Timeline for Relative coordinates -- cumulative sum
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2023 at 9:19 | answer | added | Thomas | timeline score: 2 | |
| Jun 2, 2023 at 22:45 | vote | accept | Nick | ||
| Jun 2, 2023 at 20:23 | answer | added | Nicol Bolas | timeline score: 0 | |
| Jun 1, 2023 at 20:01 | comment | added | Nick | [order of execution][1] seems to be relevant if binary tree only has one (very deep) branch. Seems to indicate lack of any parallelism :-( But what I want to do is available in SVG, so I thought there might be a way native to Vulkan. [1]: computergraphics.stackexchange.com/questions/7838/… | |
| Jun 1, 2023 at 18:51 | comment | added | Nick | By best method, I mean "Is it possible to do all this in a shader/on the GPU"? | |
| Jun 1, 2023 at 18:47 | comment | added | Nick | Nevertheless, it would be useful if the GPU can do this, as I wouldn't need to copy buffer back to CPU, calculate, then send to GPU again. | |
| Jun 1, 2023 at 18:45 | comment | added | Nick |
Thanks @Yves By cumulative sum, I mean buff_new[n] =Sum(m=1..n, buff_old[m]) or buff_new[n] = buff_old[n] buff_new[n-1] which means that calculating buff_new[n] needs to wait for buff_new[n-1] to be available. Hence, difficult to parallelise.
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| Jun 1, 2023 at 12:59 | comment | added | user1703 | Best method in what sense ? | |
| Jun 1, 2023 at 12:58 | comment | added | user1703 | "computing the sum of 1..n is difficult to parallelise": what do you mean ? a prefix sum in parallel is no real problem. Compute the local prefix sums, then globalize and adjust the local sums. | |
| S Jun 1, 2023 at 5:15 | review | First questions | |||
| Jun 1, 2023 at 11:56 | |||||
| S Jun 1, 2023 at 5:15 | history | asked | Nick | CC BY-SA 4.0 |