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$100 5x5x5 Fewest Moves Challenge

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On January 1, 2016, I will post a "last layer" position on the 5x5x5 in this section. This will be a "No Holds Barred" contest: Whatever you need to do to solve it is legal. This includes computer programs, bribing the best speed solver you know, making burnt offerings to some 5x5x5 Diety, whatever.

$100 paid by me goes to the winner.

Unlike other FMC challenges with private submissions, this contest will feature posts directly to this area while the contest is still running. That means someone might be able to use some of your solution to find a shorter solution. Tough! It's a dog-eat-dog world out there, so find the shortest solution you can before you post :)

All solutions must contain links to the Alg Cubing Net page, like this:

(Just a sample position...)

https://alg.cubing.net/?puzzle=5x5x..._3F2_D-_3L-_D_F2_D-_U_2L-_2R-_U-_D_F2_U_D-_2R

Once a solution is posted, only post subsequent solutions that are shorter. The STM metric will be used. Tiebreaks are awarded to the earlier post.

The contest ends once EDIT: 28 days elapse without an improved solution, or EDIT: 28 days go by with no solutions posted at all.
 
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#2
The contest ends once 7 days elapse without an improved solution, or 14 days go by with no solutions posted at all.
You might want to extend these time limits significantly (I'm thinking more like the record has to stand for 3 weeks or a month...) But that's just my opinion...

I think that 7 days is quite low limit to improve a complex cube solver program (especially since it's just peoples hobby and not their job...). Also even the runtime of the program could be day(s) if you start pushing the limits...

All that said, I won't compete myself because I'm not much of a programmer...
 
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#4
I was wondering if 7 days would be long enough myself, so we'll see if others also agree.
Well, of course, the more time that is given, the shorter the solutions will be, especially since (unlike the other two recent big cube reward-based big cube FMC competitions), people were not allowed to post their solutions publicly. That is, even if a programmer manages to write a program in time to be able to execute it and successfully solve the position you give, someone else might be able to use that solution or a portion of it (or "the idea behind it") to create a shorter solution.

Therefore, in my opinion, if your goal is to just test what our current knowledge is, 7 days is plenty. If you are looking to see a 5x5x5 last layer position solved in very few moves in one or more ways, then you need to allow more time. The more time you allow, the more likely you will get some interesting results.
 
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EDIT: I changed the duration to 28 days on the recommendations above.


===================

Well I don't think anyone would write a 5x5x5 solver just for this. Even using my 4x4x4 solver code as a jumping off point, it still took me about 10 days to get my 5x5x5 solver working flawlessly.

I did not exclude programs because I wanted any submitted solution to be allowed to stand, no matter what.

Why I thought of 7 days initially: I thought someone might do the following:

1. Solve the position without too much concern for move length and make a post.
2. This forces a 7-day countdown for everyone else, meanwhile the person who posts the original solution works on a better one.
3. On day 6, someone posts a better one. Solver #1 then gains 7 more days to post a better solution. If he has one, he can wait 7 more days before posting.
4. This process could repeat, extending the contest for several months, hypothetically.

If you don't imagine something like this will unfold, the duration is now 28 days.
 
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#10
I'm not sure if everyone agrees what "STM" means. You should be absolutely clear what counts as a move. I imagine you mean that any move of a single layer can count as one move, and multi-layer turns do not count as a single move. (And assuming that is what you mean, can multi-layer turns be used, but count as multiple moves?)
 
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#11
In addition to what Bruce said,

Why I thought of 7 days initially: I thought someone might do the following:

1. Solve the position without too much concern for move length and make a post.
If you are implying that you are going to show a "screen-shot" of the position instead of posting a generating move sequence (hence, the first person to post does this for you), you have to make sure that the position you want us to solve is absolutely clear. This includes preventing us from accidentally solving for the inverse of the case you show.

Therefore, I believe it's best if you actually post a generating move sequence linked to alg.cubing.net (from which we just apply moves to solve the cube from that point). Either this, or you confirm that the generating move sequence posted is the correct one somehow.
 
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#12
I have litterly no chance (I will try tho), but I would love to say 'challenge accepted'.
Hole idea is really interesting, as I think about it more big cubes FMC might be really fun.
 
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What is the motive behind this?
I keep asking people for help in the "Request an Alg" section. I decided for this particular alg, which will be tough, I will "give back" to the community.

My program has already examined 5,249,024,392,428,376,695 positions and can't solve it. Just to go one move deeper will take weeks, but I'd really like to know how to solve this.

Only 6 cubies are unsolved on the cube.

By "STM" I mean the turn of a slice counts as 1 move. The turn of a face counts as 1 move. It does not matter if it's 2 rotations, like F2, that's still one move.

Multiple slices being moved count as multiple moves.

I have two iPhone pics of the starting configuration. It will be fairly obvious to all parties what the initial position is. If I have a move sequence I would offer it. I know I can create it with 2 algs in tandem. I will post this long sequence if that would be preferred.
 
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#15
By "STM" I mean the turn of a slice counts as 1 move. The turn of a face counts as 1 move. It does not matter if it's 2 rotations, like F2, that's still one move.

Multiple slices being moved count as multiple moves.
Just to confirm, then, is the state generated by 3Rw R' considered one move or two, optimally? (AKA, what exactly is a slice to you?)
 
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#16
I'm guessing that it will be a 2-swap of midges, corners, and wings
I hope it is a combination of different piece types because if it's all wings, there really isn't going to be too much of a challenge as we can use 3x3x3 solvers such as cube explorer to quickly find close to (if not) optimal solutions in this move metric. (In other words, it would be a waste of $100 to ask for an alg that can be created with Cube Explorer in minutes).

I of course found shorter solutions than Cube Explorer for some K4 wing edge cases by hand, so it's certainly possible to achieve shorter solutions without using the moveset <U2,F2,D2,B2,l,r>, but yeah.

@unsolved, if your position is just about moving wing edges, I'll save you $100 and find one from Cube Explorer for you.

Just to confirm, then, is the state generated by 3Rw R' considered one move or two, optimally? (AKA, what exactly is a slice to you?)
unsolved has been considering that two moves (single slice half turn metric), and that's what he means here as well.
 
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Just to confirm, then, is the state generated by 3Rw R' considered one move or two, optimally? (AKA, what exactly is a slice to you?)
Technically that looks to be 4 moves. The 3Rw turns 3 slices 1 turn each and R' turns 1 face.

It can be done with 2 moves, of course, but not the way it was written.
 
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#19
Technically that looks to be 4 moves. The 3Rw turns 3 slices 1 turn each and R' turns 1 face.

It can be done with 2 moves, of course, but not the way it was written.
Sorry, my phrasing was pretty unclear despite my attempts :p I was trying to ask what the smallest number of moves (using your metric) was to solve the state generated by 3Rw R' - in other words, what is the number of moves in the optimal solution to that state - so your last sentence answers my question, I think :)
 
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Here is the puzzle: red = front, green = right, orange = back



And the back view:



I will be traveling, so I wanted to post this and give those who want to try it some bonus time. Good luck to all.
 
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