Arjun Suresh (talk | contribs) |
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<math>L(M)</math> is finite | <math>L(M)</math> is finite | ||
Any non-trivial property about the language recognized by a Turing machine is undecidable
For a property to be non-trivial, there should exist at least two Turing machines, the property holding for the language of one (<math>T_{yes}</math>) and not holding for the language of other (<math>T_{no}</math>).
Thus, as per Rice's theorem the language describing any nontrivial property of Turing machine is not recursive. It can either be recursively enumerable or not recursively enumerable. (Obviously there are also other languages which are not recursive)
Any non-monotonic property about the language recognized by a Turing machine is unrecognizable
For a property to be non-monotonic, there should exist at least two Turing machines, the property holding for the language of one (<math>T_{yes}</math>) and not holding for the language of other (<math>T_{no}</math>) and the language of the <math>T_{yes}</math> a proper subset of the language of <math>T_{no}</math>.
Examples:
<math>L(M)</math> is regular
<math>L(M)</math> is finite
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for <math>\Sigma^*</math>.
<math>L(M) = {0}</math>
We can have <math>T_{yes}</math> for <math>{0}</math> and <math>T_{no}</math> for <math>\Sigma^*</math>.
<math>L(M)</math> is regular
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for <math>\Sigma^*</math>.
<math>L(M)</math> is not regular
We can have <math>T_{yes}</math> for <math>{a^nb^n|n\ge0}</math> and <math>T_{no}</math> for <math>\Sigma^*</math>.
<math>L(M)</math> is infinite
We cannot have <math>T_{yes}</math> and <math>T_{no}</math> such that <math>L(T_{yes}) \subset L(T_{no})</math>. Hence, this is not a non-monotonic property and Rice's <math>2^{nd}</math> theorem is not applicable. Still this language is not recursively enumerable.
Any non-trivial property about the language recognized by a Turing machine is undecidable
For a property to be non-trivial, there should exist at least two Turing machines, the property holding for the language of one (<math>T_{yes}</math>) and not holding for the language of other (<math>T_{no}</math>).
Thus, as per Rice's theorem the language describing any nontrivial property of Turing machine is not recursive. It can either be recursively enumerable or not recursively enumerable. (Obviously there are also other languages which are not recursive)
Any non-monotonic property about the language recognized by a Turing machine is unrecognizable
For a property to be non-monotonic, there should exist at least two Turing machines, the property holding for the language of one (<math>T_{yes}</math>) and not holding for the language of other (<math>T_{no}</math>) and the language of the <math>T_{yes}</math> a proper subset of the language of <math>T_{no}</math>.
Examples:
<math>L(M)</math> is regular
<math>L(M)</math> is finite
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for <math>\Sigma^*</math>.
<math>L(M) = {0}</math>
We can have <math>T_{yes}</math> for <math>{0}</math> and <math>T_{no}</math> for <math>\Sigma^*</math>.
<math>L(M)</math> is regular
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for <math>\Sigma^*</math>.
<math>L(M)</math> is not regular
We can have <math>T_{yes}</math> for <math>{a^nb^n|n\ge0}</math> and <math>T_{no}</math> for <math>\Sigma^*</math>.
<math>L(M)</math> is infinite
We cannot have <math>T_{yes}</math> and <math>T_{no}</math> such that <math>L(T_{yes}) \subset L(T_{no})</math>. Hence, this is not a non-monotonic property and Rice's <math>2^{nd}</math> theorem is not applicable. Still this language is not recursively enumerable.
GATECSE