Conversion from HSI to RGB

Given $H$, $S$ and $I$, we can first find $r$, $g$, and $b$ on the RGB triangle, and the find $R$, $G$, and $B$. Consider three possible cases in terms of the hue angle $H$:

• $0< H < 120$ ($B<R$ and $B<G$, i.e., $P=(r,g,b)$ inside $\triangle P_rWP_g$)

  From $S=1-3b$, solving for $b$ we get
  

  $$b=\frac{1-S}{3}$$
  
  

  To find $r$, consider the triangle $P_rOQ_r$:

  

  Here vector $P''W$ is the projection of vector $PW$ onto the line $P_rQ_r$, and all points $W=(1/3,1/3,1/3)$, $P=(r,g,b)$, and $P''$ are on the RGB triangle. Now consider the following steps:

  • On the triangle $P_rQ_rO$, we have
    
    $$\frac{\vert P_rQ_r\vert}{\vert P_rO\vert}=\frac{\vert P_rQ_... ....}\;\;\;\;\;\; d=\frac{\vert P_rP''\vert}{\vert P_rQ_r\vert}$$
    
    
    and
    
    $$r=\vert P_rO\vert-d=1-d =1-\frac{\vert P_rP''\vert}{\vert P... ...l{*}{=}\frac{1}{3}+\frac{\vert WP''\vert}{\vert P_rQ_r\vert}$$
    
    
    (* $\vert WQ_r\vert=1/\sqrt{6}$ and $\vert P_rQ_r\vert=\sqrt{6}/2$)
  • The length of $P''W$ is $\vert WP''\vert=\vert WP\vert\cos\angle H$, and the saturation is $S=\vert PW\vert/\vert WP'\vert$, i.e., $\vert PW\vert=S\vert WP'\vert$, we have:
    
    $$\vert WP''\vert=\vert WP\vert\cos H=\vert WP'\vert S\cos H$$
    
    
    and we get
    
    $$r=\frac{1}{3}+\frac{\vert WP''\vert}{\vert P_rQ_r\vert} =\frac{1}{3}+S\cos H \frac{\vert WP'\vert}{\vert P_rQ_r\vert}$$
    
    

  • As $\vert P_rQ_r\vert/\vert WQ_b\vert=(\sqrt{6}/2)/(1/\sqrt{6})=3$, i.e., $\vert P_rQ_r\vert/3=\vert WQ_b\vert$, and also $\cos(60-H)=\vert WQ_b\vert/\vert WP'\vert$, we have
    
    $$\frac{\vert WP'\vert}{\vert P_rQ_r\vert}=\frac{1}{3}\frac{\... ...rac{\vert WP'\vert}{\vert WQ_b\vert} =\frac{1}{3\cos(60- H)}$$
    
    
    Substituting this into the expression of $r$ above, we get:
    
    $$r =\frac{1}{3}\left[1+\frac{S\cos H}{\cos(60- H)}\right]$$
    
    

  • Finally we can get
    
    $$g=1-b-r$$
    
    
    Given $r,g,b$ and $I$, we can get $R,G,B$:
    
    $$R=3Ir,\;\;\;G=3Ig,\;\;\;B=3Ib$$
    
    

• $120<H < 240$ (p inside $\triangle P_gwP_b$)

  
  

  $$H=H-120$$
  
  
  
  
  $$r=\frac{1-S}{3},\;\;\;\;\;\; g=\frac{1}{3}\left[1+\frac{S \cos H}{\cos(60-H)}\right], \;\;\;\;\;\;\;\;\;b=1-g-r$$
  
  

• $240< H < 360$ (p inside $\triangle P_bwP_r$)

  
  

  $$H=H-240$$
  
  
  
  
  $$g=\frac{1-S}{3},\;\;\;\;\;\;\;\; b=\frac{1}{3}\left[1+\frac{S \cos H}{\cos(60-H)}\right], \;\;\;\;\;\;\;\;\; r=1-g-b$$
  
  

In summary, given the H, S, and I values, we can obtain the R, G, and B components by

$$b=(1-S)/3$$

$$r =\frac{1}{3}\left[1+\frac{S\cos H}{\cos(60- H)}\right]$$

$$g=1-b-r$$

and

$$R=3Ir,\;\;\;\;\;G=3Ig,\;\;\;\;\;B=3Ib$$

Note that after the HSI to RGB conversion, we need to convert the R, G, and B values from the range of $(0,\;1)$ back to the three components for a pixel in a color image in the range of $(0,\;255)$.