cs250: discrete math for computer sciencepumping lemma for regular sets: let d = (q; ; ;q0;f) be a...
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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)∗)
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)∗)
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)∗)
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)∗)
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)∗)
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)∗)
{w ∈ {0,1}∗
∣∣ w has at least two 0’s}
= 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}∗
∣∣ w has at least two 0’s}
= 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}∗
∣∣ w has at least two 0’s}
= 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}∗
∣∣ w has at least two 0’s}
= 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}∗
∣∣ w has at least two 0’s}
= 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}∗
∣∣ w has at least two 0’s}
= 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
0
0,1
{w ∈ {0,1}∗
∣∣ w has at least two 0’s}
= 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
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.
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.
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))
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))
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))
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))
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), . . .
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), . . .
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), . . .
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), . . .
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), . . .
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), . . .
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), . . .
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), . . .
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), . . .
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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), . . .
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 = 000 ∈ L(D4), |w | ≥ 3
Split w into x · y · z s.t. 1 ∧ 2 ∧ 3 ∧ 4
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, . . . �
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, . . . �
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, . . . �
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, . . . �
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, . . . �
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, . . . �
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))
Easiest tool to prove languages not DFA acceptable
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))
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
}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
}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
}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
}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
}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
}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
}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
}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 .
F
Therefore E is 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: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
FTherefore M is not DFA acceptable. �
Prop: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
FTherefore M is not DFA acceptable. �
Prop: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
FTherefore M is not DFA acceptable. �
Prop: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
FTherefore M is not DFA acceptable. �
Prop: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
FTherefore M is not DFA acceptable. �
Prop: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
FTherefore M is not DFA acceptable. �
Prop: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
FTherefore M is not DFA acceptable. �
Prop: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
FTherefore M is not DFA acceptable. �
Prop: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
F
Therefore M is not DFA acceptable. �
Prop: M ={
w ∈ {a,b}?∣∣ #a(w) > #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 < i ≤ n
Thus xy0z = an+1−ibn ∈ M.
but an+1−ibn 6∈ M.
FTherefore M is not DFA acceptable. �
Prop: P ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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 ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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 ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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 ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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 ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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 ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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 ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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 ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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 ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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.
F
Therefore P is not regular. �
Prop: P ={
w ∈ {a,b}?∣∣ |w | is prime
}is not DFA acceptable.
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
By pumping lemma, D chooses x , y , z ∈ {a,b}∗ s.t.
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. �
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
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
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
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
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
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
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
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)
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