# 5x5x5 solver!



## Johnny.d.p (Jan 8, 2011)

PLEASE CLOSE THREAD ADMIN!


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## Kirjava (Jan 8, 2011)

doesn't exist (publically)


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## maggot (Jan 8, 2011)

stop creating threads. if you do not know how to solve it because you understand nothing about cubing, first learn to solve a regular rubiks cube and a 4x4 and then ask your questions. what is it that you are having trouble with?


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## sub_zero1983 (Jan 8, 2011)

Just follow these video's. Rob explains everything you'll need to know in a way you can understand.


Part 1 Center's: http://www.youtube.com/watch?v=44G7P-Q5E94
Part 2 Finishing centers: http://www.youtube.com/watch?v=3coU9lGXu-g&feature=list_related&playnext=1&list=PL02F429D8CF4E382A
part 3 Edge pairing: http://www.youtube.com/watch?v=Ga9z4gF3QDc&feature=channel
part 4 Parity error's: http://www.youtube.com/watch?v=KgmJg5PSifw&feature=channel


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## sub_zero1983 (Jan 8, 2011)

Johnny.d.p said:


> Uhhhhmmm... he refers to algorithms from 4x4x4. I have no idea what they are! I didn't buy a 4x4x4 or learn how to solve. Thanks anyway


 Relax...If i remember correctly, he show's you how they are done on the 5x5.


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## aronpm (Jan 8, 2011)

Johnny.d.p said:


> Uhhhhmmm... he refers to algorithms from 4x4x4. I have no idea what they are! I didn't buy a 4x4x4 or learn how to solve. Thanks anyway


 
JUST WATCH THE VIDEOS


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## maggot (Jan 8, 2011)

the omly alg from the 4x4 that you need is oll parity... it will fix an edge when it is checkerboarded. 

r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2 is the nasty alg. there are other ones, but they are just as nasy.

to solve centers: place 2 edges around its correct center piece to form a 1x3 line. without messing with that 1x3 3lne, make the other 2 1x3 blocks and then attach them to your 1st 1x3 block. once you solve 1 center conventionally most people solve the opposite center, white <> yellow.
to solve the last 4 centers can be tricky. you must fully understand break and fix method. the easiest way to describe this is breaking the cross to insert a pair and then fixing the cross again.. it is similar to that. the last 2 centers, try to get as many of the pieces onto the right center and then use a simple commutator to fix the remaining 1-2 pieces. this commutator is not really an alg, but more like a technique. look up lancetheblueknight on youtube and watch his 'last 2 centers' tutorial. 

for edge pairing, its hard to describe with text, but put the 3 dedges into the same slice with the correct orientation. sometimes you will need to flip edges (hopefully you can find a tutorial to help or you can figure this out... ) but once they are correctly oriented, you slice to put all the dedges together to form a solid 3 block bar and then R U R' to move your bar to the top amd replace it with a broken edge group and then restore your centers back. 
the last 2 edge groups can be tricky... if you are willing to memorize a bunch of algs,= memyselfandpi on youtube has a video on all the cases for the last 2 edge groups. 

good luck @[email protected] and if you dont want to pull your hair out, maybe watch a 4x4 tutorial so you can understand the general concept of reduction... the way to solve a 5x5 is only a little more complex, but it has the advantage of having fixed centers and no 3x3 parities.


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## stinkocheeze (Jan 8, 2011)

Johnny.d.p said:


> Uhhh yer.. I can solve a regular rubik's cube! I got 5x5x5 cause I wanted a challenge! I didn't want 4x4x4 If you have a problem then go somewhere else! Don't be a smartass and assume I have no idea how to solve the cube!


 


Johnny.d.p said:


> and btw I'm having trouble with the whole thing! sorry for calling u smartass! I only read half of ur post.


 


Johnny.d.p said:


> Uhhhhmmm... he refers to algorithms from 4x4x4. I have no idea what they are! I didn't buy a 4x4x4 or learn how to solve. Thanks anyway


 
TRIPLE... ALL THE WAY.... ACROSS THEE FORRUMMMMM
Edit: I can't count.


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## Bartholomew (Feb 15, 2011)

The solution of your question is 225 i.e the cube of 5,the cube is taken by 3 times multiply of a number whose cube is required.


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## MichaelP. (Feb 15, 2011)

Bartholomew said:


> The solution of your question is 225 i.e the cube of 5,the cube is taken by 3 times multiply of a number whose cube is required.



/thread


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## ben1996123 (Feb 15, 2011)

5x5 solver.

I waited the 45 seconds for you :3

download > extract > 5x5 solver > vcube5 (Virtual-cube 5. I got this program before v-cubes were released -_-) > vcube5a.htm

go to options and set the speed down to 1.


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## Forte (Feb 15, 2011)

set kinetic soldier
six sams attack into 3800 def loooool


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