Advent of Code 2024 - the home stretch - it's been an aMAZEing year
https://awful.systems/post/3123248
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It’s a wrap!
One of the easier years imho. Better than last year in any case.
I get the feeling that this is Eric’s way of saying goodbye, and that this might be the last year, but I might be wrong.
Puzzles by difficulty (leaderboard completion times)
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23!
SpoilerificGot lucky on the max clique in part 2, my solution only works if there are at least 2 nodes in the clique, that only have the clique members as common neighbours.
Ended up reading wikipedia to lift one the Bron-Kerbosch methods:
#!/usr/bin/env jq -n -rR -f reduce ( inputs / “-” # Build connections dictionary # ) as [$a,$b] ({}; .[$a] += [$b] | .[$b] += [$a]) | . as $conn | # Allow Loose max clique check # if $ARGS.named.loose == true then # Only works if there is at least one pair in the max clique # # That only have the clique members in common. # [ # For pairs of connected nodes # ( $conn | keys[] ) as $a | $conn[$a][] as $b | select($a < $b) | # Get the list of nodes in common # [$a,$b] + ($conn[$a] - ($conn[$a]-$conn[$b])) | unique ] # From largest size find the first where all the nodes in common # # are interconnected -> all(connections ⋂ shared == shared) # | sort_by(-length) | first ( .[] | select( . as $cb | [ $cb[] as $c | ( [$c] + $conn[$c] | sort ) | ( . - ( . - $cb) ) | length ] | unique | length == 1 ) ) else # Do strict max clique check # # Example of loose failure: # 0-1 0-2 0-3 0-4 0-5 1-2 1-3 1-4 1-5 # 2-3 2-4 2-5 3-4 3-5 4-5 a-0 a-1 a-2 # a-3 b-2 b-3 b-4 b-5 c-0 c-1 c-4 c-5 def bron_kerbosch1($R; $P; $X; $cliques): if ($P|length) == 0 and ($X|length) == 0 then if ($R|length) > 2 then {cliques: ($cliques + [$R|sort])} end else reduce $P[] as $v ({$R,$P,$X,$cliques}; .cliques = bron_kerbosch1( .R - [$v] + [$v] ; # R ∪ {v} .P - (.P - $conn[$v]); # P ∩ neighbours(v) .X - (.X - $conn[$v]); # X ∩ neighbours(v) .cliques ) .cliques | .P = (.P - [$v]) | # P ∖ {v} .X = (.X - [$v] + [$v]) # X ∪ {v} ) end ; bron_kerbosch1([];$conn|keys;[];[]).cliques | max_by(length) end | join(“,”) # Output password
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day 23
this is one of those days when it’s all about the right term to google right
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23-2Leaving something to run for 20-30 minutes and actually getting a valid and correct result: new positive feeling unlocked.
Now to find out how I was actually supposed to solve it.
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thanksI’ve probably learned that term at some point, so thanks for naming it. That made me realise my algorithm was too thicc and could just be greedy.
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21!
Finally managed to beat this one into submission.
P1I created this disgusting mess of a recursive search that happened to work. This problem was really hard to think about due to the levels of indirection. It was also hard because of a bug I introduced into my code that would have been easy to debug with more print statements, but hubris.
P2Recursive solution from P1 was too slow, once I was at 7 robots it was taking minutes to run the code. It didn’t take long to realise that you don’t really care about where the robots other than the keypad robot and the one controlling the keypad robot are since the boundary of each state needs all the previous robots to be on the A button. So with memoisation, you can calculate all the shortest paths for a given robot to each of the directional inputs in constant time, so O(kn) all up where n is the number of robots (25) and k is the complexity of searching for a path over 5 or 11 nodes.
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22
uhpretty straightforward. At least it’s not a grid!
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24! - Crossed Wires - Leaderboard time 01h01m13s (and a close personnal time of 01h09m51s)
SpoilersI liked this one! It was faster the solve part 2 semi-manually before doing it “programmaticly”, which feels fun.
Way too many lines follow (but gives the option to finding swaps “manually”):
#!/usr/bin/env jq -n -crR -f ( # If solving manually input need --arg swaps # Expected format --arg swaps ‘n01-n02,n03-n04’ # Trigger start with --arg swaps ‘0-0’ if $ARGS.named.swaps then $ARGS.named.swaps | split(“,”) | map(split(“-”) | {(.[0]):.[1]}, {(.[1]):.[0]}) | add else {} end ) as $swaps | [ inputs | select(test(“->”)) / " " | del(.[3]) ] as $gates | [ # Defining Target Adder Circuit # def pad: “0(.)”[-2:]; ( [ “x00”, “AND”, “y00”, “c00” ], [ “x00”, “XOR”, “y00”, “z00” ], ( (range(1;45)|pad) as $i | [ “x($i)”, “AND”, “y($i)”, “c($i)” ], [ “x($i)”, “XOR”, “y($i)”, “a($i)” ] ) ), ( [“a01”, “AND”, “c00”, “e01”], [“a01”, “XOR”, “c00”, “z01”], ( (range(2;45) | [. , . -1 | pad]) as [$i,$j] | [“a($i)”, “AND”, “s($j)”, “e($i)”], [“a($i)”, “XOR”, “s($j)”, “z($i)”] ) ), ( ( (range(1;44)|pad) as $i | [“c($i)”, “OR”, “e($i)”, “s($i)”] ), [“c44”, “OR”, “e44”, “z45”] ) ] as $target_circuit | ( # Re-order xi XOR yi wires so that xi comes first # $gates | map(if .[0][0:1] == “y” then [.[2],.[1],.[0],.[3]] end) ) as $gates | # Find swaps, mode=0 is automatic, mode>0 is manual # def find_swaps($gates; $swaps; $mode): $gates as $old | # Swap output wires # ( $gates | map(.[3] |= ($swaps[.] // .)) ) as $gates | # First level: ‘x0i AND y0i -> c0i’ and ‘x0i XOR y0i -> a0i’ # # Get candidate wire dict F, with reverse dict R # ( [ $gates[] | select(.[0][0:1] == “x” ) | select(.[0:2] != [“x00”, “XOR”] ) | if .[1] == “AND” then { “(.[3])”: “c(.[0][1:])” } elif .[1] == “XOR” then { “(.[3])”: “a(.[0][1:])” } else “Unexpected firt level op” | halt_error end ] | add ) as $F | ($F | with_entries({key:.value,value:.key})) as $R | # Replace input and output wires with candidates # ( [ $gates[] | map($F[.] // .) | if .[2] | test(“c\d”) then [ .[2],.[1],.[0],.[3] ] end | if .[2] | test(“a\d”) then [ .[2],.[1],.[0],.[3] ] end ] # Makes sure that when possible a0i comes 1st, then c0i # ) as $gates | # Second level: use info rich ‘c0i OR e0i -> s0i’ gates # # Get candidate wire dict S, with reverse dict T # ( [ $gates[] | select((.[0] | test(“c\d”)) and .[1] == “OR” ) | {“(.[2])”: “e(.[0][1:])”}, {“(.[3])”: “s(.[0][1:])”} ] | add | with_entries(select(.key[0:1] != “z”)) ) as $S | ($S | with_entries({key:.value,value:.key})) as $T | ( # Replace input and output wires with candidates # [ $gates[] | map($S[.] // .) ] | sort_by(.[0][0:1]!=“x”,.) ) as $gates | # Ensure “canonical” order # [ # Diff - our input gates only $gates - $target_circuit | .[] | [ . , map($R[.] // $T[.] // .) ] ] as $g | [ # Diff + target circuit only $target_circuit - $gates | .[] | [ . , map($R[.] // $T[.] // .) ] ] as $c | if $mode > 0 then # Manual mode print current difference # debug(“gates”, $g[], “target_circuit”, $c[]) | if $gates == $target_circuit then $swaps | keys | join(“,”) # Output successful swaps # else “Difference remaining with target circuit!” | halt_error end else # Automatic mode, recursion end # if $gates == $target_circuit then $swaps | keys | join(“,”) # Output successful swaps # else [ first( # First case when only output wire is different first( [$g,$c|map(last)] | combinations | select(first[0:3] == last[0:3]) | map(last) | select(all(.[]; test(“e\d”)|not)) | select(.[0] != .[1]) | { (.[0]): .[1], (.[1]): .[0] } ), # “Only” case where candidate a0i and c0i are in an # incorrect input location. # Might be more than one for other inputs. first( [ $g[] | select( ((.[0][0] | test(“a\d”)) and .[0][1] == “OR”) or ((.[0][0] | test(“c\d”)) and .[0][1] == “XOR”) ) | map(first) ] | if length != 2 then “More a0i-c0i swaps required” | halt_error end | map(last) | select(.[0] != .[1]) | { (.[0]): .[1], (.[1]): .[0] } ) ) ] as [$pair] | if $pair | not then “Unexpected pair match failure!” | halt_error else find_swaps($old; $pair+$swaps; 0) end end end ; find_swaps($gates;$swaps;$swaps|length)
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I did part 2 manually!
well well wellAoC, so you thought you could dredge up my trauma as an EE grad by making me debug a full-adder logic circuit? How dare you. You succeeded.
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I did end up writing a code solution.
algorithm descSo basically the problem boils down to this:
A 1 bit full adder circuit with carry in/out is described as follows:
S~i~ = X~i~ ^ Y~i~ ^ Cin~i~
Cout~i~ = (X~i~ && Y~i~) || (Cin~i~ && (X~i~ ^ Y~i~))
Where S is the output bit, X and Y are the input bits, and Cin~i~ is the carry out bit of bit i-1. For the first and last bits of the output, the circuits are slightly different due to carry in/out not mattering in the 0/last bit case.
Note that as said in the problem statement any input is correctly labelled, while outputs might be incorrect. You can then categorise each gate input/output as any of the elements in an adder circuit. You can then use set differences and intersections to determine the errors between categories. That’s all you need to do!
For example, you might have something like:
X && Y = err
if this output was used correctly, it should show up as an operand to an OR gate. So if you did:
(Set of all outputs of and gates) - (Set of all inputs to or gates), if something appears, then you know one of the and gates is wrong.
Just exhaustively find all the relevant differences and intersections and you’re done! To correct the circuit, you can then just brute force the 105 combinations of pair swaps to find what ends up correcting the circuit.
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25!
p1 tipsO(mn)/O(n^2^) is fast enough.
50 stars baby!imgur.com/a/hwEVy9H
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And done! well I had fun.
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congrats! I still have 6 stars to go, but I still think this was easier than last year.
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text/gemini