Categories

# My solution to Advent of Code 2020’s Day 8 challenge, in Python

Welcome to another installment in my Advent of Code 2020 series, where I present my solutions to this year’s Advent of Code challenges!

In this installment, I share my Python solution to Day 8 of Advent of Code, titled Handheld Halting.

Please be warned: If you want to try solving the challenge on your own and without any help, stop reading now! The remainder of this post will be all about my solution to both parts of the Day 8 challenge.

## The Day 8 challenge, part one

### The challenge

Here’s the text from part one of the challenge:

Your flight to the major airline hub reaches cruising altitude without incident. While you consider checking the in-flight menu for one of those drinks that come with a little umbrella, you are interrupted by the kid sitting next to you.

Their handheld game console won’t turn on! They ask if you can take a look.

You narrow the problem down to a strange infinite loop in the boot code (your puzzle input) of the device. You should be able to fix it, but first you need to be able to run the code in isolation.

The boot code is represented as a text file with one instruction per line of text. Each instruction consists of an operation (`acc``jmp`, or `nop`) and an argument (a signed number like `+4` or `-20`).

• `acc` increases or decreases a single global value called the accumulator by the value given in the argument. For example, `acc +7` would increase the accumulator by 7. The accumulator starts at `0`. After an `acc` instruction, the instruction immediately below it is executed next.
• `jmp` jumps to a new instruction relative to itself. The next instruction to execute is found using the argument as an offset from the `jmp` instruction; for example, `jmp +2` would skip the next instruction, `jmp +1` would continue to the instruction immediately below it, and `jmp -20` would cause the instruction 20 lines above to be executed next.
• `nop` stands for No OPeration – it does nothing. The instruction immediately below it is executed next.

For example, consider the following program:

These instructions are visited in this order:

First, the `nop +0` does nothing. Then, the accumulator is increased from 0 to 1 (`acc +1`) and `jmp +4` sets the next instruction to the other `acc +1` near the bottom. After it increases the accumulator from 1 to 2, `jmp -4` executes, setting the next instruction to the only `acc +3`. It sets the accumulator to 5, and `jmp -3` causes the program to continue back at the first `acc +1`.

This is an infinite loop: with this sequence of jumps, the program will run forever. The moment the program tries to run any instruction a second time, you know it will never terminate.

Immediately before the program would run an instruction a second time, the value in the accumulator is `5`.

Run your copy of the boot code. Immediately before any instruction is executed a second time, what value is in the accumulator?

### Importing the data

Every Advent of Code participant gets their own set of data. I copied my data and went through my usual process of bringing it into a Jupyter Notebook running a Python kernel.

This involves pasting it into a triple-quoted string and then assigning its value to the variable `main_raw_input`:

### Building the data structure

With the data imported, the next step was to build an appropriate data structure. Since today’s puzzle was based on simplified microprocessor instructions, I thought that the data structure should do the same.

I wanted my data structure to look like the table below, which shows the instructions from extracted from the first five lines of the data given to me:

Index Opcode Operand
0 acc 37
1 acc -4
2 nop 405
3 jmp 276
4 acc 39

Here’s the function that I wrote to convert the input data into the data structure:

With the function defined, here’s how I used it to create the data structure, which I assigned to a variable named `instructions`:

I then wrote `accumulator_value_at_first_repeat()`, the function that would use the data structure contained within `instructions` to solve the first challenge:

When run, it sets up the following variables:

• `program_length`: The number of instructions in the data structure.
• `program_counter`: The program’s equivalent of the “program counter (PC)” register in a microprocessor; contains the index the instruction to be executed.
• `accumulator`: The program’s equivalent of an accumulator in a microprocessor. For those of you who aren’t familiar with what happens at the microprocessor level, think of the accumulator as a “scratchpad” variable where you do all the math.
• `executed_instructions`: A set (which means that every value in it is unique) of the indices of all the instructions that have already been executed. We use it to determine if a given instruction has already been run, at which point we should stop the program and see what the value in the accumulator is.
• `repeat_instruction_encountered`: A flag that should be set to `True` when an instruction is about to be executed a second time.

The function’s main loop performs a greatly simplified version of a microprocessor’s “fetch-decode-execute” cycle:

1. Fetch the current instruction, which is the one whose index is specified by `program_counter`.
2. See if the instruction has been executed before. If this is the case, exit the loop; otherwise, recording this instruction as having been executed.
3. Decode the current instruction:
• If the instruction is jmp, add the operand value to `program_counter` and go back to the start of the loop.
• If the instruction is acc, add the operand value to `accumulator`.
• If the instruction is nop, do nothing.
• If the instruction is anything else, display an error message.
4. After the loop is done, if the `repeat_instruction_encountered` flag is set, we’ve found the value we’re looking for — display it! Otherwise, display a message saying that we’ve reached the end of the instructions without ever repeating one.

I ran the function…

…and got my result: 1801.

## The Day 8 challenge, part two

### The challenge

After some careful analysis, you believe that exactly one instruction is corrupted.

Somewhere in the program, either a `jmp` is supposed to be a `nop`or a `nop` is supposed to be a `jmp`. (No `acc` instructions were harmed in the corruption of this boot code.)

The program is supposed to terminate by attempting to execute an instruction immediately after the last instruction in the file. By changing exactly one `jmp` or `nop`, you can repair the boot code and make it terminate correctly.

For example, consider the same program from above:

If you change the first instruction from `nop +0` to `jmp +0`, it would create a single-instruction infinite loop, never leaving that instruction. If you change almost any of the `jmp` instructions, the program will still eventually find another `jmp` instruction and loop forever.

However, if you change the second-to-last instruction (from `jmp -4` to `nop -4`), the program terminates! The instructions are visited in this order:

After the last instruction (`acc +6`), the program terminates by attempting to run the instruction below the last instruction in the file. With this change, after the program terminates, the accumulator contains the value `8` (`acc +1``acc +1``acc +6`).

Fix the program so that it terminates normally by changing exactly one `jmp` (to `nop`) or `nop` (to `jmp`). What is the value of the accumulator after the program terminates?

Here’s the function I wrote to solve this challenge:

The function, `accumulator_value_and_halt_at_first_repeat()`, is an expanded version of the function from part one, `accumulator_value_at_first_repeat()`.

In addition to a set of instructions, it takes an additional parameter: the address of an instruction that should be changed — either from jmp to nop, or from nop to jmp.

The function still performs the “fetch-decode-execute” loop, and it exits the loop if it’s about to execute an instruction that’s already been executed. The main difference is that if the current instruction is the one flagged for change, it changes the instruction appropriately.

I wrote the `accumulator_value_and_halt_at_first_repeat()` function to be used by the function below:

This function goes through all the instructions in the set, looking for any jmp or nop instructions. When it finds one, it runs the program using `accumulator_value_and_halt_at_first_repeat()`, marking the jmp or nop instruction as the one to be changed.

This lets us modify the program, one jmp or nop instruction at a time, in order to find which change to a jmp or nop instruction is the one that allows the program to reach the end of the instructions.

Here’s an abridged version of what happened when I ran the function:

I entered 2060 as my answer, and step two was complete.