We say a TM decide a language L if it accepts all strings in L and rejects all strings not in L
We say a TM recognizes language L if it accepts all strings in L. It may or may not reject any string not in L- i.e., it might go into an infinite loop in case of non-acceptance. (It will never accept a string not in L)
We say a TM accepts a language (i.e., a set of strings) L, if all strings in L are accepted. For strings not in L, they can either be accepted or rejected or can cause infinite loop. (It might accept a string not in L).
(This definition is not standard but just using it to explain cases where we say TM M accepts {0,00} etc. You shouldn't define Turing acceptable language like this as this definition makes any language Turing acceptable)
Any non-trivial property of the LANGUAGE recognizable 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)
(1) <math>L(M)</math> has at least 10 strings
We can have <math>T_{yes}</math> for <math>\Sigma^*</math> and <math>T_{no}</math> for <math>\phi</math>. Hence, $L = \left\{M \mid L(M) \text{ has at least 10 strings}\right\}$ is not Turing decidable (not recursive). (Any other $T_{yes}$ and $T_{no}$ would also do. $T_{yes}$ can be any TM which accepts at least 10 strings and $T_{no}$ any TM which doesn't accept at least 10 strings )
(2) <math>L(M)</math> has at most 10 strings
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for <math>\Sigma^*</math>. Hence, <math>L = \{M\mid L(M)</math> has at most 10 strings<math>\}</math> is not Turing decidable (not recursive).
(3) <math>L(M)</math> is recognized by a <math>TM</math> having even number of states
This is a trivial property. This set equals the set of recursively enumerable languages.
(4) <math>L(M)</math> is a subset of <math>\Sigma^{*}</math>
This is a trivial property. All languages are subset of <math>\Sigma^{*}</math> and hence this set contains all languages including all recursively enumerable languages.
Any non-monotonic property of the LANGUAGE recognizable 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 <math>T_{yes}</math> must be a proper subset of the language of <math>T_{no}</math>.
(1) <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>\phi \subset \Sigma^*</math>). Hence, <math>L = \{M\mid L(M)</math> is finite<math>\}</math> is not Turing recognizable (not recursively enumerable)
(2) <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>\{0\} \subset \Sigma^*</math>). Hence, <math>L = \{M\mid L(M) = \{0\}\}</math> is not Turing recognizable (not recursively enumerable)
(3) <math>L(M)</math> is regular
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for any non-regular language. Hence, <math>L = \{M\mid L(M)</math> is regular<math>\}</math> is not Turing recognizable (not recursively enumerable)
(4) <math>L(M)</math> is not regular
We can have <math>T_{yes}</math> for <math>\{a^nb^n\mid n\ge0\}</math> and <math>T_{no}</math> for <math>\Sigma^*</math> (<math>\{a^nb^n\mid n\ge0\}\subset \Sigma^*</math>). Hence, <math>L = \{M\mid L(M)</math> is not regular<math>\}</math> is not Turing recognizable (not recursively enumerable)
(5) <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, <math>L = \{M\mid L(M)</math> is infinite <math>\}</math> is not Turing recognizable (not recursively enumerable)
(6) <math>L(M)</math> has at least 10 strings
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.
This language is in fact Turing recognizable. See here
(7) <math>L(M)</math> has at most 10 strings
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for <math>\Sigma^*</math>(<math>\phi \subset \Sigma^*</math>). Hence, <math>L = \{M\mid L(M)</math> has at most 10 strings<math>\}</math> is not Turing decidable (not recursive)
We say a TM decide a language L if it accepts all strings in L and rejects all strings not in L
We say a TM recognizes language L if it accepts all strings in L. It may or may not reject any string not in L- i.e., it might go into an infinite loop in case of non-acceptance. (It will never accept a string not in L)
We say a TM accepts a language (i.e., a set of strings) L, if all strings in L are accepted. For strings not in L, they can either be accepted or rejected or can cause infinite loop. (It might accept a string not in L).
(This definition is not standard but just using it to explain cases where we say TM M accepts {0,00} etc. You shouldn't define Turing acceptable language like this as this definition makes any language Turing acceptable)
Any non-trivial property of the LANGUAGE recognizable 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)
(1) <math>L(M)</math> has at least 10 strings
We can have <math>T_{yes}</math> for <math>\Sigma^*</math> and <math>T_{no}</math> for <math>\phi</math>. Hence, $L = \left\{M \mid L(M) \text{ has at least 10 strings}\right\}$ is not Turing decidable (not recursive). (Any other $T_{yes}$ and $T_{no}$ would also do. $T_{yes}$ can be any TM which accepts at least 10 strings and $T_{no}$ any TM which doesn't accept at least 10 strings )
(2) <math>L(M)</math> has at most 10 strings
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for <math>\Sigma^*</math>. Hence, <math>L = \{M\mid L(M)</math> has at most 10 strings<math>\}</math> is not Turing decidable (not recursive).
(3) <math>L(M)</math> is recognized by a <math>TM</math> having even number of states
This is a trivial property. This set equals the set of recursively enumerable languages.
(4) <math>L(M)</math> is a subset of <math>\Sigma^{*}</math>
This is a trivial property. All languages are subset of <math>\Sigma^{*}</math> and hence this set contains all languages including all recursively enumerable languages.
Any non-monotonic property of the LANGUAGE recognizable 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 <math>T_{yes}</math> must be a proper subset of the language of <math>T_{no}</math>.
(1) <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>\phi \subset \Sigma^*</math>). Hence, <math>L = \{M\mid L(M)</math> is finite<math>\}</math> is not Turing recognizable (not recursively enumerable)
(2) <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>\{0\} \subset \Sigma^*</math>). Hence, <math>L = \{M\mid L(M) = \{0\}\}</math> is not Turing recognizable (not recursively enumerable)
(3) <math>L(M)</math> is regular
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for any non-regular language. Hence, <math>L = \{M\mid L(M)</math> is regular<math>\}</math> is not Turing recognizable (not recursively enumerable)
(4) <math>L(M)</math> is not regular
We can have <math>T_{yes}</math> for <math>\{a^nb^n\mid n\ge0\}</math> and <math>T_{no}</math> for <math>\Sigma^*</math> (<math>\{a^nb^n\mid n\ge0\}\subset \Sigma^*</math>). Hence, <math>L = \{M\mid L(M)</math> is not regular<math>\}</math> is not Turing recognizable (not recursively enumerable)
(5) <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, <math>L = \{M\mid L(M)</math> is infinite <math>\}</math> is not Turing recognizable (not recursively enumerable)
(6) <math>L(M)</math> has at least 10 strings
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.
This language is in fact Turing recognizable. See here
(7) <math>L(M)</math> has at most 10 strings
We can have <math>T_{yes}</math> for <math>\phi</math> and <math>T_{no}</math> for <math>\Sigma^*</math>(<math>\phi \subset \Sigma^*</math>). Hence, <math>L = \{M\mid L(M)</math> has at most 10 strings<math>\}</math> is not Turing decidable (not recursive)
GATECSE