23
\$\begingroup\$

You are given the RGB values of a color. Your task is simple: to calculate the hue, in the simplest definition.

Say the channels with highest, middle and lowest value are X, Y, Z (which are either red, green or blue) and their values are x, y, z. The hue of this color is (h(X)-h(Y))*(1 + (x-y)/(x-z))/2 + h(Y), where:

h(red) = 0 (or 360 if one of X or Y is blue)
h(green) = 120
h(blue) = 240

The input consists of 3 integers from 0 to 255 which are not all equal, in any consistent order. The output can be floats, or integers rounded either up or down, which doesn't have to be consistent. If the integer part of the output is 0 or 360, you can print either of them.

You cannot call builtins for color space conversions, including implicit conversions such as while manipulating an image.

This is code-golf. Shortest code wins.

Examples

Input:  0 182 255
Output: 197 (or 198)

Input:  127 247 103
Output: 110

Input:  0 0 1
Output: 240

Input:  255 165 245
Output: 307 (or 306)

Edit

You don't have to follow the exact formula, but only have to give the same result as the above formula. I'd like to also see some answers golfing the formula itself.

\$\endgroup\$
  • \$\begingroup\$ Should we convert from sRGB to a linear scale first? I think we should, but nobody seems to have so far. \$\endgroup\$ – John Dvorak Aug 19 '15 at 13:10
  • \$\begingroup\$ @JanDvorak The task is to calculate the hue, in the simplest definition. In this case, "simplest" means you should assume the input is already in the right scale, and use the exact formula given in the question or anything that gives the same result. \$\endgroup\$ – jimmy23013 Aug 19 '15 at 13:27
  • \$\begingroup\$ But... 24 bpp usually means sRGB. If not, the format specification (you) should specify otherwise. \$\endgroup\$ – John Dvorak Aug 19 '15 at 13:33
  • \$\begingroup\$ @JanDvorak You should use this definition for RGB and the hue. \$\endgroup\$ – jimmy23013 Aug 19 '15 at 13:59
  • 9
    \$\begingroup\$ It has to be said: huehuehue. \$\endgroup\$ – TheDoctor Aug 19 '15 at 16:05

14 Answers 14

6
\$\begingroup\$

Pyth, 27 bytes

*60%+c-Ft.<QJxQKeSQ-KhSQyJ6

Demonstration. Test harness.

Fomula taken from Wikipedia.

Essentially, the steps are:

  1. .<QJxQKeSQ: Roate the largest value to the front of the list.
  2. -Ft: Take the difference of the other two values.
  3. -KhSQ: Subtract the minimum value from the maximum value.
  4. c: Divide 2 by 3.
  5. + ... yJ Add twice the index of the maximum value in the list (0 if R, 2 if G, 4 if B).
  6. % ... 6: Mod 6, to fix issues with negatives.
  7. *60: Multiply by 60 to convert to degrees, and print.
\$\endgroup\$
9
\$\begingroup\$

C#, 188 210 206 197 191 bytes

int H(int r,int g,int b){int[]a={r,g,b};System.Array.Sort(a);int x=a[2],y=a[1],c=x==g?1:(x==b?2:(y==b?3:0)),d=y==g?1:(y==b?2:(x==b?3:0));return(int)((c-d)*120*(1+(x-y)*1D/(x-a[0]))/2+d*120);}

Thanks to Sok for saving 4 bytes and to SLuck49 for saving 15!

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  • \$\begingroup\$ As you only use z once in the output calculation, and you don't use it in the preceding calculations, you do away with the variable and change the output to return(int)((c-d)*(1+(x-y)/(double)(x-a[0]))/2+d);, saving you 4 bytes. \$\endgroup\$ – Sok Aug 19 '15 at 14:47
  • \$\begingroup\$ You can factor 120 out of the c and d assignments and into the return like this c=x==g?1:(x==b?2:(y==b?3:0)),d=y==g?1:(y==b?2:(x==b?3:0)) and then return(int)((c-d)*120*(1+(x-y)/(double)(x-a[0]))/2+d*120); to save 4 bytes. \$\endgroup\$ – SLuck49 Aug 19 '15 at 20:23
  • \$\begingroup\$ Also do you really need the cast to double? If you do you can use this instead (x-a[0])*1D to save another 5 bytes. \$\endgroup\$ – SLuck49 Aug 19 '15 at 20:25
  • \$\begingroup\$ @SLuck49 Thanks! Yes, I actually need the cast, it gives inaccurate results otherwise, but that *1D multiplication is a nice trick! \$\endgroup\$ – ProgramFOX Aug 19 '15 at 20:32
  • \$\begingroup\$ Also also (just noticed) you can drop the using all together by fully qualifying System.Array for another 6 bytes. \$\endgroup\$ – SLuck49 Aug 19 '15 at 20:53
8
\$\begingroup\$

Pyth, 41 55 53 51 bytes

A.)JohN,VQ*L120?qeQhSQ3j312T+/*-HKeeJhc-GheJ-GhhJ2K

Input is expected in the form r,g,b. Here's an explanation:

                                                        Implicit: Q=eval(input()), evaluates to (r,g,b)
               ?qeQhSQ                                  Is b the smallest?
                      3j312T                            Choose [0,1,2] or [3,1,2] based on above
          *L120                                         Convert to [0,120,240] or [360,120,240]
       ,VQ                                              Pair -> [[r,0/360],[g,120],[b,240]]
   JohN                                                 Order by 1st element in each pair, store in J
A.)J                                                    Pop biggest from J, set G = x, H = h(X)
                                                        Output calculation:
                                       -GheJ                x - y
                                            -GhhJ           x - z
                                     hc                     Divide and increment
                                 KeeJ                       Set K = h(Y)
                              *-HK                          Multiply by (h(X) - h(Y))
                             /                   2          Integer division by 2
                            +                     K         Add h(Y)

Saved 4 bytes, thanks to @Jakube and @isaacg

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  • \$\begingroup\$ @jimmy23013 Fixed, thanks for the extra test case \$\endgroup\$ – Sok Aug 19 '15 at 12:59
  • 1
    \$\begingroup\$ A couple of golfs: m*120d -> *L120, save eeJ to K inline to save another byte. \$\endgroup\$ – isaacg Aug 19 '15 at 15:27
  • \$\begingroup\$ @isaacg I didn't know the L operator generated a range on an int automatically, every day is a shcool day it seems :o) Thanks! \$\endgroup\$ – Sok Aug 19 '15 at 15:47
8
+50
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Javascript (ES6), 145 115 108 100 97 90 bytes

Returns floats. Assign to a function to use.

(r,g,b)=>([x,y,z]=[r,g,b].sort((a,b)=>b-a),m=x-z,(x-r?x-g?r-g+4*m:b-r+2*m:g-b+6*m)/m%6*60)

Saved 30 bytes by inlining everything into a single ternary operator sequence and waiting until the end to normalize to 0-360.

Thanks to edc65, Vasu Adari, and ETHproductions for saving even more bytes.

JSFiddle with tests. Try in Firefox.

If removing the function declaration h= is not legal, add 2 bytes.

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  • \$\begingroup\$ You can remove 'var' and some bytes. \$\endgroup\$ – Vasu Adari Aug 26 '15 at 15:13
  • \$\begingroup\$ ES6Fiddle needs the var declaration for some reason and I didn't realize it wasn't necessary until I tried ES6 in firefox \$\endgroup\$ – DankMemes Aug 26 '15 at 17:02
  • 1
    \$\begingroup\$ You can save 6 bytes by replacing the curly braces with parentheses, the semicolon with a comma, and removing the return. I believe removing the function declaration (h=) is also legal, bringing the total down to 100. \$\endgroup\$ – ETHproductions Aug 26 '15 at 17:15
  • \$\begingroup\$ This may be obsessive (then again, aren't all good golfers? ;) ), but you could save two more bytes by getting rid of the parenthesis in %6)*60 and its partner on the other side. Also, using brute force on the addition (instead of adding 6 at the end) would actually save one byte over your current setup. (((x==r?(g-b)/m:x==g?2+(b-r)/m:4+(r-g)/m)+6)%6)*60 would become (x==r?6+(g-b)/m:x==g?8+(b-r)/m:10+(r-g)/m)%6*60. \$\endgroup\$ – ETHproductions Aug 27 '15 at 2:52
  • 1
    \$\begingroup\$ +1 for the sort, very clever, This is 90 (or 92) (r,g,b)=>([m,_,M]=[r,g,b].sort((a,b)=>a-b),C=M-m,(M-r?M-g?r-g+4*C:b-r+2*C:g-b+6*C)/C%6*60) \$\endgroup\$ – edc65 Aug 27 '15 at 15:44
6
\$\begingroup\$

Octave, 65 60 50 bytes

Edit: Saved 10 bytes thanks to pawel.boczarski

An approximate solution...

@(c)mod(atan2d(.866*c*[0;1;-1],c*[2;-1;-1]/2),360)

Test run

@(c)mod(atan2d(.866*c*[0;1;-1],c*[2;-1;-1]/2),360)
ans([0   182   255])
ans =  196.14

@(c)mod(atan2d(.866*c*[0;1;-1],c*[2;-1;-1]/2),360)
ans([127   247   103])
ans =  111.05

@(c)mod(atan2d(.866*c*[0;1;-1],c*[2;-1;-1]/2),360)
ans([0   0   1])
ans =  240.00

@(c)mod(atan2d(.866*c*[0;1;-1],c*[2;-1;-1]/2),360)
ans([255   165   245])
ans =  305.82

Octave, 107 bytes

My original (exact-ish) solution...

Code:

function H=r(c) [b,i]=sort(c);h=60*[6*(i(1)~=3),2,4](i);H=(h(3)-h(2))*(1+(b(3)-b(2))/(b(3)-b(1)))/2+h(2);

Explained:

function H=r(c)
   [b,i]=sort(c);
   h=60*[6*(i(1)~=3),2,4](i);
   H=(h(3)-h(2))*(1+(b(3)-b(2))/(b(3)-b(1)))/2+h(2);

This function takes a vector containing the R,G,B values as input c and sorts the input in ascending order

  • b contains the sorted values [z, y, x]
  • i contains the RGB plane associated with each value in b

The vector h is populated with the values

  • 60*[6, 2, 4] = [360, 120, 240] (but 3 bytes shorter)
  • unless the lowest value is in Blue (i(1) == 3), in which case the first hue value becomes zero
  • then use (i) to rearrange h into [h(Z), h(Y), h(X)] order

From there it's just a straight transcription of the formula. You can try it here.

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  • \$\begingroup\$ Hint: use anonymous function notation to squeeze more bytes: @(c)mod(atan2d(.866*c*[0;1;-1],c*[2;-1;-1]/2),360) is ten bytes shorter than definition with function keyword. \$\endgroup\$ – pawel.boczarski Aug 28 '15 at 19:22
  • \$\begingroup\$ @pawel.boczarski I was wondering if I could do away with the function header altogether, but I don't know if that's legit. But thanks for the tip! :D \$\endgroup\$ – beaker Aug 28 '15 at 19:25
  • \$\begingroup\$ @pawel.boczarski Looking back at this, I still need a r= before the anonymous function in order to call it, right? \$\endgroup\$ – beaker Aug 28 '15 at 20:01
  • \$\begingroup\$ There are many solutions where anonymous functions are posted. Moreover, you could call the so-defined function even like this: (@(c)mod(atan2d(.866*c*[0;1;-1],c*[2;-1;-1]/2),360))([127 247 103]) , or argue that you can use ans variable right after the anonymous function was defined, so that assignment is not necessary for the function definition to be complete. In one challenge (codegolf.stackexchange.com/questions/54945) a handle of existing Matlab library function was posted as full solution. \$\endgroup\$ – pawel.boczarski Aug 28 '15 at 20:51
  • \$\begingroup\$ @pawel.boczarski Wow, that's... just... evil :D I should have known Luis would be involved. I'll revert to the original code and use ans in the sample. Thanks again! \$\endgroup\$ – beaker Aug 28 '15 at 21:57
5
\$\begingroup\$

Pyth, 55

I know @Sok's answer beats mine, but since I finished mine just after he/she posted, I thought I would still post. This was my first time using Pyth so I'm sure I made some obvious mistakes.

DlZK*120ZRKJSQFNJ=Y+YxQN)=kl@Y1+k/*-leYk+1c-eJ@J1-eJhJ2

Input is expected to be r, g, b. You can try it here.

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  • \$\begingroup\$ Doesn't work for 255,165,245. \$\endgroup\$ – jimmy23013 Aug 19 '15 at 11:47
5
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PowerShell, 232 226 222 161 Bytes

See revision history for previous versions

$z,$y,$x=($r,$g,$b=$args)|sort
$c=((2,(0,3)[$y-eq$b])[$x-ne$b],1)[$x-eq$g]
$d=((2,(0,3)[$x-eq$b])[$y-ne$b],1)[$y-eq$g]
(($c-$d)*120*(1+($x-$y)/($x-$z))/2+$d*120)

Hoo boy, let's see if I can walk through this. Since \n counts the same as ; I left the line breaks in for clarity.

The first line takes input as three $args and stores them into $r, $g, $b. We're really only going to be using $b later, but we need all three so the |sort works appropriately. This makes $z, $y, $x the smallest-to-largest of the input arguments.

The next two lines setup $c and $d by using multiple index-into-an-array calls to set the numbers appropriately. Working from outside in, if $x is -equal to $g (i.e., green was the largest), we set $c=1 ... else, if $x is -notequal to $b (i.e., blue wasn't the largest) $c is either 0 or 3 depending if blue was the second largest ... else, $c=2. Similar logic sets $d.

We then calculate and print the output with the following, which is just the algorithm from the challenge golfed a little bit.

(($c-$d)*120*(1+($x-$y)/($x-$z))/2+$d*120)
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  • 1
    \$\begingroup\$ I don't know PowerShell, so correct me if I'm wrong... You don't use $z when calculating $c or $d, and you only use it once in the output calculation, so can you get rid of $z entirely and replace it with $a[0]? \$\endgroup\$ – Sok Aug 19 '15 at 15:14
4
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Ruby, 117 96 94 bytes

Code:

h=->r,g,b{z,y,x=[r,g,b].sort;v=x-z.to_f;({r=>(g-b)/v,g=>2+(b-r)/v,b=>4+(r-g)/v}[x]%6*60).to_i}
  • Saved 21 bytes by removing () and using r,g,b variables.
  • Taking modulus of 6 to convert negative value and multiplying it by 60 to convert to degrees that saved 2 bytes.

Examples:

irb(main):274:0> h.call 0,182,255
=> 197
irb(main):275:0> h.call 127,247,103
=> 110
irb(main):276:0> h.call 0,0,1
=> 240
irb(main):277:0> h.call 255,165,245
=> 306
\$\endgroup\$
3
\$\begingroup\$

SWI-Prolog, 133 bytes

a(L,H):-L=[R,G,B],max_list(L,X),min_list(L,Y),member(X:I:J:U,[R:G:B:0,G:B:R:2,B:R:G:4]),Z is 60*(U+(I-J)/(X-Y)),(Z<0,H is Z+360;H=Z).

Example: a([255,165,245],Hue). outputs Hue = 306.666666666666 .

This uses the following formula:

  • Max = max(R,G,B), Min = min(R,G,B).
  • If Max = R, U = 0. Else if Max = G, U = 2. Else U = 4.
  • If Max = R, I = G and J = B. Else if Max = G, I = B and J = R. Else I = R and J = G.
  • Z = U + (I - J)/(Max - Min)
  • Hue is either Z or Z + 360 if Z < 0.
\$\endgroup\$
  • \$\begingroup\$ Rounding is optional. \$\endgroup\$ – jimmy23013 Aug 23 '15 at 8:49
  • \$\begingroup\$ @jimmy23013 Updated, thanks. \$\endgroup\$ – Fatalize Aug 23 '15 at 8:53
3
\$\begingroup\$

Perl 5, 138 132 119 bytes

Code:

($m,$c,$M)=sort@A=($R,$G,$B)=@ARGV;print 60*(6+$M>$m?($G>$c?$B-$R:$B>$c?$R-$G:$G-$B)/($M-$m)+($G>$c?2:$B>$c?4:0):0)%360

Remarks:

Surely Perl can't win such challenge with all the Pyth'oresque golfing. But I wondered if this was possible to do with only 1 calculation step. Thanks to the modulus that worked out nicely. :)

Test:

$ perl hue.pl 0 182 255
197
$ perl hue.pl 127 247 103
110
$ perl hue.pl 0 0 1
240
$ perl hue.pl 255 165 245
307
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  • \$\begingroup\$ comparing to the middle value instead of the maximum shaved some bytes. (== versus >) \$\endgroup\$ – LukStorms Aug 28 '15 at 9:35
1
\$\begingroup\$

C++ 276 bytes

#include <iostream>
int H(int r,int g,int b){int m,n=120,o=240,l=r>g?r>b?g>b?m=r-b,o=n,n=-n,r-g:m=r-g,r-b:m=b-g,o+=n,n=-n,b-r:g>b?r>b?(m=g-b,o=0,g-r):m=g-r,n=-n,g-b:(m=b-r,o-=n,b-g);return (int)n*((float)l/m+1)/2+o;}int main(){int r,g,b;std::cin>>r>>g>>b;std::cout<<H(r,g,b);}
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  • \$\begingroup\$ A hint: you could leave H function alone in the answer, as in code-golf a standalone function is a legitimate answer, equivalent to a full program, vide meta discussion: meta.codegolf.stackexchange.com/questions/2419/… . This will make your answer more competitive (save 100 bytes in your case). You are still encouraged to leave the "full" version of program under the solution to simplify testing. \$\endgroup\$ – pawel.boczarski Aug 28 '15 at 19:40
  • \$\begingroup\$ The second test case 127 247 103 yields invalid value -120 instead of 110. \$\endgroup\$ – pawel.boczarski Aug 28 '15 at 19:46
1
\$\begingroup\$

R, 125 bytes

Very similar to beaker's Octave solution. Floating point output.

Code:

h=function(x){
  o=seq(3)[order(-x)];
  y=c(60*c(6*(o[3]!=3),2,4)[o],x[o]);
  return((y[1]-y[2])*(1+(y[4]-y[5])/(y[4]-y[6]))/2+y[2]);
}

Examples:

> h(c(0,182,255))
[1] 197.1765
> h(c(127,247,103))
[1] 110
> h(c(0,0,1))
[1] 240
> h(c(255,165,245))
[1] 306.6667
\$\endgroup\$
1
\$\begingroup\$

Python, 154 bytes

def h(c):r=c[:];c.sort();c=c[::-1];x,y,z=c;i,j=[120if n==r[1]else 240if n==r[2]else 0if z==r[2]else 360for n in[x,y]];print ((i-j)*(1+(x-y+0.)/(x-z))/2)+j

Accepts a list of values. Not sure if this can be broken down further. Here it is ungolfed:

def hue(color):
 rgb=color[:]  # copy list
 color.sort()  # sort list
 color=color[::-1]  # reverse sort
 x,y,z=color   # pull out x,y,z

 # The line 
 #   i,j=[120if n==r[1]else 240if n==r[2]else 0if z==r[2]else 360for n in[x,y]]
 # is basically the following, twice, once for x/hx and the second time for y/hy

 if x==rgb[1]: # if x is green
  hx = 120
 else:
  if x==rgb[2]: # if x is blue
   hx = 240
  else:
   if z==rgb[2]: # if z is blue and x is red
    hx = 0
   else:       # if x is red and y is blue
    hx = 1

 print ((hx-hy)*(1+(x-y+0.)/(x-z))/2)+hy  # calculate, print
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0
\$\begingroup\$

JavaScript 108

Alternate method.

function H(r,g,b){a=[r,g,b].sort(),M=a[2],c=M-a[0],h=M==r?(g-b)/c%6:M==g?(b-r)/c+2:(r-g)/c+4
return h*60|0;}

JavaScript 194

Using the example method.

Array.prototype.i=[].indexOf
function H(r,g,b,a){a=[r,g,b].sort(),i=[a.i(r),a.i(g),a.i(b)],x=[i[2]?360:0,120,240],hx=x[i.i(2)]|0,hy=x[i.i(1)]|0
return (hx-hy)*(1+(a[2]-a[1])/(a[2]-a[0]))/2+hy|0}

var input = document.getElementById("input").innerHTML;
var output = document.getElementById("output");
var html = "";

input.replace(/(\d+)\,(\d+)\,(\d+)/g, function(m, r, g, b) {
  html += H(r, g, b) + "\n";
});

output.innerHTML = html;
<pre id="input">
0,182,255
127,247,103
0,0,1
255,165,245
</pre>

<pre id="output">

</pre>

\$\endgroup\$

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