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Omnimaga
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(Moderators:
pimathbrainiac
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JWinslow23
,
c4ooo
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[ENDED] (Results Soon) Code Golf - The Reboot #2
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Topic: [ENDED] (Results Soon) Code Golf - The Reboot #2 (Read 45576 times)
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Juju
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Coder Of Tomorrow
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Extreme Addict (Next: 9001)
Posts: 5730
Rating: +500/-19
Weird programmer
Re: Code Golf - The Reboot #2
«
Reply #75 on:
June 14, 2015, 08:14:47 pm »
Quote from: pimathbrainiac on June 14, 2015, 07:02:42 pm
Actually, c4ooo is up next, then Juju.
Yep, gotta wait until next week for mine
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lirtosiast
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Posts: 44
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Re: Code Golf - The Reboot #2
«
Reply #76 on:
June 14, 2015, 11:56:43 pm »
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 10
70
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.
«
Last Edit: June 16, 2015, 11:09:52 pm by lirtosiast
»
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[ENDED] (Results Soon) Code Golf - The Reboot #2