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I have the following weighting function: $$ w(\rho) = k \cdot(\rho^\alpha)\cdot(1-\rho)^\beta,$$ where $\rho$ represents a pixel value expressed as a double, i.e., it ranges from $0$ to $1$, and $ 1\le \alpha \le \beta$.

How can I convert this function so that it shows a curve similar to the given one but works for pixel values expressed as uint8, i.e., ranging from $0$ to $255$?

In the following example of the function I take values $\alpha = 1$; $\beta =6$ and $k =17.6$.

plot for <span class=$w(\rho) $ for $0\le \rho \le 1 $">

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The simple answer is to divide your uint8 values by $255$ as a float. That gives you a value in $[0,1]$ That is almost certainly good enough for this purpose.

The subtlety comes because the float values may be equally distributed on $[0,1]$. The inverse transformation would be to multiply the floats by $255$. Since the float will never be $1$ you would never get a result of $255$. Even if you round, $0$ and $255$ will only get hit half as often as the other values. The improvement would be to multiply by $256$ and round down. Now if the floats are equally distributed, so are the uint8s. For the uint8 to float you could then divide by $256$.

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  • $\begingroup$ thanks sir, I have done the same earlier and I got zeros instead of ones during conversion of float to uint8 for the values like this (6.376246841303954e-14) and so I incremented every value by one . I know its a dumb question to ask but , I wanted to know that ,given x < P < y ranges for a variable P in function and to use it for another range x1 < P < y1 ,doing a proportional conversion just like we did earlier is fine right. $\endgroup$ – MOHANTEJA Mar 13 at 18:00
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    $\begingroup$ @MOHANTEJA I don't understand: for me it seems correct that a pixel value expressed as double is $6.376246841303954\cdot e^{-14}$ and expressed as uint8 is $0$. All doubles between $0$ and $1/256$ should be mapped to $0$ when expressed as uint8. $\endgroup$ – simonet Mar 13 at 21:21
  • $\begingroup$ @MOHANTEJA: I am surprised that you can get a value of $6.37E-14$. That is extremely small. Most pixel values have only 16 or 24 bits, but this needs 44 to represent it. I suspect you are not interpreting the input correctly. If it is a correct input, it should be mapped to 0, as should any value less than $1/256$ $\endgroup$ – Ross Millikan Mar 13 at 23:33
  • $\begingroup$ @RossMillikan I using the weight function in denominator of a division.so I had to make zeros to ones in order to prevent division by zero, sorry I didn't mentioned that in the earlier comment. $\endgroup$ – MOHANTEJA Mar 14 at 13:15
  • $\begingroup$ [the matlab code I have used ](gist.github.com/mohanteja1/775aa179df41cac21a500ebc69e6aa14 ) $\endgroup$ – MOHANTEJA Mar 14 at 13:57

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