# the pumping lemma - university of .using pumping lemma to prove non-regularity. l regular =⇒ l

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THE PUMPING LEMMA

THE PUMPING LEMMA

x

Theorem. For any regular language L there exists an integern, such that for all x L with |x| n, there exist u, v, w ,such that

(1) x = uvw

(2) |uv| n

(3) |v| 1

(4) for all i 0: uviw L.xL

THE PUMPING LEMMA

nx

Theorem. For any regular language L there exists an integern, such that for all x L with |x| n, there exist u, v, w ,such that

(1) x = uvw

(2) |uv| n

(3) |v| 1

(4) for all i 0: uviw L.xL

THE PUMPING LEMMA

nx

Theorem. For any regular language L there exists an integern, such that for all x L with |x| n, there exist u, v, w ,such that

(1) x = uvw

(2) |uv| n

(3) |v| 1

(4) for all i 0: uviw L.

xL

THE PUMPING LEMMA

n

x

(1) x = uvw

(2) |uv| n

(3) |v| 1

(4) for all i 0: uviw L.

xL

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.

xu wu 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 wu 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 wu 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 Mplus 1.

PROOF OF P.L. (SKETCH)

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

Take any x L with |x| n. Consider the path (from startstate to an accepting state) in M that corresponds to x. Thelength 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 Mplus 1.

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

Take any x L with |x| n. Consider the path (from startstate to an accepting state) in M that corresponds to x. Thelength 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 Mplus 1.

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

Take any x L with |x| n. Consider the path (from startstate to an accepting state) in M that corresponds to x. Thelength 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 Mplus 1.

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

u

vwx = uvw L

uw Luvvw Luvvvw L. . .

L regular = L satisfies P.L.L non-regular = ?L non-regular = L doesnt satisfy P.L.

Negation:

n N x L with |x| n u, v, w 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 doesnt satisfy P.L.

Negation:

n N x L with |x| n u, v, w 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 doesnt satisfy P.L.

Negation:

n N x L with |x| n u, v, w 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 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 doesnt satisfy P.L.

Negation:

n N x L with |x| n u, v, w 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 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. doesnt hold (note: showing P.L. holdsdoesnt mean regularity).

EXAMPLE 1

Prove that L = {0i1i : i 0} is NOT regular.Proof. Show that P.L. doesnt hold (note: showing P.L. holdsdoesnt 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. doesnt hold (note: showing P.L. holdsdoesnt 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. doesnt hold (note: showing P.L. holdsdoesnt 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. doesnt hold (note: showing P.L. holdsdoesnt 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) dont 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) dont 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) dont 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) dont 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) dont all hold.If (1), (2), (3) hold then x = 0n1n = uvw with |uv| n and|v| 1.

But then (4) fails for i = 0:

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) dont all hold.If (1), (2), (3) hold then x = 0n1n = uvw with |uv| n and|v| 1.

But then (4) fails for i = 0:uv0w = uw = 0s0p1n = 0s+p1n L, since s + p = n

So, u = 0s, v = 0t, w = 0p1n with

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

IN PICTURE

u, v, w such that

(1) x = uvw

(2) |uv| n

(3) |v| 1

(4) i N : uviw L.

IN PICTURE

u, v, w such that

(1) x = uvw

(2) |uv| n

(3) |v| 1

(4) i N : uviw L.

u 00000

v0 . . .

w 01111 . . . 1 L

IN PICTURE

u, v, w such that

(1) x = uvw

(2) |uv| n

(3) |v| 1

(4) i N : uviw L.

u 00000

v0 . . .

w 01111 . . . 1 L

non-empty

IN PICTURE

u, v, w such that

(1) x = uvw

(2) |uv| n

(3) |v| 1

(4) i N : uviw L.

u 00000

v0 . . .

w 01111 . . . 1 L

non-empty

If (1), (2), (3) hold then (4) fails: it is not the case that for alli, uviw is in L.

In particular, let i =

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