last day (19 days later) » 

01:18
0
A: Source permutation

Joseph SibleHaskell, score 1016 (\((.),(-))->(.)*200+mod(-39+(-))200+1).(`divMod`200).(-1+) Try it online!

@xnor Good catch. That's supposed to be outside of the mod. At some point while golfing, I must have messed up order of operations. Fixed now, thanks!
@xnor Fixed. (Sigh, there goes my score.)
@ბიმო That helped a lot, thanks!
@ASCII-only I'm not convinced that adding a `` and causing a warning makes it look better :P
@ბიმო I originally either subtracted 39 or added 161 to the remainder. Now, instead, I always subtract 39 and add 1 to the quotient in the cases when I would have added 161 before.
@ASCII-only Sure, edited.
@ბიმო Ah, I thought you were asking where the constant in my code went. Okay, fixing...
@ბიმო Fixed. (Interestingly, fixing that improved my score.)
@ASCII-only Yep, nice savings!
btw, you don't need bytecount since that doesn't matter
@ASCII-only Since . isn't in the lambdas, it can be used instead of / to save another 1 per occurrence
+ your improvements, 993
yeah
this should be pretty close to optimal for this approach, the only things with high score are mod, divMod and \
01:24
O_o
i think one of the few things that can be better would be a quine-like approach, like Jo King is doing
oh, of course, since * isn't used in that lambda
@ASCII-only :/ i'm too bad at haskell to do this though, back to things i understand i guess
 
5 hours later…
06:58
@JosephSible Sadly the xor method used in some other answers is not viable here since importing Data.Bits is too expensive. no idea if there's another easy-to-implement operation like xor that would give a permutation

  last day (19 days later) »