# 5x5x5 bld



## hait2 (Dec 17, 2007)

ok there goes me being cube-free (and ironically im posting this from a windows machine)..

so i just saw a 5x5x5 lying around in a friend's room while we were about to play risk, so of course, i had to figure out how to solve it
it took me a while (not longer than the risk game tho ), but i intuitively solved it using the method of 'turn it into a 3x3x3, then solve it normally'. 

so obviously now i want to do it blindfolded
as of now i don't really have an idea of how to do it
my initial thought was to turn it into a 3x3x3 and apply the 3x3x3 bld but i don't think that would work for various reasons

so now i'm experimenting with just solving it as i go with commutators and cycles. i still can't figure out the damn parity; i think i will have to look up the algs for that unfortunately ;( ill mess with it a bit more before giving up like that though

i don't want to look up a solution as of yet, but am i on the right track? and roughly how much harder is 5x5x5 versus a 3x3x3 in terms of bld?

sorry if im rambling its rather late here (or should i say early -.-)

thx~


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## tim (Dec 17, 2007)

hait2 said:


> i don't want to look up a solution as of yet, but am i on the right track? and roughly how much harder is 5x5x5 versus a 3x3x3 in terms of bld?



Yes, you are on the right track. Solving the 5x5x5 by turning it into a 3x3x3 and then solving it like a 3x3x3 bld works too, but in my opinion it's much harder than solving all groups of pieces independently.
I solve the 5x5x5 blindfolded this way (i used white color in order not to spoil ):

solving the outer centers with commutators (there's a tutorial in the how-to section of this forum)
solving the inner centers with commutators (the same principle as outer centers)
solving the outer edges with r2
solving the middle edge slice + corners like a 3x3x3 (in my case: M2 + Pochmann/Commutator)
And the Parity algorithm i use for the middle slice + corners(again in white):

[T perm] y (Rr)2 (Ff)2 U2 r2 U2 (Ff)2 (Rr)2
Instead of T-Perm you can use any algorithm which swaps two edges and two corners. (Rr)2 (Ff)2 U2 r2 U2 (Ff)2 (Rr)2 fixes just the outer edges.


In my opinion the 5x5x5 isn't much harder than the 3x3x3. You just have to memorize about four times as much as with the 3x3x3.


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## masterofthebass (Dec 17, 2007)

Tim,
Thanks for putting up that parity fix... It makes a lot of sense. I always did a z and then used r2 to fix the edges.


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