Computability
Construct TMs. Decidability.
Preview: next class: diagonalization
and Halting theorem.
Homework/Classwork
• Last Thursday, I showed a design for a TM that
accepts {ww | w a string in {0,1}*}. It was
decidable (always halts).
• What about
– {anbncn | n>=0}
Homework
• Video: reactions to video including reading
from one of the original Turing papers?
Enumerator
• Recall: this was a type machine that takes no input
but prints out members of a language L.
– If a language is Turing Recognizable, then there exists (we
can design) an enumerator that prints out members of the
language. If there exists an enumerator for a language L,
then there is a TM that recognizes it.
– IF a language is decidable, we can build an enumerator
that prints out the strings in lexical order. If an enumerator
prints out the machines in lexical order, then it is Turing
decidable.
• Note: we aren't saying that if a given enumerator doesn't print out
the strings in lexical order, there couldn't be another one that
does….
Enumerator Homework/classwork
• Provide formal definition of enumerator
Decidable vs Turing Recognizable
• A language is decidable if there is a TM that
accepts the language and always halts.
• A language is Turing-recognizable if there is a
TM that accepts the language and may or may
not halt on some of the strings NOT in the
language.
• All decidable languages are TM recognizable.
Hierarchy Bull's eye
Each is contained in and strictly smaller than next:
FSM  NDFSM  regular expressions
(deterministic push down automaton)
Push Down Automation  Context Free Grammars
Turing decidable  nondeterministic TM decidable
Turing recognizable  nondeterministic TM recognizable
Decidability question
• The language
{<B, w> | B is a FSM that accepts w} is decidable.
That is, there exists a TM that says yes for a
<B,w> when B accepts w, and says no
(stops and says no) if B does not accept w.
• Informal proof:
– Assume some encoding of B that a TM can check. If
the first part of input doesn't fit a valid encoding for a
FSM, reject. First states of TM check the encoding
– Add to TM the simulation for B. Also record on the
tape the state and the position scanning w. Do
simulation. When done (known by position), check if
state is accepting state.
More…
• The language
{<E,w> | E is a regular expression that
generates w} is decidable
– convert E to FSM
– Build TM to simulate the FSM. Try on w.
– Accept or reject
The language {<A> | A is a FSM and
L(A) is empty} is decidable
• Design encoding for FSM: list states and list triples
representing the arcs: pairs of states and the letter. Build TM
that checks for valid encoding. Note: alphabet includes
symbol for each state.
• Grow alphabet to include marked state for every state symbol.
• Add to the TM coding that does following marking: mark start
state. Scanning the triples, mark any state that has a transition
from a marked state. Repeat for new round. Stop when no
new state has been marked in the round. So this will take at
most N rounds when N is number of states. Are any accept
states marked? If yes, accept; otherwise reject.
Context Free Grammar
• {<B, w> | B is a CFG and generates w} is
decidable.
• Informal proof:
– Generate the CFG in Chomsky normal form that is
equivalent to B.
Claim: If length of w is n, there is a derivation of w
that is not greater than 2n + 1 steps.
– Given the limit on the size of the parse
tree/number of derivations, CLAIM we can
construct a TM that tries each one.
Parse tree
2*n -1 derivations
S
A
A1
B
B1
A2
B2
…..
w1
w2
w3
…
wn
CFG case
• The fact that the number of derivations is
bounding means that the TM knows when it
has checked enough possibilities and can
reject if there isn't a parse tree!!!
Preview: Halting problem
• Take two inputs: encoding of a TM T and input
w for that TM and simulate / run T on w.
• Will show that
{<M,w>|M is a TM and w is a string and M
accepts w} is
– Turing-recognizable but not
– Turing-decidable