Blindsolving Discussion and Help

@Penguino138:

You can get something like this: (do this as a setup move)
U' M2 U R U' M2 U R' x' M2 U' R U M2 U' R' U

Notice that the buffer is flipped and there is only a 3-cycle left. This is because (I don't know the correct terminology here) the cycle is flipped. It's UB->FR->UR->BU instead of UB->FR->UR->UB. There are still 2 unoriented edges, in this case DF and UR.

If you just break into the cycle and solve it as normal with M2, there shouldn't be a problem.
 
aronpm said:
@Penguino138:

You can get something like this: (do this as a setup move)
U' M2 U R U' M2 U R' x' M2 U' R U M2 U' R' U

Notice that the buffer is flipped and there is only a 3-cycle left. This is because (I don't know the correct terminology here) the cycle is flipped. It's UB->FR->UR->BU instead of UB->FR->UR->UB. There are still 2 unoriented edges, in this case DF and UR.

If you just break into the cycle and solve it as normal with M2, there shouldn't be a problem.
Click to expand...

How do u know during inspection, because during it, theres only one flipped edge.....
 
"If you just break into the cycle and solve it as normal with M2, there shouldn't be a problem."

You don't need to know. Solve it as normal, and your buffer will be flipped at the end.
 
aronpm said:
"If you just break into the cycle and solve it as normal with M2, there shouldn't be a problem."

You don't need to know. Solve it as normal, and your buffer will be flipped at the end.
Click to expand...

Yeah ik the buffer will be switched. (By the way is the buffer FD OR BU?) But there has to be 2 flipped pieces. And if the buffer is the only flipped piece, how do you know which edge will be flipped? (Im probaly overthinking this)
 
Er sorry, I meant that if there are no flipped edges when you start memorisation (not 'inspection' btw), then there will be no flipped pieces at the end. (if you did have a flipped edge at the start, that and the buffer would be flipped at the end)
 
AustinReed said:
When I do 4x4 BLD, it's really hard for me to tell when I've memo'ed all the edges. How do you make sure you have?
Click to expand...
What I do is NEVER assume you're finished. Just keep memoing. Then when you're sure you've finished, look as some of the edges and know their letters, ex. I see the "p" is not solved, do I have a "p" in my memo, why, yes I do. Sometimes there can be just two pieces that need swapping, and you won't notice them. Last resort is to go over the memo putting fingers on edges and stuff.
 
AustinReed said:
When I do 4x4 BLD, it's really hard for me to tell when I've memo'ed all the edges. How do you make sure you have?
Click to expand...

Count. If you have no solved pieces when you start and there is just a single cycle, there should be 23 pieces memorized. For each extra cycle, add another piece to the total needed. When you think you might be done, count the already solved pieces. If you get to the needed total, you're done; if not, you need to look for more pieces.

This is easier if you have a standard-length journey that you always use. I put 6 pieces in each location with my wings. So that means if I have no extra cycles, I'm done when I put the fifth piece in my fourth cycle. So I pretty much intuitively know when I'm done based on that.
 
aronpm said:
Er sorry, I meant that if there are no flipped edges when you start memorisation (not 'inspection' btw), then there will be no flipped pieces at the end. (if you did have a flipped edge at the start, that and the buffer would be flipped at the end)
Click to expand...

Ok, ill try to explain this in detail. So i did your scramble example you gave me. I first see ONLY ONE edge is flipped. Heres what i do: swap with RF, shoot to RU, swap with RU, then shoot to FR. Then i just have an M2 to finish. Is that how you would do it? And how would you know to do the M2?
 
This is what I would do, and I'm sure aron as well

Break into any piece. Aka UB. Just do M2. That piece belongs to FR. So do U R U' M2 U R' U'.

Now the buffer holds UR. R' U R U' M2 U R' U' R.

Now the bffer piece goes to BU. Not the same as UB. So do the BU alg. CBF to type it out
 
Last edited:
Penguino138 said:
Ok, ill try to explain this in detail. So i did your scramble example you gave me. I first see ONLY ONE edge is flipped. Heres what i do: swap with RF, shoot to RU, swap with RU, then shoot to FR. Then i just have an M2 to finish. Is that how you would do it? And how would you know to do the M2?
Click to expand...

There is no point flipping the edge at the start. COMPLETELY IGNORE YOUR BUFFER PIECE DURING MEMORISATION. When you are memorising your cube, if your buffer piece becomes solved into position REGARDLESS OF ORIENTATION, then just break into a new cycle, i.e. Shoot to somewhere that is unsolved. When you finish your cycles, your buffer will be solved unless there are other flipped edges (Edges in their correct location but oriented incorrectly).

If you do the setup as shown by aronpm and take this as an example scramble I will explain how I would do it. Your memorisation starts and the first thing you noticed is that your buffer piece is in the buffer position (Ignore the fact it is flipped, but it is in position). So you break into a new cycle. So I would start this at the UR piece, so shoot to UR, then I shoot that to BU, then that is shot to RF, which is shot to RU. You already shot your buffer piece to this location (Regardless of orientation) so that is the end of your cycle. Everything else is solved so blindfold on.

Solving. Your first piece you memorised was UR, so we start with this: R U R' U' M2 U R U' R'

Your next piece was BU (I flip the back edge, shoot to UB, and flip the edge back): B' U R' B U' M2 U B' R U' B

The next piece is RF: x' U' R2 U M2 U' R2 U x

The final piece in the cycle is RU: x' U' R U M2 U' R' U x

The cube is now solved. This is because unless there are other flipped edges on the cube, as long as you finish your cycles, the buffer piece will always be solved because it is impossible to have a single flipped edge on a cube.
 
Hi guys,

I have been getting into 5BLD again. It seems I have forgotten some things.

Solving order:
- Midges (BH)
- X-centers (BH)
- +-centers (BH)
- Wings (BH)
- Corners (BH)

When there is parity in midges, I do Ui F2 U m2 Ui F2 U. But... IT AFFECTS +-centers. DOH. Many messed up solves resulted by this. Now I do the parity-solve just before the corners. That is one more thing to remember. Is there any way around this?
 
MrMoney said:
When there is parity in midges, I do Ui F2 U m2 Ui F2 U. But... IT AFFECTS +-centers. DOH. Many messed up solves resulted by this. Now I do the parity-solve just before the corners. That is one more thing to remember. Is there any way around this?
Click to expand...
After you've practised this system for a bit, I don't think it'll take long for you to get into the habit of always asking yourself if you have midge-parity after you've solved the wings.

But the way around this is to find a way to execute the parity alg after the corners, so the very last thing. Depending on your memorising order, you may or may not want to switch your execution order. Here's one way you could execute the parity alg at the end:

-If you memorise the midges last, you'll already know (from the corners) that you have parity. You can then treat UL as if it's UB and vice-versa, this will result in all the midges solved except for UL and UB which will be swapped. By having these 2 midges swapped, it allows the midges to be executed with an even number of swaps, so the centers remain in their original positions.
-Sometime during the corners you're going to have to apply an alg that swaps 2 edges and 2 corners; the very last corner can be solved with OP.
-Unless you have an alg that swaps just 2 midges and 2 corners, you need to execute a wing swap alg to place the UL and UB wings back where they were. I don't know what alg you prefer, but here's one that could be used: B U' B' (Rw2 F2 U2 r2 U2 F2 Rw2) B U B'

I execute corners first and midges last:
That way I can swap UBL and UBR at the start, solve all the midges, then apply F2 U (T-perm) U' (Rw2 F2 U2 r2 U2 F2 Rw2) F2.

I'm sure there are better ways to fix parity, but there's a couple of options.
 
3x3 BLD using M2 + OP Corners

I just had two DNF's, both of which I felt really confident about. I ended with the exact same cube state (3 edge cycle off)for each solve, so I am wondering if I am making the same mistake often. To solve the cube, I would need UB -> DB -> DF. My M2 buffer is DF and my corner buffer is ULB. For parity, I do a "righty" j perm after corners, then (M' U2 M' U2) followed by (y T perm) after completing edges. Both of these solves had parity, but neither had any flipped edges. Does anyone have any idea what may lead to this cube state? I'm not certain, but I believe I have had this before as well. I have had plenty of successes with and without parity, so I don't believe it is anything with the method I am using, I would assume it is probably just a common stupid mistake.
 
Carson said:
3x3 BLD using M2 + OP Corners

I just had two DNF's, both of which I felt really confident about. I ended with the exact same cube state (3 edge cycle off)for each solve, so I am wondering if I am making the same mistake often. To solve the cube, I would need UB -> DB -> DF. My M2 buffer is DF and my corner buffer is ULB. For parity, I do a "righty" j perm after corners, then (M' U2 M' U2) followed by (y T perm) after completing edges. Both of these solves had parity, but neither had any flipped edges. Does anyone have any idea what may lead to this cube state? I'm not certain, but I believe I have had this before as well. I have had plenty of successes with and without parity, so I don't believe it is anything with the method I am using, I would assume it is probably just a common stupid mistake.
Click to expand...

Instead of M'U2M'U2 (y T perm), do (M2) (y L2 T perm L2).

When you're off by an M2, M'U2M'U2 (basically M U2 M' U2) will cycle UB->DB->UF. If DF-UB were already swapped (this the case for parity), then you're left with DF->UB->UF->DB and UBR-ULB, which after (y T perm) will leave DF->UB->DB.
 
aronpm said:
Instead of M'U2M'U2 (y T perm), do (M2) (y L2 T perm L2).

When you're off by an M2, M'U2M'U2 (basically M U2 M' U2) will cycle UB->DB->UF. If DF-UB were already swapped (this the case for parity), then you're left with DF->UB->UF->DB and UBR-ULB, which after (y T perm) will leave DF->UB->DB.
Click to expand...

Wow... thank you so much! Perhaps this can explain my "less than perfect" success rate. :fp
 
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