Global Nerdy

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.

Spoiler alert!

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:

nop +0
acc +1
jmp +4
acc +3
jmp -3
acc -99
acc +1
jmp -4
acc +6

These instructions are visited in this order:

nop +0  | 1
acc +1  | 2, 8(!)
jmp +4  | 3
acc +3  | 6
jmp -3  | 7
acc -99 |
acc +1  | 4
jmp -4  | 5
acc +6  |

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:

main_raw_input = """acc +37
acc -4
nop +405
jmp +276
acc +39
acc +40
acc -3
jmp +231
acc +44
acc +12
jmp +505
acc +35
jmp +282
acc +23
jmp +598
nop +392
acc +18
acc +44
acc +18
jmp +297
nop +460
jmp +152
nop +541
acc +33
jmp -11
acc -5
acc +9
jmp +327
acc +30
acc -1
acc -3
jmp +50
acc +22
acc +18
acc +33
acc +37
jmp +57
acc -17
acc -6
acc -2
jmp +535
acc -15
jmp +279
acc +34
acc +44
acc +41
jmp +349
acc +2
acc +6
nop +351
nop +252
jmp +505
jmp +1
jmp +1
nop +61
jmp +524
nop +351
jmp +399
acc +1
nop +397
acc +39
nop +141
jmp +134
acc +46
acc +14
acc +26
jmp +236
acc +7
acc -6
acc +35
jmp +397
acc +15
jmp +140
acc +3
acc -4
acc +37
acc +12
jmp +86
jmp +416
jmp +1
jmp +55
acc -19
jmp +536
jmp +1
acc -11
acc +15
jmp -61
acc +25
jmp -25
acc +50
acc +43
jmp +1
jmp +140
acc +46
nop -53
acc +1
nop +440
jmp +488
jmp +396
nop +443
acc +41
jmp +168
acc +25
nop +383
acc +12
acc -19
jmp +21
acc +29
acc +30
jmp +497
jmp +502
jmp +417
nop +351
acc -15
jmp +243
acc +21
acc +16
jmp +332
acc +28
acc +22
acc +38
jmp +476
acc +8
acc -11
jmp +458
acc +9
jmp +246
acc +40
acc +31
acc +26
jmp +218
acc +27
acc +9
nop +347
jmp +478
nop +28
nop +106
acc +25
acc -15
jmp +397
acc +31
jmp +231
acc -4
nop +136
acc +14
jmp +181
jmp +361
acc +16
acc +11
jmp -108
nop +299
acc +21
acc -2
jmp -106
jmp +246
acc +31
jmp +407
jmp +377
acc +43
acc -12
nop +142
acc +8
jmp -91
jmp +1
acc +34
acc +5
acc +31
jmp +12
acc +34
acc +7
acc +34
acc +20
jmp -45
acc -11
acc +41
acc +10
jmp +310
nop -106
jmp -36
acc +23
acc +46
acc +46
jmp +112
acc +41
nop +179
acc +17
nop +356
jmp +147
acc +42
nop +49
jmp +119
acc +0
acc +7
acc -18
acc -8
jmp +11
acc +12
acc +38
acc +39
jmp +281
nop +186
jmp +162
acc +44
acc +20
jmp +153
jmp +395
acc +49
jmp +1
acc +2
jmp +1
jmp -31
jmp +301
nop +97
jmp -102
jmp +262
acc +28
acc -15
acc +44
acc -13
jmp +191
jmp +281
acc +36
acc +1
nop +15
jmp +211
acc +6
acc -4
jmp +42
acc +34
acc +0
jmp +104
jmp +311
jmp +84
acc +43
acc -8
acc -10
acc +38
jmp -90
acc +49
jmp +303
nop +132
jmp +301
nop +60
acc +37
nop +96
jmp +182
acc +16
acc +18
nop +152
acc +19
jmp +325
jmp -63
acc +28
jmp +56
acc +18
acc +29
acc +33
jmp -115
acc +47
acc +19
jmp +1
nop +41
jmp +1
jmp -207
nop -62
acc -9
acc +42
acc -12
jmp -56
acc +28
jmp -163
acc +25
acc +17
jmp -217
acc +7
jmp +272
acc +43
acc +22
jmp +70
acc -17
jmp -117
acc +24
acc +26
nop -275
jmp -46
nop +87
acc +19
acc +28
jmp -34
acc +4
acc +9
acc +6
jmp +1
jmp +28
acc -6
nop -67
acc -10
jmp +271
acc +40
acc +25
acc -4
jmp -63
acc +46
jmp +78
acc +41
nop -126
nop +70
jmp +1
jmp +172
nop +270
jmp +30
jmp +1
acc +38
nop +68
acc +29
jmp +253
acc -18
jmp -89
acc +18
acc +30
jmp +147
acc +24
acc +11
acc +50
jmp -225
jmp -210
acc -18
acc +1
acc +38
jmp +1
jmp -79
acc +45
acc +12
jmp +209
jmp -207
acc +32
acc +4
acc +32
acc +14
jmp +83
acc +13
acc +1
acc +46
acc +38
jmp +28
nop +153
acc -17
jmp -73
acc +11
jmp +248
acc +29
acc +45
acc +16
jmp +96
jmp -273
acc +34
jmp +87
nop +99
acc -3
jmp -74
acc +12
nop -119
jmp -141
acc -18
nop -79
acc +1
acc +6
jmp +9
acc +3
acc +44
acc +39
jmp -165
acc +6
jmp +44
acc +25
jmp -133
acc +0
jmp +14
jmp +1
acc +1
jmp -223
jmp +71
nop -1
acc +22
acc +11
jmp -274
jmp -330
acc +45
jmp +1
acc +15
jmp -158
jmp -128
acc +50
acc +26
jmp -73
nop +99
jmp +71
acc +35
acc +7
jmp +192
acc +13
jmp +190
acc +4
acc -1
acc +40
acc -15
jmp +50
acc +29
jmp -337
jmp -75
acc +41
jmp +1
jmp -387
acc +28
acc +18
acc +19
jmp -62
nop -196
jmp -410
jmp +1
acc -17
jmp -267
acc +22
jmp -301
nop -98
acc -15
jmp -124
acc +45
acc -18
acc +15
acc +42
jmp -296
nop -10
acc +29
jmp -371
acc +3
jmp +1
nop +61
acc +5
jmp -361
acc -5
nop -326
jmp -379
acc -10
jmp +1
acc +44
jmp -231
acc +3
jmp -94
acc +1
jmp +113
jmp -336
acc +4
jmp -299
acc -13
jmp +1
acc +13
jmp +143
acc -11
acc -19
acc +18
nop -390
jmp -27
acc +42
jmp -232
acc +15
jmp -228
acc +21
acc +39
acc +47
acc +6
jmp +57
acc +28
acc +27
acc +50
jmp -397
acc +12
jmp -445
acc +30
jmp -352
acc -4
acc +26
acc +48
jmp +1
jmp -205
jmp +22
nop -284
acc -1
nop -361
acc +0
jmp -368
acc -17
nop -223
jmp -41
acc +4
acc +46
jmp +79
jmp -370
jmp -260
acc +42
jmp -14
acc +30
acc +50
acc +13
jmp -61
acc +46
jmp -63
nop -55
nop -320
jmp -11
acc +10
jmp -424
jmp -11
acc +3
jmp -71
acc +42
acc -13
jmp +4
nop -155
nop -138
jmp +62
acc +11
acc +19
acc +15
acc +17
jmp -73
acc -11
jmp -273
acc +8
acc +6
acc -7
acc +41
jmp -311
jmp -111
jmp -260
jmp +50
jmp -60
jmp +1
nop -89
acc +36
acc +14
jmp -220
nop -415
acc +28
jmp -402
acc +41
jmp -165
acc +9
acc -13
acc -18
acc +18
jmp -504
acc -9
acc +29
acc +44
jmp -444
acc +5
acc +47
jmp -545
acc +23
acc +7
nop -240
jmp -320
jmp -141
jmp +1
acc +28
nop -287
jmp -118
acc +44
acc -7
jmp -550
acc +10
acc +20
acc -3
jmp -401
acc +45
acc +36
jmp -375
jmp -485
acc +9
jmp -338
jmp -510
jmp -196
acc -16
jmp -372
acc +0
jmp -380
acc -3
nop -473
nop -361
jmp -311
acc +0
nop +20
jmp -436
acc +9
jmp +1
jmp -215
acc +19
jmp -451
jmp -43
acc -13
acc -10
acc -5
jmp -208
acc -11
jmp -156
acc +11
acc -2
nop -357
jmp -73
acc +21
jmp -159
acc +28
acc -16
acc +12
acc +1
jmp -282
jmp -131
acc -11
acc +45
acc +0
acc +28
jmp +1"""

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:

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

def input_to_instructions(input_data):
    instructions = []
    
    split_input = input_data.splitlines()
    
    for item in split_input:
        instruction = {}
        opcode_operand = item.split()
        instruction["opcode"] = opcode_operand[0]
        instruction["operand"] = int(opcode_operand[1])
        instructions.append(instruction)
        
    return instructions

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

>>> instructions = input_to_instructions(main_raw_input)
>>> print (instructions)

[{'opcode': 'acc', 'operand': 37},
 {'opcode': 'acc', 'operand': -4},
 {'opcode': 'nop', 'operand': 405},
 {'opcode': 'jmp', 'operand': 276},
 {'opcode': 'acc', 'operand': 39},

...

]

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

def accumulator_value_at_first_repeat(instructions):
    program_length = len(instructions)
    program_counter = 0
    accumulator = 0
    executed_instructions = set()
    repeat_instruction_encountered = False
    
    while 0 <= program_counter < program_length:
        current_instruction = instructions[program_counter]
        
        if program_counter in executed_instructions:
            repeat_instruction_encountered = True
            break
        else:
            executed_instructions.add(program_counter)
            
        if current_instruction["opcode"] == "jmp":
            program_counter += current_instruction["operand"]
            continue
        elif current_instruction["opcode"] == "acc":
            accumulator += current_instruction["operand"]
        elif current_instruction["opcode"] == "nop":
            pass
        else:
            print("Something went wrong in accumulator_value_at_first_repeated_instruction().")
            print(f"pc = {program_counter} acc = {accumulator}")
            print(f"instruction:\n{instruction}")
            
        program_counter += 1
    
    if repeat_instruction_encountered:
        print(f"The accumulator's contents at the first repeated instruction is: {accumulator}.")
    else:
        print("Reached end of instructions without repeating any of them.")

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…

>>> accumulator_value_at_first_repeat(instructions)

The accumulator's contents at the first repeated instruction is: 1801.

…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:

nop +0
acc +1
jmp +4
acc +3
jmp -3
acc -99
acc +1
jmp -4
acc +6

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:

nop +0  | 1
acc +1  | 2
jmp +4  | 3
acc +3  |
jmp -3  |
acc -99 |
acc +1  | 4
nop -4  | 5
acc +6  | 6

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:

def accumulator_value_and_halt_at_first_repeat(instructions, switch_opcode_address=-1):
    program_length = len(instructions)
    program_counter = 0
    accumulator = 0
    executed_instructions = set()
    
    while 0 <= program_counter < program_length:
        current_instruction = instructions[program_counter]
        
        if program_counter in executed_instructions:
            return {
                "halted": False,
                "program_counter": program_counter,
                "accumulator": accumulator
            }
        else:
            executed_instructions.add(program_counter)
            
        opcode = current_instruction["opcode"]
        operand = current_instruction["operand"]
            
        if program_counter == switch_opcode_address:
            print(f"Changing opcode at address {program_counter}")
            if opcode == "jmp":
                print("- Changing jmp to nop")
                opcode = "nop"
            elif opcode == "nop":
                print("- Changing nop to jmp")
                opcode = "jmp"
            
        if opcode == "jmp":
            program_counter += operand
            continue
        elif opcode == "acc":
            accumulator += operand
        elif opcode == "nop":
            pass
        else:
            print("Something went wrong in accumulator_value_at_first_repeated_instruction().")
            print(f"pc = {program_counter} acc = {accumulator}")
            print(f"instruction:\n{instruction}")
            
        program_counter += 1
            
    return {
                "halted": True,
                "program_counter": program_counter,
                "accumulator": accumulator
            }

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:

def run_instructions_with_mods(program):
    
    for index in range(len(program)):
        instruction = program[index]
        if instruction["opcode"] != "acc":
            print(f"Found jmp or nop at address: {index}")
            result = accumulator_value_and_halt_at_first_repeat(program, index)
            if result["halted"]:
                print(f"Found it! {result}")
                break
            else:
                print(f"Not the correct instruction.")

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:

>>> run_instructions_with_mods(instructions)

Found jmp or nop at address: 2
Changing opcode at address 2
- Changing nop to jmp
Not the correct instruction.
Found jmp or nop at address: 3
Changing opcode at address 3
- Changing jmp to nop
Not the correct instruction.
Found jmp or nop at address: 7
Not the correct instruction.
Found jmp or nop at address: 10
Changing opcode at address 10
- Changing jmp to nop
Not the correct instruction.

...


Found jmp or nop at address: 207
Changing opcode at address 207
- Changing jmp to nop
Not the correct instruction.
Found jmp or nop at address: 209
Changing opcode at address 209
- Changing jmp to nop
Not the correct instruction.
Found jmp or nop at address: 210
Changing opcode at address 210
- Changing jmp to nop
Found it! {'halted': True, 'program_counter': 623, 'accumulator': 2060}

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

Solutions for other days in Advent of Code 2020

• Day 1: Report Repair
• Day 2: Password Philosophy
• Day 3: Toboggan Trajectory
• Day 4: Passport Processing
• Day 5: Binary Boarding
• Day 6: Custom Customs
• Day 7: Handy Haversacks