COS3701 Assignment 3
(COMPLETE ANSWERS)
2025 - DUE 2025
For assistance contact
Email:
, Question 1: Grammars for Concatenated Languages
1. Grammars for L1=(aa)∗ and L2=(a+b)∗ab(a+b)∗Error! Filename not specified.
A grammar for a regular language is a regular grammar, also known as a right-linear grammar. It
has productions of the form A→wB or A→w, where A and B are non-terminals and w is a string
of terminals.
Grammar for L1=(aa)∗Error! Filename not specified.
The language L1 consists of strings with an even number of 'a's, including the empty string (Λ).
Start symbol: S1Error! Filename not specified.
Terminals: {a}Error! Filename not specified.
Productions: S1→aaS1 S1→ΛError! Filename not specified.
This grammar correctly generates all strings in L1. The first production allows the generation of
pairs of 'a's, and the second production allows the derivation to terminate, including the case
where no 'a's are generated (the empty string).
Grammar for L2=(a+b)∗ab(a+b)∗Error! Filename not specified.
The language L2 consists of all strings over the alphabet {a,b} that contain the substring "ab".
Start symbol: S2Error! Filename not specified.
Terminals: {a,b}Error! Filename not specified.
Non-terminals: {S2,A}Error! Filename not specified.
Productions: S2→aS2∣bS2∣abA A→aA∣bA∣ΛError! Filename not specified.
The productions for S2 first generate any arbitrary string of 'a's and 'b's (the (a+b)∗ part) before
moving to the core substring "ab". Once "ab" is generated, the non-terminal A takes over to
generate the final arbitrary string of 'a's and 'b's (the second (a+b)∗ part). The production A→Λ
allows this final part to be the empty string.
2. Using Theorem 37 to find the grammar for L1L2Error! Filename not specified.
Theorem 37 states that if a grammar G1 generates L1 and a grammar G2 generates L2, a
grammar for the concatenated language L1L2 can be constructed. A common formulation of this
theorem for regular languages involves modifying the productions of G1 that contain the empty
string (Λ).
The procedure is as follows:
1. Take all the productions of G1 and G2.
2. If the start symbol of G1, S1, has a production S1→Λ, replace this production with S1
→S2, where S2 is the start symbol of G2.
(COMPLETE ANSWERS)
2025 - DUE 2025
For assistance contact
Email:
, Question 1: Grammars for Concatenated Languages
1. Grammars for L1=(aa)∗ and L2=(a+b)∗ab(a+b)∗Error! Filename not specified.
A grammar for a regular language is a regular grammar, also known as a right-linear grammar. It
has productions of the form A→wB or A→w, where A and B are non-terminals and w is a string
of terminals.
Grammar for L1=(aa)∗Error! Filename not specified.
The language L1 consists of strings with an even number of 'a's, including the empty string (Λ).
Start symbol: S1Error! Filename not specified.
Terminals: {a}Error! Filename not specified.
Productions: S1→aaS1 S1→ΛError! Filename not specified.
This grammar correctly generates all strings in L1. The first production allows the generation of
pairs of 'a's, and the second production allows the derivation to terminate, including the case
where no 'a's are generated (the empty string).
Grammar for L2=(a+b)∗ab(a+b)∗Error! Filename not specified.
The language L2 consists of all strings over the alphabet {a,b} that contain the substring "ab".
Start symbol: S2Error! Filename not specified.
Terminals: {a,b}Error! Filename not specified.
Non-terminals: {S2,A}Error! Filename not specified.
Productions: S2→aS2∣bS2∣abA A→aA∣bA∣ΛError! Filename not specified.
The productions for S2 first generate any arbitrary string of 'a's and 'b's (the (a+b)∗ part) before
moving to the core substring "ab". Once "ab" is generated, the non-terminal A takes over to
generate the final arbitrary string of 'a's and 'b's (the second (a+b)∗ part). The production A→Λ
allows this final part to be the empty string.
2. Using Theorem 37 to find the grammar for L1L2Error! Filename not specified.
Theorem 37 states that if a grammar G1 generates L1 and a grammar G2 generates L2, a
grammar for the concatenated language L1L2 can be constructed. A common formulation of this
theorem for regular languages involves modifying the productions of G1 that contain the empty
string (Λ).
The procedure is as follows:
1. Take all the productions of G1 and G2.
2. If the start symbol of G1, S1, has a production S1→Λ, replace this production with S1
→S2, where S2 is the start symbol of G2.
