Day 17: Chronospatial Computer

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FAQ

  • Gobbel2000@programming.dev
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    10 days ago

    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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      10 days 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. :(