Timeline for Karatsuba algorithm too much recursion
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 23, 2018 at 10:25 | answer | added | Shashank Singh | timeline score: 0 | |
| Jan 17, 2012 at 23:38 | history | edited | greatwolf | CC BY-SA 3.0 |
fixed capitalization & punctuation
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| Aug 14, 2011 at 22:44 | answer | added | Francois G | timeline score: 4 | |
| Aug 14, 2011 at 22:00 | vote | accept | calccrypto | ||
| Aug 14, 2011 at 21:53 | answer | added | eugene_che | timeline score: 4 | |
| Aug 14, 2011 at 20:52 | answer | added | kjo | timeline score: 5 | |
| Aug 14, 2011 at 19:57 | history | edited | templatetypedef |
edited tags
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| Aug 14, 2011 at 19:57 | answer | added | templatetypedef | timeline score: 1 | |
| Aug 14, 2011 at 19:10 | comment | added | Francois G | 1. Shouldn't your parameter m change in some recursive call ? 2. If you work in base b, and if I give you (xb^m) with (x < b), how do you go about returning (xb^(m+1)) ? (x*b^(2m)) ? How costly is that operation ? How costly is your last line ? Remember Karatsuba is a slight improvement on fast multiplication. | |
| Aug 14, 2011 at 19:00 | comment | added | neurino |
@utdemir: if it experiences infinite recursion it means that max(x, y) < b never occur
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| Aug 14, 2011 at 18:58 | comment | added | neurino | Moreover read comments to this question | |
| Aug 14, 2011 at 18:57 | comment | added | utdemir | @neurino, at first line, if statement has return. | |
| Aug 14, 2011 at 18:56 | comment | added | utdemir |
I am not sure, maybe using x0, x1 = divmod(x, bm) would be faster.
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| Aug 14, 2011 at 18:55 | comment | added | neurino | Of course the recursion isn't stopping: where's the condition that makes the recursion to stop? | |
| Aug 14, 2011 at 18:37 | history | edited | calccrypto | CC BY-SA 3.0 |
added 41 characters in body
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| Aug 14, 2011 at 18:37 | history | edited | Oliver Charlesworth |
there is no C++ here
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| Aug 14, 2011 at 18:34 | history | asked | calccrypto | CC BY-SA 3.0 |