CS250: Discrete Math for Computer SciencePumping Lemma for Regular Sets: Let D = (Q; ; ;q0;F) be a DFA, n = jQj w 2L(D) s.t. jwj n Then 9x;y;z 2? s.t. 1. xyz = w 2. jxyj n 3. jyj>0,

CS250: Discrete Math for Computer Science

L31: Proving Languages are Not Recognized by any DFA

Which Languages are Recognized by DFAs?

So far we have seen that the following languages have DFAsthat recognize them and regular expressions that denote them:

{w ∈ {a,b}?

∣∣ #b(w) ≡ 1 (mod 2)}

L(a∗ba∗(ba∗ba∗)∗)

{w ∈ {0,1}∗

∣∣ next to last symbol is 1}

L((0|1)∗1(0|1))

{w ∈ {a,b}?

∣∣∣∣ #a(w) ≡ 1 (mod 2) ∧#b(w) ≡ 0 (mod 3)

}?

{w ∈ {0,1}∗

∣∣ w has at least two 0’s}

L(1∗01∗0(0|1)∗)

{w ∈ {0,1}∗

∣∣ w has 001 or 100}

L((0|1)∗(001|100)(0|1)∗)


So far we have seen that the following languages have DFAsthat recognize them and regular expressions that denote them:{

w ∈ {a,b}?∣∣ #b(w) ≡ 1 (mod 2)

}L(a∗ba∗(ba∗ba∗)∗)

{w ∈ {0,1}∗


L((0|1)∗1(0|1))

{w ∈ {a,b}?

∣∣∣∣ #a(w) ≡ 1 (mod 2) ∧#b(w) ≡ 0 (mod 3)

}?

{w ∈ {0,1}∗


L(1∗01∗0(0|1)∗)

{w ∈ {0,1}∗

∣∣ w has 001 or 100}

L((0|1)∗(001|100)(0|1)∗)



w ∈ {a,b}?∣∣ #b(w) ≡ 1 (mod 2)


{w ∈ {0,1}∗


L((0|1)∗1(0|1))

{w ∈ {a,b}?

∣∣∣∣ #a(w) ≡ 1 (mod 2) ∧#b(w) ≡ 0 (mod 3)

}?

{w ∈ {0,1}∗


L(1∗01∗0(0|1)∗)

{w ∈ {0,1}∗

∣∣ w has 001 or 100}

L((0|1)∗(001|100)(0|1)∗)



w ∈ {a,b}?∣∣ #b(w) ≡ 1 (mod 2)


{w ∈ {0,1}∗


L((0|1)∗1(0|1))

{w ∈ {a,b}?

∣∣∣∣ #a(w) ≡ 1 (mod 2) ∧#b(w) ≡ 0 (mod 3)

}?

{w ∈ {0,1}∗


L(1∗01∗0(0|1)∗)

{w ∈ {0,1}∗

∣∣ w has 001 or 100}

L((0|1)∗(001|100)(0|1)∗)



w ∈ {a,b}?∣∣ #b(w) ≡ 1 (mod 2)


{w ∈ {0,1}∗


L((0|1)∗1(0|1))

{w ∈ {a,b}?

∣∣∣∣ #a(w) ≡ 1 (mod 2) ∧#b(w) ≡ 0 (mod 3)

}?

{w ∈ {0,1}∗


L(1∗01∗0(0|1)∗)

{w ∈ {0,1}∗

∣∣ w has 001 or 100}

L((0|1)∗(001|100)(0|1)∗)



w ∈ {a,b}?∣∣ #b(w) ≡ 1 (mod 2)


{w ∈ {0,1}∗


L((0|1)∗1(0|1))

{w ∈ {a,b}?

∣∣∣∣ #a(w) ≡ 1 (mod 2) ∧#b(w) ≡ 0 (mod 3)

}?

{w ∈ {0,1}∗


L(1∗01∗0(0|1)∗)

{w ∈ {0,1}∗

∣∣ w has 001 or 100}

L((0|1)∗(001|100)(0|1)∗)

{w ∈ {0,1}∗


= L(1∗01∗0(0|1)∗)

Build a DFA D4 s.t. L(D4) = L(1∗01∗0(0|1)∗)

1∗D4

1

1∗01∗

0

1

qf

00,1

{w ∈ {0,1}∗


= L(1∗01∗0(0|1)∗)


1∗D4

1

1∗01∗

0

1

qf

00,1

{w ∈ {0,1}∗


= L(1∗01∗0(0|1)∗)


1∗D4

1

1∗01∗

0

1

qf

00,1

{w ∈ {0,1}∗


= L(1∗01∗0(0|1)∗)


1∗D4

1

1∗01∗

0

1

qf

00,1

{w ∈ {0,1}∗


= L(1∗01∗0(0|1)∗)


1∗D4

1

1∗01∗

0

1

qf

00,1

{w ∈ {0,1}∗


= L(1∗01∗0(0|1)∗)


1∗D4

1

1∗01∗

0

1

qf

0

0,1

{w ∈ {0,1}∗


= L(1∗01∗0(0|1)∗)


1∗D4

1

1∗01∗

0

1

qf

00,1

Which languages are recognized by DFA or denoted byregular expressions?

How can we prove a language, L, is not recognized by anyDFA?

Idea: must show that need more than a bounded size memoryto remember what we have seen so far, x , in order to decide ifextensions of x belong to L.







Pumping Lemma for Regular Sets:Let D = (Q,Σ, δ,q0,F ) be a DFA.

Let n = |Q|.

Let w ∈ L(D) s.t. |w | ≥ n.

Then ∃x , y , z ∈ Σ? s.t. the following all hold:1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))


Let n = |Q|.

Let w ∈ L(D) s.t. |w | ≥ n.


2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))


Let n = |Q|.

Let w ∈ L(D) s.t. |w | ≥ n.


2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))


Let n = |Q|.

Let w ∈ L(D) s.t. |w | ≥ n.


2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

Pumping Lemma for Regular Sets:Let D = (Q,Σ, δ,q0,F )

be a DFA, n = |Q|w ∈ L(D) s.t. |w | ≥ n

Then ∃x , y , z ∈ Σ? s.t.

1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3

Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4

x def= ε y def

= 1 z def= 0101 ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4), k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4), k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3


x def= ε y def

= 1 z def= 0101 ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4), k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4), k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3


x def= ε y def

= 1 z def= 0101 ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4), k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4), k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3


x def= ε y def

= 1 z def= 0101 ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4), k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4), k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3


x def= ε y def

= 1 z def= 0101

∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4), k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4), k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3


x def= ε y def

= 1 z def= 0101 ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4), k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4), k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3


x def= ε y def

= 1 z def= 0101 ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4),

k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4), k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3


x def= ε y def

= 1 z def= 0101 ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4), k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4), k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3


x def= ε y def

= 1 z def= 0101 ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4), k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4),

k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 10101 ∈ L(D4), |w | ≥ 3


x def= ε y def

= 1 z def= 0101 ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 0101 ∈ L(D4), k = 1 : 10101 ∈ L(D4),

k = 2 : 110101 ∈ L(D4), k = 3 : 1110101 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4), k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4), k = 3 : 00000 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4), k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4), k = 3 : 00000 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4), k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4), k = 3 : 00000 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4), k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4), k = 3 : 00000 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε

∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4), k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4), k = 3 : 00000 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4), k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4), k = 3 : 00000 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4),

k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4), k = 3 : 00000 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4), k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4), k = 3 : 00000 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4), k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4),

k = 3 : 00000 ∈ L(D4), . . .




1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

1∗D4

1

1∗01∗

0

1

qf

00,1

n = 3, w = 000 ∈ L(D4), |w | ≥ 3


x def= 00 y def

= 0 z def= ε ∀k ≥ 0 (xykz ∈ L(D4))

k = 0 : 00 ∈ L(D4), k = 1 : 000 ∈ L(D4),

k = 2 : 0000 ∈ L(D4), k = 3 : 00000 ∈ L(D4), . . .

proof: Let w ∈ L(D), |w | ≥ n, w = w1,w2, . . . ,wn · u

w = w1 w2 w3 · · · wn uq0 q1 q2 q3 · · · qn−1 qn qf

By Pigeon-Hole Principle ∃i < j ≤ n (qi = qj)

w =

x︷︸︸︷w1 . . .wi

y︷︸︸︷wi+1 . . .wj

z︷︸︸︷wj+1 . . .wnu

q0 qi qi qf

qi = δ?(qi , y) |y | ≥ 1 δ?(qi , z) = qf ∈ F

Thus, xykz ∈ L(D) for k = 0,1,2, . . . �




w =

x︷︸︸︷w1 . . .wi

y︷︸︸︷wi+1 . . .wj

z︷︸︸︷wj+1 . . .wnu

q0 qi qi qf

qi = δ?(qi , y) |y | ≥ 1 δ?(qi , z) = qf ∈ F

Thus, xykz ∈ L(D) for k = 0,1,2, . . . �




w =

x︷︸︸︷w1 . . .wi

y︷︸︸︷wi+1 . . .wj

z︷︸︸︷wj+1 . . .wnu

q0 qi qi qf

qi = δ?(qi , y) |y | ≥ 1 δ?(qi , z) = qf ∈ F

Thus, xykz ∈ L(D) for k = 0,1,2, . . . �




w =

x︷︸︸︷w1 . . .wi

y︷︸︸︷wi+1 . . .wj

z︷︸︸︷wj+1 . . .wnu

q0 qi qi qf

qi = δ?(qi , y) |y | ≥ 1 δ?(qi , z) = qf ∈ F

Thus, xykz ∈ L(D) for k = 0,1,2, . . . �




w =

x︷︸︸︷w1 . . .wi

y︷︸︸︷wi+1 . . .wj

z︷︸︸︷wj+1 . . .wnu

q0 qi qi qf

qi = δ?(qi , y) |y | ≥ 1 δ?(qi , z) = qf ∈ F

Thus, xykz ∈ L(D) for k = 0,1,2, . . . �




w =

x︷︸︸︷w1 . . .wi

y︷︸︸︷wi+1 . . .wj

z︷︸︸︷wj+1 . . .wnu

q0 qi qi qf

qi = δ?(qi , y) |y | ≥ 1 δ?(qi , z) = qf ∈ F

Thus, xykz ∈ L(D) for k = 0,1,2, . . . �


Let n = |Q|.

Let w ∈ L(D) s.t. |w | ≥ n.


2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

Easiest tool to prove languages not DFA acceptable


Let n = |Q|.

Let w ∈ L(D) s.t. |w | ≥ n.


2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

Easiest tool to prove languages not DFA acceptable

Prop: E ={

ar br∣∣ r ∈ N

}is not DFA acceptable.

proof by contradiction:Assume: E is accepted by DFA D with n states.

you (G) choose string: w ∈ E = L(D)Let w = anbn

By pumping lemma, D chooses x , y , z ∈ {a,b}?, s.t.,

1. w = anbn = xyz2. |xy | ≤ n3. |y | > 0, and4. ∀k ∈ N ( xykz ∈ E )

Since 0 < |xy | ≤ n, y = ai , 0 < i ≤ n

Thus xy0z = an−ibn ∈ E .

but an−ibn 6∈ E .

FTherefore E is not DFA acceptable. �

Prop: E ={

ar br∣∣ r ∈ N






Since 0 < |xy | ≤ n, y = ai , 0 < i ≤ n




Prop: E ={

ar br∣∣ r ∈ N



you (G) choose string: w ∈ E = L(D)

Let w = anbn



Since 0 < |xy | ≤ n, y = ai , 0 < i ≤ n




Prop: E ={

ar br∣∣ r ∈ N






Since 0 < |xy | ≤ n, y = ai , 0 < i ≤ n




Prop: E ={

ar br∣∣ r ∈ N






Since 0 < |xy | ≤ n, y = ai , 0 < i ≤ n




Prop: E ={

ar br∣∣ r ∈ N






Since 0 < |xy | ≤ n, y = ai , 0 < i ≤ n




Prop: E ={

ar br∣∣ r ∈ N






Since 0 < |xy | ≤ n, y = ai , 0 < i ≤ n




Prop: E ={

ar br∣∣ r ∈ N






Since 0 < |xy | ≤ n, y = ai , 0 < i ≤ n




Prop: E ={

ar br∣∣ r ∈ N






Since 0 < |xy | ≤ n, y = ai , 0 #b(w)

}not DFA-acceptable.

proof by contradiction:Assume: M is accepted by DFA D with n states.

you choose string: w ∈ M = L(D) Let w = an+1bn

By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.

1. w = an+1bn = xyz2. |xy | ≤ n3. |y | > 0, and4. ∀k ∈ N ( xykz ∈ M )

Since 0 < |xy | ≤ n, y = ai , 0 #b(w)






Since 0 < |xy | ≤ n, y = ai , 0 #b(w)



you choose string: w ∈ M = L(D)

Let w = an+1bn



Since 0 < |xy | ≤ n, y = ai , 0 #b(w)






Since 0 < |xy | ≤ n, y = ai , 0 #b(w)






Since 0 < |xy | ≤ n, y = ai , 0 #b(w)






Since 0 < |xy | ≤ n, y = ai , 0 #b(w)






Since 0 < |xy | ≤ n, y = ai , 0 #b(w)






Since 0 < |xy | ≤ n, y = ai , 0 #b(w)






Since 0 < |xy | ≤ n, y = ai , 0 #b(w)






Since 0 < |xy | ≤ n, y = ai , 0 < i ≤ n




Prop: P ={

w ∈ {a,b}?∣∣ |w | is prime


proof: Assume: P is accepted

by DFA D with n states.

you choose string: w ∈ P = L(D) to get contradiction

Let w = ap where p ≥ n is prime


1. w = ap = xyz2. |xy | ≤ n3. |y | > 0, and4. ∀k ∈ N ( xykz ∈ P )

y = ai , 0 < i ≤ n

Thus xyp+1z = xyzyp = apap·i = ap(i+1) ∈ P.

but p(i + 1) is not prime, so xyp+1z 6∈ P.

FTherefore P is not regular. �

Prop: P ={



proof: Assume: P is accepted by DFA D with n states.





y = ai , 0 < i ≤ n




Prop: P ={








y = ai , 0 < i ≤ n




Prop: P ={








y = ai , 0 < i ≤ n




Prop: P ={








y = ai , 0 < i ≤ n




Prop: P ={








y = ai , 0 < i ≤ n




Prop: P ={








y = ai , 0 < i ≤ n




Prop: P ={








y = ai , 0 < i ≤ n




Prop: P ={








y = ai , 0 < i ≤ n



F

Therefore P is not regular. �

Prop: P ={








y = ai , 0 < i ≤ n




Pumping Lemma for Regular SetsLet D = (Q,Σ, δ,q0,F ) be a DFA.

Let n = |Q|.

You (G) choose w ∈ L(D) s.t. |w | ≥ n.

Then D chooses x , y , z ∈ Σ? s.t. the following all hold:1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

Finally, you point out why a contradiction ensues.

Pumping Lemma for Regular SetsLet D = (Q,Σ, δ,q0,F ) be a DFA.

Let n = |Q|.

You (G) choose w ∈ L(D) s.t. |w | ≥ n.

Then D chooses x , y , z ∈ Σ? s.t. the following all hold:1. xyz = w

2. |xy | ≤ n

3. |y | > 0, and

4. ∀k ≥ 0 (xykz ∈ L(D))

Finally, you point out why a contradiction ensues.

Preview: Nondeterministic Finite Automata (NFA)

{w ∈ {0,1}∗

∣∣ w has 001 or 100}

= L((0|1)∗(001|100)(0|1)∗)

Build an NFA N5 that accepts L((0|1)∗(001|100)(0|1)∗)

ε

0,1

N5

0

0

1

1

100

000

qf

0

1 0,1


{w ∈ {0,1}∗

∣∣ w has 001 or 100}

= L((0|1)∗(001|100)(0|1)∗)


ε

0,1

N5

0

0

1

1

100

000

qf

0

1 0,1


{w ∈ {0,1}∗

∣∣ w has 001 or 100}

= L((0|1)∗(001|100)(0|1)∗)


ε

0,1

N5

0

0

1

1

100

000

qf

0

1 0,1


{w ∈ {0,1}∗

∣∣ w has 001 or 100}

= L((0|1)∗(001|100)(0|1)∗)


ε

0,1

N5

0

0

1

1

100

000

qf

0

1 0,1


{w ∈ {0,1}∗

∣∣ w has 001 or 100}

= L((0|1)∗(001|100)(0|1)∗)


ε

0,1

N5

0

0

1

1

100

000

qf

0

1 0,1


{w ∈ {0,1}∗

∣∣ w has 001 or 100}

= L((0|1)∗(001|100)(0|1)∗)


ε

0,1

N5

0

0

1

1

100

000

qf

0

1 0,1


{w ∈ {0,1}∗

∣∣ w has 001 or 100}

= L((0|1)∗(001|100)(0|1)∗)


ε

0,1

N5

0

0

1

1

100

000

qf

0

1 0,1

Goal for Friday

Kleene’s Theorem Let A ⊆ Σ? be any language. Then thefollowing are equivalent:

1. A = L(D), for some DFA D.

2. A = L(N), for some NFA N wo ε transitions.

3. A = L(N), for some NFA N.

4. A = L(e), for some regular expression e.

5. A is regular.

True by definition: (4)⇔ (5)

Goal for Friday

Kleene’s Theorem Let A ⊆ Σ? be any language. Then thefollowing are equivalent:

1. A = L(D), for some DFA D.

2. A = L(N), for some NFA N wo ε transitions.

3. A = L(N), for some NFA N.

4. A = L(e), for some regular expression e.

5. A is regular.

True by definition: (4)⇔ (5)

Documents

CS250: Discrete Math for Computer SciencePumping Lemma for Regular Sets: Let D = (Q; ; ;q0;F) be a DFA, n = jQj w 2L(D) s.t. jwj n Then 9x;y;z 2? s.t. 1. xyz = w 2. jxyj n 3. jyj>0,