Is there a way to tell if you have 3bld parity without doing memo

I’m not a BLD solver but figured that if anyone knew it would be you guys.

HTR has the same parity and the way it’s traced is basically to count how long the bld memo would be. I’m hoping there’s a faster way so I’m just wondering if maybe there’s a trick that does it quicker or you’ve noticed a pattern that reveals when you have parity.

No there isn't

There can't really be, and the nature of parity gives us a good idea of why.

If you look at all but two pieces of an orbit (e.g. just edges, or just corners), the last two could be swapped either way depending on parity. The only way to know what the remaining parity is would be to know additional information from other orbits, which involves looking at even more than 2 pieces.

Therefore, you need to look at a minimum of either 7 corners (which means you know the position of the 8th as well), or 11 edges. If you want to perform the minimum amount of work, presumably tracing CP is the easiest.

abunickabhi said:

No there isn't

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Not sure whether you ask from a BLD or sighted perspective, Lzz (you mentioned HTR in your post).

It depends on what you mean by a scramble. Scramble can be visually represented by a cube state, but it can also be textually represented by a scrambling move sequence.

By analyzing scrambling move sequence (or solving move sequence since solving move sequence can be thought of as an inverse of the scrambling move sequence), you can figure out the parity without memorizing/tracing pieces/cubies.

Using this trick in BLD, you don't need to use the memo information to determine whether you have parity or not (you need to trace pieces nonetheless).

Fun fact: if you know the scramble, there is an easy way to know if there is a parity.
Every quarter turn changes the parity. Consequently, double turns does not affect parity case; so you need to count only the pure quarter turns. For example, if your scramble is the following:
D R2 F' D' R B' L2 D2 L' U F2 D' B2 R2 U2 L2 D' F2 R2 U' F2
Count the single quarter turns (D F' D' R B' L' U D' D' U') -> there are 10 of them in this example. So no parity.

In this example:
U2 F R D2 B2 D2 B2 F2 R' F2 L' B2 R2 F' D2 R' D U B' D2 L2
There are 9 single quarter turns, so there is parity.

Of course during a regular BLD solve this would be cheating, because you shall have no information about the scrambling sequence itself, but for practice or other purposes it can be useful.