to edge and if that intensity distribution spreads over 3 or more pixels, we can possibly get subpixel resolution.
- If R = f(x) is known, we can use that information to reduce the uncertainty in ‘R’