Timeline for Bresenham circle drawing algorithm, compute the distance?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 18, 2020 at 15:41 | vote | accept | Voko | ||
| Apr 18, 2020 at 15:41 | comment | added | Voko | Thank you so much for your detailed answer, now it's clear to me | |
| Apr 18, 2020 at 14:21 | comment | added | wychmaster♦ | So, I updated my answer, after I did some lengthy, insightful calculations. | |
| Apr 18, 2020 at 14:21 | history | edited | wychmaster♦ | CC BY-SA 4.0 | added 3970 characters in body |
| Apr 17, 2020 at 19:42 | comment | added | wychmaster♦ | You are right. I have adjusted my answer for now and will further enhance it tomorrow. Running out of time today. However, even though the algorithm is not always taking the closer pixel, it only picks the wrong one if both distances are really close to each other. So you probably can't tell by just looking at the circle, that the wrong one was choosen. | |
| Apr 17, 2020 at 19:37 | history | edited | wychmaster♦ | CC BY-SA 4.0 | deleted 1046 characters in body |
| Apr 17, 2020 at 17:08 | comment | added | Voko | This explanation would have been so convincing and I had really hoped it was that simple, but if we take the square of the distance, we don't get this beautiful $x_{k+1}^2 + y_k^2 - r^2$ but instead a weird $x_{k+1}^2+y_k^2 - 2r\sqrt{x_{k+1}^2+y_k^2}+r^2$, and so I still can't see the link between this weird thing and the actual $f(N)$ that we use in the algorithm. | |
| Apr 17, 2020 at 12:14 | history | rollback | wychmaster♦ | Rollback to Revision 1 | |
| Apr 17, 2020 at 12:09 | history | edited | wychmaster♦ | CC BY-SA 4.0 | correct a typical mistake of mine -.- |
| Apr 17, 2020 at 10:44 | history | answered | wychmaster♦ | CC BY-SA 4.0 |