# Advent of Code 2020 Day 18: Rethinking Operator Precedence

12/27/20

Part 2 of AOC Day 18 asks you to evaluate some math expressions, but with a twist. `+` has a greater operator precedence than `*`. (This is opposite of how we typically read math):

``````2 * 3 + (4 * 5)
2 * 3 + 20
2 * 23
46
``````

The standard method for expression evaluation is the shunting yard algorithm. This is the right and efficient way to solve this problem. But, it’s also not very obvious or natural. It’s hard to imagine how you would think if it yourself, and doesn’t really match how people actually see mathematical expressions.

I want to show off an alternative solution demonstrating how you could think of it yourself. My code is written in Common Lisp. For this particular problem this choice goes a bit deeper than just syntax as the solution makes heavy use of a symbolic expression manipulation which is harder to do in most languages.

What does operator precedence actually mean? It tells us the order in which to evaluate operators. So naively we should just do all the additions first, and then the multiplications. Can we program that? We certainly can, but doing just one operation at a time, means at each step we produce an intermediate result which is itself a math expression, just simplified a bit. For example given the expression:

`````` (1 + 2 * 3 + 4)
``````

We first evaluate all the `+` to produce:

``````(3 * 7)
``````

And then we evaluate all the ‘*’ to produce:

``````(21)
``````

After evaluating every operator, we should get a list containing a single number.

Below is some code which does just that. The input `chain` is an expression (list of numbers and operators) and `allowed` is the operator we want to evaluate.

``````(defun simplify-chain (chain allowed)
(prog ((result '())
(op nil)
(l nil)
(r nil))

(setf l (car chain))
(setf chain (cdr chain))

loop
(setf op (car chain))
(setf chain (cdr chain))

(setf r (car chain))
(setf chain (cdr chain))

(if (eq op allowed)

; evaluate the operation.
; put the result back in the expression

(setf l (funcall op l r))

; not an operation we want to evaluate right now.
; put it's symbols back in an the expression
(progn (push l result)
(push op result)
(setf l r)))

(if (consp chain)
(go loop))

(push l result)
(return (reverse result))))
``````

Handling Parenthesis

That is quite straightforward! All the trickiness is fiddling with infix. But, there is one problem. Due to parenthesis the operands may not be “ready” to evaluate. If we have:

``````3 + (2 * 4)
``````

We can’t apply the `+` operation until the `(2 * 4)` is taken care of. But evaluating `(2 * 4)` is something we could do directly. There are only finitely many nested statements, so eventually following parenthesis must end with a list containing only numbers and operators. This is clearly a recursive problem! We need to evaluate all the parenthesized statements in a list, before we can evaluate the list itself.

The algorithm for evaluation looks like the following:

• To evaluate a number or operator symbol, just return it. (base case)
• To evaluate a list:

1. evaluate every entry in the list. (take care of parenthesis)
2. simplify the expression by applying each operator in order. (simplify chain)
3. Return the only item left in the list.

Let’s write an eval function which does this, building upon simplify chain:

``````(defun eval-expr (expr)
(if (listp expr)
(car (reduce #'simplify-chain
'(+ *)
:initial-value (mapcar #'eval-expr expr)))
expr))
``````

To summarize `simplify-chain` takes mathematical expressions and returns expressions that are simpler, while `eval-expr` always returns a number. Note that this works for any number of operators and adjusting precedence is as simple as changing the order of the `'(+ *)` list.