Suppose the ice on a server is arranged, outermost to innermost: Sensei, Wall of Static, Cell Portal, all rezzed.
If the runner does not break the subroutine on Sensei, breaks the subroutines on Wall of Static, is returned to the outermost ice by Cell Portal, and does not break the subroutine on Sensei the second time they encounter it, how many subroutines does Wall of Static have the second time they encounter it?
Three: the native subroutine, plus one from each encounter with Sensei?
Or four: the native subroutine, plus one gained during their first encounter with Wall of Static, plus two more gained during their second encounter with Wall of Static? Thanks for the question. The answer is three: one + one for each time Sensei has been encountered.
I checked some previous ruling and wanted to follow up on this. Sensei's original subroutine is still active, and so the Wall of Static would actually have four subroutines instead of three. Sorry for the confusion.
As per the current timing structure of a run, when passing a piece of ICE, if any ICE remains on the server you approach the outtermost ice on the server NOT ALREADY APPROACHED (step 2.1).
Strictly reading this, should the runner pass an Architect or Minelayer which installs a new piece of ICE on the same server, then the runner should not approach the new ICE, and if they pass this also, they'd hop over any already approached ICE (including any bypassed) as they proceed inwards.
This seems kind of unintuitive, but rules-as-written I'm finding it hard to see how it wouldn't be correct without a change to the timing structure, or an addition to the FAQ. Do you mean to say that the Runner would approach the new piece of ice? That isn’t how it works, but I agree that a strict reading of the rule could support this position. I’ll address this in the next FAQ.