Numeric Integration with OpenMP Reduction Project

CS 475/575 Project #21 of 5
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CS 475/575 — Spring Quarter 2022
Project #2
Numeric Integration with OpenMP Reduction
100 Points
Due: April 26
This page was last updated: March 24, 2022
Superquadric Surface with N = 2.5 and R = 1.2
Introduction
We all know that the equation of a sphere centered at the origin is x 2 + y2 + z2 =
R2 But, there is something called a superquadric whose equation is xN + yN + zN =
RN In this exercise, we will use parallel reduction to compute the volume of a
superquadric using
R=1.2. and N=2.5.
Using some number of subdivisions in both X and Y, NUMNODES x NUMNODES,
take NUMNODES2 height samples.
We will think of each height sample as a pin extending vertically from the middle of
a 2D tile. That really makes that height sample act as a volume where the tile is
extruded vertically from the bottom to the top.
Think about a pin rising at each dot from the floor to the top of the superquadric
surface. This pin’s contribution to the overall volume is its height times the area of
the tile underneath it.
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Note: tiles along the edges will have only 1/2 the area of tiles in the middle.
Note: tiles in the corners will have only 1/4 the area of tiles in the middle. The
volume contribution of each extruded height tile needs to be weighted accordingly.
The logic of this is for you to figure out.
Requirements
• Compute the total volume by using the given Height function. (Note that this will
give you Z=0. to Z=Height. As the superquadric has a negative half in addition to a
positive half, double the 0.-Height area as your final answer.
• Use a variety of number of subdivisions (NUMNODES). Pick at least 8 different
ones.
• Use a variety of number of threads (NUMT). You must use at least 1, 2, and 4.
• Record the data in units of something that gets larger as speed increases. Joe
Parallel used “MegaHeights Computed Per Second”, but you can use anything that
makes sense.
• From the speed-up that you are seeing, use the “Inverse Amdahl’s Law” to
determine the Parallel Fraction for this application.
• From the Parallel Fraction, determine what maximum speed-up you could ever
get, regardless of how many cores/threads you throw at this, even with a million
cores.
• Your commentary write-up (turned in as a PDF file) should include:
1. Tell what machine you ran this on
2. What do you think the actual volume is?
3. Show the performances you achieved in tables and two graphs showing:
1. Performance as a function of NUMNODES with colored lines showing
different NUMT values
2. Performance as a function of NUMT with colored lines showing different
NUMNODES values
(See the example in the Project Notes.)
4. What patterns are you seeing in the speeds?
5. Why do you think it is behaving this way?
6. What is the Parallel Fraction for this application, using the Inverse Amdahl
equation?
7. Given that Parallel Fraction, what is the maximum speed-up you could ever
get?
The Program Flow
The code below is printed in the handout to make it easy to look at and discuss. You
can get this code in the file proj02.cpp.
In this code sample, NUMNODES is the number of nodes, or dots, subdividing the floor
area. So, for example, NUMNODES=4 means that there are 4 dots on each side edge.
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Each node has some amount of tile space surrounding it:
You could use a single for-loop over all the nodes that looks like this:
#pragma omp parallel for default(none) . . .
for( int i = 0; i < NUMNODES*NUMNODES; i++ ) { int iu = i % NUMNODES; int iv = i / NUMNODES; float z = Height( iu, iv ); . . . } . . . Or, you could also use the collapse OpenMP clause, where the (2) signifies how many nested for-loops to collapse into one for-loop: #pragma omp parallel for collapse(2) default(none) . . . for( int iv = 0; iv < NUMNODES; iv++ ) { for( int iu = 0; iu < NUMNODES; iu++ ) { float z = Height( iu, iv ); . . . } } Note! The above pragma lines are not complete. You need to add more to them! The code to evaluate the height at a given iu and iv is: 4/13/22, 18:08 CS 475/575 Project #2 4 of 5 https://web.engr.oregonstate.edu/~mjb/cs575/Projects/... const float N = 2.5f; const float R = 1.2f; . . . float Height( int iu, int iv ) { float x = -1. + float y = -1. + // iu,iv = 0 .. NUMNODES-1 2.*(float)iu /(float)(NUMNODES-1); 2.*(float)iv /(float)(NUMNODES-1); // -1. to +1. // -1. to +1. float float float float if( r xn = pow( fabs(x), (double)N ); yn = pow( fabs(y), (double)N ); rn = pow( fabs(R), (double)N ); r = rn - xn - yn;