Lecture_7.md

Lecture 7

Concatenation of DFAs

• Consider $f_1: {0, 1}^* \rightarrow {0, 1}$ and $f_2: {0, 1}^* \rightarrow {0, 1}$
• Let us define $f_1 CAT f_2: {0, 1}^* \rightarrow {0, 1}$
  • $f_1CATf_2(x)$ is 1 if $f_1(x_1) =1$ and $f_2(x_2) = 1$, where $x = x_1x_2$
    • This is splitting the input and putting each split as an input - the entire concatenation function is satisfied if each split is satisfied
• 
• Example:
  • $f_1: {0, 1}^* \rightarrow {0, 1}$, where $f_1(x) = 1$ for all x
    • This function is computable by a DFA:
    • 
  • $f_2: {0, 1}^* \rightarrow {0, 1}$, where $f_2(x) = 1$ if x starts with a 1 and has a length exactly 4 and otherwise is 0
  • 
  • $f_1 CAT f_2(x)$ is 1 if the 4th bit from the end is a 1 and is 0 otherwise
    • 
• $f_1 CAT f_2$, in general, can be calculated using DFAs
  • If $f_1: {0 ,1}^* \rightarrow {0, 1}$, $f_2: {0 ,1}^* \rightarrow {0, 1}$ are computable by a DFA, then $f_1 CAT f_2: {0, 1}^* \rightarrow {0, 1}$ is also computable by a DFA
  • Pseudocode Intuition:

    • def CONCAT(f1, f2, x):
        ans = 0
        for i in range(len(x)):
          x1 = x[0:i]
          x2 = x[i:len(x)]
          if f_1(x) AND f_2(x) == 1:
            return 1
        return 0
      

      • This is not a single pass algorithm
  • Let $f_{reverse}: {0, 1}^* \rightarrow {0, 1}$ be another operation that evaluates the function but on a reversed $x$ - that is, $f_{reverse}(x) = f(reverse(x))$
    • If $f$ is computable by a DFA, then so is $f_{reverse}$
    • This cannot necessarily be done, though, by just reversing all state directions and changing the starting state to be the ending state
      • This is not feasible if there are multiple ending states
      • This is also not feasible because multiple outgoing edges will have the same bit labels (multiple zeroes) - this is nondeterministic
      • 

Non-Deterministic Finite Automata (NFA)

• A non-deterministic finite automata can have multiple outgoing edges with the same label out of a state, some edges are labeled by $\epsilon$, and some edges can go missing (i.e. they go to a dead state - this is just to make syntax easier)
• Example:
  • 
    • There are no outgoing edges from the fourth state, implying that it leads to a dead state (it does not lead back to itself)
    • There are multiple outgoing arrows labeled 1 for state zero
      • This can be interpreted as there being parallel outcomes depending on which ambiguous choice is chosen (i.e. parallel universes)
      • e.g. for Input 01001
      • 
    • If any of the parallel branches reach the end state, then one is outputted
• Example:
  • 
  • Here the $\epsilon$ transition allows for a branch to be created without needing to read any input
  • e.g. for Input 01001
  • 
    • This will output 1 because one "branch" reaches state 5
• Formal Definition:
  • $N = (T, S)$
    • $S \subseteq [C]$
    • $T: [C] \times {0, 1, \epsilon } \rightarrow Power([C])$
      • Recall the power set: $[C] = {0, 1, 2, ..., C - 1}$, $Power([C]) = {I: I \subseteq [C]}$ i.e. all possible subsets of C, including the empty set $\emptyset$ which represents a dead state
    • On an input $x$, $N(x)$ is $1$ if on any branch the final state is in $S$ and $0$ otherwise
• Formal Example:
  • 
• A function $f: {0, 1}^* \rightarrow {0, 1}$ is computable by an NFA $N$ if $\forall x$, $f(x) = N(x)$
  • 
    • This NFA checks if the 4th bit from the end is 1
  • 
    • This NFA also checks if the 4th bit from the end is 1
• If $f_1: {0, 1}^* \rightarrow {0, 1}$ is computable by a DFA $D_1$ and $f_2: {0, 1}^* \rightarrow {0, 1}$ is computable by a DFA $D_2$, then $f_1 CAT f_2: {0, 1}^* \rightarrow {0, 1}$ is computable by a NFA
  • This leverages the $\epsilon$ transitions to "split" the input - specifically, there are $\epsilon$ transitions from the accept states of $D_1$ to the start states of $D_2$; the accepting states of the NFA are the accepting states of $D_2$
  • 
• If $f: {0, 1}^* \rightarrow {0, 1}$ is computable by a DFA $D$, then $f_{reverse}$ is computable by an NFA
  • This leverages the idea of reversing the arrows, with the new accepting state being the start state
    • The new start state will be a single dummy state that has epsilon transitions to each original end state