Feedback on AoC Solution

[umma...@gmail.com]

Hello!

It's been a while since my first message last October. I was going to probably be writing more as I planned to do Advent of Code 2020 in Pure but decided to use C to destress with some nostalgia programming instead, and that's been great.

A conversation with coworkers, however, on how they solved Day 18 made me realize it would be neat to try and use Pure for a second time (the problem involves changing the precedence of arithmetic among + and * -- my colleague used JavaScript's .replace method of string objects).

Here's my Pure solution to the problem. Part one is to just have + and * have equal precedence but with left associativity. Part two is to have + bind tighter than *. If you have a few moments, could you please tell me how this could be better in a Pure idiomatic / style sense. Thanks!

using system, regex ;

leftFirst = reduce with a + b * c = (a + b) * c end ;

addFirst = reduce with

  a * b + c = prod(a, b + c) ;

  a + b * c = dorp(a + b, c) ;

  prod(a, b) + c = prod(a, b + c) ;

  a + dorp(b, c) = dorp(a + b, c) ;

  dorp(a, b) = prod(a, b) ;

  'prod(a, b) = prod(a, b) ;

  prod(prod(a, b), c) = prod(a * b, c) ;

  prod(a, prod(b, c)) = prod(a, b * c) ;

  prod(a, b) = a * b ;

end ;

fixed f a = loop a (f a) with

  loop b bb = b if same b bb ;

            = loop bb (f bb) otherwise ;

end ;

quoted s = regsub (cst "'(") "(" 0 s 0 ; // can these two be done more idiomatically?

toLong s = regsub (\s -> s!1 + "L") "[[:digit:]][[:digit:]]*" 0 s 0 ;

let input = map val . filter ("" ~=) . split "\n" . toLong . quoted . fget $ fopen "input.raw" "r" ;

run r = init . str . foldl (+) 0L . map (fixed eval . r) $ input ;

printf "Part 1: %s\nPart 2: %s\n" (run leftFirst, run addFirst) ;

[umma...@gmail.com]

Additionally, the "input.raw" file has lines of the following form (just simple expressions):

9 + 4 + (3 + (6 * 5 + 9 * 2 * 4) * 9)

((7 + 6) + 2 * 6 + 4 * 2 + 9) * 6 + 5

7 * 3 * 9 * 6 + (4 + 2 + 4 * 9) + 7

9 + (9 * 3 * 3 * (9 + 5 + 4) + 4)

(8 * 5 + (7 + 5 + 3 + 6) * 9 + 7 * (9 * 8 + 7 * 2)) * 8

3 + (6 * 2 * (9 + 2 + 5) * (9 * 2 * 6 + 8 * 5) * (2 + 2 + 3 + 7 * 9)) + 4 * 4 + (6 * 4 * (9 + 4) + 8 * 3 * (5 + 3)) * 5

2 * 8 * (8 * 2 * 4) + (9 * 3 * 6) * 9

2 + (8 * 4 * 9 * 3 + 8 * 2) + 4 * 9 * 6

((6 * 4 + 9 + 5 * 7 + 3) * 4 + 8 * (3 + 6 + 9 * 9) + 9 * 6) + (7 * 7) + 6 * 2 * 2

2 + 5 * 3

And, as shown in the code, answer is given as the sum of the evaluation of all lines.

[umma...@gmail.com]

I simplified the reduce a lot and have the following. Just updating so that no one unnecessarily types out a long email.

Happy Holidays.

using system, regex ;

leftFirst = reduce with a + b * c = (a + b) * c end ;
addFirst = reduce with
  a * b + c = prod(a, b + c) ;

  a + b * c = prod(a + b, c) ;