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Chapter 4: 27P (page 212)

Let E={M|M is a DFA that accepts some string with more 1s than 0s}. Show that E is decidable. (Hint: Theorems about CFLs are helpful here.)

Short Answer

Expert verified

E is decidable

Step by step solution

01

Definition of DFA

DFA is short for Deterministic finite automata. Automata consists of various states and edges between states to represent the transitions from one state to another. In deterministic finite automata, there is only path from current to next state for some specific input.

02

Show that E is decidable

Let A = x is a language in which x has more 1s than 0s. There exists a PDA to recognize language A. So, A is a context-free language. Now, build the TM M as follows to determine E:

Construct Bwhich is an intersection of Aand L. The language Bis also context free because Lis regular andAis context free. The language is accepted if Bis empty. Otherwise, it is rejected.

Hence, E is decidable.

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Most popular questions from this chapter

Let INFINITEPDA={(M)|MisaPDAandL(M)isaninfinitelanguage}.

Show thatINFINITEPDAis decidable.

Show that the problem of determining whether a CFG generates all strings in 1* is decidable. In other words, show that{{G}|GisaCFGover{0,1}and1*鈯�L(G)} is a decidable language.

Let PREFIX鈭�FREEREX=role="math" localid="1659933266276" {{R}|RisaregularexpressionandL(R)isprefix-free}. Show that PREFIX鈭�FREEREXis decidable. Why does a similar approach fail to show thatPREFIX鈭�FREECFG is decidable?

Let A={M|M鈥�is鈥�鈥�a鈥�鈥�DFA鈥�鈥�that鈥�鈥�doesn't鈥�鈥�accept鈥�鈥�any鈥�鈥�string鈥�鈥�contianing鈥�鈥�an鈥�鈥�odd鈥�number鈥�of鈥�鈥�1s}. Show that A is decidable.

Let ETM={M|M鈥�is鈥�a鈥�TM鈥�and鈥�LM=蠒}.Show that ETM炉, the complement of ETM, is Turing-recognizable.
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