Snick snack CPSC 121: Models of Computation 2009 Winter Term 1 DFAs in Depth Steve Wolfman, based on notes by Patrice Belleville and others

snick

snack

CPSC 121: Models of Computation2009 Winter Term 1

DFAs in DepthSteve Wolfman, based on notes by

Patrice Belleville and others

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?

• Combining DFA Languages

• CS Theory and the Halting Problem2

Lecture Prerequisites (REPEAT)

Read Section 12.2 pages 745-747 and “Designing a Finite Automaton” on 752-754 (skipping part b of the examples). We’ll reread and revisit this later; focus on getting a strong intuitive sense of how DFAs work.

Solve problems like Exercise Set 12.2 #1, #12a-c, #13a-c, and #14-19a.

Complete the open-book, untimed quiz on Vista that was due before class.

3

Learning Goals: “Pre-Class” (REPEAT)

The pre-class goals are to be able to:– Trace the operation of a deterministic finite-

state automaton (represented as a diagram) on an input, including indicating whether the DFA accepts or rejects the input.

– Deduce the language accepted by a simple DFA after working through multiple example inputs.

4

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.– Connect the graphical and formal

representations of DFAs and illustrate how each part works.

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.

Where We Are inThe Big Stories

Theory

How do we model computational systems?

Now: Establishing the profound results about the power (and lack thereof) of our DFA model and of general computational systems.

Hardware

How do we build devices to compute?

Now: Understood, in depth, a full-blown computer!

7In other words: SUCCESS!!

Outline



• Designing DFAs





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 SF the set of accepting

states; F SN : S I S is the

transition function.

a

b

b

a,b

a,ba

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).

ab

b

a,b

a,ba

Formal DFA Example (Fixed)

A DFA is a five-tuple: (I, S, s0, F, N)

a b

b

a,b

What is I? S? s0? F? N?

a,b

a

Running the DFA

How does the DFA runner work?

a b

b

a,b

a,b

a

Input: abbabOnly partly on the exam.

(You should be able to simulate a DFA, but needn’t use RunDFA.)

Outline



• Designing DFAs





Strategy for Designing a DFA

To design a DFA for some language, try:1. Should the empty string be accepted? If so, draw

an accepting start state; else, draw a rejecting one.

2. 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”.)

3. Decide for each new state whether it’s accepting.4. 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!

Problem: Design a DFA to Recognize “Decimal Numbers”

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?


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]+)?


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?

Problem: Design a DFA to Recognize “Decimal Numbers”

Problem: Design a DFA to recognize decimal numbers, defined using the regular expression:-?[0-9]+(\.[0-9]+)?

More DFA Design Problems

Design a DFA that recognizes words that have two or more contiguous sequences of vowels. (This is a rough stab at “multisyllabic” words.)

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

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.

Outline



• Designing DFAs





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.

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.

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 k as 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.)

DFAs Can’t Count

Consider the input akbk. counter must repeat (at least) one state as it processes the k as in that string.

counter accepts akbk, meaning it ends in an accepting state.

...

...

Doesn’t necessarily look like this, but similar.

a

aa a

aab

b

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 counter a(k-r)bk instead.

(Note: r > 0.)

...

...

a

aa a

aab

b

DFAs Can’t Count

Give counter a(k-r)bk instead.

counter must accept a(k-r)bk, which is not in the language anbn . Contradiction!

Thus, no DFA recognizes anbn

...

...

a

aa a

aab

b

Outline



• Designing DFAs

• Analyzing DFAs: DFAs Can’t Count




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.

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.

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.)

Decimal Number NFA... Is Easy!

s1 s2

-?[0-9]+(\.[0-9]+)?

int dot-

0-9

0-9 .

real

0-9

0-9

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).

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.

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:

DC ssCDD siisN

from labeled arcsby toconnected states ofset the),(

Decimal Number NFA... as a DFA

-

0-9

0-9 .

0-9

0-9

{s1, s2} {s2} {s2, int} {dot}

{real}{ }

0-9

.else else

else

else

LOTS of states left off that are unreachable from the start state.

Outline



• Designing DFAs


• Non-Deterministic Finite Automata: No More Powerful than DFAs



Combining Languages

Problem: Given a DFA that recognizes language A and a DFA that recognizes language B, can you build a DFA that recognizes:• Any string in language A or B• Only strings in both languages A and B• Only strings that are composed of a string

from A followed by a string from B

Worked Problem: Combining Languages

Problem: Design a combined DFA “C” for any string in BOTH language A and B.

Strategy: build a DFA that “keeps track” of where it is in both DFA “A” and DFA “B”.

Let IC = IA (which must also equal IB).Let SC = SA SB.Let sC0 = (sA0, sB0).Let FC = {(sA, sB) SC | sA FA sB FB}. (Only “double” states that accept in both DFA A and B.)Let NC((sA, sB), s) = (NA(sA, s), NB(sB, s)). (Transition to the next “double” state by tracking where

A and B would go on each part of the state.)Idea: send one “finger” tracking through A and one through B.

Outline



• Designing DFAs


• Non-Deterministic Finite Automata: No More Powerful than DFAs



Punch Line to one Big Story: Automata Theory

NFAs are no more powerful than DFAs (although we might need 2n states in a DFA to simulate an NFA).

One direction that CS theory goes is to explore what different models of computation are capable of and how they’re related.

Are all models limited?

The Halting Problem

A decision algorithm outputs only “yes” or “no” for an input (like a DFA).

If we’re really smart, we can probably write a program to solve any given decision problem (in a finite amount of time), right?

How about this one: Given a program and its input, decide whether the program ever finishes (halts). I.e., does it go into an infinite loop? This was one of the deepest questions

in math (and CS!) in 1900, posed by David Hilbert.

The Halting Problem

OK, let’s say we write the code for this inside the method doesHalt:

public static boolean doesHalt(String program,

String input) {

// Don’t know what goes here,

// but we assume (for contradiction!) it exists.

}

The Halting Problem

Now, remember Russell’s Paradox and the self-referential “set that contains all sets that do not contain themselves”?

How about a program that takes a program as input and checks whether it halts when given itself as input…

public static void main(String[] args) {

// Assume the input is args[1].

if (doesHalt(args[1], args[1]))

while(true) ;

else

return;

}Let’s call our program Russ, for short.

It halts only if its input, if run on itself, would not halt.

The Halting Problem

Now, run Russ with Russ itself as input. Does Russ halt under those circumstances or not?

Remember: Russ halts if and only if: its input does not halt when run on itself.

QED, Halt, Full Stop.

Proof (originally) thanks to Kurt Gödel.

Documents

Snick snack CPSC 121: Models of Computation 2009 Winter Term 1 DFAs in Depth Steve Wolfman, based on notes by Patrice Belleville and others