Forgive me for my ignorance as I am self-teaching myself some of this theory... I am having some trouble understanding how to systematically/algorithmically compute FOLLOW sets, given that I have computed the FIRST sets.

For the following grammar:

E -> TE'
E' -> +TE' | ε
T -> FT'
T' -> *FT' |  ε
F -> (E) | id

I found that the FIRST sets are:

FIRST(E) = { (, id }
FIRST(E') = { + ,  ε }
FIRST(T) = { (, id }
FIRST(T') = { *,  ε }
FIRST(F) = { (, id }

When it comes to computing the FOLLOW sets, I am trying to use the following rules:

1. $ goes into the set FOLLOW(S) where S is the start symbol.
2. If A -> aBb, then (FIRST(b) - { ε }) is in FOLLOW(B).
3. If A -> aB then everything in FOLLOW(A) is in FOLLOW(B)
4. If A -> aBb and ε is in FIRST(b), then everything in FOLLOW(A) is in FOLLOW(B).

Obviously, the step I am most certain here is that

FOLLOW(E) contains $, based on rule 1. But everything after that, I seem to have a hard time following a pattern on how to systematically compute the FOLLOW sets. For example, to compute FOLLOW(E), how are we supposed to approach this?

E -> TE', looks like A -> aB , so everything in FOLLOW(E) is in FOLLOW(E'). Where to go from here? Do I compute FOLLOW(E')? Then going to FOLLOW(E'):

E' -> + TE'... looks like A -> aBb. Then that means FIRST(E') - { ε } is in FOLLOW(T). So I do know that { + } is in FOLLOW(T).

As you can see, with the way I have interpreted this, I am going around all over the place computing the FOLLOW of another production. Ultimately, I end up losing track on what terminal symbols go into which follow set. :(

I am for sure misunderstanding something here.

What is the right approach/ordering to compute the FOLLOW sets?


1 Answer 1


The simple way to do it is the same as the simple way to compute FIRST sets, using a least fixed-point algorithm in which a set of inclusion rules are applied repeatedly until a fixed-point is reached:

  1. Initialise all the sets to empty.
  2. For every production in order, use the rules to add elements into the sets. (As a slight efficiency, rule 2 only needs to be applied on the first pass. I didn't apply that to the trace below.)
  3. If any set was modified during step 2, do it again.

So, here are the productions in order (I added the missing $*$ to $T'$, and inserted an augmented start rule $S$, which avoids the need for your special case rule 1):

$$\begin{align} S &\to E\; $ \\ E &\to TE' \\ E' &\to +TE' \mid \epsilon \\ T &\to FT' \\ T' &\to *FT' \mid \epsilon \\ F &\to (E) \mid id \end{align}$$

  1. Set all the $FOLLOW$ sets to $\{\}$
  2. First pass. I left out productions like $F\to id$ which don't match any of the rules. $$ \begin{array}{c|l|l} S \to E\; $ &\text{Add }\$\text{ to }FOLLOW(E) &FOLLOW(E) = \{ \$ \}\\ E \to TE' &\text{Add }FIRST(E')-\{\epsilon\}\text{ to }FOLLOW(T) &FOLLOW(T) = \{ + \}\\ &\text{Add }FOLLOW(E)\text{ to }FOLLOW(E') &FOLLOW(E') = \{ \$ \}\\ &\text{Add }FOLLOW(E)\text{ to }FOLLOW(T) &FOLLOW(T) = \{ + \$ \}\\ E' \to +TE' &\text{Add }FIRST(E')-\{\epsilon\}\text{ to }FOLLOW(T) &\text{no change}\\ &\text{Add }FOLLOW(E')\text{ to }FOLLOW(E') &\text{no change}\\ &\text{Add }FOLLOW(E')\text{ to }FOLLOW(T) &\text{no change}\\ T \to FT' &\text{Add }FIRST(T')-\{\epsilon\}\text{ to }FOLLOW(F) &FOLLOW(F) = \{ * \}\\ &\text{Add }FOLLOW(T)\text{ to }FOLLOW(T') &FOLLOW(T') = \{ + \$ \}\\ T' \to *FT' &\text{Add }FIRST(T')-\{\epsilon\}\text{ to }FOLLOW(F) &\text{no change}\\ &\text{Add }FOLLOW(T')\text{ to }FOLLOW(T') &\text{no change}\\ &\text{Add }FOLLOW(T')\text{ to }FOLLOW(F) &FOLLOW(F) = \{ * + \$\}\\ F \to (E) &\text{Add })\text{ to }FOLLOW(E) &FOLLOW(E) = \{ ) \$\}\\ \end{array} $$
  3. There were various changes, so we repeat step 2.
  4. (Step 2, second pass) $$\begin{array}{c|l|l} S \to E\; $ &\text{Add }\$\text{ to }FOLLOW(E) &\text{no change}\\ E \to TE' &\text{Add }FIRST(E')-\{\epsilon\}\text{ to }FOLLOW(T) &\text{no change}\\ &\text{Add }FOLLOW(E)\text{ to }FOLLOW(E') &FOLLOW(E') = \{ )\$ \}\\ &\text{Add }FOLLOW(E)\text{ to }FOLLOW(T) &FOLLOW(T) = \{ + )\$ \}\\ E' \to +TE' &\text{Add }FIRST(E')-\{\epsilon\}\text{ to }FOLLOW(T) &\text{no change}\\ &\text{Add }FOLLOW(E')\text{ to }FOLLOW(E') &\text{no change}\\ &\text{Add }FOLLOW(E')\text{ to }FOLLOW(T) &\text{no change}\\ T \to FT' &\text{Add }FIRST(T')-\{\epsilon\}\text{ to }FOLLOW(F) &\text{no change}\\ &\text{Add }FOLLOW(T)\text{ to }FOLLOW(T') &FOLLOW(T') = \{ + ) \$ \}\\ T' \to *FT' &\text{Add }FIRST(T')-\{\epsilon\}\text{ to }FOLLOW(F) &\text{no change}\\ &\text{Add }FOLLOW(T')\text{ to }FOLLOW(T') &\text{no change}\\ &\text{Add }FOLLOW(T')\text{ to }FOLLOW(F) &FOLLOW(F) = \{ * + ) \$\}\\ F \to (E) &\text{Add })\text{ to }FOLLOW(E) &\text{no change}\\ \end{array} $$
  5. (step 3, second pass) There were some more changes. Once again.
  6. (Step 2, third pass) $$\begin{array}{c|l|l} S \to E\; $ &\text{Add }\$\text{ to }FOLLOW(E) &\text{no change}\\ E \to TE' &\text{Add }FIRST(E')-\{\epsilon\}\text{ to }FOLLOW(T) &\text{no change}\\ &\text{Add }FOLLOW(E)\text{ to }FOLLOW(E') &\text{no change}\\ &\text{Add }FOLLOW(E)\text{ to }FOLLOW(T) &\text{no change}\\ E' \to +TE' &\text{Add }FIRST(E')-\{\epsilon\}\text{ to }FOLLOW(T) &\text{no change}\\ &\text{Add }FOLLOW(E')\text{ to }FOLLOW(E') &\text{no change}\\ &\text{Add }FOLLOW(E')\text{ to }FOLLOW(T) &\text{no change}\\ T \to FT' &\text{Add }FIRST(T')-\{\epsilon\}\text{ to }FOLLOW(F) &\text{no change}\\ &\text{Add }FOLLOW(T)\text{ to }FOLLOW(T') &\text{no change}\\ T' \to *FT' &\text{Add }FIRST(T')-\{\epsilon\}\text{ to }FOLLOW(F) &\text{no change}\\ &\text{Add }FOLLOW(T')\text{ to }FOLLOW(T') &\text{no change}\\ &\text{Add }FOLLOW(T')\text{ to }FOLLOW(F) &\text{no change}\\ F \to (E) &\text{Add })\text{ to }FOLLOW(E) &\text{no change}\\ \end{array} $$

So the $FOLLOW$ sets are: $$\begin{align} FOLLOW(S) &= \{\}\\ FOLLOW(E) &= \{ )$ \}\\ FOLLOW(E') &= \{ )$\}\\ FOLLOW(T) &= \{ + ) $\}\\ FOLLOW(T') &= \{ + ) $\}\\ FOLLOW(F) &= \{ * + ) $ \} \end{align} $$

That's a very simple grammar, of course. You might want to try it out with a more complicated one, although it is typical that it terminates after very few iterations and that many of the steps are obvious no-ops.

  • $\begingroup$ WOW! Thank you so much for this. I was able to work out the example with your explanation. The missing gap was that I had to iteratively perform the process, and had completely missed the detail about computing until no changes are seen. This is very helpful especially when you took the time to show each iteration like that. $\endgroup$ Commented Dec 13, 2018 at 0:36
  • $\begingroup$ @urbanspr1nter: That's the algorithm in the Dragon book, and it's a very simple and effective approach, although it's not the one I would recommend for a parser generator. (But, then, I wouldn't recommend writing an LL parser generator, either.) It's a useful technique, though, because least-fixed-point problems come up all the time, and not all of them can be solved simply with a transitive closure. $\endgroup$
    – rici
    Commented Dec 13, 2018 at 0:41
  • $\begingroup$ I don't have the "Dragon book". Although I have heard about it. Thinking about picking it up. I'm currently working through "Engineering a Compiler" and using a bunch of slides from the internet as a supplement. Learning curve is a bit high, but having more resources wouldn't hurt. :) Thanks. $\endgroup$ Commented Dec 13, 2018 at 0:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.