Parenthesize expressions.

Question

Can someone tell me how to add parenthesis to this expression

$a + b \cdot c \cdot d + e$.

where $a,b,c,d,e$ are algebraic variables, where addition and multiplication is as usual there. :-)

If you have any other complicated examples please explain them to me and if possible, mention the core concept which would enable me to parenthesize any complex expression.

The tag in which the question is added here maybe misleading,I could find an appropriate tag. Please explain in the context of basic math. I am mainly going to use these expressions to construct binary trees. — qwertynik, Jul 04 '14 at 12:45
@mvw I think the OP is using the word "complex" in the sense of "complicated", not the mathematical connotation of some elements $a, b, c, d \in \mathbb C$. — amWhy, Jul 04 '14 at 13:01
Soon we'll have a tag for parsley as well; after which we will have a tag for parley. — Asaf Karagila, Jul 04 '14 at 13:10

score 2 · Accepted Answer · answered Jul 04 '14 at 12:48

2

The operator $"+"$ has a lower priority than the multiplication operator, so you first do the multiplication and the the sum, and the experssion in the parenthesis has high priority so you have to write in the parenthesis the expression with the multiplication operator: $$a+(b \cdot c \cdot d) +e$$ or $$a+((b \cdot c) \cdot d)+e$$

answered Jul 04 '14 at 12:48

Mary Star

14,186
15
91
205

1

operator precedence – mvw Jul 04 '14 at 12:49

gnometorule · Answer 2 · 2014-07-04T13:08:24.427

1

If @mvw 's comment is correct, then, assuming LR, you'll parse this as $$((a+( ( b \cdot c) \cdot d) ) + e)$$ And a proper tag, if it existed, would be "recursive descent parsing" because this isn't really RegEx analysis.

If this is just standalone evaluation of simple expressions like this, you're fine using LR and top-down descent. If it is part of a larger parser, you might want to consider flattening your tree by using an operator precedence table and Dijkstra's shunting algorithm for binary expression parsing instead. The wiki page has a detailed example. For easy cases, the algorithm is straightforward; handling assignment operations ("=") and such takes a bit more thought as you must flip parsing direction.

edited Jul 04 '14 at 13:08

answered Jul 04 '14 at 12:53

gnometorule

4,670

I am going to convert this expression into a binary expression tree. Anyways thanks for your help – qwertynik Jul 04 '14 at 13:09
2

A discussion where Dijkstra shows up is always a delight. – mvw Jul 04 '14 at 13:11
@bonulti: I thought so, but the parentheses above express this (hard to "write a tree"): if (op, LHS, RHS) are the operation at the node and the children, then (+, (+, a, ((mult, (mult, b, c), d)), e). And the algorithm creates just this sub-tree. – gnometorule Jul 04 '14 at 13:15

score 0 · Answer 3 · answered Jul 04 '14 at 12:48

0

(a+b)⋅c⋅(d+e)

or

a+(b⋅c⋅d)+e

or

(a+b⋅c)⋅d+e

or

a+b⋅(c⋅d+e)

?

answered Jul 04 '14 at 12:48

BCLC

14,197

Parenthesize expressions.

3 Answers3