# Cube Exploring Competition : Kompetition 2



## kubesolver (Apr 6, 2020)

*General*

Second edition of my previous competition

The Kompetition starts now and will last until Monday 20th of April 13:00 Central Europe Summer Time. I reserve the right to finish it earlier if there is no new or updated submission nor a thread bump for more than 48 hours.

Prize for the winner 100$ cubicle gift card or similar.



Spoiler: Scoring



New scoring rules. For each problem where the higher the better your score is 100 * (your result) / (best result).In a problem where the lower is better we take do the reverse calculation so 100 * best result / your result.

For example:
In a problem 2 people got answers with a score of 8, 7, or 6 and once person gave incorrect answer.
The scores will be
8 => 100
7 => 100 * 7 / 8 = 87.56
6 => 100 * 6 / 8 = 75
wrong answer => 0





Spoiler: Problem Descriptions






Spoiler: Problem 1) Recursion number



Definition: Recursion Number (RN) of a sequence is the number of repetitions it takes to come back to starting position. E.g. R U R' U' has Recursion number 6 on nxnxn.

Design a sequence (or equivalent cube state) that has the highest RN.The answer sequence can be either in a regular notation or in some other description that tells how to construct this scramble. (e.g. you can draw the scrambled cube in paint or describe all stickers colors with some algorithm etc.) No rotations are allowed.

Note: You are expected to calculate the number yourself. I will only verify if it's correct if you're about to win a prize. If you miscalculate the number for your scramble it's considered incorrect answer and 0 points.

Score for each subproblem is a logarithm of a recursion number (no need to do FMC).

1.3) On a 3x3.
1.2) On a 2x2
1.Mega) REMOVED
1.INF) Infinite cube recursion:
What is the biggest possible number for any nxnxn cube? If you think it can be arbitrarily large please write Infinity.If it's not infinite please provide Score = sequence Recursion number / n (size of the cube),





Spoiler: Problem 2) Nemesis II



In these nemesis problems we only focus on 2x2x2 or cross on a D layer (yellow in WCA orientation).

2A) CFOP vs Petrus. Design a scramble that has a solved cross but has the toughest possible 2x2x2.
2B) Petrus vs CFOP. Design a scramble that has a solved 2x2x2, but has the toughest possible cross.

Score is a number of moves required to solve cross / 2x2x2. No need to find the shortest sequences





Spoiler: Problem 3) REMOVED



Removed





Spoiler: Problem 4) Generation



Definition generation: We will call a group of sequences a Cube-Generator if any scramble can be solved using only those sequences.



Spoiler:  "examples"



Four sequences (A,B,C,D) is a Cube-Generator if any scramble can be solved with only A, B, C, D e.g. AABCCDABCDCBACDBADCABDCABD.
The most obvious Cube-Generator is (R,L,F,B,D,U).

Another popular example comes from the common "The cube can be solved by Jperms"
The solution to 4.3 could be
(JpermA, JpermB, x, y, z) for a score of 5.03 + 0.01 * 2 * Jperm



4.3) Design a smallest set of generators for the whole cube
4.3A) Design a smallest set of generators for the whole cube without allowing the use of rotations (only RLUFDB)

Score for each subproblem is the number of sequences + (total length of all sequences in HTM) / 100.
So for example the score for sequence (R2, x y U) is 2.04
If all your sequences are longer than 25 then your score is X.25 (so it's always better to come up with very complicated 3 sequences than 4 easy ones)





Spoiler: Problem 5) FMC



Solve the following scrambles with the fewest moves using only RLUDF moves. Score is the length of a sequence in HTM metric.
5.B) B
5.B2) B2








Spoiler: Submissions



Please post your submissions as a DM to me with a title Kompetition 2 and bump this thread.



Spoiler: Please send your answer in this exact format



1.3: 6
R U R' F'
1.2: 2
R2
1.INF: 11999191293921839123
Here is how to get it on 121x121x121 blablabla.

2A: R U R' U'
2B: R U R' U'

3: Removed

4.3: 5.05
(R, U, F, D, x)
4.3A: 5.07
(R, U, F, L, D x y)


5.B: 4
R F2 U R
5.B2: 4
R F2 R' U








Spoiler: mentions



Mentioning previous participatns
@WoowyBaby @CuberStache @xyzzy @GenTheThief @TipsterTrickster @brododragon



Edit: Generation problem slighlty changed (only subproblems 4.3 and 4.3A left).
Edit: Removed problem 3
Edit: Removed problem 1.MEGA
Edit: Added example for problem 4


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## Cuberstache (Apr 6, 2020)

Wow, this looks so much more difficult than the last one! I can't wait to get started on it!


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## Cubinwitdapizza (Apr 6, 2020)

Oh goodness, I want to compete but it’s so confusing for my tiny brain
I guess I’ll try, although some things will have to be explained to me more thoroughly.
EDIT: where’d all this money to give away come from lol


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## Cubinwitdapizza (Apr 6, 2020)

Question, do I have to explain how I got the rn On the infinity problem?
And does the rn have to be 4 moves.


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## kubesolver (Apr 6, 2020)

@Cubinwitdapizza Yes, you have to explain how to get your number.
Nothing has to be 4 moves.

Edit: I added a new subproblem 4.3A)


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## Cuberstache (Apr 6, 2020)

kubesolver said:


> Design a scramble with the highest possible God Number where ALL edges are oriented using the following definition:White edge is oriented if it's in the middle layer or if it's in the U or D layer and white sticker is attached to white center or yellow center. Similar for other colors.


I'm a bit confused by your definition of "oriented". A white edge in the middle layer is always oriented? What about edges that belong in the middle layer? Is this normal EO as in ZZ or is this a different definition?


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## kubesolver (Apr 6, 2020)

Yes. According to the definition in a problem edges in the middle later are oriented.


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## Cuberstache (Apr 6, 2020)

kubesolver said:


> Yes. According to the definition in a problem edges in the middle later are oriented.


So if an edge is in the correct place in the middle layer, but flipped, it's considered oriented?


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## kubesolver (Apr 6, 2020)

Ok. I think I said it wrong. 
No edge is in the middle layer for BOTH of it's stickers. So the green-red edge is oriented if and only if
Green sticker touches Green or Blue center or Red sticker touches Red or Orange center.


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## Filipe Teixeira (Apr 6, 2020)




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## kubesolver (Apr 6, 2020)

Actually the more I think about it. the more I dislike this problem so unless someone says that he already has spend a lot of time thinking about it I consider removing the problem "oriented god number" from a competition. There seems to be already a lot of work with other problems.


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## Cuberstache (Apr 6, 2020)

kubesolver said:


> Ok. I think I said it wrong.
> No edge is in the middle layer for BOTH of it's stickers. So the green-red edge is oriented if and only if
> Green sticker touches Green or Blue center or Red sticker touches Red or Orange center.


Thank you, this clarifies it. I understand the concept of EO just fine, it was just the way you explained it that made me unsure.


kubesolver said:


> Actually the more I think about it. the more I dislike this problem so unless someone says that he already has spend a lot of time thinking about it I consider removing the problem "oriented god number" from a competition. There seems to be already a lot of work with other problems.


I would be ok with it being removed, but I now understand what the problem is asking.


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## TipsterTrickster (Apr 7, 2020)

For question 1.2 should we only use RUF moves? because if you include BDL with that you can do a "rotation" with U D', F B', or R L'.


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## ProStar (Apr 7, 2020)

Again, I'd do this but I can't get past the first problem without being confused lol


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## Cuberstache (Apr 7, 2020)

ProStar said:


> Again, I'd do this but I can't get past the first problem without being confused lol


All you need to do is find a sequence of moves that, when repeated, gets the cube back to a solved state. The example given is the sexy move, R U R' U'. When you do that 6 times on a solved cube, it returns to the solved state. So that would be a score of 6. You want to get that number as high as possible. Something that eventually restores the cube to a solved state, but with as many repetitions as possible.


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## brododragon (Apr 7, 2020)

kubesolver said:


> 4.3A) Design a smallest set of generators for the whole cube without allowing the use of rotations (only RLUFDB)


What about wide moves?


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## kubesolver (Apr 7, 2020)

The problem 3) Oriented God Number has been removed.
The reasons are that now I think it's not that good of a problem and I underestimated how much work the other problems actually are so there is no need for more problems in this competition.



brododragon said:


> 4.3A) What about wide moves?


Wide moves are forbidden. Only regular face turns R L U F D B are allowed.



TipsterTrickster said:


> For question 1.2 should we only use RUF moves? because if you include BDL with that you can do a "rotation" with U D', F B', or R L'.


RUF and BDL are both allowed in this case. But if the cube is solved in a wrong orientation it still counts as solved.


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## WoowyBaby (Apr 7, 2020)

Bump! I'll start working on it now


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## kubesolver (Apr 7, 2020)

Hi,

I have also removed the problem 1.MEGA.

I think there is more than enough to do in this comp and I will make another competition dedicated to non-cubic puzzle later.


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## Cuberstache (Apr 7, 2020)

kubesolver said:


> Hi,
> 
> I have also removed the problem 1.MEGA.


What? I spent like an hour on that and I came up with what I think is an amazingly clever solution. I can tell you my score if you want, I think you will be impressed.


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## kubesolver (Apr 7, 2020)

Keep it for the Kompetition 3  This problem will be there


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## Cuberstache (Apr 7, 2020)

kubesolver said:


> Keep it for the Kompetition 3  This problem will be there


Oh, alright. Fair enough.


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## TipsterTrickster (Apr 7, 2020)

Dang, I also did a bunch of work on it to, I guess I'll have a bit of a head start on Kompetition 3 lol.


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## WoowyBaby (Apr 7, 2020)

Bump. Submitted. Although this Kompetition is incredibly confusing and difficult so my answers to some of them were very basic and pretty dumb if I do say so myself. I know I'm not going to even come close to winning this one. GL to all.


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## TipsterTrickster (Apr 8, 2020)

Submitted! I'm pretty sure I made some errors so I'll probably resubmit later with those fixed.


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## brododragon (Apr 8, 2020)

kubesolver said:


> Score = sequence Recursion number / n (size of the cube),


Doesn't that mean the score will be higher if you prove it for a higher order nxn?


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## Cuberstache (Apr 8, 2020)

brododragon said:


> Doesn't that mean the score will be higher if you prove it for a higher order nxn?


No, you're diving by n, so the larger n is, the smaller the score will be. (Since n can only be a positive integer)


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## brododragon (Apr 8, 2020)

CuberStache said:


> No, you're diving by n, so the larger n is, the smaller the score will be. (Since n can only be a positive integer)


I meant lower. I thought lower was bad but then I realized that it isn't necessarily true.

Wait... What that_ does _mean is that the lower the recursion number, the better.


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## kubesolver (Apr 8, 2020)

I added a clarifying example to the problem 4. It seems that it was not clear that the generator sequences don't have to be simple moves (if you can solve a cube with only 2 algorithms then it's a 2 generator for the purpose of this problem.




brododragon said:


> Wait... What that_ does _mean is that the lower the recursion number, the better.


No the score for 1.INF is

```
(Recursion number)  / n
```
The result of division formula * a / b* gets bigger when *a* gets bigger, and gets smaller when *b *gets bigger.

The intuition is that it's harder to find big recursion number (hence more points for better solution i.e big recursion number) and it's harder to do so on a smaller cube (hence more points for better solution i.e. small n)


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## brododragon (Apr 10, 2020)

kubesolver said:


> number of sequences + (total length of all sequences in HTM) / 100


Just to be clear, don't you mean this: number of sequences + (total length of all sequences in HTM / 100)?


kubesolver said:


> If all your sequences are longer than 25


So all of them cumulatively?


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## WoowyBaby (Apr 10, 2020)

brododragon said:


> Just to be clear, don't you mean this: number of sequences + (total length of all sequences in HTM / 100)?




Of course, I’m almost certain that’s what he means.



brododragon said:


> so the whole's place doesn't count for anything if the combined length is > 1.00? Does X.25 = 0.25?



If you have [R, R U2 F U F’ R’ U2 F L D’ R2 U R’ U2 R U R’ U R U2 R2 F R F’ U F, R2], it would be a better score than [R, U2, F, L].

The first example only has 3 sequences, and a total HTM of 28, so the score is 3.25.
The second example has 4 sequences, and a total HTM of 4, so the score is 4.04.
Lower score is better.

I hope that clears up any confusion.


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## Cuberstache (Apr 10, 2020)

Whoops, I forgot to bump the thread when I turned in my results.


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## EasyCuber (Apr 10, 2020)

Thats Pretty Good


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## brododragon (Apr 10, 2020)

kubesolver said:


> 4.3A) Design a smallest set of generators for the whole cube without allowing the use of rotations (only RLUFDB)


Are initial rotations allowed, or does it have to be in scrambling rotation?


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## Cuberstache (Apr 10, 2020)

brododragon said:


> Are initial rotations allowed, or does it have to be in scrambling rotation?


It doesn't really matter since you have to solve all possible permutations of the cube with the generator you pick. If you have a scramble, doing a rotation just changes it to another scramble.


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## TipsterTrickster (Apr 10, 2020)

Updated my submission.


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## brododragon (Apr 12, 2020)

Bump. Finished.


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## WoowyBaby (Apr 12, 2020)

I'm feeling a bit better about this now, I am sorry for any negativity. This is really actually a ton of fun and super interesting!

I'm going to resubmit 

Edit: Also, if you want more competitors in your Kompetitions, perhaps you could make a post in the Fewest Moves Facebook group, as people that have experience with that generally know more about the cube than those who don't do it. If you don't want to post on Facebook that's your choice, it doesn't really matter, I'm just putting this out there.


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## brododragon (Apr 12, 2020)

Ya, this Kompetition is amazing! There are less questions, but am improved scoring system and I can tell you took multiple hours in each question.


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## kubesolver (Apr 14, 2020)

is anyone still working on their solutions?


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## fun at the joy (Apr 14, 2020)

kubesolver said:


> is anyone still working on their solutions?


haven't really started but I'll do it maybe today or tomorrow


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## brododragon (Apr 19, 2020)

Bump. Updated.


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## kubesolver (Apr 20, 2020)

This competition got 4 contestants. Thank you for participation. This time the winner is not clear as there are some answers that I should probably consider wrong.

The first place goes either to @TipsterTrickster or @WoowyBaby.

The problems are the following:
@TipsterTrickster gave answer to 2A which had F2L -1 solved, so it already had 2x2x2 blocks solved so he should get 0 points for problem 2A. His solution was a 7-moved to a 2x2x2 on FR slot. So I think it wouldn't break the spirit of a competition if he could resubmit, but on the other hand I feel the rules were rather clear. Would love to hear what other contestants think about it.

In this case the winner should be @WoowyBaby, but he gave an amazing short 2.x answer to both generation problems (4.3 and 4.3A), but I'm not convinced if they are correct and he didn't reply to me in PM with any explaination how to use these sequences to actually solve any state. (Edit: he did now and his answer is actually correct!)


Spoiler: His solution



4.3: 2.03
(R z, y)
4.3A: 2.05
(R L', U F D)



I think it might be correct given, that I could come up with a complex 2.x solution to this problem.


Spoiler: here is how I did it



The idea: OP-BLD style. We have one corner buffer and one edge buffer.

The idea is exactly the same for corners and edges.
1) We select one buffer piece and one target piece.
2) We come up with a sequence that cycles over all pieces except of a buffer
3) We use a swap algorithm that swaps buffer with target and we don't worry about orientation.
4) Swap algorithm also flips two edges that are neither target nor buffer so we can orientate pieces when they are all in their correct locations.


Alg1 SETUP = CORNER_SETUP EDGE_SETUP
Alg2 FLIP_AND_SWAP = JPERM + flip 2 corners + flip 2 edges = F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2


Where the exact algs are:
CORNER_SETUP = (F L D L D' F' R2 D L' D' L F L2 F' R2) //CORNER_SETUP
EDGE_SETUP = (U F2 U L2 B R2 U2 R2 B L2 D R2 U R2 L2 U2) //EDGE_SETUP

CORNER_SETUP is a 7-cycle, while EDGE_SETUP is 11-cycle so
EDGE_SETUP = (SETUP)56
CORNER_SETUP = (SETUP)22

flip(S,F) = (S2 F S2) (F S F' S' F' S F S' F S F S') (S2 F S2)'
We have two flipping algs (one for corners and one for edges):
flip(CORNER_SETUP, FLIP_AND_SWAP) = change orientation of a buffer and target corner
flip(EDGE_SETUP, FLIP_AND_SWAP) = change orientation of a buffer and target edge



To solve a cube you do 4 steps.
1) Permutate corners
2) Permutate edges
3) Orient corners and edges

1) Look what piece is in a buffer corner slot. Let's say that it's a piece X.

Call the CORNER_SETUP enough times so that the piece currently in X's spot is in target. Execute SWAP, Undo the setup.
Repeat until all Corners are in place.

2) Same but for edges.
Similar like in OP-BLD method every time we swap two edges we also swap two corners, but when all edges are in place all corners will be as well because parity.

3) Flip corners and edges.
Find a piece that has wrong orientation, bring it to the target location and execute flipping alg.

Cube Solved 


Spoiler: here is a solution for an example scramble



F U2 L2 B2 F' U L2 U R2 D2 L' B L2 B' R2 U2

[(U F2 U L2 B R2 U2 R2 B L2 D R2 U R2 L2 U2)2: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(U F2 U L2 B R2 U2 R2 B L2 D R2 U R2 L2 U2)6: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(U F2 U L2 B R2 U2 R2 B L2 D R2 U R2 L2 U2)7: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(U F2 U L2 B R2 U2 R2 B L2 D R2 U R2 L2 U2)3: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(U F2 U L2 B R2 U2 R2 B L2 D R2 U R2 L2 U2)9: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(U F2 U L2 B R2 U2 R2 B L2 D R2 U R2 L2 U2)1: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
(F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2)
[(U F2 U L2 B R2 U2 R2 B L2 D R2 U R2 L2 U2)5: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(U F2 U L2 B R2 U2 R2 B L2 D R2 U R2 L2 U2)10: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]

(F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2)

[(F L D L D' F' R2 D L' D' L F L2 F' R2)2: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(F L D L D' F' R2 D L' D' L F L2 F' R2)3: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(F L D L D' F' R2 D L' D' L F L2 F' R2)4: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(F L D L D' F' R2 D L' D' L F L2 F' R2)6: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]
[(F L D L D' F' R2 D L' D' L F L2 F' R2)5: F' U' B2 U' L U' L' F U2 L2 U2 L2 B2 D' B2 D' R2 D2 R2 B2 R2]

//at this point all pieces are in place and we can do the flipping






Other interesting problem was recursion number. The recursion number of a sequence is a lowest common multiply(LCM) of the length of all cycles.
We can have 9 * 5 cycles of corners giving 45 as a perfect answer for a 2x2x2 cube and for edges we can additionally make a cycle of length 11 with an edge twist making the max recursion number 9 * 5 * 11 * 2 = 990.

Surprisingly some people have answered with Inifinity to a 1.INF problem getting painful 0 points. All nxnxn cubes have at most 24 pieces of one kind, so the recusion number can't be higher than LCM of (1,2,3,4,...24) which is 2^4 × 3^2 × 5 × 7 × 11 × 13 × 17 × 19 × 23 = 5354228880 and it can be squezzed on a 6x6x6. @TipsterTrickster was the closest with finding the optimal solution, but packed on a 7x7x7.

Other problems turned out to be rather easy.

So current standings are like this, but this is pending explanaition from @WoowyBaby about his solutions to the 4.X problems. If @TipsterTrickster will make a claim that the problem description was misleading and will like have his solution for 2A scored 7 and nobody objects I will probably accept it.

So the final standings might change.
Here are the final standings:



Spoiler: Current standings



1. TipsterTrickster 867.00
2. WoowyBaby 790.32
3. brododragon 714.34
4. CuberStacke 692.01

Problem 1.3:
brododragon: answer: 315.0 score: 83.39835766034379
CuberStacke: answer: 210.0 score: 77.52009653651764
TipsterTrickster: answer: 990.0 score: 100.0
WoowyBaby: answer: 720.0 score: 95.38319290627571


Problem 1.2:
brododragon: answer: 36.0 score: 94.13807890996732
CuberStacke: answer: 30.0 score: 89.348540639004
TipsterTrickster: answer: 45.0 score: 100.0
WoowyBaby: answer: 45.0 score: 100.0


Problem 1.INF:
brododragon: answer: 2280.0 score: 37.799262300459965
CuberStacke: answer: 210.0 score: 26.140523935193762
TipsterTrickster: answer: 764889840.0 score: 100.0
WoowyBaby: answer: 720.0 score: 32.164132251914126


Problem 2A:
brododragon: answer: 7.0 score: 100.0
CuberStacke: answer: 7.0 score: 100.0
TipsterTrickster: answer: 7.0 score: 100.0
WoowyBaby: answer: 7.0 score: 100.0


Problem 2B:
brododragon: answer: 6.0 score: 100.0
CuberStacke: answer: 6.0 score: 100.0
TipsterTrickster: answer: 6.0 score: 100.0
WoowyBaby: answer: 6.0 score: 100.0


Problem 4.3:
brododragon: answer: 3.03 score: 67.00
CuberStacke: answer: 3.03 score: 67.00
TipsterTrickster: answer: 3.03 score: 67.00
WoowyBaby: answer: 2.03 score: 100.00


Problem 4.3A:
brododragon: answer: 5.05 score: 62.77
CuberStacke: answer: 5.05 score: 62.77
TipsterTrickster: answer: 3.17 score: 100.00
WoowyBaby: answer: 5.05 score: 62.77


Problem 5.B:
brododragon: answer: 13.0 score: 100.00
CuberStacke: answer: 13.0 score: 100.00
TipsterTrickster: answer: 13.0 score: 100.00
WoowyBaby: answer: 13.0 score: 100.00


Problem 5.B2:
brododragon: answer: 13.0 score: 69.23
CuberStacke: answer: 13.0 score: 69.23
TipsterTrickster: answer: 9.0 score: 100.00
WoowyBaby: answer: 9.0 score: 100.00



Edit: clarification from @WoowyBaby received.
Edit: As everyone agreed I gave TipsterTrickster 100 for 2A and WoowyBaby didn't provide a proof for his 4.3A answer so I gave him 50.5 there (I think it's fair to assume he would find that.


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## Cuberstache (Apr 20, 2020)

kubesolver said:


> In these nemesis problems we only focus on 2x2x2 or cross on a D layer (yellow in WCA orientation).
> 
> 2A) CFOP vs Petrus. Design a scramble that has a solved cross but has the toughest possible 2x2x2.
> 2B) Petrus vs CFOP. Design a scramble that has a solved 2x2x2, but has the toughest possible cross.
> ...


I guess it really doesn't say anything about other 2x2x2s being solved. I interpreted it to be that all yellow 2x2x2s had to be x moves to get a score of x but the rules actually aren't as clear as I thought they were...

EDIT: What was the 9-move solution to solve the B2?


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## WoowyBaby (Apr 20, 2020)

CuberStache said:


> EDIT: What was the 9-move solution to solve the B2?


R' L U2 F2 D2 R L' F2 U2 (9)
There are other solutions as well. (I believe 4 distinct)

Edit: Cube Explorer spits out this-


Spoiler: B2



Searching depth 9

U D' R2 F2 L2 U' D F2 R2 (9f*)
U2 R2 L2 D2 F2 U2 R2 L2 D2 (9f*)
U2 R2 L2 D2 F2 D2 R2 L2 U2 (9f*)
U2 F2 R' L D2 F2 U2 R L' (9f*)
U' D L2 F2 R2 U D' F2 L2 (9f*)
R L' D2 F2 U2 R' L F2 D2 (9f*)
R2 U2 D2 L2 F2 R2 U2 D2 L2 (9f*)
R2 U2 D2 L2 F2 L2 U2 D2 R2 (9f*)
R2 F2 U D' L2 F2 R2 U' D (9f*)
R' L U2 F2 D2 R L' F2 U2 (9f*)
D2 R2 L2 U2 F2 U2 R2 L2 D2 (9f*)
D2 R2 L2 U2 F2 D2 R2 L2 U2 (9f*)
D2 F2 R L' U2 F2 D2 R' L (9f*)
L2 U2 D2 R2 F2 R2 U2 D2 L2 (9f*)
L2 U2 D2 R2 F2 L2 U2 D2 R2 (9f*)
L2 F2 U' D R2 F2 L2 U D' (9f*)
Searching depth 10


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## Cuberstache (Apr 20, 2020)

WoowyBaby said:


> Edit: Cube Explorer spits out this-
> 
> 
> Spoiler: B2
> ...


I was literally about to reply with this hahaha. I'm surprised there are so many. That's really cool.


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## WoowyBaby (Apr 20, 2020)

CuberStache said:


> I was literally about to reply with this hahaha. I'm surprised there are so many. That's really cool.


Although, If you count inverses, rotations, and mirrors as the same, there are effectively only 2 different 9-move solutions to it.


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## TipsterTrickster (Apr 20, 2020)

I resubmitted my solution as <my solution> y *4. I guess I misinterpreted the rules as "hardest possible hardest possible 2x2x2" so for example if there was a 3 move 2x2 but also a 7 move 2x2 the 7 move 2x2 would be the answer. I'm ok with whatever you guys decide and think is fair.


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## brododragon (Apr 20, 2020)

TipsterTrickster said:


> "hardest possible hardest possible 2x2x2" so for example if there was a 3 move 2x2 but also a 7 move 2x2 the 7 move 2x2


That makes sense. I originally was against you resubmitting, but I'm fine with it after hearing your reasoning.


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## Cuberstache (Apr 20, 2020)

Yeah, it's pretty easy to do what he said and just set up the same case for all four 2x2x2s so I think it's fine to give him 7 points.


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## kubesolver (Apr 24, 2020)

Ok. So after some discussions and considering what is fair I have made the following calls:
@TipsterTrickster gets 100 points for 2A
Those who anwered INF to 1.INF get points for the number they provided in 1.3 instead of 0 points.
@WoowyBaby gave the answer he can't prove for the 4.3A so I changed his answer to 5.05 everyone else gave as I think 0 points would be too harsh there.
After these updates here are the final standings.
100$ goes to @TipsterTrickster.

1. TipsterTrickster 867.00
2. WoowyBaby 790.32
3. brododragon 714.34
4. CuberStacke 692.01

In the meantime I'm working on the problems for another Kompetition which will be about other puzzles than 3x3. but I will try to think a little more about problems to maybe get a higher participation


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## brododragon (Apr 24, 2020)

Almost all of my cube theory knowledge came directly by participating in the Kompetition, so I'd count it as a success.


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## WoowyBaby (Apr 24, 2020)

Congratulations to @TipsterTrickster !! After a close second on the first Kompetition, it seems right to win this one!

But wait. Let's pause for a second. Did @TipsterTrickster provide proof for his 4.3 solutions? 

I'll assume he has provided proof for those. In that case, well done and I hope you enjoy your $100 gift card prize


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## kubesolver (Apr 24, 2020)

His solution is a very elegant one kind of similar to the solutions to 5.B 
It's two center swap algorithms and U.
So e.g. R = move white center to the right face and do a U.


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## TipsterTrickster (Apr 24, 2020)

Awesome! Here are my solutions for anyone interested.


Spoiler: Solutions



1.3: 990
D' L2 B' L' U2 B' D2 F R F R2 B U' B' D2
1.2: 45
U R' F R2 F' R D B2 D2
1.INF: 764889840
On a 7x7 5354228880/7 = 764889840
Number = number of pieces cycled
F = 2 pieces flipped, both being in separate cycles
T = 2 pieces twisted, both being in separate cycles, one twisted clockwise and the other counter clockwise.








2A: F R’ F’ R2 U’ R’ U2 F’ U’ F (7 move 222) later revised to: F R’ F’ R2 U’ R’ U2 F’ U’ F y F R’ F’ R2 U’ R’ U2 F’ U’ F y F R’ F’ R2 U’ R’ U2 F’ U’ F y F R’ F’ R2 U’ R’ U2 F’ U’ F y 
2B: F R2 F U’ R’ U2 F R’ (6 move cross)


4.3: 3.03
(U, x, y)
4.3A: 3.17
(U, R L’ U D’ F’ B R L’, R L’ F’ B U’ D R L’)

5.B: 13
R’ L U2 D2 R’ L F’ R L’ U2 D2 R L’
5.B2: 9
U2 R2 L2 D2 F2 D2 R2 L2 U2


This was a lot of fun and I am super exited for the next one! Thank you for doing these.


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