Cride5
Premium Member
A description of Phasing already exists here, but I thought I'd add some extra explanation to clarify it...
Phasing is a technique for reducing the number of ZBLL cases, thus enabling completion of the LL with 'one look' and significantly less algorithms (ZZLL). However, remember that the extra inspection required to do phasing still requires a 'look' during insertion of the final F2L block, but because phasing is very lightweight it should be possible to recognise and execute it relatively quickly.
Once Phasing is complete the LL edges will be permuted so that opposite colours (eg Blue/Green or Orange/Red) are opposite each other. If two opposite coloured LL edges are phased, then the remaining two will also be opposite each other. Looking at the LL after phasing the edges will be in one of two states. Either they are SOLVED or there is PARITY, which means that adjacent edges are not correct with respect to each other. An easy way to distinguish between SOLVED and PARITY is to attempt aligning the edges by rotating the U-layer. If its only possible to align two then it is the PARITY case.
Phasing Strategy
The strategy outlined here assumes the completed front-right block is positioned in the U-Layer (in a similar manner to Winter Variation).
With this setup there are three LL edges in the U-Layer and one is in the slot. Of the three LL edges in the U-layer, only two can easily be manipulated without breaking up the block. The third edge will be 'stuck' beside the block until the block is inserted. The stuck edge is the unlabelled LL edge in the diagram below.
The idea behind this strategy is to manipulate the remaining three free edges (labelled 1 2 and 3 in the diagram) so that they are phased after block insertion. Of the three remaining edges, find the one which needs to be placed opposite the 'stuck' edge. For example, if the stuck edge is blue, you're looking for the green edge.
If this edge is in positions 1 or 2 then the solution is fairly straightforward.
With the edge in position 1, the first R turn will place it into the correct position (replacing position 2). So the solution would simply be:
R U' R'
In position 2, it is already opposite the stuck edge, so all that's required is to ensure this state is preserved during insertion. Insertion in this case would be:
U R U2 R'
If the edge to place is adjacent to the stuck edge in the U-layer (position 3), it will initially need to be taken out of the U-layer so that it can be repositioned opposite the stuck edge. It can be done in one of two ways:
(R U R' U') R U' R'
or
(U R U' R') U R U2 R'
The part in brackets really just sets up case 1 or 2, respectively. Either of these options could be chosen depending on the position of the block in the U-layer.
Move Count
For this explanation, it is assumed that the starting point is with the block to be inserted in the U-layer (similarly to WV). Assuming a particular edge is in the 'stuck' position, there is an equal probability of its opposite edge occurring in positions 1, 2 or 3. Thus the probability of each of these cases occurring is 1/3 (~33%).
In order to find the move count, the number of moves it takes for a normal insertion (without phasing) needs to be subtracted since these moves will be required in a normal solve anyway. Here we're interested the number of 'additional' moves phasing takes.
Intuition says that the insertion in the diagram takes exactly 3 moves, but this isn't necessarily the case. If the U-face is rotated by U2 or U' then an initial AUF will be required before the 3-move insertion. Thus 2/4 cases take 4 moves and the rest take 3 moves, so normal insertion takes: (3+3+4+4)/4 = 3.5 moves
Now insertion with Phasing:
With cases 1 and 2, only one U-layer position requires no initial AUF, so the move count is: (3+4+4+4)/4 = 3.75
With case 3, because we have the flexibility in having two options, two of the U-Layer positions take 7 moves and the other two take 8. Thus the move count is: (7+7+8+8)/4 = 7.5
Because all three cases happen with equal probability the average number of moves is: (3.75 + 3.75 + 7.5)/3 = 5
So the number of additional moves that phasing (from a block in the U-layer) takes on average is:
5 - 3.5 = 1.5 moves
NOTE: This is based on the assumption that the completed block is in the U-layer. If using an insertion like R U R' or some other optimised alg which doesn't result a block in the U-layer, then an alternative phasing strategy will be required. The naive approach would be to take the completed block out and use the strategy outlined above, but much better options are available: @see Michal Hordecki's phasing algs
Phasing is a technique for reducing the number of ZBLL cases, thus enabling completion of the LL with 'one look' and significantly less algorithms (ZZLL). However, remember that the extra inspection required to do phasing still requires a 'look' during insertion of the final F2L block, but because phasing is very lightweight it should be possible to recognise and execute it relatively quickly.
Once Phasing is complete the LL edges will be permuted so that opposite colours (eg Blue/Green or Orange/Red) are opposite each other. If two opposite coloured LL edges are phased, then the remaining two will also be opposite each other. Looking at the LL after phasing the edges will be in one of two states. Either they are SOLVED or there is PARITY, which means that adjacent edges are not correct with respect to each other. An easy way to distinguish between SOLVED and PARITY is to attempt aligning the edges by rotating the U-layer. If its only possible to align two then it is the PARITY case.
Phasing Strategy
The strategy outlined here assumes the completed front-right block is positioned in the U-Layer (in a similar manner to Winter Variation).
With this setup there are three LL edges in the U-Layer and one is in the slot. Of the three LL edges in the U-layer, only two can easily be manipulated without breaking up the block. The third edge will be 'stuck' beside the block until the block is inserted. The stuck edge is the unlabelled LL edge in the diagram below.
The idea behind this strategy is to manipulate the remaining three free edges (labelled 1 2 and 3 in the diagram) so that they are phased after block insertion. Of the three remaining edges, find the one which needs to be placed opposite the 'stuck' edge. For example, if the stuck edge is blue, you're looking for the green edge.
If this edge is in positions 1 or 2 then the solution is fairly straightforward.
With the edge in position 1, the first R turn will place it into the correct position (replacing position 2). So the solution would simply be:
R U' R'
In position 2, it is already opposite the stuck edge, so all that's required is to ensure this state is preserved during insertion. Insertion in this case would be:
U R U2 R'
If the edge to place is adjacent to the stuck edge in the U-layer (position 3), it will initially need to be taken out of the U-layer so that it can be repositioned opposite the stuck edge. It can be done in one of two ways:
(R U R' U') R U' R'
or
(U R U' R') U R U2 R'
The part in brackets really just sets up case 1 or 2, respectively. Either of these options could be chosen depending on the position of the block in the U-layer.
Move Count
For this explanation, it is assumed that the starting point is with the block to be inserted in the U-layer (similarly to WV). Assuming a particular edge is in the 'stuck' position, there is an equal probability of its opposite edge occurring in positions 1, 2 or 3. Thus the probability of each of these cases occurring is 1/3 (~33%).
In order to find the move count, the number of moves it takes for a normal insertion (without phasing) needs to be subtracted since these moves will be required in a normal solve anyway. Here we're interested the number of 'additional' moves phasing takes.
Intuition says that the insertion in the diagram takes exactly 3 moves, but this isn't necessarily the case. If the U-face is rotated by U2 or U' then an initial AUF will be required before the 3-move insertion. Thus 2/4 cases take 4 moves and the rest take 3 moves, so normal insertion takes: (3+3+4+4)/4 = 3.5 moves
Now insertion with Phasing:
With cases 1 and 2, only one U-layer position requires no initial AUF, so the move count is: (3+4+4+4)/4 = 3.75
With case 3, because we have the flexibility in having two options, two of the U-Layer positions take 7 moves and the other two take 8. Thus the move count is: (7+7+8+8)/4 = 7.5
Because all three cases happen with equal probability the average number of moves is: (3.75 + 3.75 + 7.5)/3 = 5
So the number of additional moves that phasing (from a block in the U-layer) takes on average is:
5 - 3.5 = 1.5 moves
NOTE: This is based on the assumption that the completed block is in the U-layer. If using an insertion like R U R' or some other optimised alg which doesn't result a block in the U-layer, then an alternative phasing strategy will be required. The naive approach would be to take the completed block out and use the strategy outlined above, but much better options are available: @see Michal Hordecki's phasing algs
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