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

Download 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

    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

    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.

    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 ,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 ,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 ,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 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

    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.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. 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.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. 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.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. 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.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. 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.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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