Combination of Recursive Functions

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Theorem

Let $f: \N^k \to \N$ and $g: \N^k \to \N$ be recursive functions (not necessarily total), where $k \ge 1$.

Let $\RR$ be a $k$-ary relation such that:

if $\map \RR {n_1, n_2, \ldots, n_k}$ holds, then $\map f {n_1, n_2, \ldots, n_k}$ is defined
if $\map \RR {n_1, n_2, \ldots, n_k}$ does not hold, then $\map g {n_1, n_2, \ldots, n_k}$ is defined.

Let $h: \N^k \to \N$ be the function defined as:

$\map h {n_1, n_2, \ldots, n_k} = \begin {cases} \map f {n_1, n_2, \ldots, n_k} & : \text{if } \map \RR {n_1, n_2, \ldots, n_k} \text { holds} \\ \map g {n_1, n_2, \ldots, n_k} & : \text{otherwise} \end {cases}$

so that $h$ is total.

Then $h$ is recursive.


Proof

Let $P_f, P_g, P_\RR$ be the URM programs computing, respectively, the functions $f$ and $g$ and the characteristic function $\chi_\RR$.

From Recursive Function is URM Computable, these programs are guaranteed to exist.

An informal algorithm for computing $h$ is as follows.

  1. Input $\tuple {n_1, n_2, \ldots, n_k}$.
  2. Use $P_\RR$ to determine whether $\map \RR {n_1, n_2, \ldots, n_k}$ holds.
  3. If it holds, use $P_f$ to compute $\map f {n_1, n_2, \ldots, n_k}$.
  4. If it does not hold, use $P_g$ to compute $\map g {n_1, n_2, \ldots, n_k}$.

The result follows from URM Computable Function is Recursive.

$\blacksquare$


Fallacious Proof

We might attempt to show that $h$ is recursive by using the equation:

\(\ds \map h {n_1, n_2, \ldots, n_k}\) \(=\) \(\ds \map f {n_1, n_2, \ldots, n_k} \map {\chi_\RR} {n_1, n_2, \ldots, n_k}\)
\(\ds \) \(+\) \(\ds \map g {n_1, n_2, \ldots, n_k} \map {\overline \sgn} {\map {\chi_\RR} {n_1, n_2, \ldots, n_k} }\)

where $\overline \sgn$ is the signum bar function, known to be primitive recursive.

Then the right hand side is obtained by substitution from known recursive functions, so is recursive, and hence $h$ is recursive.


However, this fails as follows.

Although $\map h {n_1, n_2, \ldots, n_k}$ is always defined,

\(\text {(1)}: \quad\) \(\ds \) \(\) \(\ds \map f {n_1, n_2, \ldots, n_k} \map {\chi_\RR} {n_1, n_2, \ldots, n_k}\)
\(\ds \) \(+\) \(\ds \map g {n_1, n_2, \ldots, n_k} \map {\overline \sgn} {\map {\chi_\RR} {n_1, n_2, \ldots, n_k} }\)

might not be.

$f$ and $g$ might not be total, in which case there may be some $\tuple {n_1, n_2, \ldots, n_k}$ for which at least one of $\map f {n_1, n_2, \ldots, n_k}$ and $\map g {n_1, n_2, \ldots, n_k}$ is not defined.

This means $(1)$ is not defined for those values of $\tuple {n_1, n_2, \ldots, n_k}$.

Note that $(1)$ does define a recursive function, but its domain consists only of those $\tuple {n_1, n_2, \ldots, n_k}$ for which $\map f {n_1, n_2, \ldots, n_k}$ and $\map g {n_1, n_2, \ldots, n_k}$ are both defined.

Thus we establish the result by means of the indirect approach requiring the use of the results concerning the equivalence of recursive functions and URM computable functions.