This puzzle is to implement a little computer with three op codes, add, multiply, and finish. In the first part, I’m given two starting register values, 12 and 2. In the second part, I need to brute force those values to find a given target output.

## Challenge

The puzzle can be found here. I’m given a text file with a single line of numbers, comma separated. I’m to implement a computer that runs over the input. The first number (position 0) is the op code. For this computer, there are only three:

• 2 = multiply
• 99 = exit

Add and multiply each use the next three values. The first two are the registers to take as input, and the third is the register to write the result to.

For part 1, I’m given puzzle input, but instructed to overwrite positions 1 and 2 with the values 12 and 2, and then return the output. For part two, I’m given a target output, and asked to find initial values for positions 1 and 2 that will achieve that.

## Solution

### Part 1

For part 1, I’ll read in the data and keep it as a string. Originally I converted this to a list of ints right there, but for part 2, it’s easier to have the string to get a fresh computer. I’ll also create a function to do the compute, taking in a list of ints, and returning the value in position 0 once it reaches op code 99. The function just implements a while True loop, breaking on opcode 99, and doing the operations for the other two op codes.

#!/usr/bin/env python3

import sys

def compute(program):
eip = 0
while True:
if program[eip] == 1:
program[program[eip+3]] = program[program[eip+1]] + program[program[eip+2]]
elif program[eip] == 2:
program[program[eip+3]] = program[program[eip+1]] * program[program[eip+2]]
elif program[eip] == 99:
break
else:
print("Error")
sys.exit()
eip += 4

return program[0]

with open(sys.argv[1], 'r') as f:

program = list(map(int, program_str.split(',')))
program[1], program[2] = 12, 2
print(f"Part 1: {compute(program)}")


When I run this, I get the result:

$time ./day2.py 02-puzzle_input.txt Part 1: 4330636 real 0m0.027s user 0m0.023s sys 0m0.004s  I also notice that the output is very fast. This will be useful for brute forcing in the next part. ### Part 2 In part 2, I need to find a noun (initial position 1) and verb (initial position 2) that result in the value 19690720. Given that the program ran so fast in the prior run, I decided to first try just brute forcing all possible inputs between 0 and 100. for noun in range(100): for verb in range(100): program = list(map(int, program_str.split(','))) program[1], program[2] = noun, verb if compute(program) == 19690720: print(f"Part 2: {(100*noun)+verb}") sys.exit()  This does solve the challenge, and still pretty quickly: $ time ./day2.py 02-puzzle_input.txt
Part 1: 4330636
Part 2: 6086

real    0m0.140s
user    0m0.135s
sys     0m0.004s


When I start getting to later challenges, the compute time will likely be much longer, meaning I’ll have to think about better strategies than just trying 10,000 different inputs and looking for the correct one. In this case, I can see that if either input gets larger, the output can only get larger, since I’m just using addition and multiplication. That means I could do some smarter jumping up, and then falling back down when I get larger than the target. But no need to implement that here.

## Final Code

#!/usr/bin/env python3

import sys

def compute(program):
eip = 0
while True:
if program[eip] == 1:
program[program[eip+3]] = program[program[eip+1]] + program[program[eip+2]]
elif program[eip] == 2:
program[program[eip+3]] = program[program[eip+1]] * program[program[eip+2]]
elif program[eip] == 99:
break
else:
print("Error")
sys.exit()
eip += 4

return program[0]

with open(sys.argv[1], 'r') as f: