[ENDED] (Results Soon) Code Golf - The Reboot #2

[Juju]

Actually, c4ooo is up next, then Juju.

Yep, gotta wait until next week for mine

[lirtosiast]

Here are some of my attempts at this contest, in chronological order, along with explanations.

Spoiler For 40 byte solution:

While 1
getKey
If Ans
Y-i^sum(cumSum(Ans={85,41,81→Y
Pxl-On(imag(Y),real(Y
End

This was my first real (after working for an hour or so) solution once I fixed key repeat by using WASD instead of arrow keys. I had tried using two separate variables, but it seemed that using the complex plane would be smaller, even considering the two-byte cost of the real( and imag( tokens, because raising i to a power eliminates the cost of looking up the cursor change corresponding to a particular key. You can see that up, left, down, and right correspond to W, A, S, and D keys, which are key codes 85, 41, 81, and 51 respectively.

Spoiler For 33 byte solution:

While 1
getKey
If Ans
Y-i^int(4fPart(Ans/7→Y
Pxl-On(real(Y),-imag(Y
End

I began thinking about the fact that we could assume WASD are the only pressed keys, and found I could abuse it here. The WASD key codes are all different modulo 7, so taking i to that power would give 1, i, -1, and -i; luckily, they came in the right order.

Spoiler For 31 byte solution:

While 1
getKey
If Ans
Y-i^int(4cos(8Ans→Y
Pxl-On(real(Y),imag(Y
End

Same sort of idea as the previous solution, but I used a "randomizer" function, because I knew the cosines of large interger numbers would be unpredictable. I basically searched through every token and every multiplier in the book, and finally found one that worked. These changes saved one byte in the Y-modifying line, and one from not needing to negate the imag(.

I started to think I was done golfing, but then I saw that, with a mathematical way to ignore a getKey value of zero, I could eliminate the If Ans line, and if I only used one getKey, I could include it in the algorithm and also store the cursor position in Ans instead of Y, saving five bytes.

Spoiler For 27 byte solution:

While 1
Ans-iPart(i^int(48ln(getKey-1
Pxl-On(real(Ans),imag(Ans
End

This one is honestly just magic.

Spoiler For 26 byte "solution":

If it weren't for rounding errors causing the cursor to move to the side every trillion keypresses, I would be able to eliminate the iPart(:

26 bytes:
While 1
Ans+i^int(28ln(getKey-1
Pt-On(real(Ans),imag(Ans
End

And of course replacing the While 1 with recursion would save another two bytes, but crash after a few seconds.

The 10⁷⁰ 57-byte solutions I had in mind were all the 26-byte one (before I knew it was invalid), but with the 1 in While 1 replaced with length("[some 30-byte string], of which there are about 245^30 because there are about 245 printable one-byte tokens.

EDIT: I wonder if we'll ever get the results.