# Arithmetic expressions grammar transformation

In the article Parsing Expressions by Recursive Descent by Theodore Norvell (1999) the author starts with the following grammar for arithmetic expressions:

E --> E "+" E | E "-" E | "-" E | E "*" E | E "/" E | E "^" E | "(" E ")" | v


which is quite bad, because it's ambiguous and left-recursive. So he starts from removing the left recursion from it, and his result is as such:

E --> P {B P}
P --> v | "(" E ")" | U P
B --> "+" | "-" | "*" | "/" | "^"
U --> "-"


But I can't figure out how did he get to this result. When I try to remove the left recursion myself, I'm doing it the following way:

1. Firs, I group together the productions which doesn't have left recursion in one group, and other (left-recursive) in another group:

E --> E "+" E | E "-" E | E "*" E | E "/" E | E "^" E     // L-recursive
E --> v | "(" E ")" | "-" E
2. Next, I name them and factor for easier manipulations:

E --> E B E  // L-recursive; B stands for "Binary operator"
E --> P  // not L-recursive; P stands for "Primary Expression"
P --> v | "(" E ")" | U E   // U stands for "Unary operator"
B --> "+" | "-" | "*" | "/" | "^"
P --> "-"

Now I need to deal only with the first two productions, which are now easier to deal with.

3. I rewrite those first two productions by starting from the non-L-recursive production (which is simply P, the Primary expression) and following it by the optional Tail T, which I define as the rest of the original production less the first left-recursive nonterminal (that is, just B E) followed by the Tail T, or which could be empty:

E --> P T
T --> B E T |

(note the empty alternative for the tail).

4. These two productions I can now rewrite in EBNF like this:

E --> P {B E}

which is nearly what the author get, but I have E instead of P there inside the zero-or-more repetition pattern (the Tail). The other productions I get quite the same as he have got:

P --> v | "(" E ")" | U E
B -> "+" | "-" | "*" | "/" | "^"
U -> "-"

but here too I have E instead of P in the first production for P.

So, my question is: What am I missing? What algebraic transformation on the syntax I need to proceed now to get the same exact form as the autor gets? I tried substitutions for E, but it only leads me into loops. I suspect that I need somehow to substitute P for E, but I don't know any legal transformation to justify it. Maybe you know what's the last missing step?

• Please consider using LaTeX for formatting. See here for a primer. (See here for a discussion about suitability of LaTeX in this case.)
– Raphael
Aug 8 '12 at 6:41

The missing step:

E --> P T
T --> B E T |


rewrite E in T:

E --> P T
T --> B P T T |


Simplify T:

E --> P T
T --> B P T |


Equivalent to:

E --> P T
T --> {B P}


And there you are.

• Thanks for a good answer :-) Now I see what I've missed: I substituded it the other way around and that was the problem. But still I don't understand one little piece: How do you know that you can safely merge the Ts together into one T? Is there any rule for that? (I suspect it could be somehow similar to the rule in Boolean algebraic logic that says "aa = a".)
– SasQ
Aug 6 '12 at 14:40
• BTW why have this post been moved here from cstheory.sx and what's the difference? I'd like to know to avoid mistakes in a future.
– SasQ
Aug 6 '12 at 14:49
• @SasQ CSTheory is for research-level questions in theoretical computer science only, see the FAQ of CSTheory for details.
– Juho
Aug 6 '12 at 15:00
• @SasQ: $T \to xTT \mid \varepsilon$ generates $x^*$ and so does $T \to xT \mid \varepsilon$. More general, $L^*L^* = L^*$ for $L$ any language. Note that having the empty word $\varepsilon$ as right-hand side is crucial; this does not hold for all grammar fragments, as $L^+L^+ \neq L^+$.
– Raphael
Aug 8 '12 at 6:49
• @Raphael: Does that has something to do with the idempotence rule for *? I saw in "Dragon Book" (3.3, p.91) that x** = x*. Is that the same rule you've used?
– SasQ
Aug 8 '12 at 8:23