Regular Sets

Advertisements

Previous Page
Next Page  

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*
Advertisements