Day 17: Chronospatial Computer

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FAQ

  • @SteveDinn@lemmy.ca
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    317 hours ago

    C#

    This one is mostly thanks to reading @mykl@mykl@lemmy.world’s code to understand WTF was going on for part 2, and then once I understood the basics, finally got to solving it myself. Instructions were read in as long because I didn’t want to deal with int vs. long all the time.

    using System.Collections.Immutable;
    using System.Diagnostics;
    using Common;
    
    namespace Day17;
    
    static class Program
    {
        public record State(long A, long B, long C, int InstPtr, ImmutableList<long> Output);
    
        static void Main()
        {
            var start = Stopwatch.GetTimestamp();
    
            var (sampleReg, sampleInst) = ReceiveInput("sample.txt");
            var (inputReg, inputInst) = ReceiveInput("input.txt");
    
            Console.WriteLine($"Part 1 sample: {Part1(sampleReg, sampleInst)}");
            Console.WriteLine($"Part 1 input: {Part1(inputReg, inputInst)}");
    
            (sampleReg, sampleInst) = ReceiveInput("sample2.txt");
            Console.WriteLine($"Part 2 sample: {Part2(sampleReg, sampleInst)}");
            Console.WriteLine($"Part 2 input: {Part2(inputReg, inputInst)}");
    
            Console.WriteLine($"That took about {Stopwatch.GetElapsedTime(start)}");
        }
    
        static object Part1(State state, ImmutableArray<long> instructions) =>
            Execute(instructions, state).Output.StringifyAndJoin(",");
    
        static object Part2(State state, ImmutableArray<long> instructions) =>
            RecursiveSolve(instructions, state with { A = 0 }, []).First();
    
        static IEnumerable<long> RecursiveSolve(ImmutableArray<long> instructions, State state, ImmutableList<long> soFar) =>
            (soFar.Count == instructions.Length) ? [state.A] :
            Enumerable.Range(0, 8)
                .Select(a => state with { A = (state.A << 3) + a })
                .Where(newState => newState.A != state.A)
                .Select(newState => new { newState, Execute(instructions, newState).Output, })
                .Where(states => states.Output.SequenceEqual(instructions.TakeLast(states.Output.Count)))
                .SelectMany(states => RecursiveSolve(instructions, states.newState, states.Output));
    
        static State Execute(ImmutableArray<long> instructions, State state)
        {
            while (state.InstPtr < instructions.Length)
            {
                var opcode = instructions[state.InstPtr];
                var operand = instructions[state.InstPtr + 1];
                state = Operations[opcode](state, operand);
            }
    
            return state;
        }
    
        static long ComboOperand(long operand, State state) => operand switch
        {
            >= 0 and <= 3 => operand,
            4 => state.A,
            5 => state.B,
            6 => state.C,
            _ => throw new Exception("Invalid operand."),
        };
    
        static long Adv(long op, State state) => state.A / (long)Math.Pow(2, ComboOperand(op, state));
    
        static readonly Func<State, long, State>[] Operations =
        [
            (s, op) => s with { InstPtr = s.InstPtr + 2, A = Adv(op, s) },
            (s, op) => s with { InstPtr = s.InstPtr + 2, B = s.B ^ op },
            (s, op) => s with { InstPtr = s.InstPtr + 2, B = ComboOperand(op, s) % 8 },
            (s, op) => s with { InstPtr = (s.A == 0) ? (s.InstPtr + 2) : (op <= int.MaxValue) ? (int)op : throw new ArithmeticException("Integer overflow!") },
            (s, _) => s with { InstPtr = s.InstPtr + 2, B = s.B ^ s.C },
            (s, op) => s with { InstPtr = s.InstPtr + 2, Output = s.Output.Add(ComboOperand(op, s) % 8) },
            (s, op) => s with { InstPtr = s.InstPtr + 2, B = Adv(op, s) },
            (s, op) => s with { InstPtr = s.InstPtr + 2, C = Adv(op, s) },
        ];
    
    
        static (State, ImmutableArray<long> instructions) ReceiveInput(string file)
        {
            var input = File.ReadAllLines(file);
    
            return
            (
                new State(
                    long.Parse(input[0].Substring("Register A: ".Length)),
                    long.Parse(input[1].Substring("Register B: ".Length)),
                    long.Parse(input[2].Substring("Register C: ".Length)),
                    0,
                    []),
                input[4].Substring("Program: ".Length)
                    .Split(",")
                    .Select(long.Parse)
                    .ToImmutableArray()
            );
        }
    }
    
  • @sjmulder
    link
    216 hours ago

    C

    Was looking forward to a VM! Really took my leisure for part 1, writing a disassembler, tracing, all pretty.

    Part 2 reminded me of 2021 day 24 where we also had to reverse an input. It’s been on my mind recently and I was thinking if there would be a way to backfeed an output through a program, yielding a representation like: “the input plus 3498 is a multiple of 40, and divisible by a number that’s 5 mod 8” (considering lossy functions like modulo and integer division).

    Today’s input didn’t lend itself to that, however, but analysing it having the solution ‘click’ was very satisfying.

    Code
    #include "common.h"
    
    #define MEMZ 32
    #define OUTZ 32
    #define BUFZ 64
    
    enum { ADV, BXL, BST, JNZ, BXC, OUT, BDV, CDV };
    
    struct vm {
    	int64_t mem[MEMZ]; int nm, pc;
    	int64_t out[OUTZ]; int no;
    	int64_t a,b,c;
    } vm;
    
    /* returns 0 if halted, 1 otherwise */
    static int
    step(void)
    {
    	int64_t op, ar, ac;	/* operator, lit. arg, combo arg */
    
    	assert(vm.pc >= 0);
    	assert(vm.pc+2 <= MEMZ);
    
    	if (vm.pc >= vm.nm)
    		return 0;
    
    	op = vm.mem[vm.pc++];
    	ar = vm.mem[vm.pc++];
    	ac = ar==4 ? vm.a : ar==5 ? vm.b : ar==6 ? vm.c : ar;
    
    	switch (op) {
    	case ADV: vm.a = vm.a >> ac; break;
    	case BDV: vm.b = vm.a >> ac; break;
    	case CDV: vm.c = vm.a >> ac; break;
    	case BXL: vm.b = vm.b ^ ar; break;
    	case BXC: vm.b = vm.b ^ vm.c; break;
    	case BST: vm.b = ac % 8; break;
    	case JNZ: if (vm.a) vm.pc = ar; break;
    	case OUT: assert(vm.no < OUTZ); vm.out[vm.no++] = ac%8; break;
    	default: assert(!"invalid opcode"); return 0;
    	}
    
    	return 1;
    }
    
    static int64_t
    recur_p2(int64_t a0, int pos)
    {
    	int64_t a, i;
    
    	/*
    	 * The code in the input uses up to 7 low bits of the A register
    	 * to produce a single number, then chops off the low 3 bits and
    	 * continues.
    	 *
    	 * That means bits above the current triplet influence what
    	 * number it generates, but bits below don't. To generate a
    	 * desired sequence then, we append, not prepend,  candidate
    	 * triplets.
    	 *
    	 * We may end up in a situation where, given the prefix found
    	 * for the numbers so far, no triplet yields the desired next
    	 * number. Then we have to backtrack and find another prefix,
    	 * hence the recursion.
    	 */
    
    	if (pos >= vm.nm)
    		return a0 >> 3;
    
    	for (i=0; i<8; i++) {
    		vm.a=a= a0+i;
    		vm.b=vm.c=vm.pc=vm.no= 0;
    
    		while (step() && !vm.no)
    			;
    
    		if (vm.no && vm.out[0] == vm.mem[vm.nm-pos-1])
    			if ((a = recur_p2(a << 3, pos+1)))
    				return a;
    	}
    
    	return 0;
    }
    
    int
    main(int argc, char **argv)
    {
    	char b[BUFZ], *tok, *rest;
    	int i;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	fgets(b, BUFZ, stdin); sscanf(b, "Register A: %lld", &vm.a);
    	fgets(b, BUFZ, stdin); sscanf(b, "Register B: %lld", &vm.b);
    	fgets(b, BUFZ, stdin); sscanf(b, "Register C: %lld", &vm.c);
    	fgets(b, BUFZ, stdin);
    
    	assert(vm.b == 0);	/* assumption for part 2 */
    	assert(vm.c == 0);
    
    	rest = fgets(b, sizeof(b), stdin);
    	strsep(&rest, ":");
    
    	while ((tok = strsep(&rest, ","))) {
    		assert(vm.nm < MEMZ);
    		vm.mem[vm.nm++] = atoll(tok);
    	}
    
    	while (step())
    		;
    	for (i=0; i<vm.no; i++)
    		printf(i ? ",%lld" : "17: %lld", vm.out[i]);
    
    	printf(" %lld\n", recur_p2(0, 0));
    	return 0;
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day17.c

  • @gentooer@programming.dev
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    English
    2
    edit-2
    20 hours ago

    Haskell

    Runs in 10 ms. I was stuck for most of the day on the bdv and cdv instructions, as I didn’t read that the numerator was still register A. Once I got past that, it was pretty straight forward.

    Code
    import Control.Monad.State.Lazy
    import Data.Bits (xor)
    import Data.List (isSuffixOf)
    import qualified Data.Vector as V
    
    data Instr =
            ADV Int | BXL Int | BST Int | JNZ Int | BXC | OUT Int | BDV Int | CDV Int
    type Machine = (Int, Int, Int, Int, V.Vector Int)
    
    parse :: String -> Machine
    parse s =
        let (la : lb : lc : _ : lp : _) = lines s
            [a, b, c] = map (read . drop 12) [la, lb, lc]
            p = V.fromList $ read $ ('[' :) $ (++ "]") $ drop 9 lp
        in  (a, b, c, 0, p)
    
    getA, getB, getC, getIP :: State Machine Int
    getA  = gets $ \(a, _, _, _ , _) -> a
    getB  = gets $ \(_, b, _, _ , _) -> b
    getC  = gets $ \(_, _, c, _ , _) -> c
    getIP = gets $ \(_, _, _, ip, _) -> ip
    
    setA, setB, setC, setIP :: Int -> State Machine ()
    setA  a  = modify $ \(_, b, c, ip, p) -> (a, b, c, ip, p)
    setB  b  = modify $ \(a, _, c, ip, p) -> (a, b, c, ip, p)
    setC  c  = modify $ \(a, b, _, ip, p) -> (a, b, c, ip, p)
    setIP ip = modify $ \(a, b, c, _ , p) -> (a, b, c, ip, p)
    
    incIP :: State Machine ()
    incIP = getIP >>= (setIP . succ)
    
    getMem :: State Machine (Maybe Int)
    getMem = gets (\(_, _, _, ip, p) -> p V.!? ip) <* incIP
    
    getCombo :: State Machine (Maybe Int)
    getCombo = do
        n <- getMem
        case n of
            Just 4          -> Just <$> getA
            Just 5          -> Just <$> getB
            Just 6          -> Just <$> getC
            Just n | n <= 3 -> return $ Just n
            _               -> return Nothing
    
    getInstr :: State Machine (Maybe Instr)
    getInstr = do
        opcode <- getMem
        case opcode of
            Just 0 -> fmap        ADV  <$> getCombo
            Just 1 -> fmap        BXL  <$> getMem
            Just 2 -> fmap        BST  <$> getCombo
            Just 3 -> fmap        JNZ  <$> getMem
            Just 4 -> fmap (const BXC) <$> getMem
            Just 5 -> fmap        OUT  <$> getCombo
            Just 6 -> fmap        BDV  <$> getCombo
            Just 7 -> fmap        CDV  <$> getCombo
            _      -> return Nothing
    
    execInstr :: Instr -> State Machine (Maybe Int)
    execInstr (ADV n) = (getA >>= (setA . (`div` (2^n)))) *> return Nothing
    execInstr (BDV n) = (getA >>= (setB . (`div` (2^n)))) *> return Nothing
    execInstr (CDV n) = (getA >>= (setC . (`div` (2^n)))) *> return Nothing
    execInstr (BXL n) = (getB >>= (setB . xor n)) *> return Nothing
    execInstr (BST n) = setB (n `mod` 8) *> return Nothing
    execInstr (JNZ n) = do
        a <- getA
        case a of
            0 -> return ()
            _ -> setIP n
        return Nothing
    execInstr  BXC    = ((xor <$> getB <*> getC) >>= setB) *> return Nothing
    execInstr (OUT n) = return $ Just $ n `mod` 8
    
    run :: State Machine [Int]
    run = do
        mInstr <- getInstr
        case mInstr of
            Nothing    -> return []
            Just instr -> do
                mOut <- execInstr instr
                case mOut of
                    Nothing ->           run
                    Just n  -> (n :) <$> run
    
    solve2 :: Machine -> Int
    solve2 machine@(_, _, _, _, p') = head [a | x <- [1 .. 7], a <- go [x]]
        where
            p = V.toList p'
            go as =
                let a = foldl ((+) . (* 8)) 0 as
                in  case evalState (setA a *> run) machine of
                        ns  | ns == p           -> [a]
                            | ns `isSuffixOf` p ->
                                concatMap go [as ++ [a] | a <- [0 .. 7]]
                            | otherwise         -> []
    
    main :: IO ()
    main = do
        machine@(_, _, _, _, p) <- parse <$> getContents
        putStrLn $ init $ tail $ show $ evalState run machine
        print $ solve2 machine
    
  • @mykl@lemmy.world
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    2
    edit-2
    17 hours ago

    Dart

    Part one was an exercise in building a simple OpCode machine. Part two was trickier. It was clear that the a register was repeatedly being divided by 8, and testing a few values showed that each 3-bits of the initial value defined one entry in the output, so I built a recursive routine that brute-forced each one in turn. Only <1ms to run though, so not that brutal.

    (It’s worth pointing out that at some stages multiple 3-bit values gave rise to the required value, causing a failure to resolve later on if not checked.)

    (edit: for-loop in buildQuine now avoids using a 0 for the initial triplet, as this should not be a valid value, and if it turns out to generate the required digit of the quine, you will get an infinite recursion. Thanks to SteveDinn for bringing this to my attention.)

    import 'dart:math';
    import 'package:collection/collection.dart';
    import 'package:more/more.dart';
    
    var prog = <int>[];
    typedef State = (int, int, int);
    State parse(List<String> lines) {
      var regs = lines.takeTo('').map((e) => int.parse(e.split(' ').last)).toList();
      var (a, b, c) = (regs[0], regs[1], regs[2]);
      prog = lines.last.split(' ').last.split(',').map(int.parse).toList();
      return (a, b, c);
    }
    
    List<int> runProg(State rec) {
      var (int a, int b, int c) = rec;
      combo(int v) => switch (v) { 4 => a, 5 => b, 6 => c, _ => v };
      var output = <int>[], pc = 0;
      while (pc < prog.length) {
        var code = prog[pc], arg = prog[pc + 1];
        var _ = switch (code) {
          0 => a ~/= pow(2, combo(arg)),
          1 => b ^= arg,
          2 => b = combo(arg) % 8,
          3 => (a != 0) ? (pc = arg - 2) : 0,
          4 => b ^= c, //ignores arg
          5 => output.add(combo(arg) % 8),
          6 => b = a ~/ pow(2, combo(arg)),
          7 => c = a ~/ pow(2, combo(arg)),
          _ => 0
        };
        pc += 2;
      }
      return output;
    }
    
    Function eq = const ListEquality().equals;
    Iterable<int> buildQuine(State rec, List<int> quine, [top = false]) sync* {
      var (int a0, int b0, int c0) = rec;
      if (top) a0 = 0;
      if (quine.length == prog.length) {
        yield a0;
        return;
      }
      for (var a in (top ? 1 : 0).to(8)) {
        var newState = (a0 * 8 + a, b0, c0);
        var newQuine = runProg(newState);
        if (eq(prog.slice(prog.length - newQuine.length), newQuine)) {
          yield* buildQuine(newState, newQuine);
        }
      }
    }
    
    part1(List<String> lines) => runProg(parse(lines)).join(',');
    
    part2(List<String> lines) => buildQuine(parse(lines), [], true).first;
    
    • @SteveDinn@lemmy.ca
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      320 hours ago

      I’m confused reading the buildQuine() method. It reads to me that when you call it from the top level with top = true, A0 will be set to 0, and then when we get to the 0 to 8 loop, the ‘A’ register will be 0 * 8 + 0 for the first iteration, and then recurse with top = false, but with a0 still ending up 0, causing infinite recursion.

      Am I missing something?

      I got it to work with a check that avoids the recursion if the last state’s A register value is the same as the new state’s value.

      • @mykl@lemmy.world
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        fedilink
        319 hours ago

        Oh, good catch. That’s certainly the case if an initial value of 0 correctly generates the required value of the quine. As I’d already started running some code against the live data that’s what I tested against, and so it’s only when I just tested it against the example data that I saw the problem.

        I have changed the for-loop to read for (var a in (top ? 1 : 0).to(8)) for maximum terseness :-)

        That still works for the example and my live data, and I don’t think there should be a valid solution that relies on the first triplet being 0. Thanks for your reply!

  • Rust

    First part was straightforward (the divisions are actually just right shifts), second part not so much. I made some assumptions about the input program, namely that in the end register 8 is divided by 8, then an output is made, then everything starts from the beginning again (if a isn’t 0). I found that the output always depends on at most 10 bits of a, so I ran through all 10-bit numbers and grouped them by the first generated output. At that point it’s just a matter of chaining these 10-bit numbers from the correct groups so that they overlap on 7 bits. The other 3 bits are consumed each round.

    Solution
    use rustc_hash::FxHashMap;
    
    fn parse(input: &str) -> Option<Program> {
        let mut lines = input.lines();
        let a = lines.next()?.split_once(": ")?.1.parse().ok()?;
        let b = lines.next()?.split_once(": ")?.1.parse().ok()?;
        let c = lines.next()?.split_once(": ")?.1.parse().ok()?;
        lines.next()?;
        let program = lines
            .next()?
            .split_once(": ")?
            .1
            .split(',')
            .map(|s| s.parse())
            .collect::<Result<Vec<u8>, _>>()
            .ok()?;
        Some(Program {
            a,
            b,
            c,
            out: vec![],
            program,
            ip: 0,
        })
    }
    
    #[derive(Debug, Clone, Default)]
    struct Program {
        a: u64,
        b: u64,
        c: u64,
        out: Vec<u8>,
        program: Vec<u8>,
        ip: usize,
    }
    
    impl Program {
        fn run(&mut self) {
            while self.step() {}
        }
    
        // Returns true if a step was taken, false if it halted
        fn step(&mut self) -> bool {
            let Some(&[opcode, operand]) = &self.program.get(self.ip..self.ip + 2) else {
                return false;
            };
            self.ip += 2;
            match opcode {
                0 => self.adv(self.combo(operand)),
                1 => self.bxl(operand),
                2 => self.bst(self.combo(operand)),
                3 => self.jnz(operand),
                4 => self.bxc(),
                5 => self.out(self.combo(operand)),
                6 => self.bdv(self.combo(operand)),
                7 => self.cdv(self.combo(operand)),
                _ => panic!(),
            }
            true
        }
    
        fn combo(&self, operand: u8) -> u64 {
            match operand {
                0..=3 => operand as u64,
                4 => self.a,
                5 => self.b,
                6 => self.c,
                _ => unreachable!(),
            }
        }
    
        fn adv(&mut self, x: u64) {
            self.a >>= x
        }
    
        fn bxl(&mut self, x: u8) {
            self.b ^= x as u64
        }
    
        fn bst(&mut self, x: u64) {
            self.b = x % 8
        }
    
        fn jnz(&mut self, x: u8) {
            if self.a != 0 {
                self.ip = x as usize
            }
        }
    
        fn bxc(&mut self) {
            self.b ^= self.c
        }
    
        fn out(&mut self, x: u64) {
            self.out.push((x % 8) as u8)
        }
    
        fn bdv(&mut self, x: u64) {
            self.b = self.a >> x
        }
    
        fn cdv(&mut self, x: u64) {
            self.c = self.a >> x
        }
    }
    
    fn part1(input: String) {
        let mut program = parse(&input).unwrap();
        program.run();
        if let Some(e) = program.out.first() {
            print!("{e}")
        }
        for e in program.out.iter().skip(1) {
            print!(",{e}")
        }
        println!()
    }
    
    // Some assumptions on the input:
    // * There is exactly one jump instruction at the end of the program, jumping to 0
    // * Right before that, an output is generated
    // * Right before that, register a is shifted right by 3: adv(3)
    //
    // Each output depends on at most 10 bits of a (it is used with a shift of at most 7).
    // Therefore we look at all 10-bit a's and group them by the first number that is output.
    // Then we just need to combine these generators into a chain that fits together.
    fn number_generators(mut program: Program) -> [Vec<u16>; 8] {
        let mut out = [const { vec![] }; 8];
        for a in 1..(1 << 10) {
            program.a = a as u64;
            program.out.clear();
            program.ip = 0;
            program.run();
            let &output = program.out.first().unwrap();
            out[output as usize].push(a);
        }
        out
    }
    
    fn part2(input: String) {
        let mut program = parse(&input).unwrap();
        let generators = number_generators(program.clone());
    
        let output = program.program.clone();
        // a_candidates maps from 7-bit required prefixes to the lower bits of a that
        // generate the required numbers so far.
        let mut a_candidates: FxHashMap<u8, u64> = generators[output[0] as usize]
            .iter()
            .rev() // Collects the values for each prefix
            .map(|&a| ((a >> 3) as u8, a as u64 % 8))
            .collect();
        let len = output.len();
        for (i, x) in output.iter().enumerate().skip(1) {
            let mut next_candidates = FxHashMap::default();
            for (prefix, val) in generators[*x as usize]
                .iter()
                .filter(|&a| {
                    // Take only short candidates in the end to ensure that not too many numbers are generated
                    let max_bits = (len - i) * 3;
                    (*a as u64) < (1u64 << max_bits)
                })
                // Only use generators that match any required prefix
                .filter(|&a| a_candidates.contains_key(&((a % (1 << 7)) as u8)))
                .map(|&a| {
                    let prefix = (a >> 3) as u8;
                    let val = a as u64 % 8;
                    let prev = a_candidates[&((a % (1 << 7)) as u8)];
                    (prefix, (val << (i * 3)) | prev)
                })
            {
                // Only insert first (smallest) encountered value
                next_candidates.entry(prefix).or_insert(val);
            }
            a_candidates = next_candidates;
        }
        println!("{}", a_candidates[&0]);
    
        // Verify result
        program.a = a_candidates[&0];
        program.run();
        assert_eq!(program.out, program.program);
    }
    
    util::aoc_main!();
    

    Also on github

    • @CameronDev@programming.devOPM
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      31 day ago

      Your code helped me find my bug, thanks.

      Wild that all the examples can be passed with an incorrect cdv instruction, I didnt read the last part properly and assumed it was C /= operand. :(

  • @VegOwOtenks@lemmy.world
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    English
    21 day ago

    Haskell

    Part 2 was tricky, I tried executing the algorithm backwards, which worked fine for the example but not with the input program, because it uses one-way functions .-. Then I tried to write an algorithm that would try all valid combinations of powers of 8 and but I failed, I then did it by hand.

    Solution Codeblock
    import Control.Arrow
    import Data.Bits
    import qualified Data.Char as Char
    import qualified Data.List as List
    
    replace c r c' = if c' == c then r else c'
    
    parse :: String -> ([Integer], [Int])
    parse = map (replace ',' ' ')
            >>> filter ((Char.isDigit &&& Char.isSpace) >>> uncurry (||))
            >>> words
            >>> splitAt 3
            >>> (map read *** map read)
    
    type InstructionPointer = Int
    
    adv = 0
    bxl = 1
    bst = 2
    jnz = 3
    bxc = 4
    out = 5
    bdv = 6
    cdv = 7
    
    lookupCombo _    0 = 0
    lookupCombo _    1 = 1
    lookupCombo _    2 = 2
    lookupCombo _    3 = 3
    lookupCombo regs 4 = regs !! 0
    lookupCombo regs 5 = regs !! 1
    lookupCombo regs 6 = regs !! 2
    lookupCombo regs 7 = error "Invalid operand"
    
    execute :: InstructionPointer -> [Integer] -> [Int] -> [Int]
    execute ip regs@(regA:regB:regC:[]) ops
            | ip >= length ops = []
            | instruction == adv = execute (ip + 2) [regA `div` (2 ^ comboValue), regB, regC] ops
            | instruction == bxl = execute (ip + 2) [regA, xor regB (toInteger operand), regC] ops
            | instruction == bst = execute (ip + 2) [regA, comboValue `mod` 8, regC] ops
            | instruction == jnz && regA == 0 = execute (ip + 2) regs ops
            | instruction == jnz && regA /= 0 = execute operand regs ops
            | instruction == bxc = execute (ip + 2) [regA, xor regB regC, regC] ops
            | instruction == out = (fromIntegral comboValue) `mod` 8 : execute (ip + 2) regs ops
            | instruction == bdv = execute (ip + 2) [regA, regA `div` (2 ^ comboValue), regC] ops
            | instruction == cdv = execute (ip + 2) [regA, regB, regA `div` (2 ^ comboValue)] ops
            where
                    (instruction, operand) = (ops !! ip, ops !! (succ ip))
                    comboValue             = lookupCombo regs operand
    
    part1 = uncurry (execute 0)
            >>> List.map show
            >>> List.intercalate ","
    
    valid i t n = ((n `div` (8^i)) `mod` 8) `xor` 7 `xor` (n `div` (4*(8^i))) == t
    
    part2 = const 247839653009594
    
    main = getContents
            >>= print
            . (part1 &&& part2)
            . parse
    ```haskell
    
  • lwhjp
    link
    21 day ago

    Haskell

    Woah, that was suddenly a hard one: several tricky things combined. I’m not a big fan of the kind of problems like part 2 today, but eh - you can’t please everyone.

    Solution
    import Control.Monad
    import Control.Monad.RWS
    import Data.Bits
    import Data.Foldable
    import Data.List
    import Data.List.Split
    import Data.Vector (Vector)
    import Data.Vector qualified as Vector
    
    type Machine = RWS (Vector Int) [Int] (Int, Int, Int)
    
    readInput :: String -> ((Int, Int, Int), Vector Int)
    readInput s =
      let (regs, _ : [prog]) = break null $ lines s
       in ( let [a, b, c] = map (read . last . words) regs in (a, b, c),
            let [_, s] = words prog in Vector.fromList $ map read $ splitOn "," s
          )
    
    stepMachine :: Int -> Machine Int
    stepMachine ip = do
      opcode <- asks (Vector.! ip)
      operand <- asks (Vector.! (ip + 1))
      (a, b, c) <- get
      let combo = [0, 1, 2, 3, a, b, c, undefined] !! operand
          ip' = ip + 2
          adv = a `div` (2 ^ combo)
          store 'A' v = modify (\(_, b, c) -> (v, b, c))
          store 'B' v = modify (\(a, _, c) -> (a, v, c))
          store 'C' v = modify (\(a, b, _) -> (a, b, v))
      case opcode of
        0 -> store 'A' adv >> return ip'
        1 -> store 'B' (b `xor` operand) >> return ip'
        2 -> store 'B' (combo .&. 7) >> return ip'
        3 -> return $ if a == 0 then ip' else operand
        4 -> store 'B' (b `xor` c) >> return ip'
        5 -> tell [combo .&. 7] >> return ip'
        6 -> store 'B' adv >> return ip'
        7 -> store 'C' adv >> return ip'
    
    part1 (regs, prog) =
      let (a, s, w) = runRWS (go 0) prog regs
       in intercalate "," $ map show w
      where
        go ip = when (ip < Vector.length prog) $ stepMachine ip >>= go
    
    part2 (_, prog) = minimum $ foldM go 0 $ reverse $ toList prog
      where
        go a d = do
          b <- [0 .. 7]
          let a' = (a `shiftL` 3) .|. b
              b1 = b `xor` 5
              b2 = b1 `xor` (a' `shiftR` b1)
              b3 = b2 `xor` 6
          guard $ b3 .&. 7 == d
          return a'
    
    main = do
      input <- readInput <$> readFile "input17"
      putStrLn $ part1 input
      print $ part2 input
    
  • @hades@lemm.ee
    link
    fedilink
    31 day ago

    C#

    public class Day17 : Solver {
      private List<int> program;
      private long a, b, c;
    
      public void Presolve(string input) {
        var data = input.Trim().Split("\n");
        a = long.Parse(Regex.Match(data[0], @"\d+").Value);
        b = long.Parse(Regex.Match(data[1], @"\d+").Value);
        c = long.Parse(Regex.Match(data[2], @"\d+").Value);
        program = data[4].Split(" ")[1].Split(",").Select(int.Parse).ToList();
      }
    
      public string SolveFirst() => String.Join(',', Execute(a, b, c));
    
      private List<int> Execute(long a, long b, long c) {
        List<int> output = [];
        Func<long, long> combo = operand => operand switch {
          <= 3 => operand,
          4 => a,
          5 => b,
          6 => c,
          _ => throw new InvalidDataException(),
        };
        int ip = 0;
        while (ip < program.Count - 1) {
          switch (program[ip]) {
            case 0:
              a = a >> (int)combo(program[ip + 1]);
              break;
            case 1:
              b = b ^ program[ip + 1];
              break;
            case 2:
              b = combo(program[ip + 1]) % 8;
              break;
            case 3:
              if (a != 0) {
                ip = program[ip + 1];
                continue;
              }
              break;
            case 4:
              b = b ^ c;
              break;
            case 5:
              output.Add((int)(combo(program[ip + 1]) % 8));
              break;
            case 6:
              b = a >> (int)combo(program[ip + 1]);
              break;
            case 7:
              c = a >> (int)combo(program[ip + 1]);
              break;
          }
          ip += 2;
        }
        return output;
      }
    
      public string SolveSecond() {
        Dictionary<int, List<(int, int, int)>> mapping = [];
        for (int start_a = 0; start_a < 512; start_a++) {
          var output = Execute(start_a, b, c);
          mapping.TryAdd(output[0], []);
          mapping[output[0]].Add((start_a / 64, start_a / 8 % 8, start_a % 8));
        }
        List<List<(int, int, int)>> possible_codes = [.. program.Select(b => mapping[b])];
        possible_codes.Reverse();
        List<int>? solution = SolvePossibleCodes(possible_codes, null);
        solution?.Reverse();
        long result = 0;
        foreach (var code in solution!) {
          result = (result << 3) | code;
        }
        return result.ToString();
      }
    
      private List<int>? SolvePossibleCodes(List<List<(int, int, int)>> possible_codes, (int, int)? allowed_start) {
        if (possible_codes.Count == 0) return [];
        foreach (var (high, mid, low) in possible_codes[0]) {
          if (allowed_start is null || allowed_start.Value.Item1 == high && allowed_start.Value.Item2 == mid) {
            var tail = SolvePossibleCodes(possible_codes[1..], (mid, low));
            if (tail is null) continue;
            tail.Add(low);
            if (allowed_start is null) {
              tail.Add(mid);
              tail.Add(high);
            }
            return tail;
          }
        }
        return null;
      }
    }