Math Is Fun Forum

Well, I guess I'll wait for MIF to see this thread and respond to the following, since you have answered my questions.

How long exactly (I'm not being impatient, I just feel that the specifics of the information being given is very unspecific)?  Besides, a specific answer will help those in the future who read this thread.  (Maybe it or a formal post by you or MIF or one of the moderators should be made as a sticky to explain the rules for new members...I didn't see it in the rules section.)

So in my posts, I can, for example, put a sentence at the beginning of the post asking you to convert all of my descriptions of links (including [img]image link[/img]) into links?  If so, I will do that if I really would prefer to have a link in my posts instead of a description, but I hope once MIF sees this conversation and does something so that he can save you that trouble. 

I have recently joined the wolframalpha community, and they seem to be very cautious towards spam too (but very extreme when compared to here):  they won't even let a post be displayed until they've read it, which is very annoying (because if you make a typo or forgot to add something, you have to wait hours or sometimes days before you even can see or edit your post--not to mention that people are waiting on a response already).  I think these anit-spam preventions are taking things too far.  I feel like the majority of us are being penalized because a few members have abused the system.

As for this forum's rules,
What's the difference if someone is a member for one hour or let's say 1 week (just guessing what the time limit is).  If that person wants to spam, he's going to whenever is allowed to.  Chances are, if someone joins the site, he/she is probably going to post ten things the first day he/she joins so that (in theory...which isn't reality since there is also a time limit) he/she can then post spam links.  I highly doubt those who join just for the purposes of spamming are going to have the patience to be a member (as in just joining and being a "novice") more than a minute to find out that they cannot spam.  In fact, I don't think a spammer is going to have the patience to post 10 intelligent conversation posts before he/she can "show his/her 'true' motive".

So I don't mind the post number rule (just to get rid of the "quick and easy" first post = spam rule), but I think the post number rule and the time limit rule actually have a negative effect on almost all members who join.  Reason:  it is perfectly normal in conversation on the internet to post links even in the middle of sentences.  Why should we expect to have more active members if the first thing they see is "you cannot post links because we cannot trust you yet" when they try to make a post?  That made me feel like I was just hired for a job, and I'm on probation...yeah, a very warm welcome.  In fact, it is an insult to everyone but the spammers, despite that you guys did give me (and I have noticed almost all--if not all posting members) a warm welcome in writing when we did our first post.

I hate being so blunt, especially in public (it is as if I was intentionally trying to insult or embarrass you all in front of everyone who has access to the internet), but I wasn't given any other way of contact so I hope none of this is taken the wrong way.

Perhaps if you see it this way, it might make sense to you (it helped me when I taught myself this way).

I'll show the solution to the first problem.

Note that it really helps if you make an effort to line up the equal signs when you write one equation on top of the other.  Also, keep x and y terms and the number term lined up too (a term means something you add or subtract.  For example, in the equation A + B + C = D, A, B, C, and D are terms).

Two main points about doing this process.
1) Whatever you do to one side (or even just one thing that you add or subtract with the rest), you must do to the rest, otherwise you change that equation.

2) The main idea is to end up adding

, where A is some number (or fraction) OR

.  So you only need to change one of the two equations, not both.  You can, but if you're comfortable working with fractions, then you'll only have to change one for any numbers involved.

Also, I use the

symbol which just means "implies that the following is true".  Usually people just put equal signs in between lines of work, but sometimes everything isn't equal!  So that's why I use that  instead of an equal sign.

So for two equations with two variables, there are 4 (easy) ways we can find the first variable and two ways we can find the second variable.

Thanks.  I love that version of the formal definition!

Sorry that I implied that they themselves are "stand alone" arguments, but in the proof I have seen for the product of the limits, for example, the only appearances of delta is to restate the restriction of the formal definition (the relationship between chosen epsilons and their corresponding deltas), but I think epsilons are just used in the actual arithmetic.

I stated the formal definition at the beginning of that post to state it.  So, if these arguments are correct, then I just could have added the relationship between the chosen epsilon and the corresponding delta...is that all you were implying?  (As I have said, I apologize for not mentioning that the arguments are not stand alone ones, i.e., the "background information" is provided for all arguments at the beginning of that post).

I still am not sure if these arguments are valid proofs (given now that I explained that I "abbreviated" them by providing background information for all arguments beforehand).

Here's an argument that 0^0 could be 0 or 1.

This argument may not be valid (I came up with it, and this is the first place I am placing it in for criticism), but it claims that 0^0 is both 1 and 0 in ordinary arithmetic!

The easiest way to find two points, especially when the equation is written this way, is to find the x and y intercepts.

[1] We can pretend y is zero to find the x-coordinate of the y intercept  (the y-coordinate is zero since we set it to be).

Now we solve for x.

So

is the x-intercept point.

[2] Now we set x equal to zero to find the y-coordinate of the x intercept (the x-coordinate is zero since we set it to be).

So

is the y-intercept point.

Hi everyone, this is my first post on this forum!  I tend to spend time on writing up my posts, so they are often times lengthy if I have enough to say on the subject!

I have been studying Rubik's Cubes of all sizes for several years, and I have been a part speedsolving.com forum for almost three years.  Thus I have some things to say about some of the questions and the responses to those questions.

I'm annoyed to see that I cannot post links since I'm a new member.  However, I will mention the key phrase to search for in google, and the web-page I'm referring to is the first result.

Actually, there is a simple way, but it can be tedious.  Before I explain this, I need to provide a little background information.

The 3x3x3 Rubik's Cube Pieces
It has:
12 Edges
8 Corners
6 fixed centers (they only rotate on the xyz axis core:  they don't ever move with respect to each other)

Just because the order of the Rubik's Cube Group (the diameter) is approximately 43 billion billion, there is no possible way that it can have a cycle greater than 12, because the most abundant piece type is the edges, and there are only 12 of them.

To see a powerpoint presentation I made for my university which shows how to come up with the formula for the number of possible configurations on the nxnxn Rubik's Cube, >>Link:  Look up "Christopher Mowla UNO" in Google.  It should be the first result, which is a 4.3mb powerpoint presentation.

On the Rubik's Cube, we can have more than one disjoint cycle of the same piece type of which can be different size cycles or multiples of the same cycle type.

Here is a list of all of the possible cycle types there can be for the edges of the 3x3x3 Rubik's Cube.  The cycle type is in curly braces.

But before I show the list, I want to introduce some "cubing" terminology.
Permutation - Describes when two or more pieces are cycled with each other.
Orientation - Describes when two or more pieces are cycled with themselves.  This occurs when a corner or an edge is in its solved location slot, but is twisted (common talk is a corner is twisted, and an edge is flipped).  If a corner or an edge is in some other slot (the edge can be in 11 other slots other than its own, since there are 12 edges and the corner can be in 7 other slots than its own, since there are 8 corners) and is twisted (or flipped if it's an edge), then it is NOT cycled with itself anymore:  it's only cycled with the corner (or edge) whose slot it's currently in (and there might be more corners and/or edges involved in the cycle).

So, according to the format of the list of all possible cycles below, a one-cycle is "{1}".  A 1-cycle represents either the orientation of a piece or that that piece was not affected by an algorithm, not the permutation.

It turns out that you can only have an even number of edges flipped and, if you define a frame of reference for the direction which a corner is twisted, then the sum of the orientation of the corners must be divisible by 3.

>>Link:  See "Laws of the Cube" (Ryan Heise)

>>Also, in my Powerpoint presentation, slide 74 has a great illustration for the corner law.  It will help those who don't understand to know what I am talking about.

Now, here's the list.
The Number of Possible Cycle Types for the Edges of the 3x3x3 Rubik's Cube

Notes:
*Since I mentioned a 1-cycle is not considered a "Permutation" in "cubing" terminology, I will disregard the 1-cycles.
*In order to list all possible cycles for n pieces, one must also list all cycle types for 2 to n-1 pieces as well (note that I started with 2 since I'm avoiding the mathematical "1-cycle" and separating it to be an "orientation" and not a "permutation").
*Examples:  {2} means "one 2-cycle", {4,2,2} means "one 4-cycle, and two 2-cycles", etc.
*For the corners' cycle types (classes), you can just look at the first 7 results, but all 11 are for the edges' cycle types.

2 Pieces
{2}

3 Pieces
{3}

4 Pieces
{4}, {2,2}

5 Pieces
{5}, {3,2}

6 Pieces
{6}, {4,2}, {3,3}, {2,2,2}

7 Pieces
{7}, {5,2}, {4,3}, {3,2,2}

8 Pieces
{8}, {6,2}, {5,3}, {4,4}, {4,2,2}, {3,3,2}, {2,2,2,2}

9 Pieces
{9}, {7,2}, {6,3}, {5,4}, {5,2,2}, {4,3,2}, {3,3,3}, {3,2,2,2}

10 Pieces
{10}, {8,2}, {7,3}, {6,4}, {6,2,2}, {5,5}, {5,3,2}, {4,4,2}, {4,3,3}, {4,2,2,2}, {3,3,2,2}, {2,2,2,2,2}

11 Pieces
{11}, {9,2}, {8,3}, {7,4}, {7,2,2}, {6,5}, {6,3,2}, {5,4,2}, {5,3,3}, {5,2,2,2}, {4,4,3}, {4,3,2,2}, {3,3,3,2}, {3,2,2,2,2}

12 Pieces
{12},{10,2},{9,3},{8,4},{8,2,2},{7,5},{7,3,2},{6,6},{6,4,2},{6,3,3},{6,2,2,2},{5,5,2},{5,4,3},{5,3,2,2},{4,4,4},{4,4,2,2},{4,3,3,2},{4,2,2,2,2},{3,3,3,3},{3,3,2,2,2},{2,2,2,2,2,2}

As you can see, the total number of cycle types for n pieces is the sum of the number of cycle types for n-1 pieces, and for each piece, it is the number of combinations in which integers less than or equal to itself which add up to itself.

You can compute these with Mathematica (I don't know the code for WolframAlpha) by "IntegerPartitions[n, All, {2, 3, 4, ..., n-1, n}]"

In addition, you can compute the number of each cycle type by using the formula found below.  The sum of all of the different possible cycle types adds up to n! for n objects (for this, we do consider all of the 1-cycles/pieces which are untouched).
>>Link:  Look up "Cycle classes of permutations" in Google.  It's the first result.

Now, as to why, bob bundy, I mentioned that there is an easy but tedious way to construct an algorithm which can generate any cycle type.  First of course, is to mention that there can be even permutations or odd permutations (the term "permutations" here is interchangeable with cubing terminology and mathematics).  If a cycle type (recall that in the list above, a cycle type was {#, another #,...}.  That is, the cycle type includes everything in the curly braces) has an overall odd permutation, then the algorithm we use to construct it must have an odd number of outer layer quarter turns.  If the cycle type is an overall even permutation, then the algorithm we construct must have an EVEN number of outer layer quarter turns.  When I say "outer layer quarter turns," I mean that if you turn the middle slice one quarter turn, it's actually 2 quarter turns (an even permutation).  But a quarter turn, say, R, gives an odd permutation.

Examples:
Even Permutations {3}, {3,3},{2,2}, {2,3,6}
Odd Permutations {2},{4},{6},{8},{2,3}

CUBE LAW
On slide 83 of my power-point presentation, I basically say that if there is an odd permutation in the edges, there is an odd permutation in the corners, and vice versa (an if and only if statement).

If your desired/goal cycle class type to create is an odd permutation, you need to do an odd permutation algorithm an odd number of times to the cube.  It's probably best to do a 2-cycle PLL algorithm which only affects two edges and two corners.
>>Link:  Look up "PLL speedsolving wiki" in Google and see the algorithms (they have links to an online java applet) F, J, N, R, T, V, and Y permutations.

You need to look up video tutorials on commutators and conjugates.
I made a "Derivation Video" on a 4x4x4 "Parity Algorithm" I made (which wasn't made before in the 30+ years the 4x4x4 cube was out), and in the first part, I talk a little about commutators and conjugates (conjugates are commonly called "setup moves" in cubing terminology:  where you do some moves before hand and you undo after some known algorithm).

You can watch the video starting at 4:02 to 12:00.
>>Link:  Look up "derivation of HG algorithm part 1" in Google.  It will be a You-Tube video (it's the first result).
In that video when I mention "slots" around 11:40, see slide 65 of my power-point presentation to see a good visual of what the difference is between a "slot" and a "piece".

(You can watch all 4 parts if you wish, but I think that portion of the first video and maybe the middle portion of the second video is more than enough (a lot of that material is for doing more advanced tasks).  And of course, you can watch any videos or read any guides that will help you.).
(If you like the cube software I used in the video, just see my cubetwister video for supercube stickers to get all of the information on installation).

The commutator in my video swaps only 3 pieces on the cube.  We can apply it to edges (e.g. M' U R U' M U R' U' = [M', U R U']) or to corners (R U' L' U R' U' L U = [R, U' L' U]) and just affect those pieces separately.

Now, we can make a 5-cycle, for example, by doing two applications of a 3-cycle commutator (like the one above).
Using 2 line permutation notation with the convention that the following is the identity (a subset of a solved orbit of either corners or edges):

1) Apply a 3-cycle commutator to a solved cube which affects slots A, B, and C.

2) Apply the same commutator to, for example, slots C, D, and E, so that they overlap.

, we generate a 5-cycle of either edges or corners (which ever version of the commutator we chose).

You probably will need to apply setup moves (conjugates) before the 3-cycle commutator to affect different pieces than what the commutator affects by itself.  You may also do a cyclic shift of the 3-cycle commutator you have to affect different positions.

A cyclic shift of the corner commutator I gave, for example, is:   R U' L' U R' U' L U (original) => U' L' U R' U' L U R, where you conjugate with the inverse of the first move (or the inverse of the last move) to not add more moves to the original commutator, but you affect different pieces (yay!)

A more useful commutator to generate more corner positions might be:   R' F' R U R2 U' R' F R U R2 U'  = [R' F' R U, R2].

Also, you can overlap two pieces from the previous 3-cycle commutator as well.  (You are not just limited to overlapping one piece of two 3-cycle commutators).
Just make sure that the very last 3-cycle you do also involves the first piece you affected (which is you have left alone after the first application of the 3-cycle to the solved cube) to "close" the cycle.

So once you have a commutator which swaps only three corners or only three edges, you can overlap each 3-cycle to obtain a higher order cycle.
If you have an even permutation, this is all you will need to do.
If you have an odd permutation (remember that both the corners and the edges cycle types will be odd permutations if one is), you need to do a 2-cycle PLL first, and then overlap that 2-cycle with a 3-cycle commutator (and overlap that 4-cycle "marriage" with more 3-cycle commutators).

And actually, I have modeled all of the 2-cycle PLL algorithms in the speedsolving wiki as commutators and conjugates.  A commutator, if you haven't caught on, is A B A' B' = [A,B] and a conjugate C of some algorithm X is C X C' = [C:X].  So you can see the decompositions of the 2-cycle PLL algorithms as commutators and conjugates (just so that you won't feel that these PLL algorithms are gibberish) if you look up the following:

Link>>Look up "Entire Set of Speedsolving Wiki 2-Cycle PLL Algorithms Decomposed" in Google.  It's the first result.

So that's all you really need to make any cycle class type with edges and with corners.  Now, you can actually shorten (optimize) the length of your products of commutators by inputting the entire sequence into an optimal 3x3x3 solver.  I recommend the following:

>>Link:  Look up "cube explorer 5.0" in Google.  It should be the first result.

**A way that might be easier is to make a high order cycle by just repeating the 2-cycle PLL over and over, sharing just one piece from the previous application of the algorithm at a time.  Of course, set up moves (and/or you can do a cyclic shift of the algorithm) in between each application.  This of course will scramble the edges if you do the corners first, or scramble the corners if you do the edges--unless you do what you ought to do if you choose to just use a 2-cycle PLL:  make sure you have the 2-cycle PLL affect the same two corners (or same two edges if you are permuting the corners first) every application by proper application of setup moves.

Just keep in mind that whether you use the 2-cycle PLLs once for an odd permutation purpose only OR if you use it as your only base algorithm, you take note of which two edges it affects (if you choose to do corners first) or viceversa.

This is incorrect.  We must represent the sequence R L' U as:  (urf)(ufl,rub,rbd,rdf)(dbl,fdl,ulb)(ur,br,dr,fr,uf,ul,bl,dl,fl,ub)

Which is:
[1] A 1-cycle of corners (a twist/orientation only)
[2] A 4-cycle of corners (of which some are twisted apart from their own cycle because of the Law of the cube which states that the sum of all twists must be mod 3)
[3] A 3-cycle of corners (of which ..."...")
[4] A 10-cycle of edges (permutation only)
, where [1] can have a shared orientation with [2] or with [3] AND [2] and [3] have a shared orientation.

*The cube law says that it's impossible to just have one corner twisted.  So for our case of R L' U, the urf corner's misorientation is dependent on some other corner's misorientation.

"PROOF"
Suppose that with the move sequence R L' U, ufl - rub - rbd - rdf are in a 12-cycle, since it takes 12 times of repetition to restore these corners.  Also suppose that the number of corner objects is

, because when an algorithm does a 1-cycle of a corner (twists it in its location), that algorithm must be repeated three times to restore the orientation of that corner.  (We count the number of stickers instead of the number of actual cubies).

Now, in order to compute the order of the Rubik's Cube Group, which is 43 252 003 274 489 856 000, the number of permutations which corners contribute is "only"

.

From the formula at a link I previously mentioned for calculating the number of different cycle types (first search result in Google for "Cycle classes of permutations"), we see that the number of possible 12-cycles with 24 objects (stickers) is

Note that the sum of all of the number of possible cycle classes must add up to the total number of permutations which that object contributes to the diameter of its group, and clearly only the total of one cycle class is more than

,  which is a contradiction to the order of the Rubik's Cube Group.

Q.E.D.

I could also argue (without mathematics) that your notion is false by the following example

Consider the 3-cycle corner commutator R U' L' U R' U' L U = [R, U' L' U].  [R, U' L' U]3 = the identity.
Also consider the 3-cycle corner commutator L' D R D' L D R' D' = [L', D R D'].  [L', D R D']3 = the identity.

[R, U' L' U] [L', D R D'] = [L', D R D'] [R, U' L' U], and therefore, by the results above, ([R, U' L' U] [L', D R D'])3 = ([L', D R D'] [R, U' L' U])3 = the identity.

Now what if we were to consider your argument.  Then we would assume that [R, U' L' U] is a 3-cycle because it takes 3 times of repeating it to achieve the identity.
If we twist two of the three corners it affects (with two more commutators following it):

[R, U' L' U]
(F' U' B' U F U' B U) (U R' D' R U' R' D R)  //2-corner twist

But when we repeat this 3 times, it shouldn't be solved because it's a 9-cycle now.  But ([R, U' L' U] (F' U' B' U F U' B U) (U R' D' R U' R' D R) )3 = the identity?
(we can do the same for the commutator [L', D R D'] and achieve the same result).

From this, we at least have to conclude that a disjoint 3-cycle which doesn't share an orientation with any piece outside of that cycle will stay a 3-cycle.

To verify this, let's let the two original 3-cycle corner commutators share an orientation (let's twist one corner which is in [R, U' L' U] and one that is in [L', D R D'].  It shouldn't matter which corner from each we choose.

[R, U' L' U] [L', D R D']
(D R U R' D' R U' R') (R' F L F' R F L' F')   //2-corner twist of the two front right corners.

([R, U' L' U] [L', D R D'] (D R U R' D' R U' R') (R' F L F' R F L' F')  )3

the identity, but ([R, U' L' U] [L', D R D'] (D R U R' D' R U' R') (R' F L F' R F L' F')  )9 is.

But by my "proof", that algorithm, even though it takes 9 times to achieve the identity, it is two disjoint 3-cycles which share orientation with each other, not a 9-cycle.

Actually there is.  Bruce Norskog found what is commonly called "The Devil's Algorithm", which is what you are asking about, earlier this year for the entire 2x2x2 cube group and the entire 3x3x3 cube group.

For The Devil's Algorithm for the 2x2x2 cube group (which might be better to look at first to understand how Bruce formats his algorithm)
>>Link:  Look up "Hamiltonian circuit for the entire 2x2x2 cube group" in Google.  It is the first result.

For The Devil's Algorithm for the 3x3x3 cube group
>>Link:  Look up "A Hamiltonian circuit for Rubik's Cube!" in Google.  It is the first result.