Timeline for How to implement a more efficient inverse triangular recursion?
Current License: CC BY-SA 3.0
27 events
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| Jun 16, 2020 at 9:23 | history | edited | CommunityBot |
Commonmark migration
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| Jan 17, 2017 at 14:36 | comment | added | xzczd♦ |
@faysou Actually this elegant yet abstruse function was one of the main barrier for my understanding 囧. It took me quite a while to figure out the meaning of P = (P = … today (after using Mathematica for more than 4 years! ).
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| Jan 17, 2017 at 13:50 | comment | added | faysou | The key innovation that allowed the CompileExpand function is the Step function from Mr Wizard. | |
| Jan 17, 2017 at 13:32 | comment | added | faysou | It's quite powerful when you realize what it can do. | |
| Jan 17, 2017 at 7:32 | comment | added | xzczd♦ | @faysou Actually I know (and have already upvoted) this answer for years, but was not skilled enough to understand and use that function :) | |
| Nov 11, 2016 at 9:06 | history | bounty ended | xyz | ||
| Nov 10, 2016 at 8:57 | comment | added | faysou | @xzczd you can use the CompileExpand function from the following answer instead of replacing DownValues. mathematica.stackexchange.com/a/24596/66 | |
| Aug 17, 2015 at 6:27 | comment | added | xyz |
Ok, maybe I need to consider refactor the searchSpan:)
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| Aug 17, 2015 at 6:20 | comment | added | xzczd♦ |
@ShutaoTang If the last element needs special treatment, you can use something like searchSpan2[knots_, u0_] := First@Ordering[Sign[u0 - knots], 1] + If[knots[[-1]] == u0, 1, -1]
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| Aug 17, 2015 at 5:35 | comment | added | xyz | OK, I see. The last element is a special case Thanks! | |
| Aug 17, 2015 at 5:28 | comment | added | xzczd♦ |
@ShutaoTang knots1 doesn't fit the description, right? $u_0 \in [u_i,u_{i+1})$ so $u_i<u_{i+1}$.
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| Aug 17, 2015 at 5:16 | comment | added | xyz |
Maybe a bug. knots1 = {0, 0, 0, 0, 0, 1, 2, 3, 4, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10,10}; then searchSpan2[knots1, 10] searchSpan[{2, knots1}, 10]
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| Aug 17, 2015 at 5:02 | comment | added | xyz |
For the auxiliary fuction searchSpan[], Please see my edit in here
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| Aug 17, 2015 at 4:28 | comment | added | xyz |
Thanks a lot:). I am not familiar with Compile, so your code is very helpful for me.
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| Aug 17, 2015 at 4:22 | vote | accept | xyz | ||
| Aug 17, 2015 at 4:18 | history | edited | xzczd♦ | CC BY-SA 3.0 |
trivial edit
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| Aug 17, 2015 at 4:13 | history | edited | xzczd♦ | CC BY-SA 3.0 |
correct a mistake
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| Aug 17, 2015 at 4:04 | history | edited | xzczd♦ | CC BY-SA 3.0 |
simplify the code a little
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| Aug 17, 2015 at 3:50 | history | edited | xzczd♦ | CC BY-SA 3.0 |
add a faster solution
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| Aug 17, 2015 at 3:37 | history | edited | MarcoB | CC BY-SA 3.0 |
Removed extra decimal point that had probably come from a copy/ paste too many
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| Aug 17, 2015 at 3:30 | comment | added | xzczd♦ | @ShutaoTang Have a look at my edit. | |
| Aug 17, 2015 at 3:29 | history | edited | xzczd♦ | CC BY-SA 3.0 |
add the compiled version of OP's solution
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| Aug 17, 2015 at 1:00 | comment | added | xyz |
In the function NonzeroBasis3, when I replace lst[[j, k + 1]] = (u0 - u[[j]])/(u[[j + k]] - u[[j]]) lst[[j, k]] + (1 - (u0 - u[[j + 1]])/(u[[j + k + 1]] - u[[j + 1]])) lst[[j + 1, k]] with lst[[j, k + 1]] = coeff[j, k] lst[[j, k]] + (1 - coeff[j + 1, k] ) lst[[j + 1, k]], here coeff = (u0 - u[[#1]])/(u[[#1 + #2]] - u[[#1]]) & in the Module enviroment, however, it failed. Could you tell me why?THX:)
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| Aug 17, 2015 at 0:55 | history | edited | xyz | CC BY-SA 3.0 |
added 82 characters in body
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| Aug 17, 2015 at 0:44 | history | edited | xyz | CC BY-SA 3.0 |
added 21 characters in body
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| Aug 16, 2015 at 14:59 | history | edited | xzczd♦ | CC BY-SA 3.0 |
Improve the code a little
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| Aug 16, 2015 at 14:43 | history | answered | xzczd♦ | CC BY-SA 3.0 |