CPSC 121: Models of Computation 2010 Winter Term 2

1. CPSC 121: Models of Computation2010 Winter Term 2 DFAs in Depth Benjamin Israel bisrael@cs.ubc.ca Notes heavily borrowed from Steve Wolfman’s, whose slides were based on notes by Patrice Belleville and others

2. Outline • Learning Goals • Formally Specifying DFAs • Designing DFAs • Analyzing DFAs: Can DFAs Count? • Non-Deterministic Finite Automata: Are They More Powerful than DFAs?

3. Learning Goals: In-Class By the end of this unit, you should be able to for the exam: • Build a DFA to recognize a particular language. • Identify the language recognized by a particular DFA. Discuss point of learning goals.

4. Learning Goals: In-Class By the end of this unit, you should be able to for yourself: • Use the formalisms that we’ve learned so far (especially power sets and set union) to illustrate that nondeterministic finite automata are no more powerful than deterministic finite automata. • Demonstrate through a simple contradiction argument that there are interesting properties of programs that cannot be computed in general. • Describe how the theoretical story 121 has been telling connects to the branch of CS theory called automata theory. Discuss point of learning goals.

5. Outline • Learning Goals • Formally Specifying DFAs • Designing DFAs • Analyzing DFAs: Can DFAs Count? • Non-Deterministic Finite Automata: Are They More Powerful than DFAs?

6. Reminder: Formal DFA Specification I the (finite) set of letters in the input language. S the (finite) set of states. s0 the start state; s0 S F the set of accepting states; F  S N : S  I  S is the transition function. a a,b a b b a,b

7. One Way to FormalizeDFA Operation The algorithm “RunDFA” runs a given DFA on an input, resulting in “Yes”, “No”, or (if the input is invalid) “Huh?” RunDFA((I, S, s0, F, N), in) • If in is empty, then: if s0  F, return “Yes”, else return “No”. • Otherwise, let in0 be the first element of in and inrest be the rest of in. • If in0  I, return “Huh?” • Otherwise, let s' = N(s0, in0). • Return RunDFA((I, S, s', F, N), inrest). a,b a a b b a,b

8. Formal DFA Example (Fixed) A DFA is a five-tuple: (I, S, s0, F, N) a,b a,b b a a b What is I? S? s0? F? N?

9. Running the DFA How does the DFA runner work? a,b a,b b a a b Input: abbab Only partly on the exam.(You should be able to simulate a DFA, but needn’t use RunDFA.)

10. Outline • Prereqs, Learning Goals, and !Quiz Notes • Formally Specifying DFAs • Designing DFAs • Analyzing DFAs: Can DFAs Count? • Non-Deterministic Finite Automata: Are They More Powerful than DFAs?

11. Strategy for Designing a DFA To design a DFA for some language, try: • Should the empty string be accepted? If so, draw an accepting start state; else, draw a rejecting one. • From that state, consider what each letter would lead to. Group letters that lead to the same outcome. (Some may lead back to the start state if they’re no different from “starting over”.) • Decide for each new state whether it’s accepting. • Repeat steps 2 and 3 for new states as long as you have unfinished states, always trying to reuse existing states! As you go, give as descriptive names as you can to your states, to figure out when to reuse them!

12. Problem: Design a DFA to Recognize “Decimal Numbers” Rules?

13. Problem: Design a DFA to Recognize “Decimal Numbers” • Can have a negative sign • Can have a decimal • If decimal, needs at least one digit before and one digit after the decimal.

14. Is this a decimal number? Let’s build a DFA to check whether an input is a decimal number. Which of the following is a decimal number? 3.5 2.011 -6 5. -3.1415926536 .25 --3 3-5 5.2.5 . Rules for whether something is a decimal number?

15. Regular Expressions • Syntax for defining which words to accept and which to reject • Can be converted back and forth between DFAs • Don’t worry too much about them now, they will be further introduced in lab

16. Regular Expressions Something followed by ? is optional. Anything followed by a + can be repeated. [0-9] is a digit, an element of the set{0, 1, …, 9}. – is literally –. \. means. () are used to group elements together. Rules for a decimal number: -?[0-9]+(\.[0-9]+)?

17. Problem: Design a DFA to Recognize Telephone Numbers Problem: Design a DFA to recognize telephone numbers Step 1: Design a Regular Expression

18. Problem: Design a DFA to Recognize Telephone Numbers Problem: Design a DFA to recognize telephone numbers, defined using the regular expression:((1\-)?[2-9][0-9][0-9]-)? [2-9][0-9][0-9]\- [0-9][0-9][0-9][0-9]

19. More DFA Design Problems Design a DFA that recognizes words that have two or more contiguous sequences of at least two vowels. (This is a rough stab at “multisyllabic” words.) Accepted: Sweetie, Bearee Not accepted: Queue

20. More DFA Design Problems Design a DFA that recognizes base 4 numbers with at least one digit in which the digits are in sorted order.

21. More DFA Design Problems Design a DFA that recognizes strings containing the letters a, b, and c in which all the as come before the first c and no two bs neighbour each other.

22. Outline • Prereqs, Learning Goals, and !Quiz Notes • Formally Specifying DFAs • Designing DFAs • Analyzing DFAs: Can DFAs Count? • Non-Deterministic Finite Automata: Are They More Powerful than DFAs?

23. Can a DFA Count? Problem: Design a DFA to recognize anbn: the language that includes all strings that start with some number of as and end with the same number of bs.

24. DFAs Can’t Count Let’s prove it. The heart of the proof is the insight that a DFA can only have a finite number of states and so can only “count up” a finite number of as. We proceed by contradiction. Assume some DFA recognizes the language anbn.

25. DFAs Can’t Count Assume DFA counter recognizes anbn. counter must have some finite number of states k = |Scounter|. Consider the input akbk. counter must repeat (at least) one state as it processes the kas in that string. (Because it starts in s0 and transitions k times; if it didn’t repeat a state, that would be k+1 states, which is greater than |Scounter|.) (In fact, it repeats each state after the first one it repeats up to the b.)

26. DFAs Can’t Count Consider the input akbk. counter must repeat (at least) one state as it processes the kas in that string. counter accepts akbk, meaning it ends in an accepting state. a a a ... a b ... a a b Doesn’t necessarily look like this, but similar.

27. DFAs Can’t Count Now, consider the number of as that are processed between the first and second time visiting the repeated state. Call it r. Give countera(k-r)bk instead. (Note: r > 0.) a a a ... a b ... a a b

28. DFAs Can’t Count Give countera(k-r)bk instead. countermust accept a(k-r)bk, which is not in the language anbn . Contradiction! Thus, no DFA recognizes anbn a a a ... a b ... a a b

29. Outline • Prereqs, Learning Goals, and !Quiz Notes • Formally Specifying DFAs • Designing DFAs • Analyzing DFAs: DFAs Can’t Count • Non-Deterministic Finite Automata: Are They More Powerful than DFAs?

30. Let’s Try Something More Powerful (?): NFAs An NFA is like a DFA except: • Any number (zero or more) of arcs can lead from each state for each letter of the alphabet • There may be many start states When we run a DFA, we put a finger on the current state and then read one letter of the input at a time, transitioning from state to state. When we run an NFA, we may need many fingers to keep track of many current states. When we transition, we go to all the possible next states.

31. NFAs Continued An NFA is like a DFA except: • Any number (zero or more) of arcs can lead from each state for each letter of the alphabet • There may be many start states A DFA accepts when we end in an accepting state. An NFA accepts when at least one of the states we end in is accepting.

32. Another way to think of NFAs An NFA is a non-deterministic finite automaton. Instead of doing one predictable thing at each state given an input, it may have choices of what to do. If one of the choices leads to accepting the input, the NFA will make that choice. (It “magically” knows which choice to make.)

33. Decimal Number NFA... Is Easy! . - 0-9 s1 s2 int dot 0-9 real 0-9 -?[0-9]+(\.[0-9]+)? 0-9

34. Are NFAs More Powerful than DFAs? Problem: Prove that every language an NFA can recognize can be recognized by a DFA as well. Strategy: Given an NFA, show how to build a DFA (i.e., I, S, s0, F, and N) that accepts/rejects the same strings. Hint: Formally, what is the “group of states” an NFA can be in at any given time? Build the parts of the DFA based on that (i.e., S is the set of all such things).

35. Worked Problem: Are NFAs More Powerful than DFAs? Problem: Prove that every language an NFA can recognize can be recognized by a DFA as well. WLOG, consider an arbitrary NFA “C”. We’ll build a DFA “D” that does the same thing. We now build each element of the five-tuple defining D. We do so in such a way that D faithfully simulates N’s operation.

36. Worked Problem: Are NFAs More Powerful than DFAs? C has an alphabet IC. Let ID = IC. (They operate on the same strings.) C’s states are SC. Let SD = P(SC). (Each state in D represents being in a subset of the states in C.) C starts in some subset S0C of SC. Let s0D = S0C. (D’s start state represents being in the set of start states in C.) C has accepting states FC. Let FD = {S  SC | S  FC  }. (A state in D is accepting if it represents a set of states in C that contains at least one accepting state.) C has zero or more arcs leading out of each state in SC for each letter in IC. ND(sD, i) operates on a state sD that actually represents a set of states from SC. So:

37. Decimal Number NFA... as a DFA 0-9 - . 0-9 {s1, s2} {s2} {s2, int} {dot} 0-9 else 0-9 else . {real} else { } else  0-9 LOTS of states left off that are unreachable from the start state.