In case I made any typos:

In the case of (a b)(c d) where a, b, and c denote arbitrary singular
pieces within a cycle and d denotes any number of edges within that
cycle (could be as low as 0 or as high as 9):

(a b)(c d) can be thought of as (a b c d c).

Hope this helps.

-Marcus Stuhr

--- In blindfoldsolving-rubiks-cube@yahoogroups.com, "kyuubree"
<agent_m80@...> wrote:
>
> Usually when I have a bunch of random edges to flip, I usually flip
> all 12 then flip the two I messed up.
>
> As for the cycles, when I have a bunch of doubles, I cycle them as
> normal, knowing that the FIRST edge of the next cycle will simply go
> to the back.
>
> So in your case you have (UF DF DR DL) (UL BR UB BL) (UR FL FR DB).
> Cycle (UF DF DR) as normal and you now get (UF DL) (UL BR UB BL) (UR
> FL FR DB). Instead of "keeping UF DL in mind for later," go ahead and
> cycle (UF DL UL), knowing it'll drop DL and kick UL (the end of the
> first cycle) to the end of its own cycle.
>
> (UF DL) (UL BR UB BL) (UR FL FR DB) becomes (UF BR UB BL UL) (UR FL FR
> DB). The cool thing about this approach is that it doesn't disturb
> your memory much -- all it does is finish off the second element of
> cycle 1 and kick the start of cycle 2 to the end. All you need to
> remember is what the start of the second cycle was specifically -- and
> be able to kick it to the end. So in this case you'd do (UF BR UB)
> then (UF BL UL) to finish off that mutated second cycle. Then you got
> (UR FL FR DB) to reduce after doing (UR FL FR), in which case you then
> have (UR DB) and thus a parity error.
>
> I still haven't devised an easy way to deal with parities. I am
> thinking maybe Pochmann approach would be best here? Simple two cycle
> of corners and edges using T Perms... very mindless and easy to do.
>
> --- In blindfoldsolving-rubiks-cube@yahoogroups.com, smoothcuber
> <no_reply@> wrote:
> >
> > Hey,
> >
> > I'm not good at cycle style. Here's a sample EP: Just do F2 D' B R
> > B L U' L' U B to setup.
> >
> > Now I go to memorize. I start at UF and see that the cycle goes UF-
> > >DF->DR->DL, then ends, and I have to recall that I have two to swap
> > now.
> >
> > Next my eyes go back up to my next spot, UL. I see that the cycle
> > goes UL->BR->UB->BL, then ends, and again I have to recall another
> > two to swap.
> >
> > Next my eyes go up again to my next spot, UB, but I had UB in my last
> > cycle, so I skip it, and go to my next spot, UR. I see that the
> > cycle goes UR->FL->FR->DB, ends, and once again I have two to swap.
> >
> > So I have now UF needs to swap with DL, UL with BL, and UR with DB.
> >
> > I have not yet come up with a good way to memorize this situation,
> > nor do I have a good way to solve it.
> >
> > Assuming that for a 1:20 solve I'd need to memorize this in at least
> > 30 sec, I haven't been able to do it.
> >
> > Unrelated note: What do you guys do when you have 10 edges to flip?
> >
> > I hope no typos and no mistakes.
> >
> > -Dave O.
> >
>