# odd permutation (AKA parity) - can we be unified?



## bcube (Mar 24, 2016)

Hi,

I have re-written an article about a parity.

I realize I am not entirely in agreement with some authorities, such as Jaap Scherphuis. Ryan Heise, Herbert Kociemba and other people.

On the other hand, I think I might be in agreement with another authorities, such as Chris Hardwick, Bruce Norskog (even though I still use the term "parity problem" because of (speed)cubers), Christopher Mowla, Michael Gottlieb and other people.

We can probably all agree that U move on the 3x3x3 Rubik's cube produces 1 4-cycle of corners and at the same time 1 4-cycle of edges. Now, can we be unified that 1 4-cycle + 1 4-cycle ≠ 2 4-cycles here (and similarly that for a T-perm 1 2-cycle + 1 2-cycle ≠ 2 2-cycles, while in case of H-perm or E-perm 1 2-cycle + 1 2-cycle = 2 2-cycles)?

In my opinion the term "parity" is not an issue here. From my point of view the problem is a word "permutation", despite it is well-defined mathematical term.


P.S. I think speedcubers might find more odd/shocking (or even wrong) things in my article. If you are going to read it, do not hesitate to ask (if necessary) or prove me wrong (either in this or another thread you will create).


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## Christopher Mowla (Mar 25, 2016)

Hi bcube,

I highly respect that you have written this article. I can understand how difficult it is to write on a topic like this. It took me years to write my supercube centers and odd parity paper, not just because almost all content within it was/is original, but because I rewrote it several times to make it more clear. I'm sure it's still not perfect. (I haven't changed "7x7x7" to "6x6x6" in the first example name yet, so there's one mistake!) From all of that practice, I wrote my book (which contains the pseudo god's alg solving approach and proof of my single commutator theorem) much more efficiently.

I know that English is not your first language (it's mine, and I am not perfect either), but clearly some things can be reworded to make the read better. However, I understood everything for the most part. (I guess you need to get feedback from someone who is not well-versed in parity concepts.)

So here are only my thoughts about certain things that stuck out to me in your article. (I wrote down these as I read.)

I thought the "I cycle, you cycle, we cycle" phrase was hysterical!



bcube said:


> Odd permutation is composed of odd multiples of cycles of length 1 (a cycle of length 1 will be further denoted as a 2-cycle)


This should be_ of length *2*_, correct_?_



bcube said:


> N-cycle is executed N-times in a row, the cube will return to its original configuration before executing of this algorithm.


Also be sure to add in "N-cycle is executed N-times in a row *(and not until then)*...". For example, repeating a 2 2-cycle algorithm (which affects 4 pieces) 4 times will solve it, but repeating it 2 times will also solve it. Look back at my powerpoint on the number of positions. You will see that I also said that. (And this is why I said that.)



bcube said:


> Their meaning will be "a permutation of a piece type is even (or odd)". By the way, a term "permutation of the piece types set" (in other words a puzzle permutation) doesn't make sense at all, but as a non-mathematician I can live with that .


Are you saying that when we say "the position generated by applying the move U2 is an even permutation" does not make sense? If so, we say this because we understand that if the parity of the edge orbit = parity of corner orbit. So instead of saying, for example, "The parity of the corners is even and the parity of the edges is even", we say the parity of the (3x3x3) puzzle is even because the 3x3x3 only consists of corners, edges and fixed centers, (all pieces of which their parity are equal to each other).

It's a way of "factoring out" words. For example, we can say 2+2+2+2+2+2+2 = 14 or we can say 2*7 =14.



bcube said:


> The term parity problem (or puzzle permutation) is not defined in mathematics.


We define it in mathematics by saying that a "parity problem" (as you call it) "exists when the nxnxn supercube cube is not in the commutator subgroup".



bcube said:


> Some authors are contradicting themselves when they claim: "although one move in accordance with QTM creates 1 4-cycle of edges and 1 4-cycle of corners, an overall puzzle permutation is even (since 1 4-cycle + 1 4-cycle = 2 4-cycles)". This argument can be disproved by a fact that it allows (regardless of a permutation definition) to swap a corner with an edge. In reality, however, it is not possible.


Excellent. I'm glad that you agree with me on this issue! Parity states are the core of my single commutator proof, and, unlike describing laws of the cube to non-puzzle theorists, it is _imperative_ that this be the accepted convention whenever you actually want to _apply_ or _take into account_ the laws of the cube for some further theoretical use.



bcube said:


> For more see a video whose author is Jon, a lover of combinatorial puzzles (it can be said that Jon, in a final consequence, solves occurring so-called OLL parity by 3 2-cycles of wing edges (time 22:44) - stated algorithm R2 U' R2 U R2 U D' R2 U R2 U' R2 D adapted on a 4x4x4 cube is in his realization formed by 19 moves of outer layers and 7 moves of inner layers in accordance with QTM.


YES!!! He drove me berserk when I heard he said he dodges parity by solving the 4x4x4 that way...and the ignorant believe him, just as the ignorant accepted this video claim as truth (when in fact, he's just executing a LONG and VERY complicated algorithm that is as commonly known as r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2). It's interesting that people who teach the general public rubbish are often more recognized than those that know what they are talking about. Monkey see, monkey do, I guess?



bcube said:


> For the sake of completeness you can watch analogous duo of simulators with initial odd permutation of wing edges (so either both pairs of non-paired-up wing edges or neither of them are "half-solved"). The idea of pairing (using 2 2-cycles of wing edges) is the same, as well as a principle of realization (with the only difference that a number of set-up moves has changed - instead of four, there are needed only two of them).


I mentioned on reddit that one could memorize Ed Trice's (unsolved's) 9 move checkerboard algorithm and then apply "the edge pairing algorithm" so that one would only have to memorize 9 moves. That is, if you want to only have to repair two dedges in M instead of all four with (r U2)4 r_..._
r2 U2 r' e2 r e2 r' U2 r2 //Parity Algorithm
x' z d R F' U R' F d'//Edge Pairing Algorithm



> Christopher Mowla deserves my big thanks for all the effort he shown regarding our private correspondence, not only on the topic of the permutations and analysis of a Supercube NxNxN. It wouldn't be possible to write up this article into its current form without his kind and extremely useful consultations.


You're welcome. I also appreciate that when you refer to my name, you link to that post. It really shows how I have _defined_ myself in the cubing community. (More people should post in that thread!)

Lastly, I liked how you used Gabbasoft-style supercubes! (As you know, I love using these in my parity alg videos.)


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## cuBerBruce (Mar 25, 2016)

bcube said:


> From my point of view the problem is a word "permutation", despite it is well-defined mathematical term.



Yes, you are obviously misuing the word _permutation_.

You are trying to define the "puzzle permutation" (singular term) as *multiple separate* permutations (plural).

Mathematicians are happy to talk about a permutation where the elements are constrained to certain orbits, but such a permutation of all elements does not mesh with what you want to call an odd puzzle "permutation." Perhaps you should use _odd puzzle state_ since what you are trying to define as being "even" or "odd" is not a (singular) permutation.

Also, (as cmowla noted) you are incorrectly using the term "length of a cycle." A *2*-cycle has length *2*. There are *2* elements in such a cycle. That's why it is called a *2*-cycle. You seem to be confusing the number of transpositions needed to create the cycle with the length of the cycle. The number of transpositions needed is *one less* than the length of the cycle.



> Some authors are contradicting themselves when they claim: "although one move in accordance with QTM creates 1 4-cycle of edges and 1 4-cycle of corners, an overall puzzle permutation is even (since 1 4-cycle + 1 4-cycle = 2 4-cycles)". This argument can be disproved by a fact that it allows (regardless of a permutation definition) to swap a corner with an edge. In reality, however, it is not possible.



This is just utter nonsense. It is perfectly acceptable to talk about permutations being constrained by various orbits. In fact, that's the whole point of having the term _orbit_ in the first place. Since swapping a corner piece and an edge piece violates an orbit constraint, we don't evem have to consider whether the permutation is even or odd - we know it's *not* a "valid" permutation of the puzzle because such a permutation violates an orbit constraint.


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## bcube (Mar 25, 2016)

cuBerBruce said:


> Yes, you are obviously misusing the word _permutation_.
> 
> You are trying to define the "puzzle permutation" (singular term) as *multiple separate* permutations (plural).



I agree. At least I am well aware of that (from terminology agreement: "By the way, a term "permutation of the piece types set" (in other words a puzzle permutation) doesn't make sense at all"). I will think about changing it, thank you for pointing it out. I guess a substitution of an odd permutation by a "parity problem" is not that bad in comparison with that misusage.



cuBerBruce said:


> Also, (as cmowla noted) you are incorrectly using the term "length of a cycle."



Honestly, I still don´t understand two things: 1) why the Czech mathematicians don´t use/define the term "parity" at all and 2) why I still think Czech and English definitions of a cycle differ (possibly because of my lack of understanding). I went from following: "each ordered n-tuple _(mathematical conditions here)_ we call a cycle of a permutation p of the length n-1" (I understand it as 2-cycle (n in n-tuple =2) has length 1) and "each cycle of length n can be decomposed to n transpositions" (since from the previous a 2-cycle has length 1, I see no contradiction why it couldn´t be possible to decompose it to 1 transposition (I understand a transposition as 1 2-cycle)). Therefore I see no hole in there...



cuBerBruce said:


> This is just utter nonsense.



Maybe it is possible you misunderstood? I am objecting about saying that 1 4-cycle + 1 4-cycle = 2 4-cycles in case of U move here (otherwise your _odd puzzle state_ doesn´t make sense to me).


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## bcube (Mar 25, 2016)

Hello Christopher,



Christopher Mowla said:


> clearly some things can be reworded to make the read better.



I plan to re-read it (again) and repair some sentence constructions. I translated the article literally, which was probably worse approach compared to "free" translation. 



Christopher Mowla said:


> I understood everything for the most part.



Any particular part or phrases which should be fixed with the highest priority (as misusing of a term "permutation" from the meaning point of view)?



Christopher Mowla said:


> I thought the "I cycle, you cycle, we cycle" phrase was hysterical!



Huh? Pardon me?



Christopher Mowla said:


> Also be sure to add in "N-cycle is executed N-times in a row (and not until then)...".



I checked the Czech (original) version and it says "N-cycle is executed exactly N-times in a row", so I will add "exactly" in there, hoping it will be ok.



Christopher Mowla said:


> Are you saying that when we say "the position generated by applying the move U2 is an even permutation" does not make sense?



cuBerBruce answered it for me (regarding misusing a term permutation). See my response to him regarding a length of a cycle.



Christopher Mowla said:


> For example, we can say 2+2+2+2+2+2+2 = 14 or we can say 2*7 =14.



I have mostly chosen first approach (i.e. The parity of the corners is even and the parity of the edges is even, instead of the parity of the (3x3x3) puzzle is even), believing it is correct as well. 



Christopher Mowla said:


> We define it in mathematics by saying that a "parity problem" (as you call it) "exists when the nxnxn supercube cube is not in the commutator subgroup".



It makes sense to me 



Christopher Mowla said:


> It's interesting that people who teach the general public rubbish



Oh come on, just because by "avoiding parity" Jon meant something different than you doesn´t mean he is teaching rubbish (if that´s what you meant). I find his videos amazing.


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## cuBerBruce (Mar 25, 2016)

bcube said:


> I am objecting about saying that 1 4-cycle + 1 4-cycle = 2 4-cycles in case of U move here (otherwise your _odd puzzle state_ doesn´t make sense to me).



Any two 4-cycles that are disjoint make two 4-cycles, by definition. Since the two 4-cycles here are not even in the same orbit, they are obviously disjoint. A U move on the 3x3x3 is a 4-cycle of edge pieces, and a 4-cycle of corner pieces. The total number of 4-cycles is 2. What part of that do you not understand?

It seems that your basic problem is that you have some sort of resistance to accept that permutations can include pieces from more than one orbit.

The MathWorld web site defines _orbit_ in relation to something called a group action. Basically, a group action, in some sense, describes how pieces can be moved around. For the group action to be complete, it *must* include all movable piece types, and thus, the resulting permutations must generally involve all piece types at once. But a group action also guarantees that the elements being permuted (that is, in this case, the pieces of the puzzle) will be partitioned into orbits. The elements of an orbit is a *subset* of the elements used in the permutations.

EDIT:
For example, I will illustrate using the 4x4x4 piece permutation group. (Orientation of pieces is ignored, so I simply assign some number to each of the 56 pieces of the 4x4x4. And all pieces are considered distinguishable, so we actually have a mathematical group.)

The GAP session below first defines 16 actions corresponding to a quarter turn move of each layer. Singmaster naming is used for convenvience. (2R is not a legal GAP variable name, for instance.) I then define the group determined by using those actions as generators. I then have GAP calculate the size of the group determined by those actions. I then ask GAP to tell me the orbits for this group for pieces numbered 1, 5, and 6, corresponding to the orbits for centers, corners, and edges, respectively.

Note the action U is described as four 4-cycles. GAP simply sees this as a permutation of numbers. It has no idea these numbers represent puzzle pieces, and it doesn't matter that more than one type of piece is represented by this permutation. Nevertheless, GAP correctly calculates what the orbits are. We have partitioned the 56 pieces into three disjoint sets, each set representing an orbit.


```
gap> U := (1,4,3,2)(5,14,11,8)(6,15,12,9)(7,16,13,10);
(1,4,3,2)(5,14,11,8)(6,15,12,9)(7,16,13,10)
gap> u := (17,26,23,20)(18,27,24,21)(19,28,25,22);
(17,26,23,20)(18,27,24,21)(19,28,25,22)
gap> d := (29,32,35,38)(30,33,36,39)(31,34,37,40);
(29,32,35,38)(30,33,36,39)(31,34,37,40)
gap> D := (41,44,47,50)(42,45,48,51)(43,46,49,52)(53,54,55,56);
(41,44,47,50)(42,45,48,51)(43,46,49,52)(53,54,55,56)
gap> L := (5,41,50,14)(15,17,52,38)(16,29,51,26)(27,28,40,39);
(5,41,50,14)(15,17,52,38)(16,29,51,26)(27,28,40,39)
gap> l := (1,30,56,25)(4,18,53,37)(6,42,49,13);
(1,30,56,25)(4,18,53,37)(6,42,49,13)
gap> r := (2,24,55,31)(3,36,54,19)(7,12,48,43);
(2,24,55,31)(3,36,54,19)(7,12,48,43)
gap> R := (8,11,47,44)(9,23,46,32)(10,35,45,20)(21,22,34,33);
(8,11,47,44)(9,23,46,32)(10,35,45,20)(21,22,34,33)
gap> F := (5,8,44,41)(6,20,43,29)(7,32,42,17)(18,19,31,30);
(5,8,44,41)(6,20,43,29)(7,32,42,17)(18,19,31,30)
gap> f := (1,21,54,40)(2,33,53,28)(9,45,52,16);
(1,21,54,40)(2,33,53,28)(9,45,52,16)
gap> b := (3,27,56,34)(4,39,55,22)(10,15,51,46);
(3,27,56,34)(4,39,55,22)(10,15,51,46)
gap> B := (11,14,50,47)(12,26,49,35)(13,38,48,23)(24,25,37,36);
(11,14,50,47)(12,26,49,35)(13,38,48,23)(24,25,37,36)
gap> G444p := Group (U,u,d,D,L,l,r,R,F,f,b,B);
<permutation group with 12 generators>
gap> Size (G444p);
7760717379340758543105097963403817082945536000000000
gap> Orbit (G444p, 1);
[ 1, 2, 25, 40, 3, 31, 28, 56, 24, 37, 54, 4, 19, 34, 55, 27, 53, 30, 36, 21, 22, 18, 39, 33 ]
gap> Orbit (G444p, 5);
[ 5, 14, 41, 8, 11, 50, 44, 47 ]
gap> Orbit (G444p, 6);
[ 6, 15, 42, 20, 12, 17, 51, 45, 49, 10, 43, 9, 48, 26, 52, 7, 46, 13, 35, 29, 23, 16, 38, 32 ]
```


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## bcube (Mar 26, 2016)

cuBerBruce said:


> Any two 4-cycles that are disjoint make two 4-cycles, by definition. Since the two 4-cycles here are not even in the same orbit, they are obviously disjoint. A U move on the 3x3x3 is a 4-cycle of edge pieces, and a 4-cycle of corner pieces. The total number of 4-cycles is 2. What part of that do you not understand?



I don´t understand how you can add two sets of elements.



cuBerBruce said:


> It seems that your basic problem is that you have some sort of resistance to accept that permutations can include pieces from more than one orbit.



Exactly. I understand a permutation as "mutually unambiguous (i.e. injective and onto (itself) - collectively so-called bijection) mapping of a finite set of the elements (where as an element we understand a piece of the puzzle and a set is given by an orbit, in which these pieces appear)."

That being said, the Rubik´s cube 3x3x3 consists of 3 sets, namely:
Corners = {c1, c2, c3, c4, c5, c6, c7, c8}
Edges = {e1, e2,..., e11, e12}
Centers = {ce1, ce2,..., ce5, ce6}

If I say I performed 2 2-cycles of Edges, a H-perm could be done, for example. If I say I performed 2 2-cycles (on the 3x3x3 Rubik´s cube), how do you know I didn´t swap 1 corner with 1 center and simultaneously another center with 1 edge? If I say I performed 2 2-cycles of Corners+Edges, how do you know I didn´t swap 2 edges and simultaneously 2 corners (T-perm, for instance)? 

So, how can you say 1 2-cycle + 1 2-cycle = 2 2-cycles in case of T-perm? Also, how would you define (if defined) your odd puzzle state on the 3x3x3 Rubik´s cube, if U move would perform 2 4-cycles?


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## cuBerBruce (Mar 26, 2016)

bcube said:


> I don´t understand how you can add two sets of elements.



What two sets? I have 1 set of elements of all the pieces of the puzzle. See my GAP example.



bcube said:


> Exactly. I understand a permutation as "mutually unambiguous (i.e. injective and onto (itself) - collectively so-called bijection) mapping of a finite set of the elements (where as an element we understand a piece of the puzzle and a set is given by an orbit, in which these pieces appear)."



Inside the parentheses, you needlessly and subtly impose additional conditions on what you consider a permutation to be. As a result, your definition of permutation differs from what mathematicians accept as what a permutation is.



bcube said:


> That being said, the Rubik´s cube 3x3x3 consists of 3 sets, namely:
> Corners = {c1, c2, c3, c4, c5, c6, c7, c8}
> Edges = {e1, e2,..., e11, e12}
> Centers = {ce1, ce2,..., ce5, ce6}



These three sets are subsets of the set of all puzzle pieces. The group action of the puzzle partitions the full set into these three disjoint subsets called orbits.

As an aside, I'll note that orbits are not universally determined by the types of pieces. Consider the Curvy Copter puzzle, where we consider only 180-degree turns, not any "jumbling" moves. The puzzle has 8 corner pieces, 12 edge pieces, and 24 center pieces. How many orbits does it have?

Answer:


Spoiler



17. Each edge piece can only rotate in place, so each edge piece is in its own orbit. There is one orbit of corner pieces. The center pieces (under the assumption of no jumbling moves) are in 4 orbits of 6 pieces each. 12 + 1 + 4 = 17.





bcube said:


> If I say I performed 2 2-cycles of Edges, a H-perm could be done, for example. If I say I performed 2 2-cycles (on the 3x3x3 Rubik´s cube), how do you know I didn´t swap 1 corner with 1 center and simultaneously another center with 1 edge?



Because the group action of the puzzle does not permit those operations to occur. In my GAP example, show me some composition of the "operations" { U,u,d,D,L,l,r,R,F,f,b,B } that sends piece #5 (a corner) to piece #1 (a center). I am sure you will not be able to come up with any such composition.



> So, how can you say 1 2-cycle + 1 2-cycle = 2 2-cycles in case of T-perm?


Because I simply *count* them. The count that I get is 2.

If you have a dog and a cat, how many animals do you have? By my count, you would have 2 animals. I wouldn't say you have 1 animal because a dog and a cat aren't the same type of animal.



> Also, how would you define (if defined) your odd puzzle state on the 3x3x3 Rubik´s cube, if U move would perform 2 4-cycles?



Mathematicians learn to state things in a precise manner. If a mathematician is going to claim a certain puzzle state is "odd," he will first clarify what he means for a puzzle state to be odd or even. He might, for example, say that a particular puzzle state is odd if the corner pieces are in an odd permutation. Or he might choose to say a particular puzzle state is odd if it requires an odd number of quarter turns (of face layers) to put the puzzle back into the solved state. (For this latter choice, there is an assumption which should be proved that if one way to solve the puzzle uses an odd number of quarter turns, then every path to the solved state would have to use an odd number of quarter turns.) He could also say that what he means for a puzzle state to be odd is that the overall permutation parity of all pieces is odd. Under this definition, no legal state of the puzzle would be considered odd, so it would be useless in terms of distinguishing any of the legal states of the puzzle.


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## Christopher Mowla (Mar 26, 2016)

cuBerBruce said:


> *Under this definition*, no legal state of the puzzle would be considered odd, so it would be useless in terms of distinguishing any of the legal states of the puzzle.


Exactly bcube's point.

That definition is not how it generally works for parity states on the nxnxn, and therefore bcube is not using that definition.

It should be obvious that he is referring to the nxnxn in general, because he mentioned my C(n,w,c) vector function. So what is it that he (and I) are missing?


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## cuBerBruce (Mar 26, 2016)

Christopher Mowla said:


> Exactly bcube's point.
> 
> That definition is not how it generally works for parity states on the nxnxn, and therefore bcube is not using that definition.
> 
> It should be obvious that he is referring to the nxnxn in general, because he mentioned my C(n,w,c) vector function. So what is it that he (and I) are missing?



If you read my post, my point was that there are different possible interpretations of what it would mean for a cube state to be "odd" (vs. "even"). My point is that a mathematician would make it clear precisely what he means by "odd" or "even" first, before applying it to a given cube state. What I was saying there has nothing to do with with bcube's position. bcube was *asking me* about about what I would say.

bcube seems to want his definition of "even" vs. "odd" to be universal, whereas "mathematicians" have the flexibility to discuss "even" vs. "odd" for a specific orbit, for a specific set of orbits, all orbits of the puzzle, etc. I like the flexibility to define "even" vs. "odd" differently depending on what makes sense for whatever I happen to be discussing, as long as I'm clear what I mean for that particular discussion.


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## bcube (Mar 26, 2016)

cuBerBruce said:


> What two sets? I have 1 set of elements of all the pieces of the puzzle.



Subsets then (Corners + Edges - just to be clear, I am still talking about my 3x3x3 example here). Whatever they are called, they represent the sets in "mutually unambiguous (i.e. injective and onto (itself) - collectively so-called bijection) mapping of a finite set of the elements", agreed? (if not, please explain in more detail what you are objecting about my term odd puzzle permutation)



cuBerBruce said:


> Inside the parentheses, you needlessly and subtly impose additional conditions on what you consider a permutation to be. As a result, your definition of permutation differs from what mathematicians accept as what a permutation is.



I can´t put my finger on what´s wrong with my text inside the parentheses. Could you please explain it to me in more detail (with practical example, if possible, pretty please)?



cuBerBruce said:


> orbits are not universally determined by the types of pieces.



Agreed. I believe I nowhere said they are.



cuBerBruce said:


> Because the group action of the puzzle does not permit those operations to occur.



Very well. Replace 1 edge + 1 center and 1 corner + 1 center swaps by two 1 corner + 1 corner swaps then. 



cuBerBruce said:


> If you have a dog and a cat, how many animals do you have?



That depends on a definition of animal.



cuBerBruce said:


> By my count, you would have 2 animals.



But a permutation is not about how many animals do you have, rather which animals do you have, agreed? (if not, please explain in more detail what you are objecting about my term odd puzzle permutation)



cuBerBruce said:


> ... overall permutation parity of all pieces ...



Excuse me, what is this? How is this defined? Do I understand it correctly that it is equivalent / synonym to my puzzle permutation with regard to the puzzles?


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## Christopher Mowla (Mar 26, 2016)

cuBerBruce said:


> If you read my post, my point was that there are different possible interpretations of what it would mean for a cube state to be "odd" (vs. "even").
> 
> My point is that a mathematician would make it clear precisely what he means by "odd" or "even" first, before applying it to a given cube state.


If you read the beginning of his article, you will see that he made it very clear of what he meant.



> If we keep talking about a Rubik's cube 3x3x3, then e.g. *F move is an odd permutation of edges and at the same time an odd permutation of corners*.





cuBerBruce said:


> What I was saying there has nothing to do with with bcube's position. bcube was *asking me* about about what I would say.
> 
> bcube seems to want his definition of "even" vs. "odd" to be universal, whereas "mathematicians" have the flexibility to discuss "even" vs. "odd" for a specific orbit, for a specific set of orbits, all orbits of the puzzle, etc. I like the flexibility to define "even" vs. "odd" differently depending on what makes sense for whatever I happen to be discussing, as long as I'm clear what I mean for that particular discussion.


I understand now. He said that they were "contradicting themselves", which is probably what led you to call his words "utter nonsense"



> Some authors are contradicting themselves when they claim: "although one move in accordance with QTM creates 1 4-cycle of edges and 1 4-cycle of corners, an overall puzzle permutation is even (since 1 4-cycle + 1 4-cycle = 2 4-cycles)". This argument can be disproved by a fact that it allows (regardless of a permutation definition) to swap a corner with an edge. In reality, however, it is not possible.


, but these two definitions *do contradict each other*.

But whose to say that you cannot assume one or the other *separately* when talking *just about the 3x3x3*. But then again, why would YOU even mention this other possibility if you were writing this article when clearly this definition does not fit the given context?
For those who do not understand why we cannot consider this definition to apply to the nxnxn,


Spoiler



I made a formula which "counts the number of composed 2-cycles" an outer face quarter turn does "S(n)" and the formula which counts "the number of composed 2-cycles" some number of inner slice quarter turns (w) does (on the 4x4x4 and larger cubes) "S(n,w)".

_S_(_n_,_w_), _S_(_n_)


Spoiler











That is, I made a formula which counts 3 2-cycles for each and every 4-cycle because three overlapping 2-cycles creates a 4-cycle.

Clearly, S(6) = 27 which is an odd number. If we assumed all pieces on the 6x6x6 to be in the same set, we would use this formula, it would tell us that "If we treat the 6x6x6 supercube as a 3x3x3 (turn outer face turns only), then the combined total of swaps of the "tetredges", corners, and non-fixed centers is *odd*." However, it should be _even_ since we are treating the 6x6x6 as a 3x3x3 as we are restricting moves to <U,D,F,B,L,R>.

And, in general, it looks like this formula tells us that even cube sizes which are not evenly divisible by 4 are the ones with an odd number of swaps by a single move such as U.


So maybe bcube said that some people are contradicting themselves because this definition contracts itself on the nxnxn cube. Clearly, as I and you have said before, this definition only applies to the 3x3x3, not the nxnxn, (and bcube knows this very well to not even consider it, which is _why_ he is not even considering it when you are pushing it on him for an *article which is about the nxnxn*) but yeah.

I guess I said enough.


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## bobthegiraffemonkey (Mar 26, 2016)

Christopher Mowla said:


> Clearly, S(6) = 27 which is an odd number. If we assumed all pieces on the 6x6x6 to be in the same set, we would use this formula, it would tell us that "If we treat the 6x6x6 supercube as a 3x3x3 (turn outer face turns only), then the combined total of swaps of the "tetredges", corners, and non-fixed centers is *odd*." However, it should be _even_ since we are treating the 6x6x6 as a 3x3x3 as we are restricting moves to <U,D,F,B,L,R>.



A U turn is an odd permutation of the 6x6 (nine 4-cycles in total). However, if we view it as a standard 3x3 it is an even permutation by the same definition (two 4-cycles). It's a different context so there is no reason to expect the same answer - they are different questions. Further, if we continue to deal with a supercube and look at a typical 3x3 supercube, we may have four markings on each centre which indicate the colour of adjacent faces and a U turn will make a 4-cycle of these too, so a U turn on a 3x3 supercube will be an odd permutation in that context (three 4-cycles). It all depends on the context, what's so difficult about that?


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## cuBerBruce (Mar 26, 2016)

bcube said:


> Subsets then (Corners + Edges - just to be clear, I am still talking about my 3x3x3 example here). Whatever they are called, they represent the sets in "mutually unambiguous (i.e. injective and onto (itself) - collectively so-called bijection) mapping of a finite set of the elements", agreed? (if not, please explain in more detail what you are objecting about my term odd puzzle permutation)



In my GAP example, I have 12 permutations defined, named U, u, d, etc. They all represent bijective mappings of a set unto itself - a set of 56 elements. (The permutations are specified in cycle notation.) You would seem to claim these permutations are not valid permutations because the set includes corners, edges, and centers. There is no need to make such a preposterous restriction on what constitutes a permutation.



bcube said:


> I can´t put my finger on what´s wrong with my text inside the parentheses. Could you please explain it to me in more detail (with practical example, if possible, pretty please)?



As stated above, that you can't have the set include multiple types of pieces. (For whatever reason, I don't know.)



bcube said:


> Agreed. I believe I nowhere said they are.



Nor did I actually claim that you did. But the way you simply state from the outset what the orbits are does somewhat lead to an impression that you infer the orbits based upon the piece types.

From my point of view:
1. Start with the set of all pieces
2. Define the allowed "operations" of the pieces using that set.
3. *Derive* what the piece orbits are.

You seem to start with what the orbits are, and you can only talk about how the pieces move around after that. To me, this seems kind of like putting the cart before the horse.



bcube said:


> Very well. Replace 1 edge + 1 center and 1 corner + 1 center swaps by two 1 corner + 1 corner swaps then.



It still violates the group action. You can not show me a composition of the given permutations that results in a pure swap of 2 corners. (I think that's what you meant.)



bcube said:


> That depends on a definition of animal.



This seems to be nothing but trolling... I ignore.



bcube said:


> But a permutation is not about how many animals do you have, rather which animals do you have, agreed?



No, you were asking about the number of 2-cycles, not the number of permutations. Counting the number of 2-cycles is like counting the number of animals. The 2-cycles may involve different piece types, but a 2-cycle is still just a 2-cycle regardless of the piece types. Which animals we have is like which orbits we are dealing with. You didn't ask how many orbits, you asked about how many 2-cycles. As I hope you know, cycles are like building blocks of permutations, kind of like how prime numbers are said to be the building blocks of the natural numbers.



bcube said:


> (if not, please explain in more detail what you are objecting about my term odd puzzle permutation)



I was merely answering your question about counting 2-cycles, and now you switch to talking about "odd puzzle permutation."

Well, as I said many times, you want to redefine what the term permutation means, and that you consider it nonsense that different piece types can be part of a single permutation. I don't object exactly to what what you are defining as what you mean by "even" and "odd," but I think any mathematician would interpret "odd puzzle permutation" to mean the permutation parity of all the pieces of the puzzle to be odd, and you just seem to be totally opposed to even allowing that basic concept.



bcube said:


> Excuse me, what is this? How is this defined? Do I understand it correctly that it is equivalent / synonym to my puzzle permutation with regard to the puzzles?



Permutation parity is a well-defined mathematical concept. It is the same thing you refer to as the sign of a permutation, except sign is talked about as +1 or -1 rather than even or odd. "All pieces" means we consider how each and every piece can be moved around. I don't have any idea why you object that we can have a single permutation that does this.


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## Christopher Mowla (Mar 26, 2016)

bobthegiraffemonkey said:


> A U turn is an odd permutation of the 6x6 (nine 4-cycles in total). However, if we view it as a standard 3x3 it is an even permutation by the same definition (two 4-cycles).


"Viewing it as a 3x3x3" means to considering turning only the outer layer faces, right? So you cannot possibly think of it as two 4-cycles *of physical pieces*.



bobthegiraffemonkey said:


> It's a different context so there is no reason to expect the same answer - they are different questions.


Yeah, that's what I've been saying all along. What are you trying to prove?



bobthegiraffemonkey said:


> Further, if we continue to deal with a supercube and look at a typical 3x3 supercube, we may have four markings on each centre which indicate the colour of adjacent faces and a U turn will make a 4-cycle of these too, so a U turn on a 3x3 supercube will be an odd permutation in that context (three 4-cycles). It all depends on the context, what's so difficult about that?


Are really asking me if _*I*_ think it's difficult!? No, I don't think it's difficult. I don't think I ever showed that I was confused about the matter either.


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## bobthegiraffemonkey (Mar 26, 2016)

Christopher Mowla said:


> However, it should be even since we are treating the 6x6x6 as a 3x3x3 as we are restricting moves to <U,D,F,B,L,R>
> 
> "Viewing it as a 3x3x3" means to considering turning only the outer layer faces, right? So you cannot possibly think of it as two 4-cycles *of physical pieces*.



There is a difference between a 3x3x3 and using a 6x6x6 to emulate a 3x3x3, but you seemed to say you think we should get the same answer in both cases. For the 6x6x6 allowing <U,D,F,B,L,R> a U move is odd parity, for a 3x3x3 allowing <U,D,F,B,L,R> a U move is even parity (in each case an odd/even number of 4-cycles respectively). Getting different answers is fine.


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## bcube (Mar 27, 2016)

cuBerBruce said:


> In my GAP example, I have 12 permutations defined, named U, u, d, etc. They all represent bijective mappings of a set unto itself - a set of 56 elements. (The permutations are specified in cycle notation.) You would seem to claim these permutations are not valid permutations because the set includes corners, edges, and centers. There is no need to make such a preposterous restriction on what constitutes a permutation.



Oh, I didn´t think of that. Thank you very much for giving me that counter-example (and repeating it all over again). It is sad it took me so long to understand it, but I can finally see how 1 2-cycle + 1 2-cycle in case of T-perm really is equal to 2 2-cycles. Time to re-write me article 



cuBerBruce said:


> You seem to start with what the orbits are, and you can only talk about how the pieces move around after that.



Exactly. And that´s what had caused my confusion. The problem is that I imagined a set as an orbit (i.e. actual subset) in a permutation definition from the beginning. That approach is really not bad in a puzzle solving (I would call it "solving approach" (untill now I called it mathematical definition)), but a mathematical definition (as I understand it now) is more general and allows you to imagine a set whatever legal you like (e.g. all three subsets mentioned earlier together no problem).

The idea of so-called parity states is based on that "solving approach", because it is handy to think of orbits when solving the puzzles (it feels natural for me, at least). It is just one way how to look at the permutation (i.e. one way how to imagine the sets/subsets). However, when we imagine all the pieces of the 3x3x3 Rubik´s cube as a set (because a permutatin definition allows it), we will reach different conclusion.



cuBerBruce said:


> It still violates the group action. You can not show me a composition of the given permutations that results in a pure swap of 2 corners. (I think that's what you meant.)



I meant 2 2-cycles of corners. When imaging a set as an orbit, it really is impossible to tell whether 2 edges + 2 corners or 2 corners + 2 corners or 2 edges + 2 edges have been swapped when saying that 2 2-cycles were performed (and since I imagined a set as an orbit, I claimed from the beginning generally just it is ambiguous what I just said). When imaging a set as all the puzzle pieces, it is fine to say 2 2-cycles were performed for both 2 edges + 2 corners and 2 corners + 2 corners and 2 edges + 2 edges swaps.



cuBerBruce said:


> you consider it nonsense that different piece types can be part of a single permutation



Not anymore . After realizing (because of your explanations) that a set doesn´t have to be an orbit necessarily, I don´t consider it nonsense anymore.



bobthegiraffemonkey said:


> Getting different answers is fine.



There. I am happy to say we can be unified.



Christopher Mowla said:


> So what is it that he (and I) are missing?



Thinking of a set as all puzzle pieces is the key.


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## bcube (Mar 27, 2016)

cuBerBruce said:


> "mathematicians" have the flexibility to discuss "even" vs. "odd" for a specific orbit, for a specific set of orbits, all orbits of the puzzle, etc.



Since my deduction is weak, could you please outline a way for me how to talk about the puzzle pieces/orbits without misusing the term permutation, please?



cuBerBruce said:


> You are trying to define the "puzzle permutation" (singular term) as multiple separate permutations (plural).
> 
> Mathematicians are happy to talk about a permutation where the elements are constrained to certain orbits, but such a permutation of all elements does not mesh with what you want to call an odd puzzle "permutation." Perhaps you should use odd puzzle state since what you are trying to define as being "even" or "odd" is not a (singular) permutation.



I can see how a U move creates 2 4-cycles on the 3x3x3 Rubik´s cube. That´s because a set is formed by 26 pieces. If considered 26 pieces in a set, I don´t see a way how to say a U move is an even permutation (of 3x3x3 cube), because I would be using a singular term as plural term (in other words "odd permutation of corners and odd permutation of edges" doesn´t make sense to me either if considered a set of 26 pieces).


edit: so far I perceive it as follows (please let me know if my perception is fully correct):

1) when performing a U move on the 3x3x3 cube and a set is formed by 26 pieces, I can say a permutation (of a set, i.e. the cube) is even, because 1 4-cycle + 1 4-cycle = 2 4-cycles (in general, I just sum up the N-cycles and if the number of transpositions (i.e. 2-cycles) is odd, an permutation of a set (i.e. the puzzle) is odd - this conception allows me to sum up N-cycles when N is not constant). For the 3x3x3 cube, I can not say anything about a permutation of corners, edges or centers, because I would be misusing the term permutation.

2) when performing a U move on the 3x3x3 cube and a set is formed by 8 pieces called corners (I believe a permutation definition allows me to do that) and another set is formed by 12 edges and another set is formed by 6 centers, then I can say a permutation of corners is odd, a permutation of edges is odd and a permutation of centers is even. Now I must introduce another term, let´s say "odd/even puzzle state" (otherwise I would be misusing the term permutation), if I want to talk about an equivalent ot that "permutation of a set (i.e. the puzzle)" mentioned in the previous paragraph. The permutations of piece types determine my puzzle state (odd puzzle state if at least one permutation of piece type is odd, even puzzle state if there is no odd permutation of piece type).


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## bcube (Apr 13, 2016)

New article "version" is up.

Main changes (if you are still interested):

- "puzzle permutation" mostly replaced by "puzzle status" - therefore I introduced the term puzzle status
- what is meant by a permutation is now better explained, some paragraphs added in the first half of an article
- edited a paragraph about authors´ contradiction regarding 1 move on 3x3x3 cube
- edited length of cycle "definition" (thank you Christopher and cuBerBruce for pointing it out)
- added and deleted some "a"s and "the"s  , some sentence structures changed, some typing errors fixed

I bet a mathematician can find an article to be wrong (or misused) anytime he/she wants (still I hope the meaning of used terms is now better explained than before) - if you think something should be fixed/clarified, please let me know.
There are a lot of terms used, so let me know if you got lost quickly or not.
I find some English articulation to be odd at least, however, I think I have reached my current English language grammar limits. I did my best as of now regarding an English sentence structures.

I would like to know very much what speedcubers think about that article (go ahead and criticize whatever you like  )? I am sorry Christopher and cuBerBruce, I don´t consider any of you as a typical speedcuber


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