Question

Show that a language that is described by a regular expression can also be generated by a context-free grammar. As a hint, consider each mechanism, such as using a basic form or forming the regular expression re₁ ∪ re₂ from the expressions re₁ and re₂, and show how this can be transformed into grammar rules that can be added to the grammars generating the collection of strings denoted by re₁ and re₂.

Answer 1

Explanation:

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A Regular Expression can be recursively defined as follows −

• ε is a Regular Expression indicates the language containing an empty string. (L (ε) = {ε})
• φ is a Regular Expression denoting an empty language. (L (φ) = { })
• x is a Regular Expression where L = {x}
• If X is a Regular Expression denoting the language L(X) and Y is a Regular Expression denoting the language L(Y), then
• X + Y is a Regular Expression corresponding to the language L(X) ∪ L(Y) where L(X+Y) = L(X) ∪ L(Y).
• X . Y is a Regular Expression corresponding to the language L(X) . L(Y) where L(X.Y) = L(X) . L(Y)
• R* is a Regular Expression corresponding to the language L(R*)where L(R*) = (L(R))*
• If we apply any of the rules several times from 1 to 5, they are Regular Expressions

Some RE Examples:

Regular Expressions Regular Set

(0 + 10*) L = { 0, 1, 10, 100, 1000, 10000, ... }

(0*10*) L = {1, 01, 10, 010, 0010, ...}

(0 + ε)(1 + ε) L = {ε, 0, 1, 01}

(a+b)* Set of strings of a's and b's of any length including the null string.

So L = { ε, a, b, aa , ab , bb , ba, aaa.......}

(a+b)*abb Set of strings of a's and b's ending with the string abb.

So L = {abb, aabb, babb, aaabb, ababb, ..............}

(11)* Set consisting of even number of 1's including empty string,

So L= {ε, 11, 1111, 111111, ..........}

(aa)*(bb)*b Set of strings consisting of even number of a's followed by odd number of b's ,

so L = {b, aab, aabbb, aabbbbb, aaaab, aaaabbb, ..............}

(aa + ab + ba + bb)* String of a's and b's of even length can be obtained by concatenating any combination of the strings aa, ab, ba and bb including null, so L = {aa,ab, ba, bb, aaab, aaba, ..............}

Regular Sets

Any set that represents the value of the Regular Expression is called a Regular Set.

Properties of Regular Sets

Property 1. The union of two regular set is regular.

Proof −

Let us take two regular expressions

RE1 = a(aa)* and RE2 = (aa)*

So, L1 = {a, aaa, aaaaa,.....} (Strings of odd length excluding Null)

and L2 ={ ε, aa, aaaa, aaaaaa,.......} (Strings of even length including Null)

L1 ∪ L2 = { ε, a, aa, aaa, aaaa, aaaaa, aaaaaa,.......}

(Strings of all possible lengths including Null)

RE (L1 ∪ L2) = a* (which is a regular expression itself)

Hence, proved.

Property 2. The intersection of two regular set is regular.

Proof −

Let us take two regular expressions

RE1 = a(a*) and RE2 = (aa)*

So, L1 = { a,aa, aaa, aaaa, ....} (Strings of all possible lengths excluding Null)

L2 = { ε, aa, aaaa, aaaaaa,.......} (Strings of even length including Null)

L1 ∩ L2 = { aa, aaaa, aaaaaa,.......} (Strings of even length excluding Null)

RE (L1 ∩ L2) = aa(aa)* which is a regular expression itself.

Hence, proved.

Property 3. The complement of a regular set is regular.

Proof −

Let us take a regular expression −

RE = (aa)*

So, L = {ε, aa, aaaa, aaaaaa, .......} (Strings of even length including Null)

Complement of L is all the strings that is not in L.

So, L' = {a, aaa, aaaaa, .....} (Strings of odd length excluding Null)

RE (L') = a(aa)* which is a regular expression itself.

Hence, proved.

Property 4. The difference of two regular set is regular.

Proof −

Let us take two regular expressions −

RE1 = a (a*) and RE2 = (aa)*

So, L1 = {a, aa, aaa, aaaa, ....} (Strings of all possible lengths excluding Null)

L2 = { ε, aa, aaaa, aaaaaa,.......} (Strings of even length including Null)

L1 - L2 = {a, aaa, aaaaa, aaaaaaa, ....}

(Strings of all odd lengths excluding Null)

RE (L1 - L2) = a (aa)* which is a regular expression.

Hence, proved.

Property 5. The reversal of a regular set is regular.

Proof −

We have to prove LR is also regular if L is a regular set.

Let, L = {01, 10, 11, 10}

RE (L) = 01 + 10 + 11 + 10

LR = {10, 01, 11, 01}

RE (LR) = 01 + 10 + 11 + 10 which is regular

Hence, proved.

Property 6. The closure of a regular set is regular.

Proof −

If L = {a, aaa, aaaaa, .......} (Strings of odd length excluding Null)

i.e., RE (L) = a (aa)*

L* = {a, aa, aaa, aaaa , aaaaa,...............} (Strings of all lengths excluding Null)

RE (L*) = a (a)*

Hence, proved.

Property 7. The concatenation of two regular sets is regular.

Proof −

Let RE1 = (0+1)*0 and RE2 = 01(0+1)*

Here, L1 = {0, 00, 10, 000, 010, ......} (Set of strings ending in 0)

and L2 = {01, 010,011,.....} (Set of strings beginning with 01)

Then, L1 L2 = {001,0010,0011,0001,00010,00011,1001,10010,.............}

Set of strings containing 001 as a substring which can be represented by an RE − (0 + 1)*001(0 + 1)*

Hence, proved.

Identities Related to Regular Expressions

Given R, P, L, Q as regular expressions, the following identities hold −

• ∅* = ε
• ε* = ε
• RR* = R*R
• R*R* = R*
• (R*)* = R*
• RR* = R*R
• (PQ)*P =P(QP)*
• (a+b)* = (a*b*)* = (a*+b*)* = (a+b*)* = a*(ba*)*
• R + ∅ = ∅ + R = R (The identity for union)
• R ε = ε R = R (The identity for concatenation)
• ∅ L = L ∅ = ∅ (The annihilator for concatenation)
• R + R = R (Idempotent law)
• L (M + N) = LM + LN (Left distributive law)
• (M + N) L = ML + NL (Right distributive law)
• ε + RR* = ε + R*R = R*