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 Post subject: A confession: I can't solve dihedral puzzles!Posted: Mon Aug 27, 2012 10:25 am

Joined: Mon Aug 18, 2008 10:16 pm
Location: Somewhere Else
I just went through my collection to see which puzzles I have that I can't yet solve. Most of them are dihedral: off the top of my head, I remember the list includes the Square-1, Square-2, Tower Cube, Domino, 3x3x4, Rubik's UFO, Turbo Mind Twister, Rainbow Masterball, and Standard 3-layer Pentahedron.

Several of these puzzles are usually classified as easy, but they've got me stumped. I feel really dumb.

Is there any sort of generalized advice for this class of puzzles?

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Mon Aug 27, 2012 10:54 am

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
For most dihedral puzzles it's really easy to find a 2-2 swap. I usually commutate this to get a 3-cycle.

If you have a dihedral puzzle with two part types and one is easy to make a 2-2 swap and the other part isn't then I usually solve the "hard" part first non-pure and then cycle the other part type at the end.

Here's a hint for finding the 2-2 swap. If you're able to find two axes on the puzzle that overlap by just one part group from the top to the bottom, then if you do a [1,1] commutator you'll end up with a 3-cycle of this whole group. Usually the group contains two sets of three of the part, the top three and the bottom group. When you do the [1,1] commutator you move these groups around you can look at is as two 3-cycles. Because it's dihedral though you can flip over one of these cycled groups, undo the cycle, then undo the flip for a [[1,1],1] commutator that makes a 2-2 swap. Now you can use this straight or commutate it to make a 3-cycle.

This cycle-creation recipe is generic enough that it can be adapted to almost any N-gon prism.

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Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Fri Aug 31, 2012 3:51 pm

Joined: Mon Aug 18, 2008 10:16 pm
Location: Somewhere Else
Thank you for the advice, but... what is a 2-2 swap?

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Fri Aug 31, 2012 3:58 pm

Joined: Thu Sep 24, 2009 12:21 pm
Location: Chichester, England
A 2-2 swap is two two-cycles. An example would be the H-perm on a 3x3x3.

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Fri Aug 31, 2012 4:08 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
A 3-cycle is two swaps that shared a piece so only 3 pieces are permuted. A 2-2 swap is two swaps that did not share a piece so 4 pieces are permuted.

Both are even permutations (because an even number of swaps are used to create them).

Here is a demo on the Pyraminx Crystal for a 3-cycle:
Attachment:

pc_demo_3-cycle.png [ 18.65 KiB | Viewed 2304 times ]

And here is a 2-2 swap:
Attachment:

pc_demo_2-2_swap.png [ 18.56 KiB | Viewed 2304 times ]

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Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Fri Aug 31, 2012 4:32 pm

Joined: Mon Aug 18, 2008 10:16 pm
Location: Somewhere Else
Haha, I know what a 3-cycle is. I wasn't sure if a 2-2 swap was what it sounded like.

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Sun Sep 02, 2012 10:59 pm

Joined: Sun Aug 12, 2007 8:28 pm
Location: Northern Central California
bmenrigh wrote:
For most dihedral puzzles it's really easy to find a 2-2 swap. I usually commutate this to get a 3-cycle.

If you have a dihedral puzzle with two part types and one is easy to make a 2-2 swap and the other part isn't then I usually solve the "hard" part first non-pure and then cycle the other part type at the end.

Here's a hint for finding the 2-2 swap. If you're able to find two axes on the puzzle that overlap by just one part group from the top to the bottom, then if you do a [1,1] commutator you'll end up with a 3-cycle of this whole group. Usually the group contains two sets of three of the part, the top three and the bottom group. When you do the [1,1] commutator you move these groups around you can look at is as two 3-cycles. Because it's dihedral though you can flip over one of these cycled groups, undo the cycle, then undo the flip for a [[1,1],1] commutator that makes a 2-2 swap. Now you can use this straight or commutate it to make a 3-cycle.

This cycle-creation recipe is generic enough that it can be adapted to almost any N-gon prism.

Wow! I use commutators all the time in solving all sorts of twisty puzzles. I love them. But I have no idea what you are talking about here. [1,1] ??? What does this mean in simple terms?

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Mon Sep 03, 2012 12:05 am

Joined: Sun May 29, 2011 2:56 pm
Location: New York
robertpauljr wrote:
bmenrigh wrote:
For most dihedral puzzles it's really easy to find a 2-2 swap. I usually commutate this to get a 3-cycle.

If you have a dihedral puzzle with two part types and one is easy to make a 2-2 swap and the other part isn't then I usually solve the "hard" part first non-pure and then cycle the other part type at the end.

Here's a hint for finding the 2-2 swap. If you're able to find two axes on the puzzle that overlap by just one part group from the top to the bottom, then if you do a [1,1] commutator you'll end up with a 3-cycle of this whole group. Usually the group contains two sets of three of the part, the top three and the bottom group. When you do the [1,1] commutator you move these groups around you can look at is as two 3-cycles. Because it's dihedral though you can flip over one of these cycled groups, undo the cycle, then undo the flip for a [[1,1],1] commutator that makes a 2-2 swap. Now you can use this straight or commutate it to make a 3-cycle.

This cycle-creation recipe is generic enough that it can be adapted to almost any N-gon prism.

Wow! I use commutators all the time in solving all sorts of twisty puzzles. I love them. But I have no idea what you are talking about here. [1,1] ??? What does this mean in simple terms?

Well, let's start off by defining a commutator.
X, Y, X', Y'.
If X is 1 move, and Y is 1 move, that is a [1,1] commutator.
If X is 2 moves, and Y is 1 move, that is a [2, 1] commutator.
If X is 2 moves, and Y is 2 moves, that is a [2,2] commutator.
And so on.

-Doug

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Mon Sep 03, 2012 1:17 am

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
robertpauljr wrote:
Wow! I use commutators all the time in solving all sorts of twisty puzzles. I love them. But I have no idea what you are talking about here. [1,1] ??? What does this mean in simple terms?

This is just a short-hand to describe the form the commutator takes without describing the specific details of which bits are being turned, how much, and in which direction.

There are two "operators":

Commutation:
X Y X' Y' which we describe with [X,Y]

Conjugation:
X Y X' which we describe with [X:Y]

The numbers are the short-hand and just indicate the length of X and Y.

I've written a program to dissemble move sequences into their standard-form shorthand. So for example, if I gave you:

[F', C'&2, B'&2, C&2, B'2, C'&2, B&2, C&2, B2, F, K', F', B'2, C'&2, B'&2, C&2, B2, C'&2, B&2, C&2, F, K]

You could pop it into my program and it would tell you it is of the form [[1:[[1:1],1]],1].

_________________
Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Mon Sep 03, 2012 1:42 am

Joined: Sun Aug 12, 2007 8:28 pm
Location: Northern Central California
bmenrigh wrote:
This is just a short-hand to describe the form the commutator takes without describing the specific details of which bits are being turned, how much, and in which direction.

The numbers are the short-hand and just indicate the length of X and Y.

So [1,1] indicates a sequence of 4 twists, for example: R2UR2U' ?

And [3,1] indicates a sequence of 8 twists, for example: UL2U' R2 UL2U' R2 ?

I am thinking of common sequences I use when solving a 3x3x2.

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Mon Sep 03, 2012 1:52 am

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Yep. On a cube things like R2 are self-inverse so U R2 U' R2 is a 4-move [1,1] commutator.

Your [3,1] could also be written as [[1:1],1].

I find thinking about sequences in this way very useful for finding short cycles. I never memorize whole sequences anymore. I remember the parts and re-create it on the fly. This makes inverting a routine trivial too.

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Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Mon Sep 03, 2012 2:12 am

Joined: Sun Aug 12, 2007 8:28 pm
Location: Northern Central California
bmenrigh wrote:
Yep. On a cube things like R2 are self-inverse so U R2 U' R2 is a 4-move [1,1] commutator.

Your [3,1] could also be written as [[1:1],1].
Thanks. I don't see why you would want to think about a 3 move sequence as a conjugate and 1 move sequence, but at least now I see how it works.

bmenrigh wrote:
I find thinking about sequences in this way very useful for finding short cycles. I never memorize whole sequences anymore. I remember the parts and re-create it on the fly. This makes inverting a routine trivial too.

I'm not sure I totally get what you mean here. I always needed examples in math class before I understood the concept.

Now I will go back to your previous post with a 3x3x2 in hand and see if it makes more sense.

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Mon Sep 03, 2012 12:11 pm

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
robertpauljr wrote:
Your [3,1] could also be written as [[1:1],1].
Thanks. I don't see why you would want to think about a 3 move sequence as a conjugate and 1 move sequence, but at least now I see how it works.[/quote]When I was first figuring this stuff out, I too didn't see why thinking about the 3-move sequences a conjugates was useful. It turns out to be fundamental how commutators work though.

The standard-form commutator construction principle:
For all but the really exotic twisty puzzles the shortest useful commutators are always constructed from nested commutators and conjugates that can always be reduced to sequences of 1 move.

That is, you will never have a [2,1] commutators because a 2-move sequence can not be made from a conjugate or commutator. If you have a [4,1] commutator, the 4-move sequence must itself be a [1,1] commutator.

Counterexamples can be turned into standard-form commutators through symbolic manipulation.

Once you have a feel for how a commutator must be formed, finding short ones becomes much easier.

robertpauljr wrote:
bmenrigh wrote:
I find thinking about sequences in this way very useful for finding short cycles. I never memorize whole sequences anymore. I remember the parts and re-create it on the fly. This makes inverting a routine trivial too.

I'm not sure I totally get what you mean here. I always needed examples in math class before I understood the concept.

Now I will go back to your previous post with a 3x3x2 in hand and see if it makes more sense.

I view commutators and conjugates as two discrete sequences. An X and a Y sequence. If X and Y are themselves made up of commutators and conjugates I divide them up into sub X and Y parts down to the standard-form.

In this way, if you have an 8-move commutator it'll be of the form [[1:1],1] and I only have to memorize 3 moves and the form of the commutator. So memorizing [[A:B],C] allows you to fully reconstruct the 8-move routine: A B A' C A B' A' C'. Furthermore, there is only one form an 8-move commutator can take so memorizing the form is trivial.

My point about inverting is this. [X,Y] inverted is just [Y,X]. That is: X Y X' Y' inverts to Y X Y' X'. Instead of thinking about how to do a long sequence backwards I just do the Y part first, then the X part, then undo Y, undo X.

Inverting [X:Y] is super easy too because it inverts to just [X:Y'].

If you break down your sequences into standard-form you have to memorize at most half the moves and you get the inverse for free. That's a quarter or less of what it would take to memorize your sequence and its inverse without paying attention to its form.

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Prior to using my real name I posted under the account named bmenrigh.

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 Post subject: Re: A confession: I can't solve dihedral puzzles!Posted: Mon Sep 03, 2012 3:09 pm

Joined: Sun Aug 12, 2007 8:28 pm
Location: Northern Central California
bmenrigh wrote:
The standard-form commutator construction principle:
For all but the really exotic twisty puzzles the shortest useful commutators are always constructed from nested commutators and conjugates that can always be reduced to sequences of 1 move.

Counterexamples can be turned into standard-form commutators through symbolic manipulation.

Once you have a feel for how a commutator must be formed, finding short ones becomes much easier.

I find thinking about sequences in this way very useful for finding short cycles. I never memorize whole sequences anymore. I remember the parts and re-create it on the fly. This makes inverting a routine trivial too.

I view commutators and conjugates as two discrete sequences. An X and a Y sequence. If X and Y are themselves made up of commutators and conjugates I divide them up into sub X and Y parts down to the standard-form.

In this way, if you have an 8-move commutator it'll be of the form [[1:1],1] and I only have to memorize 3 moves and the form of the commutator. So memorizing [[A:B],C] allows you to fully reconstruct the 8-move routine: A B A' C A B' A' C'. Furthermore, there is only one form an 8-move commutator can take so memorizing the form is trivial.

My point about inverting is this. [X,Y] inverted is just [Y,X]. That is: X Y X' Y' inverts to Y X Y' X'. Instead of thinking about how to do a long sequence backwards I just do the Y part first, then the X part, then undo Y, undo X.

Inverting [X:Y] is super easy too because it inverts to just [X:Y'].

If you break down your sequences into standard-form you have to memorize at most half the moves and you get the inverse for free. That's a quarter or less of what it would take to memorize your sequence and its inverse without paying attention to its form.

This really helps! Thank you thank you thank you! Working through your sequence utility with concrete examples is helping too.

Usually when I approach a new puzzle and want to find a useful commutator, I do not think in terms of R's and U's at all, since I don't want to try to figure out a way to codify the layers and slices of complex puzzles (like the Dayan Gem IV). Even the 4x4x6 is going to have u's and r's and the like. And I never can remember whether E is clockwise like U or clockwise like D. So what do I do? I just think about the 3 pieces I am cycling. Move one, replace it with another, go back, replace with the third, move it, replace it, go back, go back. So I start by looking for a sequence in which ( move, replace, go back) only changes one thing in one layer. I've just never thought of it as [move:replace] before.

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