It's clear that larger numbers will tend to require larger programs. By repeating \(B := B + A\) and \(A := B\) we can exponentiate to get to a large number quickly. We'll also need \(A := A + 1\) to get started, and to make smaller adjustments. However, \(A := A - 1\) is not nearly as useful (technically it has some utility when trying to strictly minimize the number of instructions it takes to generate a certain value, but that isn't the goal in this problem).

Since \(A := A - 1\) is not very useful, we don't need many, if any, ways to have an even number of paths at an odd distance from the start. We do, however, need to have an even number of paths at *even* distances from the start.

So our two goals when constructing a grid are to make a long enough path to fit all the instructions we need, and also maximize the number of cells at even distances from the start. For example, a grid like this:

```
@##....##....
..#.##..#.##.
.#..##.#..##.
..#.#..##.#..
.#..##.#..##.
..#.#..##.#..
.#..##.#..##.
..#.#..##.#..
.#..##.#..##.
..#.#..##.#..
.#..##.#..##.
.##.#..##.#..
....##....##.
```

This grid gives us one long path, and then many "alcoves" that are all at even distances from the start. Instruction cells along the main path will represent \(B := B + A\) and \(A := B\), and whenever we need to use \(A := A + 1\) we can add another instruction cell in one next alcove.

Given such a grid, there are multiple ways to then determine a program that fits on the grid, such as using dynamic programming where the state is the value of the two registers, or manually constructing a set of instructions where we repeatedly multiply \(A\) by 2 and optionally add \(1\) to \(A\) as needed (essentially constructing the bitstring representation of the goal value).