In Christmas 2002, I worked on a Human variant of the Thistlethwaite computer algorithm.
Below are my posts to the Yahoo! speedsolvingrubikscube mailing list which describe the method.
From: Ryan Heise <rheise@it....>
Date: Sun, 22 Jun 2003 11:30:52 +1000
To: Ron van Bruchem <rvb@chello....>
Subject: Thistlethwaite, human version
On Sat, Jun 21, 2003 at 06:13:28PM +0200, Ron van Bruchem wrote:
> Hi Ryan,
> I am very interested in the ideas you have.
> Please tell me something about the systems you came up with, and how many
> algorithms you need per stage.
Phase 1 -> <U,D,L,R,F2,B2> group
- simple, no algorithms
Phase 2 -> <U,D,L2,R2,F2,B2> group
- Direct up/down edges to up/down face (simple, no algs)
- Direct corners to up/down face (between 8 and 60 algs)
Phase 3 -> <U2,D2,L2,R2,F2,B2> group
- Corners (between 1 and 2 algs)
- Edges (between 1 and 4 algs)
Phase 4 -> place pieces
- Corners (intuitive)
- Edges (intuitive)
DETAILS OF STEPS
* PHASE 1
This is solved in 4.6 moves on average.
* PHASE 2 EDGES
This is rather simple. You can learn all 20-30 cases if you wish. I
forget the exact number. This can be solved in an average of 4 moves.
* PHASE 2 CORNERS
I used a method similar to Gaetan - first get 3 corners oriented on one
side, and then apply one of 8 algorithms. It is possible to directly
learn all 60 cases if you want (I can't remember the exact number).
I think they have an average of 8.5 moves.
* PHASE 3 CORNERS
In phase 3, it is important to do the corners first, because it is
difficult to see whether they have made it into the U2D2L2R2F2B2 group.
Just getting opposite colours on each side isn't enough. The algorithms
you learn to fix this are shorter when you don't have to worry about the
Here, I'll just describe the simplest technique that requires two
algorithms, but is very quick for the fingers and brain:
First, separate up/down colours (one colour on each side). Average 3.2
moves. There should be, for example, all red corners on top, and all
orange corners on bottom.
Now, pairs of adjacent corners will either match or mismatch. Our goal
is to make them either all match, or all mismatch. So, in this step, we
find the odd pairs out (whether they're matching or mismatching), and
fix them so they match/mismatch like all the rest. There are 4 pairs.
Either one pair is the odd one out, or two pairs are the odd ones out.
For one pair: hold the pair at UF, and do R'FR'B2RF'R. It's a
modification of the corner mover that doesn't care about the exact
positions of corners.
Two pairs: hold two pairs on F (you may need to move them there), and do
R2UF2U2R2U. (if you needed to move them there first, there's also a
trick to get it to work...)
I looked for a long time to find other methods here that used fewer
moves. I found some, but this way was definitely by far the quickest to
PHASE 3 EDGES
4 cases - simple (2,4,6 or 8 bad edges). Average 6.1 moves.
Total moves so far: 33.4. Obviously, fewer moves are necessary to
achieve an average of 40 moves overall. I worked out some shortcuts, but
I don't think they're worth it, because I could perform the longer way
PHASE 4 (the end game)
I think you already have a strategy for this. Corners, then edges. I
think it's possible to learn all cases for the edges (about 150 I think,
but easy to memorise).
A downside is the number of double turns which are more difficult to
perform. But I tried a few algorithms and they are possible to do
quickly enough. I think the main benefit of this method is fast reaction
time and no thinking. Another benefit is that it looks cool when you
solve it. None of the pieces are placed until the very end.
Above, I listed each individual step with no shortcuts. It is possible
to combine steps, or do steps in different orders depending on
opportunities. The basic method above, if you learnt all cases for each
exact step, should give an average of 45.7 moves.
From: Ryan Heise <rheise@progsoc....>
Date: Wed, 9 Jul 2003 15:42:19 +1000
Subject: [Speed cubing group] Method with the fewest non-obvious algorithms - zero
On Tue, Jul 08, 2003 at 05:09:13PM -0000, tomrokicki wrote:
> So what method requires the fewest algorithms that aren't "obvious"
> or logically clear?
Using a simplified Thistlethwaite approach, it is possible to solve the
cube without any mysterious algorithms.
Step 1 uses no algorithms.
Step 2 uses a 5 move algorithm.
Step 3 uses a 7 move algorithm and a 4 move algorithm.
Step 4 uses two 6 move algorithms.
Because all algorithms are short and simple, someone who tries this
method can say they solved the cube and understood how they did it!
It's not too easy to follow, but it's an interesting way to solve the
cube if you can do it.
** step 1 **
Put cube into <L2,R2,F,B,U,D> group, which I will call G1. That is to
say, put the cube into a position where the cube can be solved by only
using L2,R2,F,B,U,D moves. It helps in this method to think of opposite
colours as the same colour up until you get to step 4. Let's call the
left/right colours "green" (these are unimportant in step 1), the
up/down colours red, and the front/back colours white.
Once you get into G1, it is impossible to flip edges, and the only thing
we need to do to get the cube into G1 is to make sure all the edges are
flipped right. First you need to find bad edges that are flipped the
wrong way. A bad edge is one that, if moved to its home position using
only moves from G1, will be flipped the wrong way. (Of course, you don't
need to move it there to see that)
There are always an even number of bad edges, between 0 and 12, or 6 on
average. The only way to flip edges to turn the left or right sides 90
degrees, and that will flip the 4 edges on that side. To flip 4 bad
edges, you simply move them all into positions on the left side (for
example), and turn that side 90 degrees. If you have only 2 bad edges,
you can make that into 4 by moving just one of them onto the left side
and turning it. That will change one of the bad edges into 3 (plus the
other one it makes 4).
At end of step 1, you should see no red edges facing to the front or
back, and you should see no white edges facing to the top or bottom. To
the untrained eye, it will look like you have accomplished nothing so
far! Yet, these 12 edges are in a much better position now.
** step 2 **
Put cube into <L2,R2,F2,B2,U,D> group, which I will call G2. In this
group, pieces in the middle layer will stay in the middle layer without
flipping, and pieces in the top and bottom layers can only exchange
between those layers without twisting or flipping. At the end of step 2,
you should only see red on the top and bottom. In the middle layer, you
should only see green on the left/right sides, and white on the
Remember we are currently in G1, so to get to G2, we can only use moves
from G1. To start with the edges, we want to get a red cross on the top
and bottom. There are 8 red edges and because of limited space, at least
4 of those must already be forming part of the crosses. The general
strategy is to first get 3 red edges on the top and 3 edges on the
bottom, and align them so that the missing parts of both crosses are
facing to the front. The other 2 red edges must be somewhere in the
middle layer (if not, you must have 4 edges instead of 3 edges on the
top or bottom). Using double turns around the L2,R2,F2,B2 sides, move
those two remaining red edges over to the two front positions on the
middle layer, and turn the front 90 degrees. With the edges done, we
have created some symmetry so it is possible rotate the cube about the
U-D axis and change your perception of what is the front and back. Very
Now for the corners, we can twist two corners at a time. Start with one
corner at LUF (corner A) and the other corner at FDR (corner B). Then
R'D - twists the B corner
L2 - swaps the B and A corners
D'R - twists the A corner on the way back
Do this repeatedly until all corners are twisted the right way. You
should have all red on the top and bottom.
** step 3 **
Put cube into <L2,R2,F2,B2,U2,D2> group, named G3. In the previous step,
you put only opposite colours on the top and bottom (red). In this step,
you do the same for the front/back (white) and left/right (green). In
G3, each corner can only move between one of 4 positions, which I call a
circuit. There are only two circuits and each corner belongs to one of
those circuits. Within G3, the only moves you can do (double turns) will
swap two corners in one circuit, and two corners in the other circuit
that overlap. This limits the permutations that are possible, and it
means that simply getting opposite colours on all sides is not enough to
Since the corners are the tricky bit here, we will put the corners into
G3 first, then the edges. You must now distinguish between opposite
colours to know if G3 is satisfied. I will use red opposite to orange,
white to yellow, and green to blue. An easy way to see if G3 is
satisfied is to move all red corners to the top side, and all orange
corners to the bottom side. Pairs of corners around the top and bottom
will either match or mismatch. If they either all match, or they all
mismatch, the corners are in a G3 position. There is only one way for
all corners to match, and one way for all corners to mismatch (you swap
a pair of opposite corners), and so when the exact pattern occurs on the
top and bottom, the bottom corners should align exactly with the top
corners. You can switch between all matched, and none matched, by doing
R2U2R2 (that swaps a pair of opposite corners on the bottom, and a pair
of opposite corners on the top). If you're not in either of those
positions, here is a sequence that will swap a pair of opposite corners
on the bottom, and an adjacent pair of corners on the top:
R'FR' - move the 4 target corners to the back side
B2 - swap!
RF'R - move everything back
This is one case where it is more efficient to temporarily break G2. At
the end of the move, everything is restored. This move sequence will
do swapping on the bottom side, and at the rear/back. The front/top pair
of corners will be left alone. When you are looking for corners that
match or mismatch, you will usually find a pair that are not like the
rest of the pairs (eg. if all the other pairs match, this pair will
mismatch). If you find this odd pair and hold it at the top/front, doing
the above sequence will harmonise all the corners.
Now for the edges, there is nothing special here. The edges also lie on
circuits, but there are 12 edges and 3 circuits. The edges in the middle
layer were already done in step 2 and haven't left their circuit. There
are only 8 edges left do deal with, and we must move the white edges to
the white sides (front/back), ie. into their circuit, and the green
edges to the green sides (left/right). There must be an even number of
edges that are facing in the wrong direction. The easiest move to do
here is to swap two edges on the green sides for two edges on the white
sides. The edges on the greens side must be positioned on the bottom,
and the edges on the white sides must be positioned on the top. Then do:
M2 (M is the slice between L and R) - swap!
This is self explanatory. Like in step 1, if you only have 2 bad edges,
you can make it into 4 in a similar way.
At the end of this step, you should only have two (opposite) colours on
each side of the cube.
The cube can now be solved by only using double turns. However, we will
break this rule when we see an obvious short cut.
First, using only double turns, solve the corners. First, pick a side,
then join two corners on that side and the other two corners for that
side will automatically join (if you did step 3 properly). It is then
easy to join the two pairs together, and the other 4 corners will be
automatically joined (if you did step 3 properly).
To solve the edges, there are two simple moves that are easy to
F2R2 F2R2 F2R2 - Swap two pairs of edges on different circuits. The
other pieces return to normal.
F2 M2 F2 M2 - Swap two pairs of edges on the same circuit. The second
F2 returns the other pieces back to normal.
Using variations of these, it is possible to solve all the edges.
Variations of the first include:
R (F2R2 F2R2 F2R2) R'
L2B2R (F2R2 F2R2 F2R2) R'B2L2
Variations of the second include:
F2 M F2 M' (the pairs overlap so as to rotate 3 edges)
F (F2 M2 F2 M2) F'
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