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Loop interchange is a technique for rearranging a loop nest so that the right stuff is at the center. What the right stuff is depends upon what you are trying to accomplish. In many situations, loop interchange also lets you swap high trip count loops for low trip count loops, so that activity gets pulled into the center of the loop nest. It’s also good for improving memory access patterns.

Loop interchange to move computations to the center

When someone writes a program that represents some kind of real-world model, they often structure the code in terms of the model. This makes perfect sense. The computer is an analysis tool; you aren’t writing the code on the computer’s behalf. However, a model expressed naturally often works on one point in space at a time, which tends to give you insignificant inner loops — at least in terms of the trip count. For performance, you might want to interchange inner and outer loops to pull the activity into the center, where you can then do some unrolling. Let’s illustrate with an example. Here’s a loop where KDIM time-dependent quantities for points in a two-dimensional mesh are being updated:


PARAMETER (IDIM = 1000, JDIM = 1000, KDIM = 3) ...DO I=1,IDIM DO J=1,JDIMDO K=1,KDIM D(K,J,I) = D(K,J,I) + V(K,J,I) * DTENDDO ENDDOENDDO

In practice, KDIM is probably equal to 2 or 3, where J or I , representing the number of points, may be in the thousands. The way it is written, the inner loop has a very low trip count, making it a poor candidate for unrolling.

By interchanging the loops, you update one quantity at a time, across all of the points. For tuning purposes, this moves larger trip counts into the inner loop and allows you to do some strategic unrolling:


DO K=1,KDIM DO J=1,JDIMDO I=1,IDIM D(K,J,I) = D(K,J,I) + V(K,J,I) * DTENDDO ENDDOENDDO

This example is straightforward; it’s easy to see that there are no inter-iteration dependencies. But how can you tell, in general, when two loops can be inter- changed? Interchanging loops might violate some dependency, or worse, only violate it occasionally, meaning you might not catch it when optimizing. Can we interchange the loops below?


DO I=1,N-1 DO J=2,NA(I,J) = A(I+1,J-1) * B(I,J) C(I,J) = B(J,I)ENDDO ENDDO

While it is possible to examine the loops by hand and determine the dependencies, it is much better if the compiler can make the determination. Very few single-processor compilers automatically perform loop interchange. However, the compilers for high-end vector and parallel computers generally interchange loops if there is some benefit and if interchanging the loops won’t alter the program results. When the compiler performs automatic parallel optimization, it prefers to run the outermost loop in parallel to minimize overhead and unroll the innermost loop to make best use of a superscalar or vector processor. For this reason, the compiler needs to have some flexibility in ordering the loops in a loop nest.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, High performance computing. OpenStax CNX. Aug 25, 2010 Download for free at http://cnx.org/content/col11136/1.5
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