As a scoring system, I would propose a very simple approach:
Score(s) = No. of stickers solved(n) x moves remaining(m).
if m = 0, (i.e a 20 move solution) take it to be 1 when calculating score.
if m = 20, (i.e no moves applied) the organisers/WCA could consider this a DNF since no sequence would be submitted.
Analysis:
A random scramble from the FMC thread: D' R' U' F2 U F' D B2 R F2 D L' F' B2 U' D F D' U R B D2 R L D'
Full solution (1st 20 moves):
(R' B U B' U'
D B' D2 B2 D2 B'
D' F D F' D L D L'
D2) F D F2 R F R'
B R F' R' B' R F R' D
points (mn):
20 moves - 1x37 = 37
19 moves - 1x34 = 34
18 moves - 2x30 = 60
17 moves - 3x25 = 75
16 moves - 4x30 =120
It seems that the increase of the value m is too steep since we have a situation where fewer stickers are being solved, but the score is increasing. To reduce m's power, we could consider this formula: s = nx(m/2)
with: If m/2 < 1, m/2 = 1
giving:
points (mn): points(nx(m/2)):
20 moves - 1x37 = 37 1x37 = 37
19 moves - 1x34 = 34 1x34 = 34
18 moves - 2x30 = 60 1x30 = 30
17 moves - 3x25 = 75 1.5x25 = 37.5
16 moves - 4x30 =120 2x30 = 60
However, this system is not perfect, it seems that it is more rewarding to use 3 fewer moves than to solve 12 more stickers, meaning that to make using more moves worthwhile, you will need to solve on average more than 4 stickers per move.
tldr:
the score formula s = mn makes shorter solutions too op.
the score formula s = (m/2)n is better, but too make longer solutions worthwhile, you need to solve on average more than 4 stickers per move.
a better formula may be s = (m/3,4,5,etc.)n, but cannot be bothered to analyse this.
NOTE:
This is based on ONE scramble + solution not written for this event, so my assertions are not waterproof.
What are your thoughts, everyone?
