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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.
Thinking about Precedence
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:
After evaluating every operator, we should get a list containing a single number.
Below is some code which does just that.
chain is an expression (list of numbers and operators)
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))))
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.
(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 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))
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.