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The 16-color CGA palette (also known as the HTML colors) is a set of 16 colors used by early graphics adapters. The goal of this challenge is to output all 16 of them, in hex format (RRGGBB), in ascending order by hex value, separated by newlines. Thus, the output should be exactly this:
000000
000080
0000FF
008000
008080
00FF00
00FFFF
800000
800080
808000
808080
C0C0C0
FF0000
FF00FF
FFFF00
FFFFFF
A single trailing newline is allowed, but not required.
“×Ɗ¡‘ŒP»Ṫ¦209ṗ€3Fd⁴ịØHs3ṢQY
“×Ɗ¡‘ yield the code points of the characters between the quotes in Jelly's SBCS, which are 0x11 = 17, 0x91 = 145, and 0x00 = 0.
ŒP constructs the powerset of the array of code points, yielding
[[], [17], [145], [0], [17, 145], [17, 0], [145, 0], [17, 145, 0]]
The last two entries correspond to combinations that contain both 80 and FF, so we have to discard them.
»Ṫ¦209 consists of three parts:
Ṫ (tail) removes the last array of code points, i.e., [17, 145, 0].
»209 takes the maximum of each integer in the remainder of the powerset and 0xD1 = 209, replacing all of them with 209.
¦ (sparse) iterates over the elements of the remainder of the powerset. If the corresponding index is found in [17, 145, 0], the element is replaced with all 209's. If not, it is left untouched.
¦ isn't modular, so this modifies only the last array (index 0) in the remainder of the powerset. The indices 17 and 145 are too large and have no effect.
The result is as follows.
[[], [17], [145], [0], [17, 145], [17, 0], [209, 209]]
ṗ€3 computes the third Cartesian power of each array, i.e., the array of all 3-tuples of elements of each array.
Fd⁴ flattens the result and computes quotient and remainder of each integer divided by 16.
ịØH indexes (1-based) into "0123456789ABCDEF, so 0x11, 0x91, 0x00, and 0xD1 get mapped to "00", "80", "FF", and "C0" (resp.).
s3ṢQ splits the character pairs into 3-tuples, sorts the tuples, and deduplicates.
Finally, Y joins the unique tuples, separating by linefeeds.
a={00,80,FF}
eval echo $a$a$a|fmt -w9|sed '16iC0C0C0
/F0*8\|80*F/d'
{00,80,FF}{00,80,FF}{00,80,FF} gives all need combinations in the right order (excluding C0C0C0), along some extras. The extras are the ones that contain both F and 8 characters.fmt puts each entry on its own linesed expression inserts C0C0C0 in the appropriate linesed expression filters out the "extras" described above.
“mạ9ṣṃwɠƁ,¡ẓw’b4µża1$ị“08CF”s3Y
Base 250 compression of a number (“...’),
converted to base 4 (b4),
zipped (ż) with a copy of itself after a vectorised logical-and with 1 (a1$)*,
indexed (ị) into the four characters used (“08CF”),
split into chunks of length 3 (s3),
and joined with line feeds (Y).
* Thus pairing each zero digit with another zero and each of any other digits with a one. Along with the following indexed fetch this means 'F' becomes paired with another 'F' while '0','8', and 'C' each pair with a '0'.
“0FC8”, as we have 00, FF, C0, and 80.
\$\endgroup\$
– Sherlock9
Dec 10 '16 at 6:02
I think there is still a lot of golfing to do here. Golfing suggestions welcome!
Edit: -9 bytes and many thanks to Mego for helping with the string formatting. -17 bytes from figuring out a better way to print the string in the first place. -17 bytes from figuring out a better way to write the for loop in the first place. -1 byte thanks to xnor's tip to use i%3//2*"\n" instead of "\n"*(i%3<2).
for i in range(48):print(end="F0C8F000"[0x10114abf7f75c147d745d55//4**i%4::4]+i%3//2*"\n")
Ungolfing
z = 0
a = [0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 2, 2, 3, 3, 3, 1, 3, 3, 3, 1, 3, 1, 1, 3, 0, 0, 1, 1, 0, 1, 3, 3, 1, 1, 3, 1, 0, 1, 1, 3, 1, 1, 1, 1, 1]
for i in range(len(a)):
z = (z + a[i]) * 4
z //= 4 # we multiplied by 4 one too many times
for i in range(48):
m = z // 4**j % 4
s = "F0C8F000"[c::4]
if i % 3 == 2:
s += "\n"
print(s, end="")
*.
\$\endgroup\$
– user45941
Dec 10 '16 at 5:20
Saved 2 bytes, thanks to Neil
This is 7 9 bytes shorter than just returning the raw string in backticks.
_=>[...'1121173113106393'].map(v=>[4,2,0].map(x=>'08CF'[x=n>>x&3]+'000F'[x],n+=+v||21).join``,n=-1).join`
`
let f =
_=>[...'1121173113106393'].map(v=>[4,2,0].map(x=>'08CF'[x=n>>x&3]+'000F'[x],n+=+v||21).join``,n=-1).join`
`
console.log(f())
.replace(/./g) is the same length as .map().join, but .map(x=>'08CF'[x=n>>x&3]+'000F'[x],n+=+v||21) saves two bytes.
\$\endgroup\$
– Neil
Dec 10 '16 at 11:25
'777
7780
77FF
7807
78080
7FF7
7FFFF
8077
80780
80807
808080
C0C0C0
FF77
FF7FF
FFFF7
FFFFFF'-replace7,'00'
Yeah, I haven't found anything shorter than just printing the literal string... Thanks to @Martin Smith for shaving down 7 bytes using a simple replacement (which I completely overlooked). So, we're at least 7 bytes shorter than simply hardcoding it. Yay!
But that's boring!
So instead ...
[consolecolor[]](0,7+9..15)+-split'Lime Maroon Navy Olive Purple Silver Teal'|%{-join"$([windows.media.colors]::$_)"[3..8]}|sort
The [system.consolecolor] namespace defines the console colors (natively) available to the PowerShell console. If we reference them via an integer array like this, the default value is the name (e.g., Black or White or the like). We combine that with a string that has been -split on spaces, so now we have an array of strings of color names.
We loop through those |%{...} and each iteration pull out the corresponding [system.windows.media.colors] value. The default stringification for those objects is the color in #AARRGGBB format as a hex value, so we leverage that by encapsulating that call in a string with a script block "$(...)". But, since we don't want the alpha values or the hash, we take the back end [3..8] of the string, and need to -join that resulting char-array back into a string. Then, a simple Sort-Object to put them in the right order.
00 and .Replace(7,'00') is < 21 chars.
\$\endgroup\$
– Martin Smith
Dec 11 '16 at 17:07
-replace for 8 and 80 is the same byte count (saves 12 zeros, which is -replace8,80 length). Doing it for the FF is longer by two bytes because of the quotes needed around the "FF" in the -replace2,"FF" statement.
\$\endgroup\$
– AdmBorkBork
Dec 12 '16 at 15:17
Super simple base compression.
jcs@L"FC80"jC"ÿÿûÿ¿û¿ðÿ¿»¿»·wðð\0ð\0"4 6
'80FFC000'2e'3na:1Fswv1=`uIn'F4:ZaZ)6e!
'80FFC000' % Push this string
2e % Reshape with 2 rows. So 1st column contains '80', 2nd 'FF' etc
'3na:1Fswv1=`uIn' % Push this string
F4:Za % Convert from base 95 to alphabet [1 2 3 4]
Z) % Use as column indices into the first string
6e! % Reshape with 6 rows and transpose.
% Implicitly display
•P{Ætg7«r¨ëÅ,…}ù¢Ý%Vt®£8ãøÚ$0óDÛY²Zþ…ð7ê‘Ó{òìàYëØU¥•hR6ô»
What we need to output is basically (reversed and split):
FFFFFF00FFFFFF00FF0000FF0C0C0C080808000808080008000008FFFF0000FF00080800000800FF0000080000000000
Which, in decimal is:
39402003857025890357721060524755992261661062099432941475272448103296644696683709026793043150430945208910007869898752
Which in Base-214 is:
P{Ætg7«r¨ëÅ,…}ù¢Ý%Vt®£8ãøÚ$0óDÛY²Zþ…ð7ê‘Ó{òìàYëØU¥
This is the simplest solution I could come up with, because there's no way in hell I'm beating Dennis. Spent an hour trying and nothing beat his idea.
<?=strtr("000
001
002
010
011
020
022
100
101
110
111
333
200
202
220
222",["00",80,FF,CO]);
simply replacement from the digits as key in the array with the values strtr
@for %%c in (000000 000080 0000FF 008000 008080 00FF00 00FFFF 800000 800080 808000 808080 C0C0C0 FF0000 FF00FF FFFF00 FFFFFF)do @echo %%c
Yes, it's that boring. Previous 148-byte attempt:
@if "%2"=="" (call %0 000000 80&call %0 C0C0C0 FF)|sort&exit/b
@echo %1
@for %%c in (0000%2 00%200 00%2%2 %20000 %200%2 %2%200 %2%2%2)do @echo %%c
Unfortunately you can't pipe the output of a for or a call: command, so I have to invoke myself recursively.
<v"UVTYZQPefij?EDA@"
v>>9\:4/:\4/>4%:3g,0`
<^_@#:,+55$_^#!-9:,g3
F08C
The colours are encoded in the string you see on the first line, two bits per colour component, with an additional high bit set to force each value into the ASCII range (except in the case of 63, which would be out of range as 127).
The list of colours on the stack is then processed as follows:
9\ 9 is pushed behind the current colour to serve as a marker.
:4/:\4/ The colour is repeatedly divided by 4, splitting it into 3 component parts.
> The inner loop deals with each of the 3 components.
4% Modulo 4 masks out the 2 bits of the colour component that we care about.
:3g, That is used as an index into the table on line 4 to get the character to output.
0`3g, For component values greater than 0 the second char is a '0', otherwise an 'F'.
:9-! Check if the next component is our end marker.
^ If not, repeat the inner loop.
55+, Output a newline.
:_ Repeat the outer loop until there are no more colours on the stack.
void n(){string a="000000\r\n000080\r\n0000FF\r\n008000\r\n008080\r\m00FF00\r\n00FFFF\r\n800000\r\n800080\r\n808000\r\n808080\r\nC0C0C0\r\nFF0000\r\nFF00FF\r\nFFFF00\r\nFFFFFF";Console.Write(a);}
Sadly this beats, by a huge margin, the more interesting albeit incredibly convoluted (I had tons more fun writing it) C#, 270 bytes
void n(){string a,b,c,d;a="00";b="80";c="C0";d="FF";for(int i=0;i<16;i++){Console.WriteLine((i<7?a:i<11?b:i>11?d:c)+(i<3?a:i<5?b:i<7?d:i<9?a:i<11?b:i==11?c:i<14?a:d)+(i==0||i==3||i==5|i==7||i==9||i==12||i==14?a:i==1||i==4||i==8||i==10?b:i==2||i==6||i==13||i==15?d:c));}}
()=>@"string_here" (this casts to Action<string>). I also suggest using a verbatim string (@) so that you can just put the new lines directly in the string without needing to escape them.
\$\endgroup\$
– milk
Dec 12 '16 at 18:38
f(i){for(i=48;i--;)printf("%.2s%s","FFC08000"+("#&/28MNQRSV]^_ab"[i/3]-35>>i%3*2&3)*2,"\n\0"+i%3);}
After having made an attempt involving creating a list of numbers and outputting them while sorting, I compared to the naïve solution, which was sobering:
f(){puts("000000\n000080\n0000FF\n008000\n008080\n00FF00\n00FFFF\n800000\n800080\n808000\n808080\nC0C0C0\nFF0000\nFF00FF\nFFFF00\nFFFFFF");}
That one clocks in at 140 bytes compared to my try at 200 and change.
The solution was to think of it as text like any other, albeit with a small alphabet. Each colour could be thought of as a triplet of 2-bit indices into the alphabet {0xff, 0xc0, 0x80, 00}. The process of colour -> triplet -> number -> character (with offset +35 to make them all printable and avoid any need for escapes) can be illustrated as such:
000000 333 63 b
000080 332 62 a
0000FF 330 60 _
008000 323 59 ^
008080 322 58 ]
00FF00 303 51 V
00FFFF 300 48 S
800000 233 47 R
800080 232 46 Q
808000 223 43 N
808080 222 42 M
C0C0C0 111 21 8
FF0000 033 15 2
FF00FF 030 12 /
FFFF00 003 3 &
FFFFFF 000 0 #
Then it's just a matter of iterating over the resulting string and cutting out the appropriate parts of the alphabet string.
