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THE PUMPING LEMMA

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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 ∈ Σ∗,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 ∈ Σ∗,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 ∈ Σ∗,such that

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L.

xL

THE PUMPING LEMMA

n

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 ∈ Σ∗,such that

(1) x = uvw

(2) |uv| ≤ n

(3) |v| ≥ 1

(4) for all i ≥ 0: uviw ∈ L.

xL

THE PUMPING LEMMA

n

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 ∈ Σ∗,such that

(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

Theorem. For any regular language L there exists an integern, 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.

xu wu v w

L

THE PUMPING LEMMA

n

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 ∈ Σ∗,such that

(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

Theorem. For any regular language L there exists an integern, 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

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.

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

vwx = uvw ∈ L

uw ∈ L

uvvw ∈ Luvvvw ∈ 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. holdsdoesn’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. holdsdoesn’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. holdsdoesn’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. holdsdoesn’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. holdsdoesn’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

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

Proof. Show that P.L. doesn’t hold (note: showing P.L. holdsdoesn’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.If (1), (2), (3) hold then x = 0n1n = uvw with |uv| ≤ n and|v| ≥ 1.

EXAMPLE 1

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

Proof. Show that P.L. doesn’t hold (note: showing P.L. holdsdoesn’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.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

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

Proof. Show that P.L. doesn’t hold (note: showing P.L. holdsdoesn’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.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

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

Proof. Show that P.L. doesn’t hold (note: showing P.L. holdsdoesn’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.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

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

Proof. Show that P.L. doesn’t hold (note: showing P.L. holdsdoesn’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.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

v����0 . . .

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

v����0 . . .

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

v����0 . . .

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 = 0. uw �∈ L.

EXAMPLE 2Prove that L = {0i : i is a prime} is NOT regular.

EXAMPLE 2Prove that L = {0i : i is a prime} is NOT regular.

Proof. We show that P.L. doesn’t hold.

EXAMPLE 2

If L is regular, then by P.L. ∃n such that . . .

Prove that L = {0i : i is a prime} is NOT regular.

Proof. We show that P.L. doesn’t hold.

EXAMPLE 2

If L is regular, then by P.L. ∃n such that . . .

Prove that L = {0i : i is a prime} is NOT regular.

Now let x = 0m where m ≥ n + 2 is prime.

Proof. We show that P.L. doesn’t hold.

EXAMPLE 2

If L is regular, then by P.L. ∃n such that . . .

Prove that L = {0i : i is a prime} is NOT regular.

Now let x = 0m where m ≥ n + 2 is prime.x ∈ L and |x| ≥ n, so by P.L. ∃u, v, w such that (1)–(4) hold.

Proof. We show that P.L. doesn’t hold.

EXAMPLE 2

If L is regular, then by P.L. ∃n such that . . .

Prove that L = {0i : i is a prime} is NOT regular.

Now let x = 0m where m ≥ n + 2 is prime.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.

Proof. We show that P.L. doesn’t hold.

EXAMPLE 2

If L is regular, then by P.L. ∃n such that . . .

Prove that L = {0i : i is a prime} is NOT regular.

Now let x = 0m where m ≥ n + 2 is prime.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 0m is written as 0m = uvw, then 0m = 0|u|0|v|0|w|.

Proof. We show that P.L. doesn’t hold.

EXAMPLE 2

If L is regular, then by P.L. ∃n such that . . .

Prove that L = {0i : i is a prime} is NOT regular.

Now let x = 0m where m ≥ n + 2 is prime.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 0m is written as 0m = uvw, then 0m = 0|u|0|v|0|w|.If |uv| ≤ n and |v| ≥ 1, then consider i = |v| + |w|:

Proof. We show that P.L. doesn’t hold.

EXAMPLE 2

If L is regular, then by P.L. ∃n such that . . .

Prove that L = {0i : i is a prime} is NOT regular.

Now let x = 0m where m ≥ n + 2 is prime.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 0m is written as 0m = uvw, then 0m = 0|u|0|v|0|w|.If |uv| ≤ n and |v| ≥ 1, then consider i = |v| + |w|:

uviw = 0|v|0|v|(|v|+|w|)0|w|

= 0(|v|+1)(|v|+|w|) �∈ L

Proof. We show that P.L. doesn’t hold.

EXAMPLE 2

If L is regular, then by P.L. ∃n such that . . .

Prove that L = {0i : i is a prime} is NOT regular.

Now let x = 0m where m ≥ n + 2 is prime.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 0m is written as 0m = uvw, then 0m = 0|u|0|v|0|w|.If |uv| ≤ n and |v| ≥ 1, then consider i = |v| + |w|:

Both factors ≥ 2

uviw = 0|v|0|v|(|v|+|w|)0|w|

= 0(|v|+1)(|v|+|w|) �∈ L

Proof. We show that P.L. doesn’t hold.

EXAMPLE 3Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3

Again we try to show that P.L. doesn’t hold.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3

If L is regular, then by P.L. ∃n such that . . .Again we try to show that P.L. doesn’t hold.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3

If L is regular, then by P.L. ∃n such that . . .Let us consider x = 0n0n ∈ L. Obviously |x| ≥ n.

Again we try to show that P.L. doesn’t hold.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3

If L is regular, then by P.L. ∃n such that . . .Let us consider x = 0n0n ∈ L. Obviously |x| ≥ n.

Can 0n0n be written as 0n0n = uvw such that |uv| ≤ n |v| ≥ 1and that for all i: uviw ∈ L?

Again we try to show that P.L. doesn’t hold.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3

If L is regular, then by P.L. ∃n such that . . .Let us consider x = 0n0n ∈ L. Obviously |x| ≥ n.

YES! Let u = �, v = 00, and w = 02n−2.

Can 0n0n be written as 0n0n = uvw such that |uv| ≤ n |v| ≥ 1and that for all i: uviw ∈ L?

Again we try to show that P.L. doesn’t hold.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3

If L is regular, then by P.L. ∃n such that . . .Let us consider x = 0n0n ∈ L. Obviously |x| ≥ n.

YES! Let u = �, v = 00, and w = 02n−2.Then ∀i , uviw is of the form 02k = 0k0k.

Can 0n0n be written as 0n0n = uvw such that |uv| ≤ n |v| ≥ 1and that for all i: uviw ∈ L?

Again we try to show that P.L. doesn’t hold.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3

If L is regular, then by P.L. ∃n such that . . .Let us consider x = 0n0n ∈ L. Obviously |x| ≥ n.

YES! Let u = �, v = 00, and w = 02n−2.Then ∀i , uviw is of the form 02k = 0k0k.

Does this mean that L is regular?

Can 0n0n be written as 0n0n = uvw such that |uv| ≤ n |v| ≥ 1and that for all i: uviw ∈ L?

Again we try to show that P.L. doesn’t hold.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3

If L is regular, then by P.L. ∃n such that . . .Let us consider x = 0n0n ∈ L. Obviously |x| ≥ n.

YES! Let u = �, v = 00, and w = 02n−2.Then ∀i , uviw is of the form 02k = 0k0k.

Does this mean that L is regular?

NO.We have chosen a bad string x. To show that L fails theP.L., we only need to exhibit some x that cannot be “pumped”(and |x| ≥ n).

Can 0n0n be written as 0n0n = uvw such that |uv| ≤ n |v| ≥ 1and that for all i: uviw ∈ L?

Again we try to show that P.L. doesn’t hold.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 2ND ATTEMPT

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 2ND ATTEMPT

Given n from the P.L., let x = (01)n(01)n. Obviously x ∈ Land |x| ≥ n.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 2ND ATTEMPT

Q: Can x be “pumped” for some choice of u, v, w with |uv| ≤ nand |v| ≥ 1?

Given n from the P.L., let x = (01)n(01)n. Obviously x ∈ Land |x| ≥ n.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 2ND ATTEMPT

Q: Can x be “pumped” for some choice of u, v, w with |uv| ≤ nand |v| ≥ 1?

Given n from the P.L., let x = (01)n(01)n. Obviously x ∈ Land |x| ≥ n.

A: Yes! Take u = �, v = 0101, w = (01)2n−2.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 2ND ATTEMPT

Q: Can x be “pumped” for some choice of u, v, w with |uv| ≤ nand |v| ≥ 1?

Given n from the P.L., let x = (01)n(01)n. Obviously x ∈ Land |x| ≥ n.

A: Yes! Take u = �, v = 0101, w = (01)2n−2.

Another bad choice of x!

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 3RD ATTEMPT

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 3RD ATTEMPT

Given n from the P.L., let x = 0n10n1. Again x ∈ L and|x| ≥ n.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 3RD ATTEMPT

Given n from the P.L., let x = 0n10n1. Again x ∈ L and|x| ≥ n.

∀u, v, w such that 0n10n1 = uvw and |uv| ≤ n and |v| ≥ 1:

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 3RD ATTEMPT

Given n from the P.L., let x = 0n10n1. Again x ∈ L and|x| ≥ n.

∀u, v, w such that 0n10n1 = uvw and |uv| ≤ n and |v| ≥ 1:must have uv contained in the first group of 0n. Thus consider

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 3RD ATTEMPT

Given n from the P.L., let x = 0n10n1. Again x ∈ L and|x| ≥ n.

∀u, v, w such that 0n10n1 = uvw and |uv| ≤ n and |v| ≥ 1:must have uv contained in the first group of 0n. Thus consider

uv0w = 0n−|v|10n1.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.

EXAMPLE 3, 3RD ATTEMPT

Given n from the P.L., let x = 0n10n1. Again x ∈ L and|x| ≥ n.

∀u, v, w such that 0n10n1 = uvw and |uv| ≤ n and |v| ≥ 1:must have uv contained in the first group of 0n. Thus consider

uv0w = 0n−|v|10n1.Since |v| is at least 1, this is clearly not of the form yy.

Prove that L = {yy : y ∈ {0, 1}∗} is NOT regular.