Module day24 

Arithmetic Logic Unit

There are two ways to solve this problem. We can brute force emulate the ALU, but even with memoization this takes a long time to solve. Instead analyzing and reverse engineering the code shows an insight that reduces the problem to a much simpler constraint satisfaction problem.

Analysis Summary

The code consists of 14 blocks of 18 instructions, each block starting with inp w.

There are 7 “push” blocks and 7 “pop” blocks.

• Push blocks use z as a stack, multiplying by 26 and adding w plus some constant k₁. Push blocks always add to the stack.
• Pop blocks compare the top element of z plus some constant k₂ to w. If the values are equal then z is “popped” by dividing by 26. Otherwise z is pushed again with w plus some constant.

Since the push/pop blocks are equally numbered and we want z to be zero (empty stack) at the end of the program, the condition:

w₁ + k₁ + k₂ = w₂

must be true for each matching pair of push/pop blocks. Since we also know that each w is between 1 and 9 inclusive, that is enough information to determine maximum and minimum values for each of the fourteen w values.

For example:

• If k₁ + k₂ = 7 then w₁ + 7 = w₂.
• Maximum value of w₁ is 2 when w₂ is 9.
• Minimum value of w₁ is 1 when w₂ is 8.

Detailed Push block analysis

    inp w       // w = 1 to 9 inclusive
    mul x 0     // x = 0
    add x z     // x = z
    mod x 26    // x = z % 26
    div z 1     // nop
    add x 13    // x = z % 26 + 13
    eql x w     // if (z % 26 + 13) == w { x = 1 } else { x = 0 }
                // However since w is restricted to 1 to 9 this condition is always false
                // so x is always 0. Other blocks have different constants but always > 9.
    eql x 0     // x = 1 (as 0 == 0)
    mul y 0     // y = 0
    add y 25    // y = 25
    mul y x     // y = 25 * 1 = 25
    add y 1     // y = 25 + 1 = 26
    mul z y     // z = 26 * z
    mul y 0     // y = 0
    add y w     // y = w
    add y 14    // y = w + 14 (k₁ = 14)
    mul y x     // y = (w + 14) * 1 = (w + 14)
    add z y     // z = (26 * z) + (w + 14)

Detailed Push block analysis

    inp w       // w = 1 to 9 inclusive
    mul x 0     // x = 0
    add x z     // x = z
    mod x 26    // x = z % 26
    div z 26    // z /= 26 (pop)
    add x -13   // x = z % 26 - 13 (k₂ = -13)
    eql x w     // if (z % 26 - 13) == w { x = 1 } else { x = 0 }
    eql x 0     // if (z % 26 - 13) == w { x = 0 } else { x = 1 }
                // Inverts the previous conditional.
                // Unlike the push blocks, this may true or false
                // We'll split into 2 paths, depending on equals (x = 0) or
                // not equal (x = 1).
                | Equals (x = 0)        | Not Equals (x = 1)        |
    mul y 0     | y = 0                 | y = 0                     |
    add y 25    | y = 25                | y = 25                    |
    mul y x     | y = 25 * 0 = 0        | y = 25 * 1 = 25           |
    add y 1     | y = 0 + 1 = 1         | y = 25 + 1 = 26           |
    mul z y     | z = z (nop)           | z = 26 * z                |
    mul y 0     | y = 0                 | y = 0                     |
    add y w     | y = w                 | y = w                     |
    add y 4     | y = w + 4             | y = w + 4                 |
    mul y x     | y = (w + 4) * 0       | y = (w + 4) * 1           |
    add z y     | z = z                 | z = (26 * z) * (w + 4)    |

Structs

Constraint

Convert matching pairs of blocks into constraints. For the first digit value is -(k₁ + k₂) and second digit value is k₁ + k₂.

Enums

Block 🔒

Blocks are either “push” or “pop”.

Functions

parse

part1

part2