Lecture_10.md

Lecture 10

Limitations of Automata

• Given a regex $r$, we can construct a NFA $N$ such that $\forall x$, $x$ matches $r$ (that is, $N(x) = 1$)
  • Example: $r = 0(100|01)^*$
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• DFAs, NFAs, and Regular Expressions are all equivalent to each other in terms of computational power
  • A function $f: {0, 1}^* \rightarrow {0, 1}$ is regular if it is computable by a DFA or a NFA or a regex
  • These models of computations are great for string matching as well as single pass algorithms
  • From a theoretical standpoint, these models of computations work for unbounded length inputs and serve as a nice introduction to non-determinism in computations
• As discussed, DFAs (or NFAs / Regular Expressions) are limited by being only single-pass - view this as a memory limit
  • For instance, considering expressing the majority function MAJ using DFAs
    • This is not easy to using DFAs because it is essentially necessary to count the number of ones, which can be unbounded (and therefore not expressible with DFAs which have limited memory)
  • Consider the language $L_1$ containing all $x$ such that it has an equal number of 0's and 1's
    • With the same intuition as MAJ, this is not possible to do using DFAs because it is necessary to count on a potentially unbounded input
  • Consider the language $L_2$ containing all $x$ contains an equal number of 01's and 10's
    • This is actually computable via DFA's; $L_2$ is regular and can be computed by a DFA
    • This is because, for this case, a string has an equal number of 01's and 10's if the first and last bit of the string are the same
      • Even though it seems like counting is necessary, it was still possible to represent the language using a DFA via some other clever method
  • The difference between $L_2$ and $L_1$ was that $L_2$ had overlap (01 has overlap with 10)

Pumping Lemma

• Pumping Lemma: If $f$ is a regular function. Then there exists a number P such that for every string x, $f(x) = 1$ and $|x| \geq P$ can be written as $x = abc$
  • $f(a . b^i . c) = 1$ for all $i \geq 0$
    • 
  • The length of $b$ is not zero
  • The length of $ab \leq P$
• If $f$ is a regular function, this implies there exists a number $P$ such that the lemma holds (that is, $\forall x$, $f(x) = 1$, $|x| \geq P$, there exists $x = a.b.c.$ such that $f(a.b^i.c) = 1$)
  • Intuitively, every sufficiently long $x$ in a language $L$ has a piece that can be repeated while being in $L$
    • This requires that there is a sufficiently large enough $P$ - for languages where the maximum string length is less than $P$, then the lemma does not really apply (vacuously true)
      • i.e. $L = {01}$ does not violate the Pumping Lemma - it can be satisfied with $P = 3$ because there is no accepting string with legnth greater than or equal to 3
• Example: MAJ is not regular, using the Pumping Lemma as justification
  • Suppose that MAJ was regular, which means that there exists a number $P$ such that the conditions of the pumping lemma hold
  • Let $x = 0...01...1 = (0^P1^P)$
    • By the Pumping Lemma, we should be able to split it up as $x=a.b.c$ such that the the three conditions hold
    • It must be the case that $ab$ be all 0's because the $|ab| \leq P$
      • This contradicts MAJ(abbc) = 1, which is a contradiction because abbc has more 0's than 1's
      • 
  • The contradiction proves that MAJ is not a regular function
• Example: PALINDROME is not regular, using the Pumping Lemma as justification
  • Suppose that PALINDROME is regular, which implies that there exists some number $P$ such that the conditions of the Pumping Lemma hold
  • Let $x=(0^P)1(0^P) = 0...010...0$
    • The splitting into $a.b.c$ yields a contradiction because $|ab| \leq P$, implying that $ab$ is all zeroes
    • Thus, $ab^i c$ cannot be in PALINDROME, as abbc is not palinddromic
• The Pumping Lemma does not hold both ways - that is, satisfying the conditions of the Pumping Lemma is not enough to prove that a function is regular
  • The Pumping Lemma can be used to prove that a function is not regular but not able to prove that a function is regular