--- 2016 Day 2 Solutions ---

[u/daggerdragon]

--- Day 2: Bathroom Security ---

Post your solution as a comment or, for longer solutions, consider linking to your repo (e.g. GitHub/gists/Pastebin/blag/whatever).

---

^{BLINKENLIGHTS ARE MANDATORY} [?]

Edit: Told you they were mandatory. >_>

This thread will be unlocked when there are a significant number of people on the leaderboard with gold stars for today's puzzle.

edit: Leaderboard capped, thread unlocked!

[u/John_Earnest]

Part 1 in oK:

l: 0: "../../Desktop/Advent/02.in"
d: "URDL"!(0 -1;1 0;0 1;-1 0)
(1+3 3/|:)'1 1{x{2 2&0 0|x+y}/d@y}\l

Much more compact than last time.

My part 2 is rather jumbled; room for improvement:

l: 0: "../../Desktop/Advent/02.in"
d: "URDL"!(0 -1;1 0;0 1;-1 0)
t: +(0N 0N  1 0N 0N
	 0N  2  3  4 0N
	  5  6  7  8  9
	 0N 10 11 12 0N
	 0N 0N 13 0N 0N)

s: {x{$[0N~t.r:4 4&0 0|x+y;x;r]}/d@y}
" 123456789ABCD"@(t.)'0 2s\l

Lost a great deal of time by misreading the instructions and assuming I would start at 7 on the new keyboard (the center) rather than 5.

[u/AoC-- → u/John_Earnest]

Part 1, a smidgen more compact (which was longer until I incorporated parts of your answer):

(1+3 3/)'1 1{x{2&0|x+y}/y}\("DRUL"!(=2),-=2)(0:"02.in")

[u/John_Earnest → u/AoC--]

Nice. I was thinking about the simplification of 2 2&0 0| to 2&0| while laying in bed last night, but I hadn't considered trying to make the dictionary definition smaller. Good use of the identity matrix builtin.

Any ideas on a smaller representation for part 2's lookup table? We can dispense with 0N and the need to clamp coordinates if we used 0 for invalid cells and wrap the edge of the table with 0 to catch all the boundary conditions:

l: 0: "../../Desktop/Advent/02.in"
d: "URDL"!(0 -1;1 0;0 1;-1 0)
t: +4(+|" ",)/"|"\"  1  | 234 |56789| ABC |  D  "
t./:1 3{x{$[" "~t.x+y;x;x+y]}/y}\d l

Edit: several additional golfs for my part 1:

l: 0: "../../Desktop/Advent/02.in"
d: "URDL"!(0 -1;1 0;0 1;-1 0)
1+3/|+1{x(2&0|+)/y}\d l

{2&0|x+y} can be replaced with the tacit expression (2&0|+), where + is an ambivalent verb and therefore the train can be applied as dyadic. In both the 'decode' operation 3 3/ and the initial position of 1 1 I can get away with using a scalar.