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THE PUMPING LEMMA

THE PUMPING LEMMA

x

Theorem. For any regular language L there exists an integer n, such that for all x ∈ L with |x| ≥ n, there exist u, v, w ∈ Σ∗, such that

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L. x L

THE PUMPING LEMMA

n x

Theorem. For any regular language L there exists an integer n, such that for all x ∈ L with |x| ≥ n, there exist u, v, w ∈ Σ∗, such that

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L. x L

THE PUMPING LEMMA

n x

Theorem. For any regular language L there exists an integer n, such that for all x ∈ L with |x| ≥ n, there exist u, v, w ∈ Σ∗, such that

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L.

x L

THE PUMPING LEMMA

n

x

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L.

x L

THE PUMPING LEMMA

n

x

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L.

x

u v w

L

THE PUMPING LEMMA

n

x

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L.

x u w u v w

L

THE PUMPING LEMMA

n

x

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L.

x

u v wv

u w u v w

L

THE PUMPING LEMMA

n

x

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L.

x

u v wv

u w u v w

u v wv v

L

PROOF OF P.L. (SKETCH)

PROOF OF P.L. (SKETCH)

Let M be a DFA for L. Take n be the number of states of M plus 1.

PROOF OF P.L. (SKETCH)

Let M be a DFA for L. Take n be the number of states of M plus 1.

Take any x ∈ L with |x| ≥ n. Consider the path (from start state to an accepting state) in M that corresponds to x. The length of this path is |x| ≥ n.

PROOF OF P.L. (SKETCH)

Let M be a DFA for L. Take n be the number of states of M plus 1.

Since M has at most n − 1 states, some state must be visited twice or more in the first n steps of the path.

Take any x ∈ L with |x| ≥ n. Consider the path (from start state to an accepting state) in M that corresponds to x. The length of this path is |x| ≥ n.

PROOF OF P.L. (SKETCH)

Let M be a DFA for L. Take n be the number of states of M plus 1.

Since M has at most n − 1 states, some state must be visited twice or more in the first n steps of the path.

Take any x ∈ L with |x| ≥ n. Consider the path (from start state to an accepting state) in M that corresponds to x. The length of this path is |x| ≥ n.

u

vw

PROOF OF P.L. (SKETCH)

Let M be a DFA for L. Take n be the number of states of M plus 1.

Since M has at most n − 1 states, some state must be visited twice or more in the first n steps of the path.

u

vw x = uvw ∈ L

uw ∈ L uvvw ∈ L uvvvw ∈ L. . .

L regular =⇒ L satisfies P.L. L non-regular =⇒ ? L non-regular ⇐= L doesn’t satisfy P.L.

Negation:

∃ n ∈ N ∀ x ∈ L with |x| ≥ n ∃ u, v, w ∈ Σ∗ all of these hold:

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) ∀ i ≥ 0: uviw ∈ L.

USING PUMPING LEMMA TO PROVE NON-REGULARITY

L regular =⇒ L satisfies P.L. L non-regular =⇒ ? L non-regular ⇐= L doesn’t satisfy P.L.

Negation:

∃ n ∈ N ∀ x ∈ L with |x| ≥ n ∃ u, v, w ∈ Σ∗ all of these hold:

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) ∀ i ≥ 0: uviw ∈ L.

USING PUMPING LEMMA TO PROVE NON-REGULARITY

L regular =⇒ L satisfies P.L. L non-regular =⇒ ? L non-regular ⇐= L doesn’t satisfy P.L.

Negation:

∃ n ∈ N ∀ x ∈ L with |x| ≥ n ∃ u, v, w ∈ Σ∗ all of these hold:

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) ∀ i ≥ 0: uviw ∈ L.

∀ n ∈ N ∃ x ∈ L with |x| ≥ n ∀ u, v, w ∈ Σ∗ not all of these hold:

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) ∀ i ≥ 0: uviw ∈ L.

USING PUMPING LEMMA TO PROVE NON-REGULARITY

L regular =⇒ L satisfies P.L. L non-regular =⇒ ? L non-regular ⇐= L doesn’t satisfy P.L.

Negation:

∃ n ∈ N ∀ x ∈ L with |x| ≥ n ∃ u, v, w ∈ Σ∗ all of these hold:

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) ∀ i ≥ 0: uviw ∈ L.

∀ n ∈ N ∃ x ∈ L with |x| ≥ n ∀ u, v, w ∈ Σ∗ not all of these hold:

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) ∀ i ≥ 0: uviw ∈ L.

Equivalently: (1) ∧ (2) ∧ (3)⇒ not(4) where not(4) is: ∃ i : uviw �∈ L

USING PUMPING LEMMA TO PROVE NON-REGULARITY

EXAMPLE 1

Prove that L = {0i1i : i ≥ 0} is NOT regular.

EXAMPLE 1

Prove that L = {0i1i : i ≥ 0} is NOT regular. Proof. Show that P.L. doesn’t hold (note: showing P.L. holds doesn’t mean regularity).

EXAMPLE 1

Prove that L = {0i1i : i ≥ 0} is NOT regular. Proof. Show that P.L. doesn’t hold (note: showing P.L. holds doesn’t mean regularity). If L is regular, then by P.L. ∃n such that . . .

EXAMPLE 1

Prove that L = {0i1i : i ≥ 0} is NOT regular. Proof. Show that P.L. doesn’t hold (note: showing P.L. holds doesn’t mean regularity). If L is regular, then by P.L. ∃n such that . . . Now let x = 0n1n

EXAMPLE 1

Prove that L = {0i1i : i ≥ 0} is NOT regular. Proof. Show that P.L. doesn’t hold (note: showing P.L. holds doesn’t mean regularity). If L is regular, then by P.L. ∃n such that . . . Now let x = 0n1n

x ∈ L and |x| ≥ n, so by P.L. ∃u, v, w such that (1)–(4) hold.

EXAMPLE 1

Prove that L = {0i1i : i ≥ 0} is NOT regular. Proof. Show that P.L. doesn’t hold (note: showing P.L. holds doesn’t mean regularity). If L is regular, then by P.L. ∃n such that . . . Now let x = 0n1n

x ∈ L and |x| ≥ n, so by P.L. ∃u, v, w such that (1)–(4) hold. We show that ∀u, v, w (1)–(4) don’t all hold.

EXAMPLE 1

x ∈ L and |x| ≥ n, so by P.L. ∃u, v, w such that (1)–(4) hold. We show that ∀u, v, w (1)–(4) don’t all hold. If (1), (2), (3) hold then x = 0n1n = uvw with |uv| ≤ n and |v| ≥ 1.

EXAMPLE 1

x ∈ L and |x| ≥ n, so by P.L. ∃u, v, w such that (1)–(4) hold. We show that ∀u, v, w (1)–(4) don’t all hold. If (1), (2), (3) hold then x = 0n1n = uvw with |uv| ≤ n and |v| ≥ 1. So, u = 0s, v = 0t, w = 0p1n with

EXAMPLE 1

x ∈ L and |x| ≥ n, so by P.L. ∃u, v, w such that (1)–(4) hold. We show that ∀u, v, w (1)–(4) don’t all hold. If (1), (2), (3) hold then x = 0n1n = uvw with |uv| ≤ n and |v| ≥ 1. So, u = 0s, v = 0t, w = 0p1n with

s + t ≤ n, t ≥ 1, p ≥ 0, s + t + p = n.

EXAMPLE 1

x ∈ L and |x| ≥ n, so by P.L. ∃u, v, w such that (1)–(4) hold. We show that ∀u, v, w (1)–(4) don’t all hold. If (1), (2), (3) hold then x = 0n1n = uvw with

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