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)
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.
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.
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 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 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
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
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.
Top Related