Let $M$ be a NDTM (non deterministic Turing machine) which decides a certain NP-complete language, say SAT.

$M$ computes any instance $I$ of the NP-complete problem in at most $p(n)$ non deterministic steps ($p$ in a polynomial function, $n$ is the input size). The length of a computation path can be measured as the "rank" of the non deterministic step where $M$ halts. Let call $rank_M(I)$ the rank of the shortest accepting path ($rank_M(I)$ is $p(n)+1$ if $M$ rejects $I$).

Is this problem NP-complete : Given a boolean expression $I$, is it true that $rank_M(I)=k$ ? ($k$ is polynomially bounded in $n$)

Edit --- $rank_M(I)$ has been modified : it defines now the shortest accepting path (instead of just the rank of the step where $M$ halts). The question is then to know if the problem above is (many-one polytime) reducible to the NP-complete known "Bounded halting problem" cited in the @Bruno's answer/comment.

Thank you.

Let $M$ be a NDTM (non deterministic Turing machine) which decides a certain NP-complete language, say SAT.

$M$ computes any instance $I$ of the NP-complete problem in at most $p(n)$ non deterministic steps ($p$ in a polynomial function, $n$ is the input size). The length of a computation path can be measured as the "rank" of the non deterministic step where $M$ halts. Let call $rank_M(I)$ the rank of the shortest accepting path ($rank_M(I)$ is $p(n)+1$ if $M$ rejects $I$).

Is this problem NP-complete : Given a boolean expression $I$, is it true that $rank_M(I)=k$ ? ($k$ is polynomially bounded in $n$)

Edit --- $rank_M(I)$ has been modified : it defines now the shortest accepting path (instead of just the rank of the step where $M$ halts).

Thank you.

Let $M$ be a NDTM (non deterministic Turing machine) which decides a certain NP-complete language, say SAT.

$M$ computes any instance $I$ of the NP-complete problem in at most $p(n)$ non deterministic steps ($p$ in a polynomial function, $n$ is the input size). The length of a computation path can be measured as the "rank" of the non deterministic step where $M$ halts. Let call $rank_M(I)$ this rank ($rank_M(I)$ is $p(n)+1$ if $M$ rejects $I$).

Is this problem NP-complete : Given a boolean expression $I$, is it true that $rank_M(I)=k$ ? ($k$ is polynomially bounded in $n$)

Thank you.

Let $M$ be a NDTM (non deterministic Turing machine) which decides a certain NP-complete language, say SAT.

$M$ computes any instance $I$ of the NP-complete problem in at most $p(n)$ non deterministic steps ($p$ in a polynomial function, $n$ is the input size). The length of a computation path can be measured as the "rank" of the non deterministic step where $M$ halts. Let call $rank_M(I)$ the rank of the shortest accepting path ($rank_M(I)$ is $p(n)+1$ if $M$ rejects $I$).

Is this problem NP-complete : Given a boolean expression $I$, is it true that $rank_M(I)=k$ ? ($k$ is polynomially bounded in $n$)

Edit --- $rank_M(I)$ has been modified : it defines now the shortest accepting path (instead of just the rank of the step where $M$ halts). The question is then to know if the problem above is (many-one polytime) reducible to the NP-complete known "Bounded halting problem" cited in the @Bruno's answer/comment.

Thank you.