It is possible in a CFG that the derivation of strings does not require the use of all the production rules and symbols. Additionally, there might be some unit and null productions. The term “simplification of CFGs” refers to the removal of certain productions and symbols. Essentially, simplification consists of the following steps:
- CFG Reduction
- Unit Productions Removal
- Null Productions Removal
In this article, we will look more into the Simplification of CFG according to the GATE Syllabus for (Computer Science Engineering) CSE. We will read ahead to find out more about it.
Table of Contents
- What is the Simplification of CFG?
- Removal of Useless Symbols
- Elimination of ε Production
- Removing Unit Productions
What is the Simplification of CFG?
As we’ve seen, context-free grammar can effectively represent a variety of languages. Because not all grammar is streamlined, some grammars may contain superfluous symbols (non-terminal). Adding extra pointless symbols lengthens the grammar. The grammar that has been reduced by the removal of symbols is said to be simpler. Below are some reduced characteristics of grammar:
- A word in L is derived from each variable (i.e., non-terminal) and every terminal in G.
- Where X and Y are not terminal, there should be no production as X → Y.
- There need not be a production of X → ε if it is not in the L language.
Let us study the reduction process in detail.
Removal of Useless Symbols
If a symbol does not exist on the RHS of the production rule and does not participate in the derivation of any string, it may be considered to be useless. Similar to this, if a variable is not used to create any strings, it may be worthless.
Example:
T → xxY | xbX | xxT
X → xX
Y → xy | y
Z → xz
In the example given above, the variable ‘Z’ will never occur in any string’s derivation. Thus, the production Z → xz is useless. Therefore, let’s eliminate it. This way, all the other productions would be written in a way that the variable Z can never reach from the ‘T’ starting variable.
Production X → xX would also be useless because it cannot terminate it in any way. But if it never terminates, it can never ever produce a string. Thus, such a production can not participate in any derivation ever.
For the removal of this useless production X → xX, we will find all the variables first that never lead to a terminal string, like the variable ‘X’. We will then remove all the productions where the variable ‘B’ occurs.
Elimination of ε Production
All the productions that are of type S → ε are known as ε productions. These types of productions can be removed only from the grammars that do not happen to generate ε.
Step 1: Find all nullable non-terminal variables which derive ε first.
Step 2: For all the productions of X → a, construct all production X → a, where a is obtained from x by removing non-terminals from the first step.
Step 3: Combine the result of the second step with the original production. Remove all the ε productions.
Example:
Let us remove the production from the CFG given below by preserving its meaning.
S → ABA
A → 0A | ε
B → 1B | ε
Solution:
While removing ε production, the rule A → ε and B → ε would be deleted. To preserve the CFG’s meaning, we are placing ε actually at the RHS whenever A and B have appeared.
Let us take
S → ABA
If the first A at RHS is ε. Then,
S → BA
Similarly, if the last A in RHS = ε. Then,
S → AB
If B = ε, then
S → AA
If B and A are ε, then
S → A
If both A is replaced by ε, then
S → B
Now,
S → AB | BA | AA | A | B
Now, let us consider
A → 0A
If we place ε at RHS for A, then
A → 0
A → 0A | 0
Similarly, B → 1B | 1
We can rewrite the CFG collectively with removed ε production as
S → AB | BA | AA | A | B
A → 0A | 0
B → 1B | 1
Removing Unit Productions
The productions that result in the giving of one non-terminal to another are known as unit productions. To get rid of unit production, take the following actions:
Step 1: To remove A → B, add production A → x to the grammar rule whenever B → x occurs in the grammar.
Step 2: Delete A → B from the grammar.
Step 3: Repeat the first and the second steps until all the unit productions are removed.
Example:
S → 0X | 1Y | Z
X → 0S | 00
Y → 1 | X
Z → 01
Solution:
S → Z is a unit production. However, while removing S → Z, we have to consider what Z gives. So, we can add a rule to S.
S → 0X | 1Y | 01
Similarly, Y → X is also a unit production, so we can modify it as
B → 1 | 0S | 00
Thus, we can write CFG finally without the unit production, as follows:
S → 0X | 1Y | 01
X → 0S | 00
Y → 1 | 0S | 00
Z → 01
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