The monkeys that stole all our stuff awhile ago are back — and they are just standing around shouting. Well, at least they aren’t stealing stuff. A nearby elephant let me know that they saw I was lost and thought I should solve a really complicated algebra problem to get their help.
Well, sure!
The puzzle input is a symbol dictionary — like
root: vtsj + tfjf
mqjw: 5
qplt: tpfl * vmpn
dpsb: 2
It starts at root, and then you substitute that with vtsj + tfjf, and then substitute those terms with their expansions until you have a super long expression to solve. My first instinct was to use a stack parser, which is the usual way to do these things. My second instinct was just to make the whole thing into an expression that Python could evaluate, and just let Python worry about it. This is exactly what I used to solve Part 1 before work this morning.
I do occasionally peek in on Reddit while at work, and though I didn’t look at anyone’s code, I saw people mention several times that the sympy Python library could handle these complicated algebra problems easily.
So this is my moral dilemma: I was going to solve Part 2 (which introduces a variable for which we should solve) by just writing that stack parser, and just preserving the variable, and then simplifying the resulting equation by hand. I could also solve Part 1 this way, first, which would make it easier to port this to, say, Java, or other languages without such a large number of useful libraries.
But then, I thought I should really learn sympy, as it looks incredibly useful. And since I am writing up these AoC solutions, maybe someday in the future, someone will stumble upon this post and find out about it, too. Nobody would find my stack parser helpful — but maybe this would be good to know.
And so here we are. I rewrote Part 1 to use sympy, and then did Part 2 in it as well.
Python 3.11
From sympy, symbols is how you define symbols (x in this case), solve solved equations for a symbol, and simplify simplifies equations. Part 1 uses simplify, Part 2 uses solve.
There is a little bit of magic to set up Part 2, but this is one of the shortest, if not the shortest, solution, and I feel a little guilty that I let a library do the heavy lifting when I could have done it myself with a little more effort.
from sympy import symbols, solve, simplify
def expand(s):
if s.isnumeric() or s == 'x': return s
toks = s.split()
return '(' + expand(symbols_dict[toks[0]]) + ' ' + toks[1] + ' ' + expand(symbols_dict[toks[2]]) + ')'
with open(r"2022\puzzle21.txt") as f:
symbols_dict = {l[:4]: l[6:] for l in f.read().splitlines()}
print ("Part 1:", simplify(expand(symbols_dict['root'])))
symbols_dict['humn'] = 'x'
s = symbols_dict['root'].split()
symbols_dict['root'] = s[0] + ' - ' + s[2]
print ("Part 2:", solve(expand(symbols_dict['root']), symbols('x'))[0])
