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Re: Please me with the Rubik's cube group elements order problem!
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 In speedsolvingrubikscube@yahoogroups.com, "danix800"
<danix800@...> wrote:>
Rubik's
> I'm a big rubik's cube fan. It's well know that for a normal
> cube the cube group has its maximum element
take
> order 1260, say, the maneuver UR'UF'D2 needs 1260 repeations to
> the cube back
it.
> to its original configuration. But the problem is, I can't prove
> I've searched the web
with
> extensively, no results! I guess the problem has something to do
> the cycles of the
Could
> edges or corners. And the cubie orientations are also important.
> you please give me
grateful.
> some introductions or maybe whatever helps. I would be very
> Also, some people around me said, for the picture cube, the
R' F
> maximum order changes
> to at least 1980, still, he can't prove it. Here's an example, F'
> D' U' F' L' D2 L' B L B2 R' D2 F',
you
> it takes 990 times to recover the normal cube, but for the picture
> cube, we need to recover
> the orientations of the centerpieces, so it rises to 1980. Could
> say something about this?
here a site where you can fillin your 1260 alg
> Any help would be appreciated!
>
http://www.solvethecube.co.uk/tools.html and then go to order
greets
~A.O. 0 Attachment
 In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
<stefan.pochmann@...> wrote:> Incomplete because you rely on definitions of orientations that you
But I don't need the definition of an arbitrary "correct" orientation.
> omit. Nonobvious because it's not at all clear to me that your
> statements are correct. And cumbersome compared to sticker cycles.
A pure orientation of cubies is any equivalence class of legal
sequences acting on the Rubik's Cube by twisting/flipping any number
of cubies in place, not relative to predefined orientation. And Bruce
provided even less detail than me. Maybe the sticking point is the
distinction between the cubie itself and the cubie position?
More details:
View the cube group elements as equivalence classes of legal sequences
of moves on a Rubik's Cube, with the obvious equivalence relation
(disregarding rotated centers), so that the cube group act on a
Rubik's Cube in the obvious way. Any cube group element can be
represented by a pure permutation of cubie _positions_ followed by a
pure orientation of cubie _positions_. (That's clear enough. Don't
make me write out the components of the semidirect product and exhibit
an isomorphism.) By abuse of terminology, we will say that the cube
group element _is_ a pure permutation followed by a pure orientation.
Associate the permutation subgroup with a subgroup of S_12 X S_8 and
write out the disjoint cycles in each direct product component. Take
an ncycle in this decomposition. Restrict the action of the cube
group element to the cubies moved by this cycle. The restriction is
now written as a pure permutation of n cubie positions followed by a
pure orientation of the n cubie positions.
Starting at any position on the cycle, number the positions 1 through
n along the cycle. An ncycle (1, ..., n) sends the cubie in position
1 to position 2, etc. An orientation of ncycle (1, ..., n) is a
ntuple (t_1, ..., t_n), where t_i is 0 if the corner in position i is
left untwisted, 1 if twisted clockwise, and 2 if twisted
counterclockwise, where the cycle is written as a pure permutation
followed by a pure orientation. The sum of orientations of the ncycle
is t_1 + ... + t_n. For example, a 3cycle of edges (e_1, e_2, e_3,
e_4) with orientation (1, 1, 1, 0) looks as follows, where "e_1 e_2
e_3 e_4," for example, means that e_1 is in position 1, etc. We keep
track of flipped edges using *.
Starting configuration: e_1 e_2 e_3 e_4 (after renumbering, if necessary)
Permute: e_4 e_1 e_2 e_3. Flip: e_4* e_1* e_2* e_3.
Permute: e_3 e_4* e_1* e_2*. Flip: e_3* e_4 e_1 e_2*.
Permute: e_2* e_3* e_4 e_1. Flip: e_2 e_3 e_4* e_1.
Permute: e_1 e_2 e_3 e_4*. Flip: e_1* e_2* e_3* e_4*.
Note that each cubie has been flipped three times, once each time it
was in positions 1 through 3. The argument I provided below (a) and
(b) should make more sense now.
> In essence, you're checking every single piece of a cycle against an
An "obvious natural" way? You're such a Pochmann! ;) Sticker cycles
> unspecified artificial arbitrary orientation scheme, instead of
> simply checking the cycle once as a whole for orientation in an
> obvious natural way.
alone don't give you enough structure to figure out the maximum bound,
though.
macky 0 Attachment
At first I didn't think that had anything to do with this proof, but I
realize you can put exponents on sequences with that tool.
It only suffices for showing/proving that 1260 is a lowerbound. This
is the existence claim/assertion Macky made. But to do this you have
to also selectively try certain factors of it, by choosing from its
prime factorization.
Between Bruce and Macky they've got all the abstract math questions
covered! I'm impressed. Good reads.
> here a site where you can fillin your 1260 alg
> http://www.solvethecube.co.uk/tools.html and then go to order
>
> greets
> ~A.O.
> 0 Attachment
 In speedsolvingrubikscube@yahoogroups.com, "mackymakisumi"
<mackymakisumi@...> wrote:>
Maybe, though I'm not sure what you mean by that. But I think I
> Maybe the sticking point is the distinction between the cubie
> itself and the cubie position?
understand you now (but whoa was reading your message hard). Earlier
you wrote:
> (a) An ncycle of edges has order
Only now do I realize that you mean "are flipped" in the sense of
> n if an even number of edges are flipped
> 2n if an odd number of edges are flipped
"get flipped", i.e., the process rather than the result. I had
misunderstood this, thinking about your orientation scheme for
blindcubing where you determine which edges "are flipped".
But I should've realized that that interpretation doesn't make sense
here. Sloppy thinking by me, I apologize and walk away in shame.
Although, I really recommend writing "get" instead of "are" to make
it clearer. I believe I've come across this subtlety a few times
myself when writing something and intentionally chose "get" to hint
towards change rather than state.
> View the cube group elements as equivalence classes of legal
Not trying to nitpick or poke holes, but can you explain this a bit?
> sequences of moves on a Rubik's Cube, with the obvious equivalence
> relation (disregarding rotated centers), so that the cube group act
> on a Rubik's Cube in the obvious way.
It's not obvious to me at all what that relation is supposed to be. I
pretty much understand the rest of your message, but this part has me
totally stumped, and I'd like to understand it.
> Sticker cycles alone don't give you enough structure to figure out
Yeah, I only meant it for the lower level part of the proof, i.e.,
> the maximum bound, though.
the 2n/3n statement. And now that I understand what you mean, there's
no big difference anymore. I was really acting noob, sorry again.
Cheers!
Stefan 0 Attachment
you don't want to know about these things. possession of such
knowledge may result in spending copious amounts of money on cheap
plastic.
 In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
>
> Where can I find a site on cube math? What's a corner cycle?
What's parity?
> 0 Attachment
 In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
<stefan.pochmann@...> wrote:>
Earlier
>  In speedsolvingrubikscube@yahoogroups.com, "mackymakisumi"
> <mackymakisumi@> wrote:
> >
> > Maybe the sticking point is the distinction between the cubie
> > itself and the cubie position?
>
> Maybe, though I'm not sure what you mean by that. But I think I
> understand you now (but whoa was reading your message hard).
> you wrote:
Alright, I think I just doubleconfused myself. I did have your
>
> > (a) An ncycle of edges has order
> > n if an even number of edges are flipped
> > 2n if an odd number of edges are flipped
>
> Only now do I realize that you mean "are flipped" in the sense of
> "get flipped", i.e., the process rather than the result. I had
> misunderstood this, thinking about your orientation scheme for
> blindcubing where you determine which edges "are flipped".
blindcubing orientation scheme in mind, but actually only as part of
the picture. I did understand your "are flipped" as change rather
than state, but change *according to* a certain fixed orientation
scheme like you use for blindcubing, where you say an edge is or
isn't flipped if it moves from here to there. It just wasn't clear to
me that you meant it in a more general way where each piece "carries
its own orientation" with itself, so to speak. I'm not sure I've
thought about orientations this way before. I might've, but right now
I'm in a thoroughly confused state of mind.
Cheers!
Stefan 0 Attachment
Last I checked, you could buy a Rubik's Cube for less than $20,
including shipping to everywhere. If you wanted to put it together
yourself, you can get them for less than half that price. I think
it's fairly easy to see the incredible price gulf between that and V
Cubes (at the previously quoted amount of $80 + shipping).
2008/7/2 Bob Burton <rubikscubewhiz@...>:> you don't want to know about these things. possession of such
> knowledge may result in spending copious amounts of money on cheap
> plastic.
>
>  In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
>>
>> Where can I find a site on cube math? What's a corner cycle?
> What's parity?
>>
>
>
>
> 
>
> Yahoo! Groups Links
>
>
>
> 0 Attachment
Ok, I read that proof again, paying more attention. Improvement ideas
below.
 In speedsolvingrubikscube@yahoogroups.com, "mackymakisumi"
<mackymakisumi@...> wrote:>
edge
> A cycle with order divisible by 11 must be an 11cycle of edges,
> leaving a 1cycle. The total order for edges is maximized by giving
> both cycles an odd number of edge flips: lcm(22,2)=22. Since this
> permutation is odd, the restriction on permutation parity forces an
You could shorten this part by saying that even the product
> even corner permutation. We can easily list all distinct cycle types
> of A_8:
> Only even cycles: 1,1,1,1,1,1,1,1 3,1,1,1,1,1 5,1,1,1 7,1
> 3,3,1,1 5,3
> Two odd cycles: 2,2,1,1,1,1 4,2,1,1 6,2 4,4
> Two odd cycles and even cycle(s): 3,2,2,1
> Four odd cycles: 2,2,2,2.
>
> Ignoring the restriction on corner orientation, the total order for
> corners with cycle type l_1,...,l_k is maximized by multiplying each
> length by 3, according to (b). This at most lcm(3l_1, ...,
> 3l_k)=3lcm(l_1, ..., l_k), which takes the maximum 45 for cycle type
> 5,3. Thus these cube group elements have order at most lcm(22,45) =
> 2*3^2*5*11 = 990 < 1260.
maximizing partition of the eight corners, which is (8/e)^e=18.8,
isn't enough to reach the necessary (1260/22)/3=19.09.
For the productmaximizing partition:
http://mathforum.org/library/drmath/view/52441.html
http://www.jstor.org/pss/2690532
> Case 2: One factor of 3
Aren't you already done at the point where you need 12 corners?
> We must show that there cannot be four factors of 2. This means a
> flipped 8cycle, which forces corners 7 and 5. The 3 must come from
> a 3cycle of the edges, but again this violates the permutation
> parity.
> Done.
Cheers!
Stefan 0 Attachment
 In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
<stefan.pochmann@...> wrote:> You could shorten this part by saying that even the product
Nice!
> maximizing partition of the eight corners, which is (8/e)^e=18.8,
> isn't enough to reach the necessary (1260/22)/3=19.09.
> > Case 2: One factor of 3
Haha, yeah.
> > We must show that there cannot be four factors of 2. This means a
> > flipped 8cycle, which forces corners 7 and 5. The 3 must come from
> > a 3cycle of the edges, but again this violates the permutation
> > parity.
> > Done.
>
> Aren't you already done at the point where you need 12 corners?
> > View the cube group elements as equivalence classes of legal
I'm not sure how much you know about equivalence relation or group
> > sequences of moves on a Rubik's Cube, with the obvious equivalence
> > relation (disregarding rotated centers), so that the cube group act
> > on a Rubik's Cube in the obvious way.
>
> Not trying to nitpick or poke holes, but can you explain this a bit?
> It's not obvious to me at all what that relation is supposed to be. I
> pretty much understand the rest of your message, but this part has me
> totally stumped, and I'd like to understand it.
action, so I'll try to be as detailed as I can. Sorry if any part
bores you.
We begin with the free group G on six letters, U, D, R, L, F, B. Its
elements are any finite products of these six letters and their
inverses (denoted U', etc, for convenience), and since I haven't
specified that these are actually moves on a Rubik's Cube, any two
distinct "words" (e.g. RRLL and LLRR; the identity, RRRR, R'R'R'R',
RRRRRRRR, etc) are considered distinct in this group ("a free group
has no relation"). The group operation is concatenation, or stringing
together the words one after another. The groups axioms are easily
checked.
Definition (based on Wiki http://en.wikipedia.org/wiki/Group_action):
If G is a group and X is a set, then a group action of G on X is a
function from X x G (all ordered pairs (x, g), where x is in X and g
is in G) into X, where x*g denotes the image of (x, g), that satisfies
the following two axioms:
1. x*(gh) = (x*g)*h for all g, h in G and x in X
2. x*e = x for every x in X (where e denotes the identity of G)
We say that G acts on X.
[Wiki uses left group action and writes x*g. Right group action works
better for us. Also, you usually use a dot instead of an *.]
Let X be the set of all legal configuration of the cube, with the
centers fixed. Then the free group G constructed above acts on X if we
assign the six letters their normal meanings as moves on a cube.
Intuitively, any finite sequence g of those six letters and their
inverses corresponds to a sequence of moves on the cube, and so we can
define x*g as the legal cube configuration that results when we apply
the moves corresponding to g to the legal configuration x. Axioms 1
holds as the group operation in G is concatenation (note that this
wouldn't have worked if we had used left group action instead) and
axiom 2 is obvious since the identity of G is the empty word,
corresponding to doing nothing.
Definition: An equivalence relation is a binary relation ~ on some set
G that satisfies the following for all a, b, and c in G.
Reflexivity: a ~ a for all a in G
Symmetry: If a ~ b, then b ~ a
Transitivity: If a ~ b and b ~ c, then a ~ c.
The equivalent class of a under ~, denoted [a], is the subset of G
whose elements b satisfy a ~ b.
When G is a group, ~ is said to be Ginvariant if whenever g ~ u and h
~ v, we have gh ~ uv.
Two examples of equivalence relations:
Equality on any set
Congruence modulo n on the integers.
It's easy to show that an equivalence relation on set G partitions G
into equivalent classes; that is, each element of G is in exactly one
equivalent class. For example, in the second example, there are n
congruence classes, [0] through [n1], and every integer is congruent
to exactly one of 1 through n1 mod n.
The idea is to introduce some equivalent relation on our free group G
so as to "collapse" it to the cube group. We want consider all
sequences with the same effect on the solved cube to be the same in
the cube group. So we use the action of G on X to define our
equivalence relation "~":
For g, h in G, g ~ h if and only if (solved)*g = (solved)*h, where
(solved) is the solved configuration in X.
It's easy to see that this is an equivalence relation. Moreover,
Claim 1: ~ is Ginvariant.
Proof: This is easy to see for our example.
The set of equivalence classes, denoted by G/~, now has the desired
set structure of the cube group. It remains to place a group structure
by defining a group operation between two equivalent classes of G/~.
Again, we use the group action:
Claim 2: Let [g], [h] be equivalent classes in G/~ (so g, h are in G).
The binary operation [g][h] = [gh] makes G/~ into a group.
Proof: We first need to show that the operation is well defined. That
is, whatever elements of G we pick from the equivalent classes [g] and
[h], the operation returns the same equivalent class [gh]. But this is
precisely the Ginvariance of ~; if g ~ u and h ~ v, meaning [g] = [u]
and [h] = [v], then gh ~ uv, so [gh] = [uv]. Closure, associativity,
and the existence of identity and inverse follow from the respective
properties of the group operation in G.
Introducing an equivalent relation is a standard way of "collapsing" a
group. The other, related, method uses quotient groups, which need a
bit more development. Ginvariance is related to the concept of normal
subgroups and is the exact condition needed for us to be able to
induce the group structure on the smaller set G/~. It's a pretty
strong condition, but using action made it very easy in our example.
> Although, I really recommend writing "get" instead of "are" to make
OK, I can see the source of confusion.
> it clearer. I believe I've come across this subtlety a few times
> myself when writing something and intentionally chose "get" to hint
> towards change rather than state.
> And now that I understand what you mean, there's
No problem! I know you're a very reasonable person, and it's cool that
> no big difference anymore. I was really acting noob, sorry again.
you want to understand the little details.
macky 0 Attachment
I really don't think it's a linear scale.
On Wed, Jul 2, 2008 at 5:00 PM, Bart <banaticus@...> wrote:
> Last I checked, you could buy a Rubik's Cube for less than $20,
> including shipping to everywhere. If you wanted to put it together
> yourself, you can get them for less than half that price. I think
> it's fairly easy to see the incredible price gulf between that and V
> Cubes (at the previously quoted amount of $80 + shipping).
>
> 2008/7/2 Bob Burton <rubikscubewhiz@... <rubikscubewhiz%40yahoo.com>
> >:
>
> > you don't want to know about these things. possession of such
> > knowledge may result in spending copious amounts of money on cheap
> > plastic.
> >
> >  In speedsolvingrubikscube@yahoogroups.com<speedsolvingrubikscube%40yahoogroups.com>,
> Bart <banaticus@...> wrote:
> >>
> >> Where can I find a site on cube math? What's a corner cycle?
> > What's parity?
> >>
> >
> >
> >
> > 
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
>
>
[Nontext portions of this message have been removed] 0 Attachment
bart yeah you're pretty much a dipshit, and no one cares about your extremely ignorant opinion on this matter. if you had even a clue as to the amount of money and effort that goes into producing a toy on a worldwide scale (its not JUST the plastic) then you might be able to appreciate these for what they are  probably the most complex twisty puzzles ever built and well worth the prices being asked. you're ignorant and uninformed, kindly quit trying to be a troll, it hardly works and will just make you look like a bigger dumbass (if that's even possible). if you think its unfair or not reasonable, produce 6x6 and 7x7 cubes for a price less than theirs of the same quality and i'll eat my hat. until then....cork it, asshat.
 On Wed, 7/2/08, Bart <banaticus@...> wrote:
From: Bart <banaticus@...>
Subject: Re: [Speed cubing group] Re: Please me with the Rubik's cube group elements order pr
To: speedsolvingrubikscube@yahoogroups.com
Date: Wednesday, July 2, 2008, 4:00 PM
Last I checked, you could buy a Rubik's Cube for less than $20,
including shipping to everywhere. If you wanted to put it together
yourself, you can get them for less than half that price. I think
it's fairly easy to see the incredible price gulf between that and V
Cubes (at the previously quoted amount of $80 + shipping).
2008/7/2 Bob Burton <rubikscubewhiz@ yahoo.com>:
> you don't want to know about these things. possession of such
> knowledge may result in spending copious amounts of money on cheap
> plastic.
>
>  In speedsolvingrubiksc ube@yahoogroups. com, Bart <banaticus@. ..> wrote:
>>
>> Where can I find a site on cube math? What's a corner cycle?
> What's parity?
>>
>
>
>
>    
>
> Yahoo! Groups Links
>
>
>
> 0 Attachment
 In speedsolvingrubikscube@yahoogroups.com, Clancy Cochran
<perscription_death@...> wrote:
i'll eat my hat. until then....cork it, asshat.>
Not the slayer hat!
 0 Attachment
Sure, I'd moved on to asking about a different topic, until Bob threw
that back up in my face again. If you'd care to actually answer the
question so that I can learn more about Rubik's Cubes, I'd love to
hear it. 0 Attachment
Well, you can get a cube in the store for less than $10, and shipping
on DIY's usually end up being more than $10 total, but those are
facts, so I'm sure you're not too concerned about them.
Anyways, if a Rubik's Cube is 3x3x3, wouldn't a 6x6x6 be the
equivilant of 8 Rubik's cubes?
Anyways, if you're trying to figure out corner cycle and parity, why
not try Google?
I'd suggest getting a new Yahoo ID and try not to be an idiot in the
future.
 In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
>
> Last I checked, you could buy a Rubik's Cube for less than $20,
> including shipping to everywhere. If you wanted to put it together
> yourself, you can get them for less than half that price. I think
> it's fairly easy to see the incredible price gulf between that and V
> Cubes (at the previously quoted amount of $80 + shipping).
> 0 Attachment
A simple Google search would yield the answer to each of your three
quesitons.
If you want a website about "cube math," Google 'Rubik's Cube Math'
If you want to know about a "corner cycle," Google 'Rubik's Cube
corner cycle'
If you want to know about "parity," Google 'Rubik's Cube parity'
The first link for each of those three Google searches yields the
answer to your question.
...however, I'd stay away from the Rubik's Cube math. You seem to
have the mathematical ability and reasoning of a goldfish.
Bob
 In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
>
> Sure, I'd moved on to asking about a different topic, until Bob threw
> that back up in my face again. If you'd care to actually answer the
> question so that I can learn more about Rubik's Cubes, I'd love to
> hear it.
> 0 Attachment
Thanks. A guy asked what my thoughts were about the price. I
answered. People jumped all over my case and I defended my reasoning.
Then those people that, I presume, have a financial interest in
seeing the company succeed started using profanity and slinging other
insults. Whatever floats their boat, I guess. Maybe they should be
upset with the guy who asked to hear people's opinion in the first
place. 0 Attachment
 In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...>
wrote:>
reasoning.
> Thanks. A guy asked what my thoughts were about the price. I
> answered. People jumped all over my case and I defended my
> Then those people that, I presume, have a financial interest in
other
> seeing the company succeed started using profanity and slinging
> insults. Whatever floats their boat, I guess. Maybe they should be
Nobody minds you saying they cost too much for you. The problem
> upset with the guy who asked to hear people's opinion in the first
> place.
>
started when you claimed a ridiculous x1000 markup and insulted the V
cubes people.
Stefan 0 Attachment
 In speedsolvingrubikscube@yahoogroups.com, "mackymakisumi"
<mackymakisumi@...> wrote:>
And it even works when done right. Should be e^(8/e)=18.97.
>  In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
> <stefan.pochmann@> wrote:
> > You could shorten this part by saying that even the product
> > maximizing partition of the eight corners, which is (8/e)^e=18.8,
> > isn't enough to reach the necessary (1260/22)/3=19.09.
>
> Nice!
> I'm not sure how much you know about equivalence relation or group
I know equivalence relations, but again misread your post. I missed
> action, so I'll try to be as detailed as I can. Sorry if any part
> bores you.
that you actually went down to the level of move sequences, as in the
whole thread I was solely thinking in terms of their effects (cube
group elements). I agree it's completely obvious when read correctly.
And I'd actually say this obviousness is what made me misread your
message. I need to be more careful.
Group actions however still confuse me. I've read about them and
similar stuff before but I have trouble grasping/remembering these
things. I'll read your message again later when I have more time, and
will hopefully understand it better then. Thanks for taking the time
to explain, I imagine it's interesting/useful for others as well.
Cheers!
Stefan 0 Attachment
I don't think any of us have a financial interest in VCubes. We
simply want them to succeed so they put out more cubes. But
presumptions like this just make you look more ignorant.
 In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
>
> Then those people that, I presume, have a financial interest in
> seeing the company succeed started using profanity and slinging other
> insults. 0 Attachment
VCubes' profits would be the catalyst for 11x11x11 engineering.
Our investment goes further than what we directly pay for. 0 Attachment
 In speedsolvingrubikscube@yahoogroups.com, "Bruce Norskog"
<brnorsk@...> wrote:>
I built such a case yesterday and then today found your above message
> In this group of "permutable centers," the maximum order is 2520.
> This can be achieved by:
>
> corner cycles: 5*3, 3*3
> edge cycles: 7*2, 4*2, 1*1
> center cycles: 4, 1, 1
>
> LCM(15,9,14,8,1,4,1,1) = 2520
again. Here's an algorithm in four steps:
x
U L D' L' U' L D L'
M' U M' U M' U M' U
U B' U' F2 M2 F2 M2 U B U'
The steps are smalleffect algs with specific intentions:
1) Create several 4cycles (and change edges+corners parity)
2) Create the misoriented corner 3cycle and 5cycle
3) Create the misoriented edge 4cycle (in the L layer)
4) Create the edge 7cycle
You can see steps 24 simply combine the 4cycles created in step 1
in order to build the desired cycles.
Cheers!
Stefan
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