Machines That Can’t Count

CS 15-251Lecture 15 Lecture 15 b

ba

b

a

a

a

ba

b

Let me teach you a programming language so simple

that you can learn it in

less than a minute.

Meet “ABA” The Automaton!Meet “ABA” The Automaton!

b

ba

ba

a

a

bab

Input StringInput String ResultResult

abaaba AcceptAccept

aabbaabb RejectReject

aabbaaabba AcceptAccept

AcceptAccept

The Simplest Interesting Machine:The Simplest Interesting Machine:

Finite State MachineFinite State Machine OR OR

Finite Automaton Finite Automaton

Finite set of Finite set of statesstates

A start stateA start state

A set of A set of accepting accepting statesstates

A finite A finite alphabetalphabet

a b #a b #

x 1x 1State State transition transition instructionsinstructions

1 2{ , , , , }o kQ q q q q

Finite AutomatonFinite Automaton

oq

1 2, , ,

ri i iF q q q

:

( , )i j

Q Q

q a q

iq jq

a

How it worksHow it works

•At any point in time, it’s in one of its states (or it has crashed).

•It starts in its start state.

•It “reads” one letter at a time from the input string (going from left to right).

If it is in state qi and reads a then

If it has the transition

Then it changes to state qj

Else it crashes

( , )i jq a q

M accepts the string x if when M reads x it ends in an accepting state.

M rejects the string x if when M reads x it ends in a non-accepting state.

M crashes on x if M crashes while reading x.

Let M=(Q,,F,) be a finite automaton.

The set (or language) accepted by The set (or language) accepted by M is:M is:

x

*|

...

M accepts x

All length k strings over the alphabet

k

* 0

n s

1 2 3

Notice that this is

What is the language What is the language accepted by this machine?accepted by this machine?

L = {a,b}* = all finite strings of a’s and b’s

a,b

What is the language What is the language accepted by this machine?accepted by this machine?

L = all even length strings of a’s and b’s

a,b

a,b

What machine accepts this What machine accepts this language?language?

L = all strings in {a,b}* that contain at least one a

aa,bb

What machine accepts this What machine accepts this language?language?

L = strings with an odd number of b’s and any number of a’s

b aa

b

What is the language What is the language accepted by this machine?accepted by this machine?

L = any string ending with a b

b ba

a

What is the language What is the language accepted by this machine?accepted by this machine?

L = any string with at least two a’s

b ba,ba a

What machine accepts this What machine accepts this language?language?

L = any string with an a and a b

b

ba,b

a

a

b a

What machine accepts this What machine accepts this language?language?

L = strings with an even number of ab pairs

a

bb

a

a

b a

b

L = all strings containing ababb as a consecutive substring

b

a,ba a

b

b

bb

aa

a

a ab aba abab

If I am in state “aba” then the longest prefix of ababb that forms a suffix of the string I have read until now is aba.

Problem:Does the string S appear inside the text T ?

The “grep” ProblemThe “grep” Problem

Input:• text T of length t• string S of length n

Cost: O(nt) comparisons1 2 3, , , , ta a a a

symbols

n Naïve method:

Automata SolutionAutomata Solution

•Build a machine M that accepts any string with S as a consecutive substring.

•Feed the text to M.

•Cost: t comparisons + time to build M.

•As luck would have it, the Knuth, Morris, Pratt algorithm builds M quickly.

•By the way, it can be done with fewer than t comparisons in the worst case!

Real-life uses of finite state Real-life uses of finite state machinesmachines

•grep

•coke machines

•thermostats (fridge)

•elevators

•train track switches

•lexical analyzers for parsers

Let L be a language.

L is called a regular language if there is some finite automaton that accepts L.In this lecture we have seen many regular languages.•

• even length strings

• strings containing ababb

Theorem: Any finite langage is regular.

Proof: Make a machine wth a “path” for each string in the language.

Example: L = {a, bcd, ac, bb}

b

d

a

c b c

Are all languages regular?

NO!NO!Counting Argument

•There are uncountably many L

•There are countably many automata (each automaton can be specified by a finite string)

NO!NO!Halting Argument

•K = { P | P(P) } is undecidable for ideal computers with unlimited memory, i.e., K is not recursive.

•Finite automata can be simulated by computers. (they are a weak sub-class)

The fact that not all languages are regular can be argued more directly (without resorting to macho techniques).

, , , ,n na b ab aabb aaabbb

Consider the language

i.e., a bunch of a’s followed by an equal number of b’s

No finite automaton accepts this language.

Can you prove this?

APE 0835

anbn is not regular. No machine has enough states to keep track of the number of a’s it might encounter.

That is a fairly weak argument. Consider the following example…

Can’t be regular. No machine has enough states to keep track of the number of occurrences of ab.

L = strings where the # of occurrences of the pattern ab is equal to the number of occurrences of the pattern ba

Remember “ABA”?Remember “ABA”?

b

ba

ba

a

a

bab

ABA accepts only the strings with an equal number of ab’s and ba’s!

Professional Strength ProofProfessional Strength Proof

Theorem: anbn is not regular.

Proof: Assume that it is. Then M with k states that accepts it.

For each 0 i k, let Si be the state M is in after reading ai.

i,j k s.t. Si = Sj, but i j

M will do the same thing on aibi and ajbi . (to Si ) (to Si )

But a valid M must reject ajbi and accept aibi.

MORAL:MORAL:

Finite automata can’t count.

AdvertisementAdvertisement

You can learn much more about these creatures in the FLAC course.

Formal Languages, Automata, and Computation

• There is a unique smallest automata for any regular language

• It can be found by a fast algorithm.

• ……

Cellular AutomataCellular Automata•Line up a bunch of identical finite automata in a straight line.

( , , )k i j l

Q Q Q Q

q q q q

kq lq

,i jq q

iq jqkq

•Transitions are based on the states of the machine’s two neighbors or an indicator that a neighbor is missing. (There is no other input.)

•All cells move to their next states at the same time:synchronous transition

The Firing Squad ProblemThe Firing Squad Problem•Five “soldiers” all start in the sleep state. You change the one on the left to the wake state.•All five must get to the fire state at the same time (for the first time).

sleepwakefire

sleepwakefire

sleepwakefire

sleepwakefire

sleepwakefire

sleepwakefire

sleepwakefire

sleepwakefire

sleepwakefire

sleepwakefire

ShorthandShorthand

, , ,Q a b d

Means use this transition when your left neighbor is in any state at all and your right neighbor is in state a,b, or d.

11 22 33 44 55

11 wakewake sleepsleep sleepsleep sleepsleep sleepsleep

22 wakewake22

sleepsleep sleepsleep sleepsleep sleepsleep

33 wakewake33

wakewake33

wakewake33

sleepsleep sleepsleep

44 wakewake44

wakewake44

wakewake44

wakewake44

sleepsleep

55 firefire firefire firefire firefire firefire

sleep

wake wake2 wake3 wake4

fire!

{w2},Q{w},Q {w3},Q

Q,Q

{w4},Q

{sleep,fire,end},Q

Q,Q Q,Q Q,Q

QuestionQuestion

No. Finite automata can’t count!

Can you build the soldier’s finite automaton brain before you know how many soldiers will be in the line?

Don’t jump to conclusions! It is possible to design a single cellular automaton that works for any number of soldiers!

Useful subroutine:method to find the center

after leftmost soldier is awakened.

Define the center of the line to be:

•The guy in the middle if the # of soldiers is odd.

•The two guys in the middle if the the # of soldiers is even.

Your neighbor “notices” that you are in a signaling state.

An automaton can “signal” to its neighbors.

Certain subsets of its states will correspond to signaling.

signal

normaloperation

A Method to Find the CenterA Method to Find the Center• Send a signal back and forth from “end” to “end”.

• Each time it gets to an “end”, send it back, but tell the neighbor of the “end” that it is the “new end”.

•The center(s) will get a “new end” signal from both sides.

• # of steps: 2O n

Slick Center AlgorithmSlick Center Algorithm

• Use two signals: one fast and one slow

• When you get the fast signal, pass it on (or bounce it back) on the next time step.

• When you get the slow signal, hold it for two steps, then pass it on the next time step.

• When you wake and you are leftmost, consider yourself to hold both the slow and fast signal.

• If you get one signal from each side, shout, “I am the center!”

Firing Squad ProblemFiring Squad Problem

•Keep finding the middle recursively: If you’re the center, mark yourself as an “end” and find the centers on each side of you.

•If you are an “end” and your neighbor is starting both signals, send back a “fire next step”.

Firing Squad AnalysisFiring Squad Analysis

3 3 3 3

2 2 2 2 4 2 83 1 1 1

12 2 4 8

32 3

2

n n nn

n

n n

How many steps to find the center?

How many steps to fire?

3 steps

2n

n

2

n

Lowed BoundLowed Bound

Can you see why any solution must use at least 2n steps?

There is a clever way to do it in 2n steps.

Other ResearchOther Research

• Different topologies

• Self-stabilizing clocks

• all cells fire simultaneously at regular intervals

• even if you mess with the states of all the cells they get back on schedule (eventually)