AoC 2021 D24: Arithmetic Logic Unit

| Problem statement | Source code | Tags: VMManual inspectionPuzzle

Part 1

This is a question type I enjoy a lot: there's more pen-and-paper (or, for me, iPad and typing in a text file) than coding.

I tried to formally work out the required input range, using Hoare logic to symbolically deduce the weakest precondition for each line of code, given that the postcondition is z = 0. This is actually feasible given that no loops exist. I even have a solution that I believe works; the problem is that the branching still leads to an exponential number of paths, and Node OOMs. So I was forced back to manual inspection.

The program has 14 chunks, each of the form:

 1 | inp w
 2 | mul x 0
 3 | add x z
 4 | mod x 26
 5 | div z <n0>
 6 | add x <n1>
 7 | eql x w
 8 | eql x 0
 9 | mul y 0
10 | add y 25
11 | mul y x
12 | add y 1
13 | mul z y
14 | mul y 0
15 | add y w
16 | add y <n2>
17 | mul y x
18 | add z y

Among the 14 chunks, only lines 5, 6, 16 may be different in their constant. In particular, <n0> can either be 1 or 26; <n1> can be between -15 and 15; <n2> can be between 1 and 14. Furthermore, n1 <= n0 iff n0 == 26.

If we convert this to TypeScript:

function chunk(w: number, n0: number, n1: number, n2: number) {
  x = (z % 26) + n1 === w ? 0 : 1; // Lines 2, 3, 4, 6, 7, 8
  z = Math.floor(z / n0); // Line 5
  y = 25 * x + 1; // Lines 9, 10, 11, 12
  z *= y; // Line 13
  y = (w + n2) * x; // Lines 14, 15, 16, 17
  z += y; // Line 18
}

More compactly:

function chunk(w: number, n0: number, n1: number, n2: number) {
  if (z % 26 === w - n1) {
    z = Math.floor(z / n0);
  } else {
    z = Math.floor(z / n0) * 26 + w + n2;
  }
}

Note that while n1 can be negative, all other numbers are always positive, including z.

This is using z as a stack in base-26:

If z.top() == w - n1 and n0 == 26, then z.pop().
If z.top() == w - n1 and n0 == 1, then do nothing.
If z.top() != w - n1 and n0 == 1, then z.push(w + n2).
If z.top() != w - n1 and n0 == 26, then z.replaceTop(w + n2).

function chunk(w: number, n0: number, n1: number, n2: number) {
  if (n0 == 1) {
    z.push(w + n2);
  } else if (z[z.length - 1] == w - n1) {
    z.pop();
  } else {
    z[z.length - 1] = w + n2;
  }
}

The goal at the end is to have z empty, and we have 7 n0 == 1 chunks and 7 n0 == 26 chunks, so we need to pop every time we can. We can pair up each push-chunk and pop-chunk, such that w_push + n2_push == w_pop - n1_pop. This creates 7 equations, each of the form w[i] - w[j] = n where i > j.

function getConstraints(data: string[]): [number, number, number][] {
  const chunks = data.join("\n").split("inp w\n");
  const nVals = chunks.filter(Boolean).map((chunk) => {
    const lines = chunk.split("\n");
    return [lines[3], lines[4], lines[14]].map((line, i) =>
      Number(line.split(" ")[2]),
    );
  });
  const stack: [number, number][] = [];
  // Each constraint of the form w[i] - w[j] = n (i > j)
  const constraints: [number, number, number][] = [];
  for (const [i, [n0, n1, n2]] of nVals.entries()) {
    if (n0 === 1) {
      // Push-chunk
      stack.push([i, n2]);
    } else {
      // Pop-chunk
      const [iPush, n2Push] = stack.pop()!;
      constraints.push([i, iPush, n2Push + n1]);
    }
  }
  return constraints;
}

Because we want to maximize w, we can just greedily maximize each w[i], since these equations are independent of each other. If w[i] - w[j] = n, then when n is positive, set w[i] = 9 and w[j] = 9 - n; when n is negative, set w[j] = 9 and w[i] = 9 + n.

const optW = Array(14).fill(0);
for (const [i, iPush, n] of constraints) {
  optW[i] = Math.min(9, 9 + n);
  optW[iPush] = Math.min(9, 9 - n);
}
console.log(optW.join(""));

Part 2

Exactly the same, except we minimize w instead of maximizing it. In this case we want w[i] and w[j] as close to 1 as possible.

const optW = Array(14).fill(9);
for (const [i, iPush, n] of constraints) {
  optW[i] = Math.max(1, 1 + n);
  optW[iPush] = Math.max(1, 1 - n);
}
console.log(optW.join(""));

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