Timeline for Computing even huger Fibonacci numbers in Java - follow-up
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 12, 2018 at 14:46 | comment | added | CiaPan | Additionally, as you're 'manually' multiply matrices instead of using some general, three-loop routine or matrix multiplication, you can get rid of a 2×2 matrix and use a 3–items vector instead, which could save some overhead on a second dimension indexing operations. | |
| Jan 12, 2018 at 13:52 | comment | added | CiaPan | By recursion (or mathematical induction) all your matrices are symmetric, so you can save approximately 1/4 of work by not calculating the item at [1][0] and using item [0][1] instead. | |
| Jan 12, 2018 at 13:41 | vote | accept | coderodde | ||
| Jan 12, 2018 at 13:36 | answer | added | Eric Duminil | timeline score: 1 | |
| Jan 12, 2018 at 11:40 | comment | added | coderodde | @EricDuminil I have already rolled the Java port, but be my guest. | |
| Jan 12, 2018 at 11:35 | comment | added | coderodde | @EricDuminil That's a nice result! | |
| Jan 12, 2018 at 11:03 | comment | added | coderodde | @EricDuminil You are wrong. My solution performs \$\Theta(\log N)\$ multiplications of a \$2 \times 2\$ matrices, and the one in the link \$\Theta(N)\$ additions. For instance, my version computes the millionth Fibonacci number in half a second; the one behind the link takes almost 40 seconds. | |
| Jan 12, 2018 at 1:47 | answer | added | greybeard | timeline score: 0 | |
| Apr 13, 2017 at 12:40 | history | edited | CommunityBot | replaced http://codereview.stackexchange.com/ with https://codereview.stackexchange.com/ | |
| Nov 1, 2015 at 7:58 | history | asked | coderodde | CC BY-SA 3.0 |