Theory of Computing

Lecture 23MAS 714

Hartmut Klauck

The game

• Two players, Alice and Bob• Alice receives an input x2§*• Bob receives y2§*• Alice sends a message to Bob, who decides

whether x±y2 L– ±: concatenation, we will also just write xy

• Cost of the game: number of messages used

DFA and the game

• Theorem [DFA vs. game]:– If L is regular, then the optimal number of

messages in the game is equal to the smallest number of states in a DFA.

– If L is not regular, then the smallest number of messages is infinite

Application

• How to show that we need many messages?• Consider the infinite matrix M, rows labeled

with x2§* and columns labeled with y2§* and M[x,y]=1 iff xy2 L

• Call this matrix the communication matrix of L• Lemma: The minimum number of messages in

the game is equal to the number of distinct rows in the matrix

Proof

• Two rows labeled x, x’ are distinct if there is a column y where M[x,y]=1 and M[x’,y]=0– or vice versa

• If Alice sends the same message on x and x’ Bob cannot distinguish x and x’, so there is an error on xy or on x’y

• On the other hand it is enough for Alice to send the same message for rows that are not distinct

Examples

• Parity: M has two distinct rows– one for x with odd number of 1’s and for even

• {0n 1n}:– Consider the rows labeled 0k for all k=1,…,1– The rows are all different

• Conclusion: {0n 1n: n natural} is not a regular language• Note: We can consider subsets of the rows and columns of

M and show that this submatrix has infinitely many distinct rows. Then M also has infinitely many distinct rows

Example

• L={xx: x2{0,1}* } is not regular• Proof: we need to show the number of distinct

rows is infinite• Consider the rows and columns labeled by 0n1

for all natural n• Take two rows labeled by x=0p1 and x’=0r1• Clearly there is a 1 in M[x,y] for y=0p1 but there

is no 1 in M[x’,1]• Again M contains an infinite identity matrix

Proof [DFA to game]

• Assume L is regular. There is a DFA M for L.• Alice simulates M on x, and sends the state of

M reached after reading x to Bob, who continues the simulation

• Bob accepts iff M accepts xy• Clearly this is a correct protocol, and the

number of messages is the size of the set of states

Proof [game to DFA]• Now assume that there is a protocol with m messages• We can assume that Alice uses one message for each distinct row of

the matrix, otherwise the protocol is not optimal– i.e., for two rows that are not distinct she uses the same message

• We construct a DFA with m states corresponding to the m messages• q0 [the starting state] is the message that Alice sends on the empty

string• We use m-1 other states, corresponding to messages

• Set of accepting states: messages/states, such that Bob accepts on the message when y is empty string

• Remains to define the transition function

Proof

• Assume that on some input x Alice sends message q, and on input xa she sends message q’

• Then we include the transition q,a q’• Clearly this defines transitions, but are they consistent?• I.e., If Alice sends q on x and q’ on xa, and Alice also sends q

on x’, then she should send q’ on x’a• But x and x’ have the same row, hence xa and x’a also have

the same row [if not, there is some y such that xay2 L and x’ay not in L, hence rows of x and x’ not the same]

• Hence the transition function is well-defined.• We get a DFA with m states for L

Note

• The minimum number of messages in our communication game is usually called the Myhill-Nerode index

2-way DFA

• Theorem: 2-way DFA can recognize only regular languages

• Note:– Sometimes it is easier to show that L is regular by

using a 2-way DFA– Sometimes 2-way DFA need fewer states– Proof: We show that the Myhill-Nerode index is

finite.

Examples

• L={w: third last symbol of w is 1}• 2-way DFA: move to the right end of the input,

move 3 steps left, accept if there is a 1

• Lrev for a regular language L is the set of w, such that w read in reverse is in L– Move to the right end and then simulate the DFA

for L while reading the input backwards

Proof (theorem)

• Proof: We show that the Myhill-Nerode index is finite.• Language L is decided by a 2-way DFA M, where M has n

states• Consider words w=xy• Suppose the tape-head of M comes back from y into x

while M is in state q, and the same happens later in state q’– qq’ otherwise there is an infinite loop

• Hence the states on which M crosses from y to x are distinct, and there are at most n such crossings

• Simulation in the communication game:

Proof• We describe the communication protocol• Let S=[q1,...,qm] be a sequence of any m· n distinct states of M• Alice simulates M on x, assuming that when M first leaves x it later

returns to x in state q1 – Etc: If M leaves x for the k-th time it returns to x in state qk

• Alice’s message: for every possible S Alice sends the m states in which M leaves x (for even m), and additionally if she would accept, for odd m

• Bob finds the sequence S that is consistent with M’s behavior on y and Alice’s message (S is the correct sequence occurring on M and xy)

• Bob accepts if M accepts• It is easy to see that this protocol is correct, and uses no more than

O(n2n) messages, which is finite

2-way NFA

• Definition: a two-way NFA is a nondeterministic Turing machine that cannot write and cannot leave the tape area that contains the input.

• Theorem: Two-way NFA can decide only regular languages

• Proof: Same as for two-way DFA, Alice and Bob check consistency with respect to the nondeterministic A– Is there a computation that leads to the communicated

sequence of states and accepts?

Open Problem

• Is there a language, for which 2-way NFA need much fewer states than 2-way DFA?

Size of Automata?

• We know L is regular, if there is a DFA with a finite number of states for L

• Minimum number of states is the Myhill-Nerode Index

• Questions:1. How can we find a minimal DFA for L?2. Is the minimal DFA unique (up to renaming states)?3. Are NFA sometimes smaller than minimal DFA?4. Are two-way DFA sometimes smaller than DFA?

Question 4

• Consider the language L={xi : x2{0,1}n and i2{0,1} log n and xn-i=1}– n is fix, this is a finite language, hence regular

• It is easy to see that there are at least 2n rows in the communication matrix– Consider only columns for the different i=n-1,…,0– Rows labeled x contains the string x at entries i=n-1…0

• Hence any DFA for L has at least 2n states• Can give a 2-way DFA with O(n2) states:– go to the right end of the input, read i into the state, move i

steps left, accept if there is a 1

Question 3

• Consider L={xy: x,y 2 {0,1}n, x y}– This is a FINITE language, n is fixed

• Exercise: there is an NFA with O(n2) states for L• DFA-size: the matrix has more than 2n distinct

rows– DFA size is exponential

• Hence NFA can be exponentially smaller than DFA for some languages

• Example where they are not: complement of L

2) Uniqueness

• Uniqueness of the minimal DFA (up to vertex names) follows from the Myhill-Nerode characterization

• The optimal DFA has exactly one state for each set of equal rows of the communication matrix

• Edges/the transition function are also determined uniquely, i.e., if x is in set q and xa is in set q’ then there must be an edge labeled a from q to q’