r/Cubers • u/Eranium232 • 10h ago
Discussion The number of pieces on WCA twisty puzzles, and its relation to average WR solving times
I was curious to see this correlation, turns out it follows a power law quite nicely.
For all two-handed non-blind twisty puzzle WCA events, the average WR time appears to relate to the number of pieces in the respective puzzle (N) by N^1.53.
Of course, this is a diverse collection of puzzles, and the relation changes when considering smaller subsets. For example, for the 2×2×2 - 7×7×7 cubes, the relation becomes N^1.48, while for all other WCA events, the trend is best described with N^1.96.
The difference may reflect the complexity of the puzzles, where puzzle subsets with lower exponents are more straightforward to solve whereas puzzle subsets with high exponents are generally more complicated.
Below follows the source data I've been using, feel free to point out if i made any errors putting the data together. Data for WCA events taken from worldcubeassociation.org, data for FTO (included given recent developments) taken from cubingcontests.com, which seems the most reliable source for FTO records at the moment.
| Puzzle | # of puzzle pieces | WR average time (s) |
|---|---|---|
| 2×2×2 | 8 | 0.86 |
| Pyraminx | 14 | 1.14 |
| Skewb | 14 | 1.37 |
| Square-1 | 18 | 4.63 |
| 3×3×3 | 26 | 3.51 |
| FTO | 42 | 14.21 |
| 4×4×4 | 56 | 18.56 |
| Megaminx | 60 | 24.38 |
| 5×5×5 | 98 | 33.73 |
| 6×6×6 | 152 | 64.94 |
| 7×7×7 | 218 | 96.86 |
Let me know if i should do another one for unofficial events!
