1 / 22

Introduction to CUDA Programming

Introduction to CUDA Programming. Scan Algorithm Explained Andreas Moshovos Winter 2009. Reading. You are strongly encouraged to read the following as it a contains a more formal treatment of the algorithm, plus an overview of various applications of scan.

tayte
Download Presentation

Introduction to CUDA Programming

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Introduction to CUDA Programming Scan Algorithm Explained Andreas Moshovos Winter 2009

  2. Reading • You are strongly encouraged to read the following as it a contains a more formal treatment of the algorithm, plus an overview of various applications of scan. • Guy E. Blelloch. “Prefix Sums and Their Applications”. In John H. Reif (Ed.), Synthesis of Parallel Algorithms, Morgan Kaufmann, 1990. http://www.cs.cmu.edu/afs/cs.cmu.edu/project/scandal/public/papers/CMU-CS-90-190.html

  3. Two phases • Up-Sweep • Essentially a reduction • Produces many partial results • Down-Sweep • Propagating the partial results to all relevant elements

  4. Up-Sweep • Just a reduction: 1 2 2 5 6 3 8 2 4 1 5 2 7 9 3 5 1 3 2 7 6 9 8 10 4 5 5 7 7 16 3 8 1 3 2 10 6 9 8 19 4 5 5 12 7 16 3 24 1 3 2 10 6 9 8 29 4 5 5 12 7 16 3 36 1 3 2 10 6 9 8 29 4 5 5 12 7 16 3 65

  5. Up-Sweep • Now let’s see this is a tree 1 2 2 5 6 3 8 2 4 1 5 2 7 9 3 5 3 7 9 10 5 7 16 8 10 19 12 24 29 36 • Notice we only have these nodes left in our array: • the rest were partial results 65 1 3 2 10 6 9 8 29 4 5 5 12 7 16 3 65

  6. Up-Sweep • So, this is what’s left • nodes without values don’t exist, they were partial results 1 2 6 8 4 5 7 3 3 9 5 16 10 12 29 65

  7. Down-Sweep • For the second phase we need to think: • The edges in reverse • The empty nodes as placeholders for partial results 1 2 6 8 4 5 7 3 3 9 5 16 10 12 29 65

  8. Down-Sweep • Now let’s view the tree as a collection of nsubtrees • The root of each sub tree, where it’s still present contains the reduction of all subtree elements • i.e., the sum of all subtree elements 1 2 6 8 4 5 7 3 3 9 5 16 10 12 29 65

  9. Down-Sweep • Let’s focus on the rightmost subtree: 1 2 6 8 4 5 7 3 3 9 5 16 10 12 29 65

  10. Down-Sweep • Before the last step of the down-sweep phase the yellow element will contain the sum (57) of all elements to the left of the subtree. 3 57 • The last step will take the following two actions • 3+ 57 = 60, this goes on the rightmost element • This is the sum of all elements including 3 but excluding the right most one • overwrite 3 with 57 • This is the sum of all elements left of 3

  11. Down-Sweep • In terms of the array stored in memory the aforementioned actions look like this: 57 61 3 57 • Where: • the dark arrows represent addition • the red dotted arrow represents a move

  12. Down-Sweep • Let’s now focus at the rightmost subtree that contains the last four nodes: • This will be processed at the step before the previous subtree we just discussed 7 3 16

  13. Down-Sweep • Before the previous to the last step of the down-sweep phase the green element will contain the sum (41) of all elements to the left of the subtree. 7 3 16 41

  14. Down-Sweep • The actions that will be taken at this step are: • 16 + 41 = 57 will be written as the root of the rightmost subtree • As we saw before this is the sum of all element left of the rightmost subtree • 41 will replace 16 • This is the sum of all elements left of the subtree rooted by 16 7 3 41 57 41

  15. Down-Sweep • In terms of the array stored in memory the aforementioned actions look like this: 7 41 3 57 7 16 3 41 • Where: • the dark arrows represent addition • the red dotted arrow represents a move

  16. Down-Sweep • Now let’s go a step back looking at the complete right subtee (in green) 4 5 7 3 5 16 12

  17. Down-Sweep • Before this step the root node will contain the sum (29) of all elements of the left subtree 4 5 7 3 5 16 12 29

  18. Down-Sweep • As before we’ll do two things: • 29+12 = 41 and this becomes the root of the rightmost subtree • This should be the sum of all elements to the left of that subtree for the next step (which we saw previously) • 29 replaces 12 4 5 7 3 same reason: 29 is the sum of all elements left of the subtree rooted by what was 12. 5 16 29 41 29

  19. Down-Sweep • Let’s try to generalize what happens at every step of the down-sweep phase • Let’s look at step 1: • There is only one subtree shown in purple 1 2 6 8 4 5 7 3 3 9 5 16 10 12 29 65

  20. Down-Sweep • Before we process this tree as described before the root node must contain the sum of all elements to the left of the tree • There are no elements • Hence the root must be 0 1 2 6 8 4 5 7 3 3 9 5 16 10 12 29 0

  21. Down-Sweep • Now repeat the steps we saw before • 29 + 0 = 29 and this becomes the root of the right subtree • 29 gets replaced by 0 1 2 6 8 4 5 7 3 3 9 5 16 10 12 0 29 0

  22. Down-Sweep • In terms of the array stored in memory the aforementioned actions look like this: 1 3 2 10 6 9 8 0 4 5 5 12 7 16 3 29 1 3 2 10 6 9 8 29 4 5 5 12 7 16 3 0 • Where: • the dark arrows represent addition • the red dotted arrow represents a move

More Related