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Magazine Chapter 3: 6E (page 188)
In Theorem 3.21 we showed that a language is Turing-recognizable iff some enumerator enumerates it. Why didn鈥檛 we use the following simpler algorithm for the forward direction of the proof? As before, s1,s2,... is a list of all strings in
E = 鈥淚gnore the input.
1. Repeat the following for
2. Run M on si.
3. If it accepts, print out si.鈥�
Short Answer
We need to simulate on each of the strings for a fixed length of time so that no looping can occur.
Step by step solution
Turing machine.
A Turing Machine (TM) is a mathematical model which consists of an infinite length tape divided into cells on which input is given. It consists of a head which reads the input tape. A state register stores the state of the Turing machine.
A Turing machine consists of a tape of infinite length on which read and writes operation can be performed. The tape consists of infinite cells on which each cell either contains input symbol or a special symbol called blank. It also consists of a head pointer which points to cell currently being read and it can move in both directions.
Solution.
The problem with the proof is that M on si might loop forever. If it loops forever, then E0 doesn鈥檛 print out si.
More importantly, E isn鈥檛 going to move on to test the next string. Therefore, it won鈥檛 be able to enumerate any other strings in L.
E = 鈥淚gnore the input.
1. Repeat the following for
2. Run M on si.
3. If it accepts, print out si.鈥�
For this reason, we need to simulate M on each of the strings for a fixed length of time so that no looping can occur.
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