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

Here R 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 1st production to the last.

If A→ x1x2…xn is one of these productions and A is

flagged, then flag x1,x2,…,xn (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→ x1x2…xn is any such production,

then if x1,x2,…,xn 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 1st production to

last. If A→ x1x2…xn is any such production, then

if x1,x2,…,xn 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 equivalentto the one below but without ε-productionsS → ABaCA → BCB → b | εC → D | εD → 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

 B D E

  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

EXAMPLEConsider 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 | BB → A | bbA → a | bc | B

ANSWERS → Aa | bb | a | bc since S => B and S => AB → 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).