Pumping Lemma

Automata

Author

Prof. Calvin

Sketch

  • Pumping Lemma
    • Statement
    • Examples

Preface

Avoid this trap

  • Suppose we wish to prove a language is not regular.
  • We must prove there is no DFA that recognizes the language.
  • It may be tempting to conclude:
    • I thought about it really hard.
    • I could find no DFA.
    • Therefore the language is irregular.

Example

  • Take \(\Sigma = \{0,1\}\)
    • Take \(B\) to be the language for which there are equal numbers of 0s and 1s
    • Take \(C\) to be the language for which there are equal numbers of 01s and 10s.
      • 0101 \(\notin C\)
      • 0110 \(\in C\)
  • One of these is regular.
  • Left as an exercise to the student.

Pumping Lemma

Statement of Lemma

\[ \begin{aligned} &\forall A :\exists p \in \mathbb{N} : \\ &\exists xyz \in A : |xyz| \geq p \implies \\ &\forall i \in \mathbb{N} : xy^iz \in A \land \\ &|y| > 0 \land \\ &|xy| \leq p \end{aligned} \] - That is, \(\{xz, xyz, xyyz\} \in A\)

  • We “pump up” the number of occurances of y

Definitions

  • Take \(M | L(M) = A\)
  • Take \(p = |M| = |Q|\)
  • Take \(s \in A = xyz : |s| \geq p\)

Sketch of Proof

  • We note that \(s\) visits states \(p\) times.
    • We make no claims about which states.
  • Define sequence of states \(S = (q_0,s_1,s_2,\ldots,s_i,s_{i+1},\ldots,s_p)\)
  • \(\forall s_i \in S : s_i \in Q\)
  • \(p = |S| > |Q|\)
  • \(\therefore \exists i,j : s_i = s_j\)

Find x,y,z

  • Let \(x = (a_o,\ldots,a_{i-1})\)
    • \(M\) is in state \(s_j\) after reading \(x\)
  • Let \(y = (a_i,\ldots,a_{j-1})\)
    • \(M\) is in state \(s_j\) after reading \(y\) when beginning in \(s_i\)
    • \(s_i = s_j\)
  • Let \(z = (a_{j},\ldots,a_p)\)
    • \(M\) accepts after reading \(z\) when beginning in \(s_j\)

Pump on \(y\)

  • We note
    • \(xz\) is accepted by \((q_0,s_1,\ldots,s_i,s_{j+1},\ldots,s_p)\)
      • Denote as \([0,i][j+1,p]\)
    • \(xyz\) is accepted by \((q_0,s_1,\ldots,s_p)\)
      • Denote as \([0,p] = [0,i][i+1,j][j+1,p]\)

Pump on \(y\)

  • We note
    • \(xyyz\) is accepted by \((q_0,s_1,\ldots,s_{i+1},\ldots,s_{j-1},s_i\ldots,s_j, s_{j+1},\ldots, s_p)\)
      • Denote as \([0,i][i+1,j][i+1,j][j+1,p]\)
    • \(xy^nz\) is accepted by \([0,i][i+1,j]^n[j+1,p]\)

Python

# assume S as a list of states.
S : List[states]
x : str
y : str
z : str
assert(states(x + z)         == S[:i] + S[j:]) 
assert(states(x + y + z)     == S    == S[:i]+S[i:j]+S[j:])
assert(states(x + y + y + z) == S[:i] + S[i:j] + S[i:j] + S[j:])
assert(states(x + y * 2 + z) == S[:i] + S[i:j] * 2      + S[j:])
# This can't run - it's infinite
assert(all((states(x+y*n+z) == A[:i]+A[i:j]*n+A[j:]) for n in count()))
  • Much easier with Pythonic slices.

Graphically

finite_automata d0 q0 q 0 d0->q0 qn q n q0->qn ?

Show \(s_i=s_j\)

finite_automata d0 q0 q 0 d0->q0 qi s i q0->qi ? qi->qi ? qn q n qi->qn ?

Label Edges

finite_automata d0 q0 q 0 d0->q0 qi s i q0->qi x qi->qi y qn q n qi->qn z

  • This is valid GNFA that reduces to a DFA.

Pumping Lemma

\[ \begin{aligned} &\forall A :\exists p \in \mathbb{N} : \\ &\exists xyz \in A : |xyz| \geq p \implies \\ &\forall i \in \mathbb{N} : xy^iz \in A \land \\ &|y| > 0 \land \\ &|xy| \leq p \end{aligned} \]

Proof

finite_automata d0 q0 q 0 d0->q0 qi s i q0->qi x qi->qi y qn q n qi->qn z

Easy to read

  • \(s \in A \land |s| > p\) requires
    • \(xy^nz \in A\)
    • \(|y| > 0\)
    • \(|xy| \leq p\)

Examples

Irregular 0

  • Take \(D = \{ 0^k1^k | k \in \mathbb{N}\}\)
  • Show \(D\) is not regular.
  • We proceed by contradiction.
    • Assume \(D\) is regular.
    • Apply the pumping lemma.
    • Derive a contradiction.

Strategy

  • We take string \(s = 0^p1^p\)
  • Pumping lemma requires \(\exists xy : |xy| \leq p\)
  • \(\forall xy \not\exists 1 \in xy\)
  • So \(xz = 0^{p-|y|}1^p\)
  • Pumping lemma requires \(|y| > 0\)
  • \(D\) requires \(p - |y| = p\)
  • Contradiction \(\blacksquare\)

On Graphics

  • We note we cannot show this graphically.
  • We have proven no DFA accepts \(D\)
  • How could we draw a DFA!
  • This is also why we can’t rely on graphics!

Irregular 1

  • Take \(F = \{ ww | w\ \in \Sigma*\}\)
  • Show \(F\) is not regular.
  • We proceed by contradiction.
    • Assume \(F\) is regular.
    • Apply the pumping lemma.
    • Derive a contradiction.

Avoid this

  • Say we chose \(0^p0^p\)
  • We take, say, \(x = y = 0^{\frac{p}{2} - 1}\)
  • Well we actually can pump \(y\) if \(|y|\) is odd.
  • Have to be careful!

Strategy

  • We take string \(s = 0^p10^p1\)
  • Pumping lemma requires \(\exists xy : |xy| \leq p\)
  • \(\forall xy \:\not\exists 1 \in x\)
  • So \(xz = 0^{p-|y|}10^p1\)
  • Pumping lemma requires \(|y| > 0\)
  • \(F\) requires \(p - |y| = p\)
  • Contradiction \(\blacksquare\)

Combined Tactics

  • We have used the assumptions of the pumping lemma.
  • We have proven other results!
  • We now explore a result using closure properties (Theorems 1-4)

Irregular 2

  • Recall \(B\)
    • “…equal numbers of 0s and 1s”
  • Assume by contradiction \(B\) is regular.
  • Since \(B\) is regular, \(B \cap 0*1*\) is regular
    • Closure under intersection
  • We note that \(D = B \cap 0*1*\)
    • \(D = \{ 0^k1^k | k \in \mathbb{N}\}\)
    • \(D\) is irregular!
  • \(B\) is irregular.
  • \(\blacksquare\)

Stretch Break