This example also illustrates why adding non-determinism to the original deterministic model \(\mathfrak{T}\) does not enlarge the class of decidable problems. For if \(N \in \mathfrak{N}\) decides \(X\), then it is possible to construct a machine \(T_N \in \mathfrak{T}\) which also decides \(X\), by successively simulating each of the finitely many possible sequences of non-deterministic choices \(N\) might have made in the course of its   computation.[12]   It is evident that if \(N\) has time complexity \(f(n)\), then \(T_N\) must generally check \(O(k^{f(n)})\) sequences of choices (for fixed \(k\)) in order to determine the output of \(N\) for an input of length \(n\). While the availability of such simulations shows that the class of languages decided by \(\mathfrak{N}\) is the same as that decided by \(\mathfrak{T}\) – i.e. exactly the recursive ones – it also illustrates why the polynomial time decidability of a language by a non-deterministic Turing machine only guarantees that the language is decidable in exponential time by a deterministic Turing machine.