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When the Runner encounters Komainu, it gains "SubroutineDo 1 net damage." for each card in the Runner's grip for the remainder of this run.



Related FAQsEdit

Related RulingsEdit

  • The "SubroutineEnd the run." added by Sensei is located after the subroutines Komainu gains by itself.[1]:Ruling
  • Komainu loses its subroutines when derezzed during the run.[2]:Ruling


SourcesEdit

  1. Sensei's Added Subroutine Location and Timing Ruling
    Suppose the runner encounters Sensei and does not break the subroutine, and then encounters Ice Wall, which has been chosen as the target of Femme Fatale. Does it cost 1 credit to bypass Ice Wall with Femme Fatale (because these are simultaneous conditional abilities, and the runner may trigger Femme Fatale before Ice Wall gains a subroutine), or 2 credits (because 'for the remainder of this run, each piece of ice encountered except Sensei gains "SubroutineEnd the run" after all its other subroutines' is a constant ability)? If the previous answer is 2 credits, then suppose instead the runner encounters Sensei and does not break the subroutine, and then encounters Komainu. Does the end-the-run subroutine come first, because it is already present, as the first subroutine, when Komainu gains its net damage subroutines, or last, because Sensei says the end-the-run is after all other subroutines? Or something else entirely?
    The Sensei is creating a constant ability that gives each piece of ice an additional subroutine. So when you initially encounter the Ice Wall, it costs 2 credits to use Femme Fatale in order to bypass it. With the Komainu, the subroutine will be by itself, unless the Runner has cards in hand. If the Runner has cards in hand, then the subroutine drops to the last position as soon as the Komainu gains its subroutines.
  2. Komainu Loses Subroutines When Derezzed Ruling
    Does Komainu keep the subroutines for that run even after it got derezzed? Or the derez will reboot it as a new ice?
    Komainu will lose its subroutines. Doesn't keep them when derezzed.

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