Download Report

CS 235
Automata etc
Rhys Price Jones
Introduction to Turing Machines
Some Questions are Decidable
Given a CFG G and a regular expression r is L(G) ⊂ L(r)?
___
L(G) ∩ L(r) is CF (why?)
We can decide if a CF language is empty (how?)
L(G) is contained in L(r) if and only if
___
there is empty intersection between L(G) and L(r) (why?)
Some Questions are Decidable
Given a CFG G, does G generate any odd-length
strings?
is decidable
How?
Some Questions are Decidable
Given a CFG G, does G generate any odd-length
strings?
Lodd = L((0+1)(00+01+10+11)*) is regular
L(G) ∩ Lodd is CF (why?)
G generates an odd-length string if and only if
L(G) ∩ Lodd ≠ ∅
and we can decide that (how?)
Some Questions are Undecidable
Given CFL L, is L = Σ*?
Given CFL L, is the complement of L CF?
Given CFL L, is L regular?
Given two CFLs L1 and L2 is L1 = L2?
Given CFLs L1, L2 is L1 ⊆ L2?
Given CFLs L1, L2 is L1 ∩ L2 = ∅?
Is the CFL L inherently ambiguous?
Is the context-free grammar G ambiguous?
Let’s build the Math
so we can prove things are undecidable?
“That is undecidable” is a much better answer
than “I don’t know how to do that”
We need a formal notion of computability
Turing machines!
As defined by Sipser
A Turing Machine is a 7-tuple (Q ,Σ,Γ,δ,q0,qa,qr)
Q is a finite set of states
Σ is a finite input alphabet not including the blank
symbol ☐
Γ is a finite tape alphabet with ☐∊Γ and Σ⊆Γ
δ:Q×Γ → Q×Γ×{L,R} is the transition function
q0,qa,qr ∊ Q are the start, the accept and the reject states
TM def Continued
Initially M receives input from Σ* on the left end of
its one-way infinite tape.
The rest of the tape is filled with the blank symbol
☐. The head starts on the left end of the tape.
The initial state is q0, after which the behavior of
M is governed by the transition function δ.
but the head cannot move off the left end.
TM as recognizer
TMs should halt in states qa and qr
accept or reject
(Rhys
! and JFlap: TM can also reject by crashing)
The collection of strings that M accepts is the
language recognized by M, denoted L(M)
A language is Turing-recognizable if some TM
recognizes it
TMs don’t have to halt
A TM can
accept by halting in state qa
reject by halting in state qr
(Reject by crashing — RPJ and JFlap)
loop forever
A TM that halts on all inputs is a decider
A language is Turing decidable if it’s the language of a decider.
Sipser Example 3.7
TM exercises
Build TM to
recognize {anbmcn+m |n,m >= 0}
Look at checkadditionTM.jff
try many inputs: aabbcccc, etc
TM exercises
Build TM to
recognize {anbmcn+m |n,m >= 0}
Look at checkadditionTM.jff
try many inputs: aabbcccc, etc
Now try ababcccc
ooopps! (checkadditionTMfixed.jff)
TM exercises
Build TMs to:
recognize {anbmcn*m |n,m >= 0}
recognize {anbncn |n >= 0}
recognize {anbn! |n >= 0}
TM as calculator
Start with 111+11 on tape, end in qa with 11111
unaryaddition.jff
Similar machines for *, -, /, ...
For a calculator of a different flavor, try Sipser’s
Example 3.12
Variants
You will consider some variants in Lab 9
2-way infinite tape
multitrack tape
...
Wile E Coyote wants to sell me an enhanced Sipser
TM with a “Stay Option”
Will I buy it?
TM as enumerator
Start with blank tape to be used as work tape
Another blank tape (the printer)
The class of all strings that are eventually printed out is the
language enumerated by M
Note Theorem 3.21 and its proof
L is Turing-recognizable iff an enumerator enumerates it
Given enumerator E, create recognizer M:
Whenever E prints, compare output to w, if == accept
Why is the converse direction so complicated?
“Less powerful” TMs
I can replace the infinite tape with two stacks
and still have the same power.
I can replace the infinite tape with a single queue
and still have the same power.
I can replace the infinite tape with a finite
random-access memory (as long as I have
unlimited external storage and a good
replacement algorithm -- “virtual memory”
What means “the same power”
Busy TMs
How many TMs are there with n states and a two
element tape alphabet (1 and ◻)?
(4(n+1))2n
For each of n states there are two transitions out
giving 2n transition arrows
Each has two choices 0 or ◻ to write
Each has two choices for movement L or R
and n+1 possible target states (including halt)
For n=1 there are 64 TMs
Some run forever when started on blank tape:
Of the 64 one-state TMs
Some make one move and halt
Of these, some write a non-blank symbol
Some don't
There are 20,736 2-state TMs
(4(n+1))2n|n=2 gives 124 = 20,736
Of these
some run forever when started on blank tape
some halt and never write a non-blank
some halt and write some non-blanks
Tibor Rado (1962): How many?
n-state Busy Beaver
started on blank tape will halt and write maximal
number of non-blank symbols.
2-state busy beaver:
makes 6 moves
and prints 4 1s
1-state busy beaver?
makes one move and writes one 1.
Proof ?
3-state busy beaver?
Can you improve on?
makes 11 moves
leaves 4 non-blank
(at one stage had 5)
Known busy beavers
A 3-state busy beaver writes 6 1s
A 4-state busy beaver writes 13 1s
Nobody knows about the 5-state busy beaver
but it does write at least 4,098 1s
Nobody knows the 6-state busy beaver
but it does write at least 1018267 1s