Olypmicube #6a

[Scott]

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?

 

[dougreed]

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

 

[Richard]

What exactly is it?

 

[Scott]

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

Pic1
Pic2
Pic3
Pic4

Video of the 6x6x6

 

[gillesvdp]

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

 

[Richard]

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

 

[dougreed]

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

 

[CraigBouchard]

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

 

[Scott]

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

 

[Richard]

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?

 

[pjgat09]

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.

 

[dougreed]

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

 

[Ravi]

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

-Ravi

 

[Alexander]

my bet is Stefan would be the first one

 

[pjgat09]

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

 

[Scott]

Haha, i cant even do 3!

 

[pjk]

Do you guys know the potential costs of these?

 

[CraigBouchard]

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

 

[Scott]

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.

 

[BillT]

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