Random Cubing Discussion

Is there any point in using the term "forced"? Sometimes you see people saying things like I did R' F R F' instead of U R U' R' for my last pair to skip OLL. In those cases, it's more likely that the person did that to orient all edges and got a OCLL skip.

What do people really mean when they say that they "forced" a skip?
 
Is there any point in using the term "forced"? Sometimes you see people saying things like I did R' F R F' instead of U R U' R' for my last pair to skip OLL. In those cases, it's more likely that the person did that to orient all edges and got a OCLL skip.

What do people really mean when they say that they "forced" a skip?
If they knew MW and did a sledgehammer to skip OLL, then I'd say that was forced. If they did it to orient edges, whether they knew it would give an OCLL skip or not, then that isn't a forced OLL skip, it's a forced OELL skip; the OCLL skip was luck.
 
I think that makes sense. Another random example: Let's say you've learnt full PLL. And let's say you are doing a solve and you get a T perm for PLL. I guess you could say that you forced an EPLL skip or CPLL skip by doing T perm, although you probably wouldn't ever hear this, but I guess it's true :p
 
Here's my attempt:

Where X is any step, or even any other part of a step i.e. could be OLL, or just OELL.
For X to be forced, you must
1. Know at least 2 different algs for completing the particular step you are doing
2. Use an alg that is not the outright fastest for completing that step, but with the aim of getting a better result overall
3. Know that X will be done after doing the alg
 
Here's my attempt:

Where X is any step, or even any other part of a step i.e. could be OLL, or just OELL.
For X to be forced, you must
1. Know at least 2 different algs for completing the particular step you are doing
2. Use an alg that is not the outright fastest for completing that step, but with the aim of getting a better result overall
3. Know that X will be done after doing the alg

I like these. So like if I did a mirror of a COLL, knowing it would cause an EPLL skip that would be forced.

My general definition: An action performed as an alternative to another, usually more common/fast/etc. action in order to create a better/easier solution.
 
I think that makes sense. Another random example: Let's say you've learnt full PLL. And let's say you are doing a solve and you get a T perm for PLL. I guess you could say that you forced an EPLL skip or CPLL skip by doing T perm, although you probably wouldn't ever hear this, but I guess it's true :p
Only if using 4lLL; otherwise you're not doing EPLL or CPLL, so you don't have either of those steps to skip :p
2. Use an alg that is not the outright fastest for completing that step, but with the aim of getting a better result overall
What if you know all the algs of that set and it just happens to be the easiest of them? Then would it still be a forced skip (e.g. if you used ZZ and knew WV and had R U' R' as WV, would that be a 'forced OCLL skip')?
 
Only if using 4lLL; otherwise you're not doing EPLL or CPLL, so you don't have either of those steps to skip :p

What if you know all the algs of that set and it just happens to be the easiest of them? Then would it still be a forced skip (e.g. if you used ZZ and knew WV and had R U' R' as WV, would that be a 'forced OCLL skip')?

If you use ZZ and full WV, than OCLL isn't even a step for you anymore; your method is now [EOLine -> F2L-1 -> WV -> PLL]. So your first sentence applies: there's no OCLL step to skip. EDIT: wait there's still R U R' insert never mind

Now if you use CFOP with WV, OLL is still actually a step for you that you can't 'skip' every single time, and I think your point applies a little better. I don't have an answer to that question, though. Even knowing WV, I'd feel a little silly claiming a forced O(C)LL skip with a R U' R' insert.
 
If you use ZZ and full WV, than OCLL isn't even a step for you anymore; your method is now [EOLine -> F2L-1 -> WV -> PLL]. So your first sentence applies: there's no OCLL step to skip. EDIT: wait there's still R U R' insert never mind

Now if you use CFOP with WV, OLL is still actually a step for you that you can't 'skip' every single time, and I think your point applies a little better. I don't have an answer to that question, though. Even knowing WV, I'd feel a little silly claiming a forced O(C)LL skip with a R U' R' insert.

I would call it "predicted"
 
If you use ZZ and full WV, than OCLL isn't even a step for you anymore; your method is now [EOLine -> F2L-1 -> WV -> PLL]. So your first sentence applies: there's no OCLL step to skip. EDIT: wait there's still R U R' insert never mind

Now if you use CFOP with WV, OLL is still actually a step for you that you can't 'skip' every single time, and I think your point applies a little better. I don't have an answer to that question, though. Even knowing WV, I'd feel a little silly claiming a forced O(C)LL skip with a R U' R' insert.
Yeah, I did contradict myself, you're right. It's hard to say whether that could be called a forced skip or not... or if it even a skip if that's your method.
I don't know what to call a forced skip any more...
 
Here's my attempt:

Where X is any step, or even any other part of a step i.e. could be OLL, or just OELL.
For X to be forced, you must
1. Know at least 2 different algs for completing the particular step you are doing
2. Use an alg that is not the outright fastest for completing that step, but with the aim of getting a better result overall
3. Know that X will be done after doing the alg

for me, 1) and 3) suffice (because as TDM said, sometimes you just get the short case anyway)

i would go with something like 'deliberately combine 2 steps or sub-steps of a method (eg f2l-4 and OLL in CFOP, CMLL and 4a in Roux), using a known algorithm'
 
I would say a forced skip is when for two steps X and Y, you choose a solution to X that also solves Y, but is different to what you would normally do to solve X alone.
 
I would say a forced skip is when for two steps X and Y, you choose a solution to X that also solves Y, but is different to what you would normally do to solve X alone.
Let X be the last F2L pair, and let Y be the last layer. Instead of doing U R U' R', I do R' F R F' and skip the last layer. With this definition, I have forced a last layer skip. Is that sensible?
 
How can 1 and 3 suffice? So every time I get an OLL skip and I see it coming it's forced?

I think it depends when you saw it coming :P
Say U' R U' R' solves a last pair and OLL. If you see that this is the case before you do the alg, then i would consider that forced. If you work out halfway through the step that you're going to get a skip, then that's not forced because you already decided to put the pair in for some other reason (EOLL skip, whatever). i don't see how omitting 2) implies this :s
 
Let X be the last F2L pair, and let Y be the last layer. Instead of doing U R U' R', I do R' F R F' and skip the last layer. With this definition, I have forced a last layer skip. Is that sensible?
Last layer isn't generally an individual step, but if you had learned 1LLL and knew you were forcing a skip, yes. Let Y = OLL or OELL and it makes more sense.

Now let Y = OLL and let Z = PLL. I do R' F R F' expecting to skip Y and fortuitously skip Y+Z, the whole last layer. I could say that I forced OLL skip, but skipping PLL too was lucky.
 
I think it should be possible to force a skip in every solve, which means sometimes the fastest alg should be the one you end up using. So I would say you forced a skip if you (a) know multiple algs for a case, and you choose one of them (c) specifically in order to skip part or all of a future step. So if you always or almost always use one alg, and happen to see that it skips the next step, you didn't force it because condition (b) isn't satisfied - you did that alg because you always do, not because it causes the skip.
 
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