y' // Inspection
In this orientation, we see that the DL corners are solved, so we will use this orientation. We will use beginner tracing for this example. In this case, "5" is in UBL and "6" is in DBR. These are of the same parity. Thus we mentally swap UFL-DBR to make "6" friends with "5", and we mentally swap UFR-DFR to preserve CP. Now, since 5 is in an even position, we will read the thread as DBR-DFR-UFR-UBR-UBL-UFL. This gives us 4-1-3-2-5-6. Ignoring 5 and 6, we get 4-1-3-2. Reading cyclically from 4, we find that the CP case is 13. This CP case can be solved by swapping either "2"-"3", or "1"-"6" or "4"-"5". In this particular case, "2"-"3" look like a good option to swap, since we can set them up in UFL-UFR with a U' and use the F' U' F trigger to solve CP. While doing so, we can also simultaneously solve the DL edge by making the F' an f'.
U' f' U' F // CP-line
Now we will use the intuitive way to solve pEO-extension. Here, we see that the two first block edges are already aligned with each other, and we just need to bring the centre in the same line before connecting them all. This can be done by bringin the centre in the top face without moving either of the edges (i.e. using r' R). From there, we can connect the three pieces while making centres proper and insert the extension.
r' R u r' R' u // pEO-extension
Note that right before the second r', the DF edge was bad. Thus, we simultaneously happened to orient the DB edge and we do not need to worry about it anymore, thus completing the step.
For the next step, we have 2 bad edges in UL and UR. These can be oriented followed by the solving of DB-DF as follows:
R U' r U R2 U' r R U' r2 // EOBF
Now we proceed with the 2-gen finish
U R' U R2 U' R U R U' R' // F2L
z U R U' R U' R2 U2 R U2 R U2 R' U' R' // 2GLL
y' x // Inspection
In this orientation we see that the DL corners are only an F move away from being solved. "1" is in the DFL spot, so the DFL corner (which is currently in DFR) will be called "1" Now, we use the beginner tracing. "5" is in DBR and "6" is in UFL. These are of different parity. Thus, we mentally swap UFL-DFR to make "6" friends with "5", and we mentally swap UFR-UBR to preserve CP. Now, since 5 is in an even position, we will read the thread as DBR-DFR-UFR-UBR-UBL-UFL. This gives us 5-6-4-2-3-1. Ignoring 5 and 6, we get 4-2-3-1. Reading cyclically from 4, we find that the CP case is 23. Because "1" is stuck in DFL we recall that in the 23 CP case "1" swaps with "4". The DFL corner is already an F away from being solved so we can keep it as it is, and we need to put "4" in UBR. Luckily, "4" is already in UBR. Thus we can simply go ahead and do the F move to solve CP and DL corners. However, doing an f instead will actually make it easier to insert the DL edge.
f U2 S' // CP-line
Now for pEO extension, we will use the algorithmic approach. Note, doing u' R will attach a first block edge with the orange centre. Then we can bring the centre to the right face while ensuring the other edge stays in the top layer and a non-white-yellow centre comes to the top.
u' R r' u' R // setup
UL is good, UB is bad, UR is good. Thus we do the corresponding algorithm to make DB good, make centres proper, and preserve some good edges.
U' R u2' U' r // pEO extension
We have 4 bad edges, which are in UF, UR, FR and RD. Thus we can solve them as written below. Note that we end EO with U r' to ensure that we can solve DB-DF efficiently.
r2 U' R' U' r' U r' U' r2 // EOBF
Now we simply proceed with the 2-gen finish
U' R' U R U' R' U R // Square
U R U' R' U2' R U2' // F2L
R2 U' R U' R' U2' R' U2' R' U' R U' R2 U2' // 2GLL
z2 x' // Inspection
In this orientation the DL corners are solved, thus we shall use this orientation. We will use advanced tracing for this solve. "5" is in UBR and "6" is in UFL. This is the letter N case. "4" is in UFR, thus following the circlet in the direction determined by the position of "5", we need to read DFR-UBL, doing which we find that the CP case is 23. We can place the DL edge without disturbing the corners, and then we can proceed to swap "3" and "6" since they are in convinient spots.
S2 U' F' U' F // CP-line
Note, the above can also be executed as y' r2 R2 U' R' U' R y.
Now we will use intuitive pEO-extension to solve this case. Note that we can align all 3 pieces using u U'; then we can attach them and insert the extension. This will leave a non-white-yellow centre on top, so we will find a bad edge on top layer and put it in UB, then do r.
u U' r2 R u r // pEO extension
It so happened that UB was already bad, so we directly inserted the edge in BD while making the centres proper. This leaves us we 2 bad edges in UF and UL. These can be solved followed by DB-DF as follows.
U' r U R' U r R2 U' r2 // EOBF
Now we continue with F2L and 2GLL
R' U' R' U R' U R2 U R U2 R // F2L
U2 R2 U R2 U2 R' U' R U' R U' R U' R2 U2 R' // 2GLL
z' y’ // Inspection
In this orientation the DL corners are 2 moves from being solved. "6" is stuck in DFL, thus the actual DFL corner (currently in UFL) will be numbered "6". We will use advanced tracing for this solve. "5" is in UFR and "6" is in UFL. This is an almost-friends case. "4" is in UBL, thus following the circlet in the direction determined by the position of "5", we need to read UBR-DFR, doing which we find that the CP case is 12. This is the special case (i.e. where DL corners are unsolved but CP is solved), thus we will try to put the DFL corner an F2 away from being solved. But before that, we can connect the DL edge with the DFL corner by doing r. Thus, the solution will be:
r u’ f U f // CP line
Here both algorithmic and intuitive pEO extension will give the same solution since it is a simple case. One edge is already connected to the centre, thus we can bring it to the right face, and then attach the other edge while orienting DB and making centres proper. Then we can insert the extension.
u2 r’ R’ U’ r’ u2 // pEO extension
We now have 4 bad edges. These can be oriented along with solving of DB-DF as follows:
r’ U2 r’ U r’ U’ r U r2 // EOBF
Now we continue with F2L, but on the side face
R U R2 U // square
R’ U R’ U’ R2 U R2 U’ // F2L
We can now rotate and get a 2GLL case.
z’ y R U R’ U R U’ R’ U R U2 R’ U2 // 2GLL
Movecount: 45 - 55
Average non-[r, R, U] moves: 5
The above example solves try to cover almost every technique, beginner and advanced, upto EO-CP-2x2x3. The average movecount of the above solves is in the low 40's, however during a speedsolve one may not always identify the most efficient way to do things, especially in a method that is designed to rely on ergonomics and TPS; and a better estimate of the average movecount during speedsolves would be close to 50.