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One-Answer Puzzle Theory Question Thread

Roman

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God's number for solving 1 side on the 3x3 Cube?

I have a feature request for my mosaicbuilding tool to output "scrambles" to apply to solved cubes and make mosaics even faster. I'm thinking of implementing a breadth first search to find shortest solutions for each cube. I've done this already in C++ but here we're talking about JS and a few hundreds of cubes so I'm trying to come up with the best approach atm in case I have free time for this stuff which I always happen to find because I love doing stupid things that literally only a few people on the planet may find useful
 

Sub1Hour

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God's number for solving 1 side on the 3x3 Cube?

I have a feature request for my mosaicbuilding tool to output "scrambles" to apply to solved cubes and make mosaics even faster. I'm thinking of implementing a breadth first search to find shortest solutions for each cube. I've done this already in C++ but here we're talking about JS and a few hundreds of cubes so I'm trying to come up with the best approach atm in case I have free time for this stuff which I always happen to find because I love doing stupid things that literally only a few people on the planet may find useful
@pjk can we get a big brain react for stuff like this?
 

EngiNerdBrian

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I want to nerd out on 3x3 and general puzzle theory and mechanics. What are some foundational elements, references, topics, and tutorials you can recommend? I've been on a non-WCA puzzle kick lately and have found myself reverting back to my old school big cube last 2 center commutators or making parallels to the 3x3 algs i know but don't intuitively understand to find solutions to these puzzles. While just struggling through these puzzles has been fun i know i would benefit from having more tools in the tool belt!

I want to gain the fundamental knowledge of how and why the puzzles behave the way the do. Where should i start?
 
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ray5

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Is there an explanation why you (without affecting anything else):
  1. Can cycle 3 edges
  2. Can cycle 3 corners
  3. Can not swap any two edges
  4. Can not swap any two corners
  5. Can swap 2 edges and 2 corners
I don't mind it being explained with group theory if that gives a good description of why.

For (1) and (2) we can build algs with commutators, but for (5) I didn't manage to come up with algs and I wonder why we would expect this to be possible before finding an alg.
 

xyzzy

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I don't mind it being explained with group theory if that gives a good description of why.
For #3 and #4:

For any \( n\ge2 \), there's a (surjective) group homomorphism from \( S_n \) (symmetric group of degree \( n \), i.e. the permutations of \( n \) elements) to \( \mathbb Z/2 \) (integers mod 2) given by the parity of the permutation.

Consider the permutation of the corner pieces and the permutation of the edge pieces (completely ignoring orientation). These live in \( S_8 \) and \( S_{12} \) respectively, and we can define the product map \( \mathsf{parity}:S_8\times S_{12}\to(\mathbb Z/2)\times(\mathbb Z/2) \) by taking the above parity map component-wise. This is also a group hom.

The six clockwise face moves on a 3×3×3 (U, D, L, R, F, B) generate every possible cube state with the centres in their original locations. (Note: this is why this reasoning fails on a void cube, which does not have centre pieces and it's truly possible to swap only two edge pieces there. You need either a rotation or a slice move to truly generate the full set of states.) The permutations induced by these six face moves generate some subgroup \( G \) of \( S_8\times S_{12} \), and it's a priori not easy to determine exactly what this subgroup is. However, by pushing it through the \( \mathsf{parity} \) map defined above, since \( \mathsf{parity}(U)=\mathsf{parity}(D)=\mathsf{parity}(L)=\mathsf{parity}(R)=\mathsf{parity}(F)=\mathsf{parity}(B)=(1,1) \) (they're all a 4-cycle of corners and a 4-cycle of edges, both odd permutations) we have that:

\( \begin{aligned}\mathsf{parity}(G)&=\mathsf{parity}(\langle U,D,L,R,F,B\rangle)\\
&=\langle\mathsf{parity}(U),\mathsf{parity}(D),\mathsf{parity}(L),\mathsf{parity}(R),\mathsf{parity}(F),\mathsf{parity}(B)\rangle\\
&=\langle(1,1),(1,1),(1,1),(1,1),(1,1),(1,1)\rangle\\
&=\{(0,0),(1,1)\}\end{aligned} \)

In other words, the permutation parities of the corners and of the edges must be either both even or both odd. Swapping only two edges corresponds to an odd permutation in the edges and an even permutation in the corners; swapping only two corners corresponds to an odd permutation in the corners and an even permutation in the edges; either way, these are both impossible.

This reasoning generalises easily to big cubes: you just have to care about more piece types.

For #5: Do a U move, then use a corner 3-cycle to solve two corners and an edge 3-cycle to solve two edges. You'll be left with one of the 2-cycle PLL cases (J perm, R perm, T perm, F perm).

Heuristically, as long as you've covered the parity constraints, permutation groups that show up in twisty puzzles tend to follow the rule of "anything not forbidden by parity is allowed". This is because picking a few random elements in \( S_n \) is already very likely to generate at least \( A_n \) (where "at least \( A_n \)" really means "either \( A_n \) or \( S_n \)"); the generating elements need to be quite "structured" in order to generate smaller subgroups.
 
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Christopher Mowla

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Is there an explanation why you (without affecting anything else):
.
.
.
Can swap 2 edges and 2 corners

I don't mind it being explained with group theory if that gives a good description of why.
I can see why you're confused about this, as the same knowledge to derive an algorithm for the single dedge flip!

I show in this video (Part 1 of my (and the first-ever) derivation of the "Red Bull"/Old standard 15 move algorithm how the two ideas are directly related (with images/visuals/animations) how they are related and the logic behind it exactly the way I think you're looking for. You can watch Part II while you're at it to get the rest of it (just in case YouTube's recommendations doesn't show you that there is a second part). If like parity algorithm theory (after watching that video), perhaps you would like to see this video (and Part 2 of that series) to see how to use the same logic to derive wide-turn based edge flip algorithms! (It's ironic, but I discovered THAT before I came up with a derivation for the Red bull alg!)

(And no, "group theory language" is not really needed to explain (or justifiy) why any of these situations are true. It's rather just an "exercise" for those who are fluent in higher mathematics to relate math to the cube, but that type of math really isn't needed to solve the cube. Conjugates are a "natural" way to do "minimal damage" to a cube, and commutators are nothing more than two conjugates (which have a special relationship to each other.)
 
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Wgg

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So i had a question
If you would have a coll
Doesnt that mean youd have to orient all four edges in f2l?
Because otherwise youd get another oll case which you can just do with those 50 algs
In that case every last f2l which ends up in an all edges oriented case must be a wv case.right?
Basicaly meaning youd never get a coll???
Let me know if coll is relevant even if you know wv
Because this question is annoying me in my head for a while and i dont know whether to learn coll or not
 
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COLL is a first step for learning ZBLL but by itself, it actually is not that great. WV on the other hand is actually great because you don't get OLL skips often and learning the easy cases helps and the odds of an OLL skip (With CFOP)is reduced from 1/216 to 1/8.
So, from the above facts, it is indeed WV FTW.
Moreover, COLL Recognition is hard at first and WV recognition is somewhat easier.
What's still good is WV is less algs. Of course 40 algs with harder recognition sucks in front of 27 algs with somewhat easier recognition.
On top of all this, OLL skips are a big boost to solves if you have good PLL algs and fingertricks as they cut down at least a second or two from your times.
(I don't know full WV and I'm in the process of learning it. I know somewhere north of 15 cases but still it helps and all the stuff I've written above is all my experience. Btw I know almost all of COLL too.)
 

xyzzy

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What's still good is WV is less algs. Of course 40 algs with harder recognition sucks in front of 27 algs with somewhat easier recognition.
I don't think this is a fair comparison. On the surface, it looks like WV has only 27 algs and COLL has 42 (40 not counting PLL), but:

(i) Most people skip the S/AS subsets of COLL, bringing that down to 28 algs. Or really, 28 − 5 = 23 new algs on top of OLL/PLL. Contrast with 27 − 2 = 25 new algs for WV.

(ii) WV is 25 algs for one slot in one specific orientation only, and 50 if you want to be able to do it with two slots. If you want to count mirrors as the same cases, then you should do the same for COLL as well, which brings the alg counts from
50 (WV) versus 23 (COLL sans S/AS) versus 33 (full COLL)
to
25 (WV) versus 18 (COLL sans S/AS) versus 23 (full COLL).
 

Wgg

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So could i
COLL is a first step for learning ZBLL but by itself, it actually is not that great. WV on the other hand is actually great because you don't get OLL skips often and learning the easy cases helps and the odds of an OLL skip (With CFOP)is reduced from 1/216 to 1/8.
So, from the above facts, it is indeed WV FTW.
Moreover, COLL Recognition is hard at first and WV recognition is somewhat easier.
What's still good is WV is less algs. Of course 40 algs with harder recognition sucks in front of 27 algs with somewhat easier recognition.
On top of all this, OLL skips are a big boost to solves if you have good PLL algs and fingertricks as they cut down at least a second or two from your times.
(I don't know full WV and I'm in the process of learning it. I know somewhere north of 15 cases but still it helps and all the stuff I've written above is all my experience. Btw I know almost all of COLL too.)
So should i just learn wv and zbll?

I don't think this is a fair comparison. On the surface, it looks like WV has only 27 algs and COLL has 42 (40 not counting PLL), but:

(i) Most people skip the S/AS subsets of COLL, bringing that down to 28 algs. Or really, 28 − 5 = 23 new algs on top of OLL/PLL. Contrast with 27 − 2 = 25 new algs for WV.

(ii) WV is 25 algs for one slot in one specific orientation only, and 50 if you want to be able to do it with two slots. If you want to count mirrors as the same cases, then you should do the same for COLL as well, which brings the alg counts from
50 (WV) versus 23 (COLL sans S/AS) versus 33 (full COLL)
to
25 (WV) versus 18 (COLL sans S/AS) versus 23 (full COLL).
Ok thanks but does learning wv mean id never get a coll?
 

PapaSmurf

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Learn OLL/PLL if you're doing CFOP. Then learn the 4 or 5 worthwhile WV cases that are easy (R U' R', R U2 R', R U R' U' R U' R', L' U R U' R' L, L' U2 R U R' U2 L). Then learn the COLL cases that are worth it (all of TUL, all but one of H and Pi). Full of either set isn't worth it.

Yes, WV means you never get COLL.
 

Cubing5life

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Learn OLL/PLL if you're doing CFOP. Then learn the 4 or 5 worthwhile WV cases that are easy (R U' R', R U2 R', R U R' U' R U' R', L' U R U' R' L, L' U2 R U R' U2 L). Then learn the COLL cases that are worth it (all of TUL, all but one of H and Pi). Full of either set isn't worth it.

Yes, WV means you never get COLL.
Actually you could get COLL with some sledge inserts or advanced F2L solutions for last pair that don‘t use R U‘ R‘ inserts.
 

Gerry

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Something I noticed made me wonder about how many ways there are to "solve" a 3x3. From the perspective of the logo there is at least 2 different solves. I noticed this while watching beginner tutorials that when the blue side is towards me and white side up, the logo is correctly oriented. Then I scrambled it and solved it. Now when the white side is up, and blue towards me, the logo is upside down. If I drew a picture on each side so you could still solve with the colors, would the images be scrambled? Or would just the centers of the images spin? Originally my guess was just the centres would spin, with 3 axis. For 64 different ways. (4x4x4) Now I'm thinking that the centers can spin separately from each other for 4x4x4x4x4x4 for 4096 different ways. Can anyone confirm or deny my guess?
 

bcube

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Now I'm thinking that the centers can spin separately from each other for 4x4x4x4x4x4 for 4096 different ways. Can anyone confirm or deny my guess?

Actually, it´s (4^6)/2 = 2048. One quarter turn rotates the center piece by 90 degrees, but also makes corners and edges unsolvable while keeping that center piece rotated - so only 1/2 of combinations which can be reached by executing quarter turns are solvable.

What you are describing is a supercube, if you are only interested in centers, click here.
 

PetrusQuber

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Yeah, certain things can rotate the centre in a 3x3, while it is not considered different solved states generally.
Actually, it´s (4^6)/2 = 2048. One quarter turn rotates the center piece by 90 degrees, but also makes corners and edges unsolvable while keeping that center piece rotated - so only 1/2 of combinations which can be reached by executing quarter turns are solvable.

What you are describing is a supercube, if you are only interested in centers, click here.
^
Doing two T Perms one after the other rotates the centre by 180 degrees for example
 
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