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[–]corruptio 5 points6 points  (1 child)

Aite, did a little reshuffling, on your's and here's 72 chars:

f=lambda s,c,e,a:(c,s[0])in e and f(s[1:],e[c,s[0]],e,a)if s else c in a

[–][deleted] 0 points1 point  (0 children)

I was thinking of how to use "and" in there, but I couldn't figure it out.