subject

Questions refer to the following variant of Turing machines. A linear bounded automaton (LBA) is like an ordinary one-tape Turing machine, except for the fol lowing modification: its tape alphabet contains an extra distinguished symbol - the night endmarker, and the machine is constrained never to move to the right of the right endmarker nor overwrite it with a different symbol. The input string is initially enclosed between the left and right endmarkers with no blank cells, and the machine starts in its start states scanning the left endmarker ; that is, the start configuration on input res is (s, 30). 1. State formally in terms of the transition function 6: Qr+Qxrx (L, R) what it means to say, "The machine is constrained never to move to the right of the right endmarker nor overwrite it with a different symbol." Take care to use proper quantification E, M) in your formal statement.
2. Prove that the halting problem for LBAS is decidable.
3. Prove that it is undecidable whether a given LBA runs in polynomial time. (Hint. Encode the halting problem for arbitrary Turing machines Given Me, build an LBA with M and encoded in its finite control. On any input the LBA ce its input and do something interesting with Afand).

ansver
Answers: 1

Another question on Computers and Technology

question
Computers and Technology, 22.06.2019 15:10
David is in week 3 of his current ashford course and has a paper due by monday night at midnight. he has finished everything but the concluding paragraph. as he boots up his computer to work on it, he sees a flash across the screen and then the screen goes black. he begins to panic as he tries desperately to turn the laptop back on. david should have saved his work on what kind of portable device?
Answers: 2
question
Computers and Technology, 22.06.2019 20:00
What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }
Answers: 3
question
Computers and Technology, 23.06.2019 22:30
What would be the address of the cell, which is at the intersection of the second row and the third column in a worksheet?
Answers: 1
question
Computers and Technology, 24.06.2019 11:00
Under the home tab, where can a user find options to change the bullet style of an outline? in the slides group in the font group in the paragraph group in the drawing group
Answers: 1
You know the right answer?
Questions refer to the following variant of Turing machines. A linear bounded automaton (LBA) is lik...
Questions
question
Mathematics, 01.03.2021 14:00
question
Biology, 01.03.2021 14:00
question
Mathematics, 01.03.2021 14:00
question
Social Studies, 01.03.2021 14:00
Questions on the website: 13722360