Re: 5x5x5 help

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• ... apply steps 6 and 7 to ... Hi, Thanks for your interest in my method. Step3. (Middle edges.) and Step6. (Permute last U edges) may be the most difficult
Message 1 of 23 , May 3, 2004
wrote:
> in masayuki akimoto's solution to the professor cube, how do you
apply steps 6 and 7 to
> the 5x5x5?
Hi,

Thanks for your interest in my method.
Step3. (Middle edges.) and Step6. (Permute last U edges)
may be the most difficult parts in my method.
I worry if you may not understand without pictures.

Step6. (Permute last U edges)
Before explanation, I would define each edge as El, Ec, Er.

Ec :MD2M'
El, Er :rl'D2r'l
El,Ec, Er:rMl'D2r'M'l
Those algorithms should not mix up center pieces.
Use these algs effectively.

If you have pair of El-Ec, Ec-Er, or El-Ec in same orientation, solve
this pair first.
El or Er left. You can apply Step 6-2 of 4x4x4.
Ec left. MD2M' or S'U'SDS'USD'

In the other case, solve either Ec or Er (El) pretending l, M, and r
as one slice set, you can solve at least one edge either El, Ec, Er
as same as 3x3x3 but you may need extra Ds to flip the rest of edges.
If you solved Ec first, and if you could keep either Er or El which
can be solved by rl'D2r'l, you can solve the last edge as same as
4x4x4.

If you solved Er or El first, you can also place Ec which can be
solved by MD2M'. you can solve the last edge easily.

I normally don't use but in case you have edge pair in different
orientation that means Er-Ec or El-Ec, solve Ec with flipped Er or
El.
Then you can apply 7-2-2iv to solve Er and El simultanously.

Step7 (Last Layer)
7-1 (corners) Must be as same as smaller cubes.

7-2-1,7-2-2
You can use all the algorithms.

7-2-3
This is Chris's algorithm. You will have middle edges problem if you
directly apply this for 5x5x5.
You can fix the centers by E'L2Er2E'L2Er2.
Or you can solve by combintaion of 7-2-2iii and 7-2-2-iv

7-2-4
You can directly apply them to 5x5x5.

Good luck.

Masayuki Akimoto
• Does anyone have a list of the 469 algs used in Watermans( i think) CF method? I d like to see them if possible... thanks jake
Message 2 of 23 , May 3, 2004
Does anyone have a list of the 469 algs used in Watermans( i think)
CF method? I'd like to see them if possible... thanks

jake
• Where did you get the 469 figure from ? Wayne
Message 3 of 23 , May 4, 2004
Where did you get the 469 figure from ?

Wayne

--- In speedsolvingrubikscube@yahoogroups.com, j_rueth
> Does anyone have a list of the 469 algs used in Watermans( i think)
> CF method? I'd like to see them if possible... thanks
>
> jake
• I remember someone saying that there is a corners first method uses 469 algs, i think dan gosbee was saying that this was the method he was using, or
Message 4 of 23 , May 4, 2004
I remember someone saying that there is a corners first method uses 469
algs, i think dan gosbee was saying that this was the method he was using,
or developed... I just wanted to see what this method is like

:) jake

--- In speedsolvingrubikscube@yahoogroups.com, "Wayne" <
mylib_2000@y...> wrote:
> Where did you get the 469 figure from ?
>
> Wayne
>
>
> --- In speedsolvingrubikscube@yahoogroups.com, j_rueth
> > Does anyone have a list of the 469 algs used in Watermans( i think)
> > CF method? I'd like to see them if possible... thanks
> >
> > jake
• His method may require that many but the Waterman system is probably around 150. Wayne ... 469 ... was using, ... think)
Message 5 of 23 , May 4, 2004
His method may require that many but the Waterman system is probably
around 150.

Wayne

--- In speedsolvingrubikscube@yahoogroups.com, j_rueth
> I remember someone saying that there is a corners first method uses
469
> algs, i think dan gosbee was saying that this was the method he
was using,
> or developed... I just wanted to see what this method is like
>
> :) jake
>
> --- In speedsolvingrubikscube@yahoogroups.com, "Wayne" <
> mylib_2000@y...> wrote:
> > Where did you get the 469 figure from ?
> >
> > Wayne
> >
> >
> > --- In speedsolvingrubikscube@yahoogroups.com, j_rueth
> > > Does anyone have a list of the 469 algs used in Watermans( i
think)
> > > CF method? I'd like to see them if possible... thanks
> > >
> > > jake
• Dan Gosbee s CF method requires 463 algorithms. Kelley
Message 6 of 23 , May 4, 2004
Dan Gosbee's CF method requires 463 algorithms.

Kelley
• ... I don t know any method requiring 469 algorithms , but if the method you re looking for belongs to the CF class of approaches, you could try the
Message 7 of 23 , May 4, 2004
wrote:
> Does anyone have a list of the 469 algs used in Watermans( i think)
> CF method? I'd like to see them if possible... thanks
>
> jake

I don't know any method requiring 469 "algorithms", but if the method
you're looking for belongs to the CF class of approaches, you could
try the cornersfirst Yahoo group.

Gilles.

PS: Josef Jelinek is building a page dealing with the method used by
Marc Waterman (work in progress). See
http://www.rubikscube.info/waterman/index.html
• Thanks for the link Gilles, I ll check it out. I thought watermans solution was a lot bigger. Thanks jake
Message 8 of 23 , May 4, 2004
Thanks for the link Gilles, I'll check it out. I thought watermans
solution was a lot bigger. Thanks

jake
• ... think) ... There are a total of 141 algs used in Waterman s method, provided that you don t run into any complications. This total includes a lot of
Message 9 of 23 , May 4, 2004
> Does anyone have a list of the 469 algs used in Watermans( i
think)
> CF method? I'd like to see them if possible... thanks
>
> jake

There are a total of 141 algs used in Waterman's method, provided
that you don't run into any complications. This total includes a
lot of mirrored algs.

Roughly, the method involves the following steps:

1. solve one face

2. orient and permute the remaining corners (1 of 43 algs)

Next, hold the solved face in your left hand, with the corners
solved in step 2 in the R face. The next steps are to solve the
right edge pieces (called "redges"), and orient all the middle edge
pieces (called "midges") in two algs. The M slice is referred to as
the "ring".

3. Solve 2 redges such that you leave 1 or 0 redges in the ring.
Do this by picking an alg which solves 1 redge in the ring and 1 in
the R face (1 of 6 algs), OR pick an alg which solves 2 redges in
the ring (1 of 6 algs)

4. Solve the last 2 redges, and orient all the midges in 1 alg. If
both redges are in the R face, then you have to pick 1 of 14 algs.
If one redge is in the R face, and one is in the ring, then pick one
of 48 algs.

5. solve the midges, 1 of 3 algs, which are trivial.

additional algs are included to handle situations where one or more
redges are solved by the end of step 2.

flip one redge and orient midges: 2 algs

solve one redge in the ring and orient midges: 16 algs (8 algs and
their mirrors)

if all redges are solved, orient midges: 3 algs

I think that the algs from step 3 are borrowed from the algs in step
4, but listed separately.

In addition to this set of algs, there are additional components to
the system designed to handle ugly situations such as when you have
3 or 4 redges cycled in the r face, or when you have all 4 redges in
the ring.

If you can master all of the bad situations in addition to the basic
(!) system, you should be able to solve the R face and orient the M
slice in 2 algs or processes, no matter what.

since you pretty much use only U, R, and M, you don't have to regrip
with your left hand too much (there are some algs with F moves).

Here are a few example algs from step 3 to show the basic idea. Try
doing the inverse of the algs on a solved cube to set these up.

To solve 2 redges in the ring, one at FU and one at FD (FU means
that the R facelet of the redge is on the F face as opposed to UF,
which puts the R facelet on the U face) , do this:

U2 (R) M U2 M' U M2 U (R)' U2

When you execute this alg, the redge at FU will get placed in the
hole at RU, and the redge at FD will get placed in the hole in the R
face which is turned up to the RU position with the turn (R). (R)'
is just the inverse of whatever R move you did in (R).

To solve 2 redges where one is in the ring at DF (R facelet of this
redge is in the D face), and one is in the correct hole in the R
face at the position RU but flipped, do this:

U2 (R) U' M U2 M2 U' (R)' U2

in this case, the move (R) means turn the R face so that the hole
that the redge in the ring belongs to gets moved to the RU position.

In any case its a neat but scary system.

Lucas
• Only several more notes. To count exact number of algorithms in Waterman s method is quite difficult because each algorithm using (R) and (R) symbols is a
Message 10 of 23 , May 6, 2004
Only several more notes.

To count exact number of algorithms in Waterman's method
is quite difficult because each algorithm using (R) and (R)'
symbols is a shortcut for more algorithms and should
be learnt that way to get really fast in performing them...

The original Waterman's description of "bad luck cases"
and "complications" is too complicated and in fact can be
handled simpler with less algorithms with the same efficiency.

The algorithms mentioned in the previous message (see below)
are not the most efficient (and this is true for more algs
in the Waterman's method). You can use the following
replacements (found by myself):

U M (R) U' M U M' (R)' U'
(instead of U2 (R) M U2 M' U M2 U (R)' U2)
- shorter, without half moves

U M' U' (R) U' M' U2 M2 U'
(instead of U2 (R) U' M U2 M2 U' (R)' U2)
- less half moves, only one (R) in alg., 1 more M turn,
there is a hole in RU and not the flipped edge (!)

In addition, I have found replacements for all algorithms
that use F turns (hard to perform quickly) and now
all algorithms use only U M R moves (!) and are of the
same length or even shorter (except for one or two algs.)

Josef

> There are a total of 141 algs used in Waterman's method, provided
> that you don't run into any complications. This total includes a
> lot of mirrored algs.
>
> Roughly, the method involves the following steps:
>
> 1. solve one face
>
> 2. orient and permute the remaining corners (1 of 43 algs)
>
> Next, hold the solved face in your left hand, with the corners
> solved in step 2 in the R face. The next steps are to solve the
> right edge pieces (called "redges"), and orient all the middle
edge
> pieces (called "midges") in two algs. The M slice is referred to
as
> the "ring".
>
> 3. Solve 2 redges such that you leave 1 or 0 redges in the ring.
> Do this by picking an alg which solves 1 redge in the ring and 1
in
> the R face (1 of 6 algs), OR pick an alg which solves 2 redges in
> the ring (1 of 6 algs)
>
> 4. Solve the last 2 redges, and orient all the midges in 1 alg.
If
> both redges are in the R face, then you have to pick 1 of 14
algs.
> If one redge is in the R face, and one is in the ring, then pick
one
> of 48 algs.
>
> 5. solve the midges, 1 of 3 algs, which are trivial.
>
> additional algs are included to handle situations where one or
more
> redges are solved by the end of step 2.
>
> flip one redge and orient midges: 2 algs
>
> solve one redge in the ring and orient midges: 16 algs (8 algs and
> their mirrors)
>
> if all redges are solved, orient midges: 3 algs
>
> I think that the algs from step 3 are borrowed from the algs in
step
> 4, but listed separately.
>
> In addition to this set of algs, there are additional components
to
> the system designed to handle ugly situations such as when you
have
> 3 or 4 redges cycled in the r face, or when you have all 4 redges
in
> the ring.
>
> If you can master all of the bad situations in addition to the
basic
> (!) system, you should be able to solve the R face and orient the
M
> slice in 2 algs or processes, no matter what.
>
> since you pretty much use only U, R, and M, you don't have to
regrip
> with your left hand too much (there are some algs with F moves).
>
> Here are a few example algs from step 3 to show the basic idea.
Try
> doing the inverse of the algs on a solved cube to set these up.
>
> To solve 2 redges in the ring, one at FU and one at FD (FU means
> that the R facelet of the redge is on the F face as opposed to UF,
> which puts the R facelet on the U face) , do this:
>
> U2 (R) M U2 M' U M2 U (R)' U2
>
> When you execute this alg, the redge at FU will get placed in the
> hole at RU, and the redge at FD will get placed in the hole in the
R
> face which is turned up to the RU position with the turn (R).
(R)'
> is just the inverse of whatever R move you did in (R).
>
> To solve 2 redges where one is in the ring at DF (R facelet of
this
> redge is in the D face), and one is in the correct hole in the R
> face at the position RU but flipped, do this:
>
> U2 (R) U' M U2 M2 U' (R)' U2
>
> in this case, the move (R) means turn the R face so that the hole
> that the redge in the ring belongs to gets moved to the RU
position.
>
> In any case its a neat but scary system.
>
> Lucas
• ... Nicely done! I should have guessed that there was room for improvement with the original algs. On closer inspection of the algs in Waterman s method, I see
Message 11 of 23 , May 6, 2004
--- In speedsolvingrubikscube@yahoogroups.com, "Josef Jelinek"
<gloom@e...> wrote:
> Only several more notes.....
>
> The original Waterman's description of "bad luck cases"
> and "complications" is too complicated and in fact can be
> handled simpler with less algorithms with the same efficiency.
> (about 4 + 2 algs)
>
> The algorithms mentioned in the previous message (see below)
> are not the most efficient (and this is true for more algs
> in the Waterman's method). You can use the following
> replacements (found by myself):
>
> U M (R) U' M U M' (R)' U'
> (instead of U2 (R) M U2 M' U M2 U (R)' U2)
> - shorter, without half moves
>
> U M' U' (R) U' M' U2 M2 U'
> (instead of U2 (R) U' M U2 M2 U' (R)' U2)
> - less half moves, only one (R) in alg., 1 more M turn,
> there is a hole in RU and not the flipped edge (!)
>
> In addition, I have found replacements for all algorithms
> that use F turns (hard to perform quickly) and now
> all algorithms use only U M R moves (!) and are of the
> same length or even shorter (except for one or two algs.)
>
> Josef
>

Nicely done!

I should have guessed that there was room for improvement with the
original algs.

On closer inspection of the algs in Waterman's method, I see that
there are similar algs in table 4 which are similar to your second
example: U M' U' (R) U' M' U2 M2 U'.

Table 4, alg 17a: U M' U' (R) U M2 U2 M U

This makes me wonder why they published Table 2 with less efficient
algs borrowed from table 4 which accomplish the same thing!

How many algs have you found which are more efficient for the method?
I would love to see them.

I am sort of glad that I never had the guts to try to master the
original algs, given that I would have to spend a lot of time
unlearning the algs with more efficient alternatives.

Lucas
• Hi guys, I need an alg to switch two opposite edge centers without affecting any other center pieces. I feel dumb because I can t figure this out on my own.
Message 12 of 23 , Oct 25, 2005
Hi guys,

I need an alg to switch two opposite edge centers without affecting
any other center pieces. I feel dumb because I can't figure this out
on my own. =/ Help please!

-Jason
• Nevermind, I figured it out. Probably not the most efficient way of doing it but at least it gets the job done. :P -Jason
Message 13 of 23 , Oct 25, 2005
Nevermind, I figured it out. Probably not the most efficient way of
doing it but at least it gets the job done. :P

-Jason

--- In speedsolvingrubikscube@yahoogroups.com, "Jason Baum"
<speedrunningcuber@y...> wrote:
>
> Hi guys,
>
> I need an alg to switch two opposite edge centers without affecting
> any other center pieces. I feel dumb because I can't figure this out
> on my own. =/ Help please!
>
> -Jason
>
• anyone have any tips for making a 5x5x5 loose. i have used silicon and it helped a little but it is still stiff. Patrick
Message 14 of 23 , Jun 10, 2007
anyone have any tips for making a 5x5x5 loose. i have used silicon and
it helped a little but it is still stiff.

Patrick
• Wearing it down is the best way, imo. Jonathan Choi
Message 15 of 23 , Jun 10, 2007
Wearing it down is the best way, imo.

Jonathan Choi

--- In speedsolvingrubikscube@yahoogroups.com, "Patrick Jameson"
<poker19@...> wrote:
>
> anyone have any tips for making a 5x5x5 loose. i have used silicon and
> it helped a little but it is still stiff.
>
> Patrick
>
• Cube with it for maybe 2 or 3 months keeping the lubrication minimal, and then take it apart, claen it all out, lube it, and put it all back together. The lack
Message 16 of 23 , Jun 10, 2007
Cube with it for maybe 2 or 3 months keeping the lubrication minimal,
and then take it apart, claen it all out, lube it, and put it all back
together. The lack of lube smooths out all the rough edges nicely.
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