Histogram matching

Consider a grayscale input image X. It has a probability density function p_r(r), where r is a grayscale value, and p_r(r) is the probability of that value. This probability can easily be computed from the histogram of the image by

${\textstyle p_{r}(r_{j})={n_{j} \over n}}$

Where n_j is the frequency of the grayscale value r_j, and n is the total number of pixels in the image.

Now consider a desired output probability density function p_z(z). A transformation of p_r(r) is needed to convert it to p_z(z).

Each pdf (probability density function) can easily be mapped to its cumulative distribution function by

${\displaystyle S(r_{k})=\sum _{j=0}^{k}p_{r}(r_{j}),\qquad k=0,1,2,3,\ldots ,L-1}$

${\displaystyle G(z_{k})=\sum _{j=0}^{k}p_{z}(z_{j}),\qquad k=0,1,2,3,\ldots ,L-1}$

Where L is the total number of gray level (256 for a standard image).

The idea is to map each r value in X to the z value that has the same probability in the desired pdf. I.e. S(r_j) = G(z_i) or z = G⁻¹(S(r)).^[3]

The following input grayscale image is to be changed to match the reference histogram.

The input image has the following histogram

It will be matched to this reference histogram to emphasize the lower gray levels.

After matching, the output image has the following histogram

And looks like this

Given two images, the reference and the target images, we compute their histograms. Following, we calculate the cumulative distribution functions of the two images' histograms – ${\displaystyle F_{1}()\,}$ for the reference image and ${\displaystyle F_{2}()\,}$ for the target image. Then for each gray level ${\displaystyle G_{1}\in [0,255]}$, we find the gray level ${\displaystyle G_{2}\,}$ for which ${\displaystyle F_{1}(G_{1})=F_{2}(G_{2})\,}$, and this is the result of histogram matching function: ${\displaystyle M(G_{1})=G_{2}\,}$. Finally, we apply the function ${\displaystyle M()}$ on each pixel of the reference image.

In typical real-world applications, with 8-bit pixel values (discrete values in range [0, 255]), histogram matching can only approximate the specified histogram. All pixels of a particular value in the original image must be transformed to just one value in the output image.

Exact histogram matching is the problem of finding a transformation for a discrete image so that its histogram exactly matches the specified histogram.^[4] Several techniques have been proposed for this. One simplistic approach converts the discrete-valued image into a continuous-valued image and adds small random values to each pixel so their values can be ranked without ties. However, this introduces noise to the output image.

Because of this there may be holes or open spots in the output matched histogram.

The histogram matching algorithm can be extended to find a monotonic mapping between two sets of histograms. Given two sets of histograms ${\displaystyle P=\{p_{i}\}_{i=1}^{k}}$ and ${\displaystyle Q=\{q_{i}\}_{i=1}^{k}}$, the optimal monotonic color mapping ${\displaystyle M}$ is calculated to minimize the distance between the two sets simultaneously, namely ${\displaystyle \operatorname {argmin} _{M}\sum _{k}d(M(p_{k}),q_{k})}$ where ${\displaystyle d(\cdot ,\cdot )}$ is a distance metric between two histograms. The optimal solution is calculated using dynamic programming.^[5]

• Histogram equalization
• Image histogram
• Color mapping