Solving the same scramble

Disagree.

Oops, you're right, it's not guaranteed that the chosen set of 18-movers contains algs for all 17-movers. But since there are 2.4 times as many 18-movers as 17-movers and you can take both the first 17 moves or the last 17 moves, I suspect it's possible to at least cover almost all 17-movers. No idea how to determine this, though.

Edit: Oh and I forgot about the Devil's algorithm, so you just have to practice *one* algorithm. Surely you have enough time for that.
 
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Oops, you're right, it's not guaranteed that the chosen set of 18-movers contains algs for all 17-movers. But since there are 2.4 times as many 18-movers as 17-movers and you can take both the first 17 moves or the last 17 moves, I suspect it's possible to at least cover almost all 17-movers. No idea how to determine this, though.

Edit: Oh and I forgot about the Devil's algorithm, so you just have to practice *one* algorithm. Surely you have enough time for that.


Are there any positions which are 17 moves far from solved state, but you cannot get to an 18-move position with only 1 move?
 
Are there any positions which are 17 moves far from solved state, but you cannot get to an 18-move position with only 1 move?

Gah, another thing I missed. Don't know about 17, but there's a 6-mover that can't be turned into a 7-mover with only one turn (R2 L2 U2 D2 F2 B2).

Edit: Also does happen at least at distances 9, 10 and 11:
http://cubezzz.dyndns.org/drupal/?q=node/view/69
 
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