CS52 - Fall 2015 - Class 23

Lecture notes

  • Admin
       - Assignment 9 & 10
       - Office hours might change on Thursday
       - Mentors for next semester (applications due Wednesday!)

  • deterministic finite automata (DFA) review
       - basic idea
          - we have a set of states (indicated by circles)
          - we have a start state where computation (indicated by an arrow)
          - we have a collections of final states (indicate by states with an inner circle)
          - for each state and each letter in our alphabet, we have a transition to another state

       - computing
          - we have a string as input on a tape
          - we start at the beginning of the string
          - read a symbol from the tape and transition to the state indicated by the model
          - if:
             - we end in a final state (i.e. get to the end of the string) we accept the string
             - otherwise, if when we get to the end of the string/tape we're in a non-final state we reject

  • DFAs over numbers
       - we can use any alphabet we want
       - if we use 1's and 0's we can interpret them as binary numbers!

       - greater_5 (6): determines if the input string, when interpreted as a binary number, is greater than 5

       - write a DFA that determines if a number is odd
          - look at odd_number

  • non-deterministic finite automata (NFA)
       - almost identical definition to DFA except:
          - for a given state and input, can go to zero, one or *more* states (rather than just a single one for DFAs)

          - can have epsilon (or sometimes called lambda) transitions from one state to another
             - doesn't read anything from the input, just transitions
          
          
          - do not require that there is a transition for every alphabet letter for every state
             - if you encounter a state without a transition for a particular letter, it does *not* accept that path

       - tend to be a bit easier to create than DFAs

  • some NFA examples (found in NFA_examples)
       - start_end_a (1): start and end with a
          - a(a|b)*a

       - 2_or_3_a (2): strings of a's that have lengths divisible by 2 OR 3
          - (aa)*|(aaa)*

       - end_aa (3): ends in two a's
          - (a|b)*aa

       - aa_bb (4): has either aa or bb as a substring
          - (a|b)*(aa|bb)(a|b)*

       - no_bb (5): any string of a's and b's that doesn't have two adjacent b's
          - a*b(aa*b)*a*

  • Do NFAs give us more power, i.e. are there some languages that we can recognize with NFAs that we cannot recognize with DFAs?
       - how would we show that NFAs are more powerful?
          - find a language that can be represented by an NFA, but cannot be represented by a DFA

       - how would we show that they're not more powerful?
          - if we can show that for any NFA there is an equivalent DFA and vice versa, then we can show that they are equivalent, i.e. have the same representative power

       - Given an DFA, how can we create an equivalent NFA?
          - Easy... don't do anything!

       - Given an NFA, how can we create an equivalent DFA?

       - Consider the end_aa NFA (found in NFA_examples)
          - Is aaaaa in the language?
             - what states could we be in after reading the first a?
                - q_0 or q_1
             - what states would we be in after reading the second a?
                - we could start in either q_0 *or* q_1
                   - if we were in q_0, we'd end up in q_0 or q_1
                   - if we were in q_1, we'd end up in q_2
                - therefore, after reading two a's we could end up in any of: q_0 or q_1 or q_2
             - the third a?
                - we could be in q_0 or q_1 or q_2
                   - if we were in q_0 -> q_0 or q_1
                   - if we were in q_1 -> q_2
                   - if we were in q_2 -> reject
                      - does this matter?
                      - No. We only need to find *one* path through the state transitions that ends in an accepting state
             - the fourth a?
                - q_o or q_1 or q2
             - the fifth a?
                - q_o or q_1 or q2
                - since q2 is *an* option, then there's a set of transitions between the states that gets us to an accepting state

          - Is ababaa in the language?
             - what states would we be in after reading the first a?
                - q_0 or q_1
             - second b?
                - q_0
             - third letter, a?
                - q_0 or q_1
             - fourth letter, b?
                - q_0
             - fifth letter, a?
                - q_0 or q_1
             - sixth letter, a?
                - q_0 or q_1 or q_2
                - since q_2 is *an* option, then there's a set of transitions between the states that gets us to an accepting state

          - is ababa in the language?
             - a:
                - q_0 or q_1
             - b:
                - q_0
             - a:
                - q_0 or q_1
             - b:
                - q_0
             - a:
                - q_0 or q_1
                - reject: no way to get to an accepting state

  • Constructing a DFA from an NFA
       - the basic idea is that we're going to create NFA states that represent one or more of the NFA states
          - DFA state [Q] (where Q is 1 or more NFA states) will transition to DFA state [Q'] on letter l if there exists a transition to every q' \in Q' on letter l from *some* q \in Q
          - start state is [q_0]
          - accepting states are any [Q] where at least one q \in Q is an accepting state in the NFA

       - how many DFA states can we have at most?
          - 2^k - 1 where k is the number of NFA states
             - think of each DFA state like a k bit number
                - 1 if it represents the original NFA state
                - 0 otherwise
             - can't have all zeros, so 2^k - 1

       - one algorithm
          - create the start state (q_0)
          - add [q_0] to process queue
          - as long as process queue isn't empty:
             - remove state s from process queue
             - new_s = []
             - for each letter l in the alphabet
                - if any "old state", q_i, in s has a transition q_i: l -> q_j
                   - add q_j to new_s

             - if new_s doesn't exist already
                - create state new_s
                - add new_s to process q
          - if any states don't have transitions for all letters in the alphabet, create a "sink" state that transitions to itself on all letters and have states transition to here for any remaining alphabet letters

  • A few examples (in NFA examples):
       - We can construct a DFA from the end_aa NFA: end_aa_DFA

       - We can construct a DFA from the start_end_a NFA: start_end_a_DFA

  • Does this show that DFAs and NFAs are equivalent?
       - Yes, given either one (a DFA or NFA) we can create through a deterministic process a corresponding machine of the other type
       - Therefore, they can process/accept the same set of languages

  • We can handle lambda/epsilon transitions in a similar way
       - I'll let you figure out/investigate for those that are curious :)

  • regular language
       - any language that can be described by a DFA (or an NFA, remember, they're equivalent)
       - any language that can be described by a regular expression!
          - how would you prove this?

  • What languages are *not* regular?
       - 0^n 1^n for any n
          - i.e. the language of some number of zeros followed by the *same* number of 1s
       
       - why not?
          - can you come up with a regular expression for this language?
             - seems hard, since there's no tool for us to count
          - can you come up with a DFA or NFA for this language?
             - would have to have 2^(n+1) states
                - states are the only way we can count
             - only problem is that n isn't finite!
                - consider any DFA that recognizes strings of 0^n 1^n for some fixed n
                - won't recognize string O^(n+1) 1^(n+1)

       - This is a bit of a "hand-wavy" proof
          - see the pumping lemma (or take CS81) to see more concrete proof
    •