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Chapter 5: 3E (page 239)

Find a match in the following instance of the Post Correspondence Problem.{[ababab],[ba],[abab],[aaa]}

Short Answer

Expert verified

The required match isaaaabab

Step by step solution

01

Post Corresponding Problem

Post Corresponding Problem is undecidable problem where we have more than one tile which contains multiple strings. The goal of this problem is to arrange the order of strings such that the numerator and denominator are identical.

02

Finding match

It is required to find the matching strings of Post Correspondence Problem in such a way that we will concatenate the strings in manner so that upper string and lower string are equal.

Let鈥檚 give number to each instance such as:

1=ababab, 2=ba, 3=abab, 4=aaa

So, pattern is:4,4,2,1

We have,4,4,2,1=aaaababaaaabab=aaaababaaaabab

Hence this is the required match.

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Most popular questions from this chapter

Show that 猢�m is a transitive relation?

Let 螕=0,1,蟽be the tape alphabet for all TMs in this problem. Define the busy beaver function BB:N鈫�N as follows. For each value of K, consider all K-state TMs that halt when started with a blank tape. LetBBk be the maximum number of1s that remain on the tape among all of these machines. Show thatBB is not a computable function.

Show that EQCFG is undecidable.

Define a two-headed finite automaton (2DFA) to be a deterministic finite automaton that has two read-only, bidirectional heads that start at the left-hand end of the input tape and can be independently controlled to move in either direction. The tape of a 2DFA is finite and is just large enough to contain the input plus two additional blank tape cells, one on the left-hand end and one on the right-hand end, that serve as delimiters. A 2DFA accepts its input by entering a special accept state. For example, a 2DFA can recognize the languageanbncn|n鈮�0 .

  • a. Let A2DFA={<M,x>|Mis a 2DFA and M acceptsx} . Show that A2DFA is decidable.
  • b. Let E2DFA={<M>|Mis a 2DFA and LM=鈭�}. Show that E2DFA is not decidable.

Question: Let螕={0,1,鈯�} be the tape alphabet for all TMs in this problem. Define the busy beaver function BB:N鈫�Nas follows. For each value of k, consider all K-state TMs that halt when started with a blank tape. LetBB(k) be the maximum number of 1s that remain on the tape among all of these machines. Show that BB is not a computable function.
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