Given Color codes in CMYK, convert it to RGB value.
Input:
string of 4 integers(ranging from 0-100) separated by space
86 86 0 43
28 14 0 6
0 41 73 4
Output:
#141592
#ABCDEF
#F49043
Shortest code wins!
HINT: For converting CMYK to RGB you may use formula such as:
Red = 255 x (1 - Cyan/100) x (1 - Black/100) Green = 255 x (1 - Magenta/100) x (1 - Black/100) Blue = 255 x (1 - Yellow/100) x (1 - Black/100)and use these three variables to get the value in
#RRGGBBformat
<?="#";for($c=explode(" ",$argn);$i<3;)printf("%02X",255*(1-$c[+$i++]/100)*(1-$c[3]/100));
-2 bytes thanks to Rod.
lambda s:'#'+''.join('%02X'%int(.0255*(100-int(i))*(100-int(s.split()[3])))for i in s.split()[:3])
ḲV÷ȷ2ạ1×Ṫ$×255ḞṃØHṙ1¤ṭ”#
A full program which prints the result.
Note: rounding rather than flooring may be used by inserting the two bytes of code +. between 255 and Ḟ.
ḲV÷ȷ2ạ1×Ṫ$×255ḞṃØHṙ1¤ṭ”# - Main link: list of character, s
Ḳ - split at spaces (makes a list of lists of characters)
V - evaluate as Jelly code (makes a list of the decimal numbers)
ȷ2 - literal 100
÷ - divide (vectorises to yield [C/100, M/100, Y/100, K/100])
ạ1 - absolute difference with 1 -> [1-(C/100),...]
$ - last two links as a monad:
Ṫ - tail (this is 1-(K/100))
× - multiply (vectorises across the other three)
×255 - multiply by 255 (vectorises)
Ḟ - floor to the nearest integer
¤ - nilad followed by link(s) as a nilad:
ØH - hex-digit yield = "0123456789ABCDEF"
ṙ1 - rotate left by 1 -> "123456789ABCDEF0"
ṃ - base decompress (use those as the digits for base length (16))
”# - literal character '#'
ṭ - tack
- implicit print
_.Ċ instead of +.Ḟ...but the latter is maybe more widely used.
\$\endgroup\$
Commented
Jul 1, 2017 at 7:47
\d+
$*1;100$*
(1*);\1
1(?=.* (1*))|1
$1
1
51$*
(1{32000})*(1{2000})*1*.
;$#1;$#2
T`d`L`1\d
;B\B|;
^
#
Try it online! Note: This code is very slow, so please don't hammer Dennis's server. Explanation:
\d+
$*1;100$*
(1*);\1
Convert each number to unary and subtract from 100.
1(?=.* (1*))|1
$1
Multiply all the numbers by the last number, which is deleted.
1
51$*
Multiply by 51, so that once we divide by 2000, we get 100 * 100 * 51 / 2000 = 255 as desired.
(1{32000})*(1{2000})*1*.
;$#1;$#2
Divide by 32000 and floor divide the remainder by 2000, thus generating a pair of base 16 values, although sadly themselves still written in base 10.
T`d`L`1\d
;B\B|;
Convert from base 10 to base 16.
^
#
Insert the leading #.
s->{int i=0,c[]=java.util.Arrays.stream(s.split(" ")).mapToInt(Byte::new).toArray();for(s="#";i<3;)s+=s.format("%02X",(int)(.0255*(100-c[i++])*(100-c[3])));return s;}
f=
(s,z=s.split` `,k=z.pop())=>'#'+z.map(x=>('0'+(.0255*(100-x)*(100-k)+.5|0).toString(16)).slice(-2)).join``<input id=i value="28 14 0 6"/><button onclick="o.innerHTML=f(i.value)"/>Go</button>
<pre id=o></pre>
#import<iostream>
#import<iomanip>
#define F(x)int(.0255*(100-x)*(100-k))
int main(){
int c,m,y,k;
std::cin>>c>>m>>y>>k;
std::cout<<"#"<<std::hex<<F(c)<<F(m)<<F(y);
}
Using the optimized formula. Added which is not necessary.+.5 to convert CMYK=0 0 0 0 correct to RGB=0xffffff
range[3] as [0,1,2], unwanted [] removedn=input().split()
print('#'+''.join(hex(int(.0255*(100-int(n[i]))*(100-int(n[3]))))[2:]for i in[0,1,2]))
gsub(/(.+) (.+) (.+) (.+)/){'#%X%X%X'%[$1,$2,$3].map{|n|255*(1-n.to_i/1e2)*(1-$4.to_i/1e2)}}
s=>"#"+[0,1,2].map(n=>("0"+((255-2.55*s[n])*(1-s[3]/100)|0).toString(16)).slice(-2),s=s.split` `).join``
Example code snippet:
f=
s=>"#"+[0,1,2].map(n=>("0"+((255-2.55*s[n])*(1-s[3]/100)|0).toString(16)).slice(-2),s=s.split` `).join``
console.log(f("86 86 0 43"))
console.log(f("28 14 0 6"))
console.log(f("0 41 73 4"))
printf"#%2X%2X%2X",map{.0255*(100-$_)*(100-$F[3])}@F
Solution:
"#",raze{(last($)0x0 vs)each"h"$.0255*x[3]*x 0 1 2}100-
Examples:
q)"#",raze{(last($)0x0 vs)each"h"$.0255*x[3]*x 0 1 2}100-86 86 0 43
"#141491"
q)"#",raze{(last($)0x0 vs)each"h"$.0255*x[3]*x 0 1 2}100-28 14 0 6
"#adcef0"
q)"#",raze{(last($)0x0 vs)each"h"$.0255*x[3]*x 0 1 2}100-0 41 73 4
"#f59042"
Explanation:
Fairly straightforward, stole the 0.0255 trick from other solutions (thanks!). Evaluation is performed right to left.
"#",raze {(last string 0x0 vs) each "h"$ .0255 * a[3] * a 0 1 2}100- / ungolfed
{ } / lambda function
100- / subtract from 100 (vector)
a 0 1 2 / index into a at 0, 1 and 2 (CMY)
a[3] / index into at at 3 (K)
* / multiply together
.0255 * / multiply by 0.255
"h"$ / cast to shorts
( ) each / perform stuff in brackets on each list item
0x0 vs / converts to hex, 1 -> 0x0001
string / cast to string, 0x0001 -> ["00", "01"]
last / take the last one, "01"
raze / join strings together
"#", / prepend the hash
Notes:
Rounds numbers by default, would cost 3 bytes (_) to floor instead before casting to short.
$#т/-¤s¨*255*hJ'#ì
-1 thanks to kalsowerus.
Has floating-point inaccuracies, so results might be off-by-one, but the formula in the question is used.
$ is just the same as 1¹
\$\endgroup\$
Commented
Jul 7, 2017 at 12:45
input when there would be multiple
\$\endgroup\$
Commented
Jul 7, 2017 at 12:48
q=(1-).(/100)
x!y=h$ceiling$q x*(q y)*255
f c m y k=concat["#",c!k,m!k,y!k]
h x|x<16=[s!!x]|0<1=(h((x-m)`quot`16))++[s!!m] where m=x`mod`16
s=['0'..'9']++['a'..'f']
PROGRAM C
REAL,DIMENSION(4,3)::d
READ(*,*)((d(i,j),i=1,4),j=1,3)
WRITE(*,'((A,3(Z2)))')(35,(INT(.0255*(100-d(i,j))*(100-d(4,j))),i=1,3),j=1,3)
END PROGRAM C
CMYKvalues as decimals from0to1or is it required to do0to100? \$\endgroup\$