Presentation on theme: "SIMPLIFYING GRAMMARS Definition: A useless symbol of a context-free grammar is one which does not occur in the derivation of any sentence of that grammar."— Presentation transcript:
SIMPLIFYING GRAMMARS Definition: A useless symbol of a context-free grammar is one which does not occur in the derivation of any sentence of that grammar. For example:G→ RT R→ Ra T→b HereR is useless.
Clearly a symbol is useless if and only if either: a)we cannot derive any string containing it from the goal symbol and/or b)we cannot derive a terminal string from that symbol Notation: a) is expressed by saying that the symbol is not reachable from the goal symbol. b) is expressed by saying that the symbol does not derive a terminal string
Algorithm: To find all those symbols that are not reachable from the goal symbol. 1) Make a list of all the grammar symbols, all initially unflagged. 2) Flag the goal symbol. 3) Go through the grammar from the 1 st production to the last. If A→ x 1 x 2 …x n is one of these productions and A is flagged, then flag x 1,x 2,…,x n (those ones not already flagged). 4) Where any new symbols flagged during the iteration of step 3? If so, repeat step 3 again, otherwise stop. Any symbol that has not been flagged at this stage is not reachable from the goal symbol.
EXAMPLE Grammar 1 z → b e a → a e | e b → c e | a f c → c f d → f d is not reachable f √ f c √ c d is not reachable
Grammar 2 Z → E + T E → E | S + F | T F → F | F P | P P → G G → G | G G | F T→ T * i | i Q → E | E + F | T | S S → i Q is not reachable Q is not reachable
Grammar 3 G → A Q → P R P→Q Q, P, R are not reachable
Algorithm: To determine which symbols do not derive a terminal string. 1) Make a fresh list of all the symbols, initially unflagged. 2) Flag all the terminals. 3) Go through the grammar from the first production to the last. If A→ x 1 x 2 …x n is any such production, then if x 1, x 2, …,x n are all flagged, flag A. 4) Were any new symbols flagged in step 3? If so, go back to step 3. If not, all symbols not flagged at this stage do not derive a terminal string.
TRY THE ABOVE ALGORITHM ON GRAMMARS 1- 3 ABOVE
Definitions: 1) A α means you can derive α from A or α=A 2) A symbol A is said to vanish if A ε 3) A production of the form χ→ε is called an ε-production Note that the textbook uses λ to denote the empty string, whereas these slides employ ε for this purpose
Algorithm: To determine which symbols of a grammar vanish. 1) Make a list of symbols, initially unflagged. 2) Flag all the left hand sides of ε-productions. 3) Go through the grammar from 1 st production to last. If A→ x 1 x 2 …x n is any such production, then if x 1, x 2, …,x n are all flagged, then flag A. 4)Were any new symbols flagged in step 3? If so, go back to step 3, else stop. The flagged symbols are those which vanish.
Example: Try the algorithm on the following Grammar. Grammar G4 A → b Y D | A Y c Y → E F | ε D → g h i F → N O | Y N N → ε O → Y N E → Y O N Y
Defns: An - production is one of the form A -> . If A, in this case, is the goal symbol, the production is referred to as a null goal production Theorem: For every cfg G, there exists a cfg G’, such that L(G’) = L(G), and G’ has no -productions with exception that if L(G), then G’ contains a null goal production.
Proof. G’ can be formed from G as follows: 1. Discard all the -productions. 2. For each production of G, add to the grammar all possible productions that can be formed from it by omitting from its rhs some subset of those symbols (if any) that vanish.. 3. Remove all productions with useless symbols. 4. If the goal symbol of G vanishes, add a null goal production.
Example 1 G -> AVw A -> aA | a V -> rUcW | U -> W -> First of all, determine which symbols vanish: U, V, W.
1) Remove -productions, gives: G -> AVw A -> aA | a V -> rUcW 2) Considering G -> AVw in step 2 of the algorithm, we add to the grammar G -> Aw Considering V -> rUcW, we add V -> rc V -> rUc V -> rcW 3) W, U are now useless symbols, so leaving out all productions with W, U, we get: G -> AVw | Aw A -> aA | a V -> rc
EXAMPLE. Provide a grammar equivalent to the one below but without ε-productions S → ABaC A → BC B → b | ε C → D | ε D → d Try working this out for yourself, before consulting the answer on the next slide. Note carefully that the symbol A is one of those that vanishes.
ANSWER S → ABaC | ABa | AaC | Aa | BaC |Ba | aC | a A → BC | B | C B → b C → D D → d
Defn. A unit production of a grammar is one of the form A -> B where A, B are both non-terminals. Theorem.For any context-free grammar G, a cfg G’ s.t. L(G’) = L(G) and G’ does not contain any unit productions.
Proof. G’ can be formed from G as follows 1.Eliminate -productions from G to form G* (with possibly a null goal symbol) 2.If A is the left hand side of a unit production and B is any symbol that can be derived from A, and B -> is any production with B as left hand side where is not a single non-terminal, then add to grammar A -> . By step 1, any derivation of B from A must consist entirely of a sequence of non-terminals. Do step 2 for all symbols which are the left hand side of a unit production
To find all single symbols that can be derived from a symbol A, consider the derivation tree in which no symbol occurs more than once, e.g.: A BDE C F N M If say M B, we do not include it, as B already occurs in the tree. Hence the depth of the tree is < = the number of unit productions
3. Now discard all unit productions
EXAMPLE Consider the grammar: E → E + T | T T → T * F | F F → ( E ) | a Since E => T and T → T * F, we add to the grammar E → T * F and since E => F and F → ( E ) | a, we add E → ( E ) and E → a Also since T => F, we add T → ( E ) | a
Discarding all unit productions, then gives us: E → E + T | T * F | ( E ) | a T → T * F | ( E ) | a F → ( E ) | a
EXAMPLE 3. “Remove” unit productions from: S → Aa | B B → A | bb A → a | bc | B ANSWER S → Aa | bb | a | bc since S => B and S => A B → bb | a | bc since B => A A → a | bc | bb since A => B But B is a useless symbol, so discard the production involving B
EXAMPLE 4. “Remove” unit productions from S → Aa | bb | a | bc | B B → bb | a | bc A → a | bc | bb
ANSWER S → Aa | bb | bc | a B → bb | a | bc A → a | bc | bb Again, B is a useless symbol, and so the productions involving it should be discarded
Defn. A nice context free grammar is one: a) without useless symbols, b) without -production except possible for a null goal production, and c) without unit productions Notation. cfg stands for context free grammar, and ncfg stands for nice context free grammar Corollary. For every cfg G, a ncfg G ’, such that L(G ’ ) = L(G).