Computation of pi in CUDA

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# Computation of pi in CUDA - PowerPoint PPT Presentation

Computation of pi in CUDA. Measure circumference of the circle by counting pixels on the edge of the circle. Compute value of pi using this circumference. Motivation. Say I have a digital camera and magnification system which gives me pixel size of 1 unit on object plane

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Presentation Transcript
Computation of pi in CUDA
• Measure circumference of the circle by counting pixels on the edge of the circle.
• Compute value of pi using this circumference
Motivation

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.

Algorithm

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

Implementation in CUDA

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.

Some important parts of the CUDA code

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;

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 d = 0;

int flag = 0;

{ 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; }

Kernel function

The function computes contour length

of a circular segment with radius ‘R’

Some important parts of the CUDA code

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

Sources of error

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)

Reducing uncertainty in ‘R’ estimation
• 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’