# Olypmicube #6a



## Scott (Mar 19, 2006)

Ok, the "comming in the next couple of months" thing has been up there for like 6 months.

Does anyone know how close we are to acualy having this great puzzle?


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## dougreed (Mar 19, 2006)

No clue. From the videos, the thing looks hella awesome. Whenever it comes out, I hope to be able to buy from the first batch. I can't wait to get my hands on one of those things.

-Doug


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## Richard (Mar 19, 2006)

What exactly is it?


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## Scott (Mar 19, 2006)

it's a working 6x6x6 rubik's cube, and acualy looks to be very smooth.

Pic1
Pic2
Pic3
Pic4

Video of the 6x6x6


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## gillesvdp (Mar 19, 2006)

Anybody going to Greece soon ?
If so, please try to contact those OlympicCube guys !


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## Richard (Mar 19, 2006)

Oh man! That thing is amazing. Good luck solving that thing...I think i'll stick to my 3X3X3 a while longer....


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## dougreed (Mar 19, 2006)

BTW, just because you can't physically buy one of these things now doesn't mean you can't practice them!

For more high-stress, high-order cube-solving fun:
http://www.puzzlingaddiction.com/Cube/applet/

-Doug


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## CraigBouchard (Mar 20, 2006)

I'm gunna go with a no comment on this...Lemme just say I'm hoping in the summer???


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## Scott (Mar 20, 2006)

> _Originally posted by CraigBouchard_@Mar 19 2006, 09:01 PM
> * I'm gunna go with a no comment on this...Lemme just say I'm hoping in the summer??? *


 Hmmm, you just commented after saying you wouldn't

lol


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## Richard (Mar 21, 2006)

When new cubes like this come up, how do find the new algorthims for it? Do you make a computer program for it or what?


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## pjgat09 (Mar 21, 2006)

When I solve any cube above 4 I really do it off the top of my head. The only alg i really use cycles 3 edges around. The rest is like a 4x4 solve: centers, edges, 3x3. That was my method for the 40x40.


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## dougreed (Mar 21, 2006)

> _Originally posted by Richard_@Mar 20 2006, 09:04 PM
> * When new cubes like this come up, how do find the new algorthims for it? *


You can use commutators to come up with algorithms for any number of puzzles, not just cubes. You can learn more about their 3x3x3 applications at Joel's Speedcubing Page, and then scale them up to work with nxnxn cubes using the same techniques.

Feel free to make a new thread to get more information about commutators if you're interested.

HTH,
Doug


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## Ravi (Mar 21, 2006)

I wonder who'll be the first to BLD one of those...

-Ravi


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## Alexander (Mar 21, 2006)

my bet is Stefan would be the first one


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## pjgat09 (Mar 21, 2006)

That would be scary to BLD a +6 order cube!! I can't even do 4!


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## Scott (Mar 21, 2006)

Haha, i cant even do 3!


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## pjk (Mar 21, 2006)

Do you guys know the potential costs of these?


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## CraigBouchard (Mar 22, 2006)

Yes I did comment...but w/e...No clue on potential cost, I'm still hoping around summer...Maybe they'll be sold for the first time at US Nationals...hehehe and not sold anywhere else for a month  That would be intense...

Craig


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## Scott (Mar 22, 2006)

Well, if it follows the pattern, it'll cost around 42 USD.

x = cube layers.
p = price.

p = 2x + x

*Aproximate Prices*
2x2x2 cube, 2(2) + 2 = $6 USD
3x3x3 cube, 3(3) + 3 = $12 USD
4x4x4 cube, 4(4) + 4 = $20 USD
5x5x5 cube, 5(5) + 5 = $30 USD
6x6x6 cube, 6(6) + 6 = $42 USD
7x7x7 cube, 7(7) + 7 = $49 USD
8x8x8 cube, 8(8) + 8 = $74 USD
9x9x9 cube, 9(9) + 9 = $90 USD
10x10x10 cube, 10(10) + 10 = $110 USD

Yes i made that formula, no it's not exact, and no... you dont bash it.


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## BillT (Mar 22, 2006)

> _Originally posted by Scott+Mar 21 2006, 10:02 PM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>*QUOTE* (Scott @ Mar 21 2006, 10:02 PM)</td></tr><tr><td id='QUOTE'>you dont bash it. [/b]_


_
I'm afraid I must.  
<!--QuoteBegin-Scott_@Mar 21 2006, 10:02 PM
*x = cube layers.
p = price.

p = 2x + x

3x3x3 cube, 3(3) + 3 = $12 USD
4x4x4 cube, 4(4) + 4 = $20 USD
5x5x5 cube, 5(5) + 5 = $30 USD
*[/quote]
If we go with your original formula, these would be the prices:

3x3x3 cube, 2(3) + 3 = $9 USD
4x4x4 cube, 2(4) + 4 = $12 USD
5x5x5 cube, 2(5) + 5 = $15 USD :blink: 

It should be: p = x^2 + x  

And yes, I realise I'm being rather picky. (I was bored  )

-Bill


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## pjk (Mar 22, 2006)

I wouldnt be suprised to see it at $42.


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## Scott (Mar 22, 2006)

yea, it seams reasonable


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## gillesvdp (Mar 22, 2006)

I'm sorry to disagree with this 42$ thing.
Adding one layer to a cube requires to have a much more complicated rotating system than the previous system.
So I think it's more an exponential that linear relaitonship between the number of layers of a cube and its price. ;-)

Gilles.


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## pjgat09 (Mar 22, 2006)

I think it will start at more like $60 dollars, and as they are mass produced, drop down to around $40-45


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## FrankMorris (Mar 22, 2006)

whatever the cost.... it will be worth it right?


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## pjgat09 (Mar 22, 2006)

As long as the price isnt $100 or above, the sides turn very well, and the puzzle is stable, I would buy one! I would love to own one, just toy play with it, even if i dont speedsolve it.


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## Scott (Mar 22, 2006)

haha, speedsolving it would be rediculous. you'd have to like turn it into a 4x4, then turn the 4x4 into a 3x3 or something like that.


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## pjk (Mar 22, 2006)

Well, it will be quite popular. We will have to see when I comes out.


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## CraigBouchard (Mar 22, 2006)

Gilles: Notice how it is exponential?!?!?!

Everyone else: It will be solved the same as any other puzzle bigger than 3x3...Centers, edges, solve as 3x3, as you get to bigger cubes you almost have to do this...Edges first starts to fail...And yes...I need one...


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## pjgat09 (Mar 23, 2006)

I would solve it like this:

2 opposite centers
2 other opposite centers
2 last centers
Put all the edges together
Solve 3x3
Fix Parity

I know it works, how do you think I solved a 40x40?


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## Scott (Mar 23, 2006)

how long did that take? lol


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## gillesvdp (Mar 23, 2006)

> _Originally posted by CraigBouchard_@Mar 22 2006, 11:16 PM
> * Gilles: Notice how it is exponential?!?!?!
> 
> Everyone else: It will be solved the same as any other puzzle bigger than 3x3...Centers, edges, solve as 3x3, as you get to bigger cubes you almost have to do this...Edges first starts to fail...And yes...I need one... *


 Why would Per's cage method be worse for bigger cubes than the traditional centers first method ?


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## Scott (Mar 23, 2006)

because you'd have ALOT more center piece to work with i think, i dunno.


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## dougreed (Mar 25, 2006)

> _Originally posted by Scott_@Mar 23 2006, 05:49 AM
> * because you'd have ALOT more center piece to work with i think, i dunno. *


 Yeah, I think you would be right if Per relied only on 3-cycles of single pieces to solve the center. But considering he doesn't 3-cycle single blocks all the time, he also cycles blocks of pieces, I think his method woudl still be very efficient.

-Doug


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## pjk (Mar 25, 2006)

The price of the 6x6 is going to be between $42-$100, I got that straight from the makers


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