# Chomsky normal form

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Chomsky normal form

In computer science, a context-free grammar is said to be in Chomsky normal form if all of its production rules are of the form: $A \rightarrow BC$ or $A \rightarrow \alpha$ or $S \rightarrow \varepsilon$

where A, B and C are nonterminal symbols, α is a terminal symbol (a symbol that represents a constant value), S is the start symbol, and ε is the empty string. Also, neither B nor C may be the start symbol.

Every grammar in Chomsky normal form is context-free, and conversely, every context-free grammar can be transformed into an equivalent one which is in Chomsky normal form. Several algorithms for performing such a transformation are known. Transformations are described in most textbooks on automata theory, such as (Hopcroft and Ullman, 1979). As pointed out by Lange and Leiß, the drawback of these transformations is that they can lead to an undesirable bloat in grammar size. Using | G | to denote the size of the original grammar G, the size blow-up in the worst case may range from | G | 2 to 22 | G | , depending on the transformation algorithm used (Lange and Leiß, 2009).

## Alternative definition

Another way to define Chomsky normal form (e.g., Hopcroft and Ullman 1979, and Hopcroft et al. 2006) is:

A formal grammar is in Chomsky reduced form if all of its production rules are of the form: $A \rightarrow\, BC$ or $A \rightarrow\, \alpha$

where A, B and C are nonterminal symbols, and α is a terminal symbol. When using this definition, B or C may be the start symbol.

## Converting a grammar to Chomsky Normal Form

1. Introduce S0
Introduce a new start variable, S0 and a new rule $S_0 \rightarrow S$ where S is the previous start variable.
2. Eliminate all $\epsilon$ rules $\epsilon$ rules are rules of the form $A \rightarrow \epsilon$ where $A \not= S_0$ and $A \in V$ where V is the CFG's variable alphabet.
Remove every rule with $\epsilon$ on its right hand side (RHS). For each rule with A in its RHS, add a set of new rules consisting of the different possible combinations of A replaced or not replaced with $\epsilon$. If a rule has A as a singleton on its RHS, add a new rule $R = A \rightarrow \epsilon$ unless R has already been removed through this process. For example, examine the following grammar G: $S \rightarrow AbA | B$ $B \rightarrow b | c$ $A \rightarrow \epsilon$
G has one epsilon rule. When the $A \rightarrow \epsilon$ is removed, we get the following: $S \rightarrow AbA | Ab | bA | b | B$ $B \rightarrow b | c$
Notice that we have to account for all possibilities of $A \rightarrow \epsilon$ and so we actually end up adding 3 rules.
3. Eliminate all unit rules $A \rightarrow B \ni A,B \in V$
After all the $\epsilon$ rules have been removed, you can begin removing unit rules, or rules whose RHS contains one variable and no terminals (which is inconsistent with CNF.)
To remove $A \rightarrow B$ $\forall B \rightarrow U$ add rule $A \rightarrow U$ unless this is a unit rule which has already been removed.
4. Clean up remaining rules that are not in Chomsky normal form.
Replace $A \rightarrow u_1,u_2, ... u_k, k \ge 3, u_1 \in V \cup \Sigma$ with $A \rightarrow u_1 A_1 , A_1 \rightarrow u_2 A_2 , ... , A_{k-2} \rightarrow u_{k-1} u_k$ where Ai are new variables.
If $u_i \in \Sigma$, replace ui in above rules with some new variable vi and add rule $v_i \rightarrow u_i$.

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