Function Obtained by Primitive Recursion from URM Computable Functions
Theorem
Let the functions $f: \N^k \to \N$ and $g: \N^{k+2} \to \N$ be URM computable functions.
Let $h: \N^{k+1} \to \N$ be obtained from $f$ and $g$ by primitive recursion.
Then $h$ is also URM computable.
Proof
From the definition:
- $\forall n \in \N: h \left({n_1, n_2, \ldots, n_k, n}\right) = \begin{cases} f \left({n_1, n_2, \ldots, n_k}\right) & : n = 0 \\ g \left({n_1, n_2, \ldots, n_k, n-1, h \left({n_1, n_2, \ldots, n_k, n-1}\right)}\right) & : n > 0 \end{cases}$
Let $P$ and $Q$ be normalized URM programs which compute $f$ and $g$ respectively.
Let $s = \lambda \left({P}\right)$ and $t = \lambda \left({Q}\right)$ be the number of basic instructions in $P$ and $Q$.
Let $u = \max \left\{{\rho \left({P}\right), \rho \left({Q}\right), k+2}\right\}$.
Thus registers $R_{u+1}, R_{u+2}, \ldots$ are neither used by $P$ nor $Q$ nor for input.
So they can be used for storing values.
The following algorithm can be followed to create a URM program $H$ to compute $h$.
We assume that the input is $\left({n_1, n_2, \ldots, n_k, n}\right)$, which is held in $R_1, R_2, \ldots, R_k, R_{k+1}$.
The plan is that the program will compute $h \left({n_1, n_2, \ldots, n_k, i}\right)$ for $i = 0, 1, 2, \ldots, n$ in order.
At the end of the last iteration, $h \left({n_1, n_2, \ldots, n_k, n}\right)$ will be left in $R_1$.
We are to use the following registers:
- $R_{u+1}, R_{u+2}, \ldots, R_{u+k+1}$ will be used to store the input $\left({n_1, n_2, \ldots, n_k, n}\right)$ so it does not get overwritten. We note that $R_{u+k+1}$ will hold the value of $n$.
- $R_{u+k+2}$ will hold the current value of the recursion variable $i$.
When $r_{u+k+2} = r_{u+k+1}$, the computation will have ended.
Step | Process | Notes | $\lambda \left({H}\right)$ | |
---|---|---|---|---|
$1$ | Append a Block Copy Program $C \left({1, u+1, k+1}\right)$ to $H$^{[1]}. | This stores the input somewhere safe so it can be accessed again later. | $k + 1$ | |
$2$ | Append $Z \left({k+1}\right)$ to $H$. | This clears $R_{k+1}$ so that $R_1, R_2, \ldots, R_{k+1}$ holds the input for calculating $f$. | $k + 2$ | |
$3$ | Increment the Jumps in $P$ by $k+2$ 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 $P'$. | ||
$4$ | Append $P'$ to $H$. | This will compute $h \left({n_1, n_2, \ldots, n_k, n}\right) = f \left({n_1, n_2, \ldots, n_k}\right)$. | $k + s + 2$ | |
$5$ | Append the command $J \left({u+k+1, u+k+2, k+s+t+8}\right)$ to $H$. | If the iterator equals the value of $n$ that was input, this Jumps $H$ to a non-existent line to force $H$ to terminate. | $k + s + 3$ | |
$6$ | 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 $h \left({n_1, n_2, \ldots, n_k, i-1}\right)$. | $k + s + 4$ | |
$7$ | Append the command $C \left({u+k+2, k+1}\right)$ to $H$. | This moves the current contents of $R_{u+k+2}$, the current iteration count, into the last but one input location for function $g$, that is, $i-1$ itself. | $k + s + 5$ | |
$8$ | Append a Block Copy Program $C \left({u+1, 1, k}\right)$ to $H$. | This restores the input that we saved in step $1$ so it will be ready for program $Q$. | $2 k + s + 5$ | |
$9$ | Append the command $S \left({u+k+2}\right)$ to $H$. | This increments the iterator $i$. | $2 k + s + 6$ | |
$10$ | Calculate $l = \lambda \left({H}\right)$. | This is the length of $H$ so far. | ||
$11$ | Increment the Jumps in $Q$ by $k + l$ lines. Call this amended version $Q'$. | As $Q$ 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 $Q'$. | ||
$12$ | Append $Q'$ to $H$. | This will compute $h \left({n_1, n_2, \ldots, n_k, n}\right) = f \left({n_1, n_2, \ldots, n_k}\right)$. | $2 k + s + t + 6$ | |
$13$ | Append the command $J \left({1, 1, k + s + 3}\right)$ to $H$. | This makes the program jump back to the command added at step $6$. | $2 k + s + t + 7$ |
The program $H$ finally looks like this:
Line | Command | Comment | ||
---|---|---|---|---|
$1$ | $C \left({1, u+1}\right)$ | Start of block copy of input to store. | ||
$\vdots$ | ||||
$k+1$ | $C \left({k+1, u+k+1}\right)$ | End of block copy of input to store. | ||
$k+2$ | $Z \left({k+1}\right)$ | Clear the iterator to zero. | ||
$k+3$ | Start of $P'$. | |||
$P'$ | ||||
$k+s+2$ | End of $P'$. | |||
$k+s+3$ | $J \left({u+k+1, u+k+2, k+s+t+8}\right)$ | If all the iterations have finished, jump to the exit line. | ||
$k+s+4$ | $C \left({1, k+2}\right)$ | Copy the output of $P'$ into input of $Q'$. | ||
$k+s+5$ | $C \left({u+k+2, k+1}\right)$ | Copy the current iteration count input location for function $Q'$. | ||
$k+s+6$ | $C \left({u+1, 1}\right)$ | Start of block copy of stored input back to input of $Q'$. | ||
$\vdots$ | ||||
$2k+s+5$ | $C \left({u+k, k}\right)$ | End of block copy of stored input back to input of $Q'$. | ||
$2k+s+6$ | $S \left({u+k+2}\right)$ | Increment the iterator. | ||
$2k+s+7$ | Start of $Q'$. | |||
$Q'$ | ||||
$2k+s+t+6$ | End of $Q'$. | |||
$2k+s+t+7$ | $J \left({1, 1, k + s + 3}\right)$ | Jump back to start the next iteration through $Q'$. | ||
$2k+s+t+8$ | Empty | Exit line. |
It can easily be determined that $H$ computes $h$.
Hence $h$ is URM computable.
$\blacksquare$
Comment
Note that there are two things going on here. The first table is an algorithm for creating the program, the second table is the program itself.
Footnotes
- ↑ $H$, at this point, is a null URM program.
- ↑ 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 $J \left({m, n, q+r}\right)$.