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Olypmicube #6a

Scott

Member
Joined
Mar 17, 2006
Messages
207
Location
Glastonbury CT, USA
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

Member
Joined
Mar 17, 2006
Messages
287
Location
Austin, TX
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
 

pjgat09

Member
Joined
Mar 16, 2006
Messages
87
Location
Hiding, and waiting
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

Member
Joined
Mar 17, 2006
Messages
287
Location
Austin, TX
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
 

CraigBouchard

Member
Joined
Mar 14, 2006
Messages
274
Location
Kingston, ON, Canada
WCA
2005BOUC01
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 :p That would be intense...

Craig
 

Scott

Member
Joined
Mar 17, 2006
Messages
207
Location
Glastonbury CT, USA
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. :p
 

BillT

Member
Joined
Mar 20, 2006
Messages
23
Location
Reading, PA, USA
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. :p[/b]

I'm afraid I must. :D
<!--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 :D

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

-Bill
 
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