Odd edges and even corners in blind solve

(Oliver Christensen) #1

So I’ve been trying to learn how to solve the cube blindfolded, but when I wrote down the letters, I noticed that there were an odd number of edges and an even number of corners. Surely this is a mistake, I thought and tried solving it, yet it came out solved!
Does anyone know why this is?

You can try backtracking the sequence:

Edges:
udafxq nkn ow rpcr
Corners:
wmso dfgd

#2

That shouldn’t be possible, because you always get either both odd or both even… did you do it blindfolded or just one after another? There should have been a parity fix in there after edges since you had an odd number. Did you do an R-perm between edges and corners (assuming you use OP as a beginner’s method.)

What do you use? OP, M2, Turbo, 3-cycle? Got the original scramble rather than the letters you’ve memorized?

(Oliver Christensen) #3

I did it with my eyes closed and I use OP. And no, I didn’t do any R-perm since I had no idea if I needed to use it in that case. I don’t have the scramble.

#4

Well, in 3BLD, an odd number of edge targets always means you have an odd number of corner targets… that’s just the math that goes along with it.

As far as the R-perm goes, it is needed after an odd number of edges because every time you do a T-perm (or J-perm for that matter) you swap URB and UFR. Do it once, it’s off… do it again, good, once more, it’s off again, etc. So any odd number of edges requires an R-perm before going into corners, so that the pieces in URB and UFR are in fact the pieces you memorized.

Sorry, but without the scramble it’s just guesswork what happened there… it’s a good idea to write down your scrambles during practice for cases like this.

#5

It looks like you have the n target twice. Is this a flipped edge perhaps?
I found this information on the speed solving community helpful.

Corner Targets = 7 + (number of cycle breaks) - (permuted pieces)
Edge Targets = 11 + (number of cycle breaks) - (permuted pieces)

However, if you twist corners/flip edges by targeting the same piece twice, then it is:

Corner Targets = 7 + (number of cycle breaks) - (solved pieces) + 2*(twisted corners)
Edge Targets = 11 + (number of cycle breaks) - (solved pieces) + 2*(flipped edges)