Non-Cubers say the darndest things!

after miserably failing a 5x5 solve
Are you in MENSA?
No.
Do you hold any world records.
I don't believe you.

also a noob commenting on my videos

cubing, or blindfolding cubing, is only good, if it is used or applied somewhere else, its not, you people must not realize how stupid you really are because your blinded by being good at ONE thing, has it helped you in life?, impressed any girls?, helped you get a job, gain you money. . anything, no, center fail. . ?(reffering to when i said wow center fail(on 4x4)), haha, you and everyone else. . . (which doesnt look like many, which is a good thing) are fai8ling at life, you wonder why you have only a few views and subscribers,?(no not really)
 
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after miserably failing a 5x5 solve
Are you in MENSA?
No.
Do you hold any world records.
I don't believe you.

also a noob commenting on my videos

cubing, or blindfolding cubing, is only good, if it is used or applied somewhere else, its not, you people must not realize how stupid you really are because your blinded by being good at ONE thing, has it helped you in life?, impressed any girls?, helped you get a job, gain you money. . anything, no, center fail. . ?(reffering to when i said wow center fail(on 4x4)), haha, you and everyone else. . . (which doesnt look like many, which is a good thing) are fai8ling at life, you wonder why you have only a few views and subscribers,?(no not really)

Anyone want to "epic noob" this guy?
He's just showing off his abs and ****.
Also, this guy's the guy who commented on CN4's vid.
Here's the link of the epic noob's video:
http://www.youtube.com/watch?v=uSnKvCPbeoM

Also, I got this one today:
My record is like 3 sides, but none were opposite.

^ I lol'd at this, and thought of the simple setup and U-perm that could be done.
 
Funniest youtube quote ever (on one of nakaji's vids):

The First 2 Layers and Pendulum Last Layers and Octulum last Layers are all easy, what's reeally hard is the 7x7, and more challenging, 12x12.. this is just 3x3


LOL FAIL

PENDULUM IS SO HARD YOU DO NOT UNDERSTAND :mad:

lol I don't even know how one could come up with Pendulum and Octulum
 
To all that say there is no One algorithm to solve any cube, there is... it would just have to be 43,000,000,000,000,000 (fourty-three-quintillion) moves long.

No, there are 43,000,000,000,000,000 algorithms, but you only need one per solve.

One? Well, if you know all of ZB, I guess.
 
To all that say there is no One algorithm to solve any cube, there is... it would just have to be 43,000,000,000,000,000 (fourty-three-quintillion) moves long.

No, there are 43,000,000,000,000,000 algorithms, but you only need one per solve.

One? Well, if you know all of ZB, I guess.

Right? One algorithm that solves the cube in one step, but you have to memorize all 43 quintillion algorithms for all 43 quintillion possible combinations.
 
To all that say there is no One algorithm to solve any cube, there is... it would just have to be 43,000,000,000,000,000 (fourty-three-quintillion) moves long.

No, there are 43,000,000,000,000,000 algorithms, but you only need one per solve.

... I'm acting as if this algorithm is sort of like a beginner doing the RUR'U' trigger, tell them after a bit it will solve itself, they'll check after every move. If you did the same for one long algorithm/scramble, this scramble will give you every possible case.
 
To all that say there is no One algorithm to solve any cube, there is... it would just have to be 43,000,000,000,000,000 (fourty-three-quintillion) moves long.

No, there are 43,000,000,000,000,000 algorithms, but you only need one per solve.

... I'm acting as if this algorithm is sort of like a beginner doing the RUR'U' trigger, tell them after a bit it will solve itself, they'll check after every move. If you did the same for one long algorithm/scramble, this scramble will give you every possible case.

http://games.groups.yahoo.com/group/speedsolvingrubikscube/message/41581
 
Anyone want to "epic noob" this guy?
He's just showing off his abs and ****.
Also, this guy's the guy who commented on CN4's vid.
Here's the link of the epic noob's video:
http://www.youtube.com/watch?v=uSnKvCPbeoM

Also, I got this one today:
My record is like 3 sides, but none were opposite.

^ I lol'd at this, and thought of the simple setup and U-perm that could be done.

please do epic noob this guy
for the second one it could also be an F-perm
 
Anyone want to "epic noob" this guy?
He's just showing off his abs and ****.
Also, this guy's the guy who commented on CN4's vid.
Here's the link of the epic noob's video:
http://www.youtube.com/watch?v=uSnKvCPbeoM

Also, I got this one today:
My record is like 3 sides, but none were opposite.

^ I lol'd at this, and thought of the simple setup and U-perm that could be done.

please do epic noob this guy
for the second one it could also be an F-perm

True.
Now that I think about it, it could be out of a lot of permutations.
 
My friend whose a newer cuber said He saw a video on youtube where they said there were enough combinations on the petaminx to fill up 15 universes. How many things are wrong with that?
 
My friend whose a newer cuber said He saw a video on youtube where they said there were enough combinations on the petaminx to fill up 15 universes. How many things are wrong with that?

There are more?

Let's just look at the teraminx which I found a firm number on the # of combinations for (petaminx obviously has more but I haven't seen a good looking calculation)
1.7989 x 10^571 possible positions. Drewseph says his teraminx fits in an 11x11x11cm cube. A dodecahedron's edge length is approximately half its cubic dimensions so I'll go with 5.5cm per edge. A dodecahedron's volume is IIRC approximately equal to 7.66*(edge length^3) so the volume of a teraminx is approximately 1274.4325 cubic centimeters. Going with 13.7 billion light years for the radius of the universe, light traveling at 3*10^8m/s, we get 4/3*pi*((1.37*10^10)*(3600*24*365)*(3*10^8)*(100))^3 cubic centimeters in the universe.

1274.4325*1.7989*10^571
= 2.29257662*10^574 cm^3 for all possible teraminx positions, assuming we can seamlessly pack teraminxes together.
4/3*pi*((1.37*10^10)*(3600*24*365)*(3*10^8)*(100))^3
= 1.4043993*10^85 cm^3 in the universe


Obviously the petaminx would fill even more universes.
 
My friend whose a newer cuber said He saw a video on youtube where they said there were enough combinations on the petaminx to fill up 15 universes. How many things are wrong with that?

There are more?

Let's just look at the teraminx which I found a firm number on the # of combinations for (petaminx obviously has more but I haven't seen a good looking calculation)
1.7989 x 10^571 possible positions. Drewseph says his teraminx fits in an 11x11x11cm cube. A dodecahedron's edge length is approximately half its cubic dimensions so I'll go with 5.5cm per edge. A dodecahedron's volume is IIRC approximately equal to 7.66*(edge length^3) so the volume of a teraminx is approximately 1274.4325 cubic centimeters. Going with 13.7 billion light years for the radius of the universe, light traveling at 3*10^8m/s, we get 4/3*pi*((1.37*10^10)*(3600*24*365)*(3*10^8)*(100))^3 cubic centimeters in the universe.

1274.4325*1.7989*10^571
= 2.29257662*10^574 cm^3 for all possible teraminx positions, assuming we can seamlessly pack teraminxes together.
4/3*pi*((1.37*10^10)*(3600*24*365)*(3*10^8)*(100))^3
= 1.4043993*10^85 cm^3 in the universe


Obviously the petaminx would fill even more universes.

First off, the size of the universe is infinite, second, solution is not an object, so it has no volume. Third there is only one universe.
 
My friend whose a newer cuber said He saw a video on youtube where they said there were enough combinations on the petaminx to fill up 15 universes. How many things are wrong with that?

There are more?

Let's just look at the teraminx which I found a firm number on the # of combinations for (petaminx obviously has more but I haven't seen a good looking calculation)
1.7989 x 10^571 possible positions. Drewseph says his teraminx fits in an 11x11x11cm cube. A dodecahedron's edge length is approximately half its cubic dimensions so I'll go with 5.5cm per edge. A dodecahedron's volume is IIRC approximately equal to 7.66*(edge length^3) so the volume of a teraminx is approximately 1274.4325 cubic centimeters. Going with 13.7 billion light years for the radius of the universe, light traveling at 3*10^8m/s, we get 4/3*pi*((1.37*10^10)*(3600*24*365)*(3*10^8)*(100))^3 cubic centimeters in the universe.

1274.4325*1.7989*10^571
= 2.29257662*10^574 cm^3 for all possible teraminx positions, assuming we can seamlessly pack teraminxes together.
4/3*pi*((1.37*10^10)*(3600*24*365)*(3*10^8)*(100))^3
= 1.4043993*10^85 cm^3 in the universe


Obviously the petaminx would fill even more universes.

First off, the size of the universe is infinite, second, solution is not an object, so it has no volume. Third there is only one universe.

Proof of universe being infinite?
 
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