Talk:xkcd: volume 0
How the pages are counted[edit]
The pages are counted in something akin to a trinary number system: the only digits used are 0, 1, and 2. But 2 always rolls over, even if there are digits behind it. So:
| Number (xkcd v0) | Number in denary |
| 1 | 1 |
| 2 | 2 |
| 10 | 3 |
| 11 | 4 |
| 12 | 5 |
| 20 | 6 |
| 100 | 7 |
| 101 | 8 |
| 102 | 9 |
| 110 | 10 |
| 111 | 11 |
| 112 | 12 |
| 120 | 13 |
| 200 | 14 |
| 1000 | 15 |
How would someone do a base conversion between "xkcd trinary" and denary? I was thinking to do something akin to (value in "normal trinary")-(value of the previous digit in the same trinary) for each magnitude in xkcd trinary, but I can't figure out how rigorous it would be, much less accurate. Any ideas? 108.162.241.141 23:33, 19 June 2024 (UTC)
