Clear Registers Program
From ProofWiki
URM Program
Let $a, b \in \N$ be natural numbers such that $0 < a$.
Then we define the URM program $Z \left({a, b}\right)$ to be:
| Line | Command | |
|---|---|---|
| $1$ | $Z \left({a}\right)$ | |
| $2$ | $Z \left({a + 1}\right)$ | |
| $3$ | $Z \left({a + 2}\right)$ | |
| $\vdots$ | $\vdots$ | |
| $b - a + 1$ | $Z \left({b}\right)$ |
This program clears (that is, sets to $0$) all the registers from $R_a$ through to $R_b$.
If $a > b$ then $Z \left({a, b}\right)$ is the null URM program.
The length of $Z \left({a, b}\right)$ is:
- $\lambda \left({Z \left({a, b}\right)}\right) = \begin{cases} 0 & : a > b \\ b - a + 1 & : a \le b \end{cases}$
Proof
Self-evident.
$\blacksquare$