Leftmost and Rightmost Derivations

Now consider the grammar G = ({S, A, B, C}, {a, b, c}, S, P) where:

S ABC

A aA | λ

B bB | λ

C cC | λ

With this grammar, there is a choice of variables to expand. Here is a sample derivation:

S ABC aABC aABcC aBcC abBcC abBc abbBc abbc

If we always expanded the leftmost variable first, we would have a leftmost derivation:

S ABC aABC aBC abBC abbBC abbC abbcC abbc

Conversely, if we always expanded the rightmost variable first, we would have a rightmost derivation:

S ABC ABcC ABc AbBc AbbBc Abbc aAbbc abbc

There are two things to notice here:

1. Different derivations result in quite different sentential forms, but 2. For a context-free grammar, it really doesn't make much difference in what order we expand

the variables.

Chomsky normal form

Transform the grammar with productions

S abAB

A baB|λ

B Baa|A|λ

into Chomsky normal form.

Solution:

1. Removing λ-productions: Here A and B are nullable variables

S abAB|abA|abB|ab

A baB|baB Baa|A|Ba|Aa|a

2. Removing unit-productionsS abAB|abA|abB|abA baB|baB Baa|baB|ba|Ba|Aa|a

There are no useless productions in the grammar.3. Introduce variables C and D to substitute terminals a and b.

S CDAB|CDA|CDB|CDA DCB|DCB BAC|DCB|DC|BC|AC|aC aD b

4. Introduce variables to shorten the right sides of the production.S EF|EA|EB|CDA GB|DCB BH|GB|DC|BC|AC|aE CDF ABG DCH ACC aD b

Now the grammar is in CNF.

Show that the following grammar is ambiguous, by producing one string in the language that hastwo different parse trees:

S aS | aSbS | λ

Solution:Consider the string aaba and the two parse trees in the figure below:

S

a S

a S b S

λ a S

λ

S

a S b S

a S a S

λ λ

Consider the grammar:

S aS | aSbS | λ

Show that the grammar is ambiguous, by giving two parse trees for the

string aab.

Solution:

S -> aSbS -> aaSbS -> aabS -> aabS -> aS -> aaSbS -> aabS -> aab

Construct PDAs that accept each of the following language.a) {aibj | 0 ≤ i ≤ j}

aλ|A bA|λ , bλ| λ

λλ|λ q0

q1