I never followed up with this! A little while after, I refined this significantly and then a year or two ago, I implemented it in my z80float z80float library (link is directly to sqrtHLIX sqrt32) and that's how I made square roots faster than multiplication.

Today I was excited to learn that this is how GMP does it, it just looks scarier there because they are optimizing the notation to make it easy to implement in code without necessarily understanding how it works (which is cool, they present the pseudocode in a very easy-to-digest manner). But that's not what I'm here for, I'm here to show how it works!

Here is the TL;DR if you don't want to go through the derivation:
##
\begin{array}{rcl}
x_{0} & \approx & \sqrt{c} \\
a_{0} &=& c - x_{0}^{2} \\
q_{n} + \frac{r_{n}}{2x_{n}} &=& \frac{a_{n}}{2x_{n}} \\
x_{n+1} &=& x_{n} + q_{n} \\
a_{n+1} &=& r_{n} - q_{n}^{2} \\
\end{array}
##
(The third line means "divide ##\frac{a_{n}}{2x_{n}}##, but truncate at some precision to get the quotient and the remainder")

---

Let's start. I like the "Babylonian Method" where you start with a guess ##x_{0}## as an approximation to the square root of ##c##, then you run this iteration a few times until you have the precision you want:
##
\begin{array}{l}
x_{n+1} &=& \frac{\left(x_{n}+\frac{c}{x_{n}}\right)}{2}
\end{array}
##
All you are doing is averaging the guess, with ##c## divided by that guess. If your guess was an overestimate, ##\frac{c}{x_{n}}## is going to be an underestimate, so their average will be closer to the actual answer.


And that's just a simplification of the Newton Iteration:
##
\begin{array}{l}
x_{n+1} &=& x_{n} + \frac{c-x_{n}^{2}}{2x_{n}}
\end{array}
##

We are going to build off of the Newton Iteration, but make it a little more complicated by turning ##c-x_{n}^{2}## into it's own recurrence:
##
\begin{array}{l}
a_{n} &=& c-x_{n}^{2} \\
x_{n+1} &=& x_{n} + \frac{a_{n}}{2x_{n}}
\end{array}
##

And actually, in practice we are going to truncate that division so that ##q_{n} + \frac{r_{n}}{2x_{n}} = \frac{a_{n}}{2x_{n}}## where ##q_{n}## is the (truncated) quotient, and ##r_{n}## is whatever the remainder was after dividing by ##2x_{n}##. So what we'd really be doing is:
##
\begin{array}{rcl}
a_{n} &=& c-x_{n}^{2} \\
q_{n} + \frac{r_{n}}{2x_{n}} &=& \frac{a_{n}}{2x_{n}} \\
x_{n+1} &=& x_{n} + q_{n}
\end{array}
##

Now that everything is more complicated, let's take a look at what happens to ##a_{n+1}##:
##
\begin{array}{l}
a_{n+1} &=& c-x_{n+1}^{2} \\
&=& c-(x_{n} + q_{n})^{2} \\
&=& c-(x_{n}^{2} + 2x_{n}q_{n} + q_{n}^{2}) \\
&=& c - x_{n}^{2} - 2x_{n}q_{n} - q_{n}^{2} \\
&=& a_{n} - 2x_{n}q_{n} - q_{n}^{2} \\
\end{array}
##

Already, we have an optimization. Those two multiplications are half the precision of a full ##x_{n+1}^{2}## if you do it right, and that is strictly faster. But we're not done! Remember that division? Let's re-write it:
##
\begin{array}{rcl}
q_{n} + \frac{r_{n}}{2x_{n}} &=& \frac{a_{n}}{2x_{n}} \\
2x_{n}q_{n} + r_{n} &=& a_{n} \\
r_{n} &=& a_{n} - 2x_{n}q_{n} \\
\end{array}
##

Well ain't that awfully convenient? Let's rewrite our ##a_{n+1}## recurrence:
##
\begin{array}{l}
a_{n+1} &=& a_{n} - 2x_{n}q_{n} - q_{n}^{2} \\
&=& r_{n} - q_{n}^{2} \\
\end{array}
##

So now we've significantly improved that step by replacing the squaring operation with one at half the precision! Let's look at the recurrence relations now:
##
\begin{array}{rcl}
q_{n} + \frac{r_{n}}{2x_{n}} &=& \frac{a_{n}}{2x_{n}} \\
x_{n+1} &=& x_{n} + q_{n} \\
a_{n+1} &=& r_{n} - q_{n}^{2} \\
\end{array}
##

Neat!

---

Now let's retry our example of finding the square-root of 21:
##
\begin{array}{rcl}
x_{0} & \approx & \sqrt{c} \\
&=& 4\\
a_{0} &=& c - x_{0}^{2} \\
&=& 21-16 \\
&=& 5 \\
q_{0} + \frac{r_{0}}{2x_{0}} &=& \frac{a_{0}}{2x_{0}} \\
&=&\frac{5}{8} \\
&=&.5 + \frac{1}{8} \\
x_{1} &=& x_{0} + q_{0} \\
&=& 4.5 \\
a_{1} &=& r_{0} - q_{0}^{2} \\
&=& 1 - .5^{2} \\
&=& .75 \\

q_{1} + \frac{r_{1}}{2x_{1}} &=& \frac{a_{1}}{2x_{1}} \\
&=&\frac{.75}{9} \\
&=&.083 + \frac{.003}{9} \\
x_{2} &=& x_{1} + q_{1} \\
&=& 4.583 \\
a_{2} &=& r_{1} - q_{1}^{2} \\
&=& .003 - .083^{2} \\
&=& -.003889 \\

q_{2} + \frac{r_{2}}{2x_{2}} &=& \frac{a_{2}}{2x_{2}} \\
&=&\frac{-.003889}{9.166} \\
&=&-.0004243 + \frac{.0000001338}{9.166} \\
x_{3} &=& x_{2} + q_{2} \\
&=& 4.5825757 \\

\end{array}
##

EDIT: updates z80float link to point to the routine that actually implements this

You were on the right track-- "getKey" on its own returns a single keypress, but you can also use "getKey(n)" to check if key "n" is pressed (it returns 0 for not pressed, 1 for pressed).

So if you want to check for both Enter and the Down arrow, simultaneously:
If getKey(9)
//Enter is pressed
End

If getKey(1)
//Down is pressed
End

You can also do some useful arithmetic, like if you have (X,Y) coordinates, you might want:
X+getKey(3)-getKey(2)→X
Y+getKey(1)-getKey(4)→Y
And that has a bonus of letting the user press multiple arrow keys simultaneously

Hopefully that helps!

EDIT: Also, welcome to Omninaga, you should Introduce Yourself!

Necropost: this came up elsewhere and I made a python version if anybody is interested. I think it could be implemented really well with logic gates!
https://gist.github.com/Zeda/98aab98fd32a231b878772a435ffb306

#!/usr/bin/python3

# This is ported from:
#    https://www.omnimaga.org/asm-language/sine-approximation-(z80)/
# (Xeda112358  a.k.a.  Zeda  a.k.a.  ZedaZ80)
#

from math import sin

def sineb(x):
    """Approximates 128*sin(x*2*pi/256)"""

    a1 = int(x)
    a0 = a1 << 1
    a2 = a1 >> 1

    # iteration 1
    if a1 & 64:
        l = ~a0 & 127
    else:
        l =  a0 & 127

    # iteration 2
    if a1 & 32:
        l += ~a1 & 31
    else:
        l +=  a1 & 31

    # iteration 3
    if a1 & 16:
        l += ~a2 & 7
    else:
        l +=  a2 & 7

    # check if it needs to be negated
    if a1 & 128:
        return -l
    else:
        return l

# Plot a graph of the approximation vs the actual
for x in range(0, 256, 2):
    y = sineb(x)
    z = int(sin(x*3.1415926535/128)*128)

    # translate and scale for "graphing"
    y += 128
    z += 128
    y >>= 1
    z >>= 1

    # "graph"
    #    X - Approximation
    #    O - Actual
    #    8 - used when the graphs overlap
    if y == z:
        print(" "*y + "8")
    elif y > z:
        print(" "*z + "X" + " "*(y-z-1) + "O")
    else:
        print(" "*y + "O" + " "*(z-y-1) + "X")


The code is difficult to follow, but it looks like you are mixing 0-indexing and 1-indexing.
You loop J from 1 to Rooms, but then you are checking if J>0 in the same loop (which is always true).

J+(J*((J>0)*4))-4→Y
Simplifies to:
J+(J*(1*4))-4→Y
J+(J*4)-4→Y
J*5-4→Y

Maybe there is something there?

Could it be that I used WabbitSign?  Should I use something else to sign the app?

I haven't used wabbitsign in a long time and I'm not sure if that is the issue. I generally use spasm and I've got it configured to automatically sign apps.

I know Wabbitemu is very generous with loading apps (it bypasses the calc's security which is great for streamlining development, but it hides signing issues).

I'm not much help with this stuff these ays .__.

Holy heck that looks so good :O

Awww yiss, that looks pretty cool!
I definitely like the monochrome graphics (I also find that grayscale is just too rough to look at on actual hardware and with moving images).