Lecture_11.md

Lecture 11

Pumping Lemma and Proof

• Example: Prove $L = {1^{n^2}: n \geq 1 }$, or strings with only 1's and the number of 1's is a perfect square (i.e. 1, 1111, ...), is not regular
  • Suppose L is regular, then there is some number p such that the Pumping Lemma holds
  • Let $X = 1^{p^2} = 1 ... 1$
    • This challenge string cannot be broken up into $abc$ such that the conditions of the Pumping Lemma holds
    • This is because $x = abbc \notin L$
    • $|abbc| = p^2 + |b| \leq p^2 + p$
      • The length of $ab$ is at most $p$
    • $p^2 + p$ is not necessarily a perfect square, thus proving that $L$ is not regular
      • $p^2 < |abbc| < p^2 + p < (p+1)^2 = p^2 + 2p + 1$

Proof

• If $L$ is a regular language, there is some number such that for any $x \in L$ of length $\geq p$, it can be written as $x = abc$ where: (i) $ab^ic \in L$, (ii) $b$ is not empty, (iii) $|ab| \leq P$
  • Since $L$ is a regular language, there is a DFA $D$ that computes $L$
  • Let $p$ be the number of states in the DFA $D$
  • Let $x = x[0]x[1]x[2]...x[p-1]x[p]...x[n-1]$
    • $N \geq p$
  • 
    • There are a sequence of $p + 1$ state indices $S_0, S_1, S_2, ... S_p$
      • These indices are in ${0, 1, ..., p - 1}$
    • By the Pigeonhole Principle, one state index must repeat because the length of $x$ is greater than or equal to $p$
      • That is, there exists some $i, j \leq p$ such that $S_i = S_j$
      • If there is some repeated state index, then all states between the repeated indices can be repeated
        • Let the states between the repeated indices be $b$ then, if repeating $b$ leads to the same end state then $abbc \in L$
      • 
• Let $S_0 = 0$ be the starting state and $S_i = T_D(S_{i - 1}, x[i - 1]), i = 1, .., N$
  • As $N \geq P$, we must have two indices $i \neq j$, $i, j \leq p$ such that $s_i = s_j$
  • By the Pigeonhole Principle, some two elements $S_0, S_1, S_2, ..., S_p$ must be identical
  • Let $a = x[0] ... x[i - 1]$
  • Let $b = x[i]x[i + 1] ... x[j - 1]$
  • Let $c = x[j] ... x[n-1]$
  • Intuitively, think of $b$ representing a loop of states - that is, following the DFA from $b$ will result in the same end state, so repeating $b$ will have the same effect as simply including it once
    • 

Turing Machines

• Turing machines are another model of computation that can deal with an arbitrary length of inputs
  • They are as powerful a model as there can be because they extend the limitations of DFAs (which are restricted to single-pass algorithms and constant memory)
  • To be more precise, Turing machine have variable memory and allow the "pointer" to the current input bit to move left or right (think of DFAs or NFAs as a "pointer" that can only move right)
    • Aside from these two modifications, Turing machines function similarly to DFAs
• Anatomy of a Turing Machine:
  • Imagine an infinite "tape" (array) with cells that are initially loaded with the input and the rest of the cells are empty
  • There is a head pointer that starts at the very beginning of the input
    • This head can move left, right, or stay
  • Although memory is infinite, the number of states $k$ is finite
    • This can be intuitively thought of as making the description of the machine inherently finite
    • State is separate from memory
  • Operation:
    • If at state $i$ and head reads symbol $a$, then:
      • Change the state to some other state
      • Write something at the current head position
      • Move the head (left, right, stay, or halt/terminate)
• Formal Definition:
  • Let there be $k$ states
  • Let there be an alphabet $\Sigma \supseteq {0, 1, \Delta, \emptyset }$
    • $\Delta$ is notation for the start of the tape
    • $\emptyset$ is notation for nothing in the cell (empty)
  • Let there be a transition function $\delta: {0, 1, ..., k - 1} \times \Sigma \rightarrow {0, 1, ... k - 1} \times \Sigma \times {L, R, S, H}$
    • $\delta(state[i], a) = (state[j], b, movehead)$