# New event idea: 20 moves



## Robert-Y (May 29, 2015)

You take a scrambled 3x3x3 and you have 20 moves (OBTM) to solve as many stickers as possible. Simple. You do not need to use all of your moves. I haven't thought about the time limit much. You can use sheets similar to the ones that you use for FM.

Discuss.


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## Username (May 29, 2015)

This is actually a pretty cool idea. Have you tried it yourself yet? Any "example solves"?

E: How would this be judged/how would results be ranked?


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## cashis (May 29, 2015)

Would it be changed to 19 moves if the FMC record gets broken? (joking)


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## Robert-Y (May 29, 2015)

Nope I only thought about this idea today for like 10 minutes lol...

How would this be judged? Similar to FM. Trusted checkers apply the written string of moves for each competitor then count up the number of stickers solved, and number of moves used.

How would results be ranked? I was thinking that it's number of stickers, followed by number of moves used.


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## Meneghetti (May 29, 2015)

Cool idea!
But what would define a sticker as "solved"? Would it be its relative position to the centers?
What if you use a Corners First method, which disregards the centers in the first steps, and you "solve" every corner in the "wrong" position?


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## Robert-Y (May 29, 2015)

Yes it would be its relative position to the centres. It's just more simple like this I think. I don't want to make things "complicated"


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## StachuK1992 (May 29, 2015)

I've been thinking about this as well, but my internally-proposed format was something like

Micro-FMC:
- 1hr total time limit
- 5 scrambles given
- 10 moves allowed max per scramble

the points for each scramble are based on how many stickers (or pieces) are solved

take out best/worst, average middle 3.


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## OrigamiCuber1 (May 29, 2015)

I like the idea of solving the stickers as opposed to pieces. This seems like a cool event


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## United Thought (May 29, 2015)

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?


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## Phinagin (May 29, 2015)

For a scoring system how about s=2m+n where m,s, and n are defined the same was previous post ^^.


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## cashis (May 29, 2015)

As much as I think this would be cool, I dont really think there's enough time at competitions for another long event like this.


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## OLLiver (May 29, 2015)

sounds good, i like all these new ideas!


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## tseitsei (May 29, 2015)

I actually don't like this idea. I find the idea that you don't actually solve the cube be somehow..."wrong"

You don't even have to solve a single piece correctly :/ 

IMO goal of all events should ultimately be to solve the puzzle...


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## TDM (May 29, 2015)

I don't think this is a suggestion for a new official event, just an unofficial one, like match the scramble.


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## scottishcuber (May 29, 2015)

tseitsei said:


> IMO goal of all events should ultimately be to solve the puzzle...



But that's boring.


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## tseitsei (May 29, 2015)

scottishcuber said:


> But that's boring.



Then maybe you shouldn't be practising speedSOLVING?


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## giorgi (May 29, 2015)

This way non cubers will have a chance to show their skills


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## primarycuber (May 29, 2015)

Nice idea!


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## adimare (May 29, 2015)

I would propose an even simpler approach for scoring:

Score = amount of solved stickers.
Move count could be used as a tie-breaker, but it seems too complicated and unnecessary to come up with a formula to integrate both numbers into the final score.


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## United Thought (May 30, 2015)

Yeah, I think I too this too seriously...


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## IAmAPerson (Jun 2, 2015)

adimare said:


> I would propose an even simpler approach for scoring:
> 
> Score = amount of solved stickers.
> Move count could be used as a tie-breaker, but it seems too complicated and unnecessary to come up with a formula to integrate both numbers into the final score.



I like this.
Also, if a local comp had this as an event, I would do it.


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## qqwref (Jun 2, 2015)

I want to see this as a linear, head-to-head event.

Explaining a bit more: choose some number of participants and randomly place them into a single-elimination tournament format. In each match-up, the two people are seated across from each other and given the same scramble. They have 15 seconds to look at it, then every few seconds (3? 5?) the judge says "turn" and each competitor performs up to one turn, HTM. This happens exactly 20 times. At the end, the competitor with more solved stickers wins. If there is a tie, they go again with a new scramble. If a competitor turns more than once per "turn" or turns at some other time, they forfeit the round. Podium is decided like with any other head-to-head event.


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## mark49152 (Jun 2, 2015)

What does "solved sticker" mean? Would a red sticker adjacent to a red centre count as solved even if its piece is in the wrong place? Why not count solved pieces instead?


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## pdilla (Jun 2, 2015)

Huh, so a correctly solved corner piece counts as 3 solved stickers?


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## notfeliks (Jun 2, 2015)

mark49152 said:


> What does "solved sticker" mean? Would a red sticker adjacent to a red centre count as solved even if its piece is in the wrong place? Why not count solved pieces instead?



That is starting to sound too much like FMC. I think Rob's intention with it being stickers as opposed to pieces was to have some differentiation between this and FMC.


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## IAmAPerson (Jun 2, 2015)

pdilla said:


> Huh, so a correctly solved corner piece counts as 3 solved stickers?



I guess. I suppose (because of that) that FMCetrus::20 Moves:Corners-First.


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## molarmanful (Jun 2, 2015)

IAmAPerson said:


> I guess. I suppose (because of that) that FMCetrus::20 Moves:Corners-First.



If this becomes a real event, then people will start using CF more often... Waterman will be extremely well-suited for this event (if you can block build efficient FL's). FL + CLL = at least 32 stickers solved, most likely more if some of the edge stickers happen to fall in place while doing CLL.

Side note: do center stickers count? Because if so, then with Waterman, after finishing FL + CLL, if you still have moves to go, then you can do slice turns (or OBTM equivalents) to get some more edges into place. Displaces the 4 center stickers, but it's all right if you can get at least 4 edge stickers into place.


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## IAmAPerson (Jun 2, 2015)

molarmanful said:


> Side note: do center stickers count?



I wouldn't think so. I believe the stickers have to be solved in relation to the centers, so a solved side minus the center would be null.


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## Animorpher13 (Jun 2, 2015)

tseitsei said:


> IMO goal of all events should ultimately be to solve the puzzle...



I disagree. It is easy to just memorize algorithms and train your fingers and eyes to recognize patterns and perform executions, but this trains you (just like FMC) to develop and understand how the cube works and not just memorize. thinking > memorizing.

Side note: How did skewb become an official event? That was before my time (as a speedcuber). Was it purely the WCA who announced it or did cubers petition for a new event or what?


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## Bilbo7 (Jun 2, 2015)

Yes I think this would be a great idea!
However, maybe increase the moves number as it may get the same results out of different competitiors!
Otherwise I would definately have this as an event!


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## Rnewms (Jun 2, 2015)

qqwref said:


> This happens exactly 20 times.



I think head-to-head FMC would be better. Do what you described until solved instead of stopping at 20 moves to count stickers. I like both ideas.


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## Phinagin (Jun 2, 2015)

Rnewms said:


> I think head-to-head FMC would be better. Do what you described until solved instead of stopping at 20 moves to count stickers. I like both ideas.


That sounds really cool actually.


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## IAmAPerson (Jun 2, 2015)

Bilbo7 said:


> Yes I think this would be a great idea!
> However, maybe increase the moves number as it may get the same results out of different competitiors!
> Otherwise I would definately have this as an event!



I think 20 moves was chosen since it makes it possible to completely solve the cube (perfect score). However, that's insanely difficult. 25-30 would be cool. Enough moves for the pros to solve the cube but not enough for the majority of people.


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## molarmanful (Jun 3, 2015)

First attempt at 20 moves: here

28 stickers solved, which is more than half the cube!


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## Brest (Jun 3, 2015)

molarmanful said:


> First attempt at 20 moves: here



L' B2 F2 U2 L' B2 L2 R B2 L2 F2 U' R F2 R B D2 L' U2 B' U2

B' R U2 B D' B' L'
R U' R' U R2 U R' U2
F2 L F' L' F'
(39)
View at alg.cubing.net


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## Phaint (Jun 3, 2015)

I see this just turning into who ever can made the biggest most efficient block.


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## molarmanful (Jun 3, 2015)

Okay, decided to give this a second try... 38 stickers!


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## Ranzha (Jun 3, 2015)

Animorpher13 said:


> Side note: How did skewb become an official event? That was before my time (as a speedcuber). Was it purely the WCA who announced it or did cubers petition for a new event or what?



My take on it was that skewb got popular enough, had developed solving methods, was easy to integrate in the WCA database and Regulations, and (I think the biggest challenge) had a random-state scrambler integrated in TNoodle.
Sarah Strong did a whole ton of work in order to petition the Board for skewb's approval.


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## Brest (Jun 3, 2015)

molarmanful said:


> Okay, decided to give this a second try... 38 stickers!



I count 44


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## ottozing (Jun 3, 2015)

D2 B2 L R2 B2 L B2 D2 B2 D2 U2 B' D' R2 U' B' U' B' U F2

z'
L U' L2 U
F' L2 U R U2 L
U' F B' R F' B
R' U R U2

33

yay


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## obelisk477 (Jun 3, 2015)

I would think a corners first-ish approach would be ideal for this, since corners are the pieces with the highest sticker 'density' (3 per piece). You get 30 stickers just solving corners alone


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## adimare (Jun 3, 2015)

EO might be worth it as a first step since unsolved edges at the top layer will at least be oriented and count as one point:

Scramble: L' B2 F2 U2 L' B2 L2 R B2 L2 F2 U' R F2 R B D2 L' U2 B' U2
B R2 L' F // EO
B2 // DB
L2 U L2 // Block
R U F2 // DF 
L' U2 L // Pair
R U2 R // Edges
alg.cubing.net

17 moves, 43 stickers


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## Robert-Y (Jun 3, 2015)

Has anyone found some decent strategies yet? I tried 20 moves a few times but got a bit bored because I don't really know how to improve


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## ottozing (Jun 3, 2015)

Robert-Y said:


> Has anyone found some decent strategies yet? I tried 20 moves a few times but got a bit bored because I don't really know how to improve



I tried doing corners first because 2x2 skillz+Corners have more stickers but the first time I tried that it failed miserably. I'll still probably look at it some more though.


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## mark49152 (Jun 3, 2015)

obelisk477 said:


> You get 30 stickers just solving corners alone


Are centres counted? Aren't they the point of reference for other stickers?

An interesting variant might be to aim for the lowest possible score.


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## molarmanful (Jun 3, 2015)

Brest said:


> I count 44



Not counting centers, it's 38.



mark49152 said:


> Are centres counted? Aren't they the point of reference for other stickers?
> 
> An interesting variant might be to aim for the lowest possible score.



I'm guessing that centers aren't counted. That's how I judge my solves, anyhow.



Robert-Y said:


> Has anyone found some decent strategies yet? I tried 20 moves a few times but got a bit bored because I don't really know how to improve



I usually build a 2x2x3 block, which usually requires less than or equal to 10 moves, then I see if I can finish F2L-1 along with several corners in 10 moves. If not, I just try to solve corner-edge pairs and finish some LL corners. When I finish my attempt, I usually end up with 6 corners solved and a bunch of edges solved or oriented relative to a center. Example with 40 stickers solved (not including centers): here.

The first time I tried 20 moves, I went with a Waterman-esque approach, building an FL-1 and seeing if the FL could be finished in 3 moves. Then I moved onto CLL, and I used an 8-move commutator to finish up some of the remaining corners.


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## IAmAPerson (Jun 3, 2015)

First ever attempt: 33 stickers (excluding centers)! I used a corners-first-style solve.
Here's the solve.

Scramble: *B' L' B R B' F2 L2 R2 D' F L2 B' D U' B2 F' D2 U2 F' U'*
2x2: *z2 U2 R' U B' L' D R' B' R* (9 moves! wow!)
Edges: *x2 M' F' M' B' M B' E B2 E'* (9 moves. I didn't use the last two.)

If I _have_ to use the last two moves, add a U U' to that solve.


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## TDM (Jun 3, 2015)

IAmAPerson said:


> First ever attempt: 33 stickers (excluding centers)! I used a corners-first-style solve.
> Here's the solve.
> 
> Scramble: *B' L' B R B' F2 L2 R2 D' F L2 B' D U' B2 F' D2 U2 F' U'*
> ...


The metric used stated in the OP is OBTM, so slice moves count as 2 moves rather than one.

(also x2 y2 = z2, if you want to know, but that's not important )


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## IAmAPerson (Jun 3, 2015)

TDM said:


> The metric used stated in the OP is OBTM, so slice moves count as 2 moves rather than one.



I spent a literal 1-2 hours on that solve. Can this one solve count? :'(

Also, if slice moves count 2 moves, then my corners-first method I used will do down the drain. 50% of turns past the 2x2 stage are slice turns.


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## molarmanful (Jun 3, 2015)

IAmAPerson said:


> I spent a literal 1-2 hours on that solve. Can this one solve count? :'(
> 
> Also, if slice moves count 2 moves, then my corners-first method I used will do down the drain. 50% of turns past the 2x2 stage are slice turns.



I suggest using a hybrid method (Roux? Waterman?) if you want to use CF. Solve corners and edges at the same time.


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## IAmAPerson (Jun 3, 2015)

molarmanful said:


> I suggest using a hybrid method (Roux? Waterman?) if you want to use CF. Solve corners and edges at the same time.



On my last competition, I tried FMC just for the experience. 63 moves, last place (Petrus method). I guess that's what you get for doing literally _one_ FMC solve _ever_ before the comp.


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## leeo (Jun 10, 2015)

I've been contemplating the 20-move event with a scoring system similar to the card-game Bridge. Here's what I cam up with so far:
A sticker is consider solved if it matches the center face color. 2 points for each corner sticker that matches the center face color. 3 points for each edge sticker that matches the center face color. Thus each sub-cube has a total point contribution of 6 points. For placing and orienting all 20 sub-cubes, 120 points.

for each unused move +4 points. As a tie-breaker, and a consideration of the quarter-turn metric, count the number of double faceturns (U2, F2, D2, etc.), and subtract one point for each two double faceturns submitted. Here is the table for this:

double face turns | penalty
0 | 0
1 | 0
2 | -1
3 | -1
4 | -2
5 | -2
6 | -3
7 | -3
...

I additionally feel that this should be a speed-type event. Thus, for instance, each contestant is given five scrambles, and each contestant has one hour to submit the best partial solution to all five scrambles on the 20-move budget. The total points of all five partial (or even complete  scrambles is then compared across all contestants in the event.


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