Function Obtained by Minimization from URM Computable Functions

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Theorem

Let the function $f: \N^{k+1} \to \N$ be a URM computable function.

Let $g: \N^k \to \N$ be the function obtained by minimization from $f$ thus:

$g \left({n_1, n_2, \ldots, n_k}\right) \approx \mu y \left({f \left({n_1, n_2, \ldots, n_k}\right) = 0}\right)$.


Then $g$ is also URM computable.


Proof

Let $f: \N^{k+1} \to \N$ be a URM computable function.

Let $P$ be a URM program which computes $f$.

Let $u = \rho \left({P}\right)$ be the number of registers used by $P$.

We can use:

  • the registers $R_{u+1}, R_{u+2}, \ldots, R_{u+k}$ to store the input $\left({n_1, n_2, \ldots, n_k}\right)$;
  • the register $R_{u+k+1}$ to store the current value of the recursion variable $y$.

We check whether or not $f \left({n_1, n_2, \ldots, n_k}\right) = 0$ by comparing the output from running $P$ (appearing in register $1$ at the end of the running of $P$) with the number in register $R_{u+k+2}$, which remains at $0$ throughout.


The following algorithm can be followed to create a URM program $H$ to compute $h$.

We assume that the input is $\left({n_1, n_2, \ldots, n_k}\right)$, which is held in $R_1, R_2, \ldots, R_k$.


We are to use the following registers:

  • $R_{u+1}, R_{u+2}, \ldots, R_{u+k}$ will be used to store the input $\left({n_1, n_2, \ldots, n_k}\right)$ so it does not get overwritten.
  • $R_{u+k+1}$ will hold the value of $y$.
  • $R_{u+k+2}$ will hold the $0$ throughout.

When $r_{u+k+2} = r_{u+k+1}$, the computation will have ended.

We also define $v = \lambda \left({H}\right)$ to be the number of basic instructions in $H$.


Step Process Notes $\lambda \left({H}\right)$
$1$ Append a Block Copy Program $C \left({1, u, k}\right)$ to $H$[1]. This stores the input somewhere safe so it can be accessed again later. $k$
$2$ Increment the Jumps in $P$ by $k$ lines[2]. Call this amended version $P'$. As $P$ was written so as to start from line 1, we need to move all the Jumps so as to point to the same lines relative to the start of $P'$.
$3$ Append $P'$ to $H$. This will compute $f \left({n_1, n_2, \ldots, n_k, y}\right)$. $k + s$
$4$ Append the command $J \left({1, u+k+2, v}\right)$ to $H$. This Jumps $H$ to the end of the program if the output of $P$ is zero. $k + s + 1$
$5$ Append the command $S \left({u + k + 1}\right)$ to $H$. Increment $y$. $k + s + 2$
$6$ Append a Block Copy Program $C \left({u+k+1, 1, k+1}\right)$ to $H$. Recall the input values, and put $y$ in $R_{k+1}$. $2 k + s + 3$
$8$ Append the command $J \left({1, 1, k + 1}\right)$ to $H$. This makes the program jump back to the start of $P'$. $2 k + s + 4$
$7$ Append the command $C \left({1, k+2}\right)$ to $H$. This copies the output of $P'$ into the last input location for function $g$, that is, the output from $h \left({n_1, n_2, \ldots, n_k, i-1}\right)$. $2 k + s + 5$
$8$ Append the command $C \left({u+k+1, 1}\right)$ to $H$. Put $y$ into the output register. $v = 2 k + s + 6$


It can easily be determined that $H$ computes $g$.

Hence $g$ is URM computable.

$\blacksquare$

Footnotes

  1. $H$, at this point, is a null URM program.
  2. To increment the Jumps by $r$ for any normalized URM program is done by changing all Jumps of the form $J \left({m, n, q}\right)$ to $J \left({m, n, q+r}\right)$.