Exact Histogram Specification for Digital Images Using a Variational Approach

Abstract

We consider the problem of exact histogram specification for digital (quantized) images. The goal is to transform the input digital image into an output (also digital) image that follows a prescribed histogram. Classical histogram modification methods are designed for real-valued images where all pixels have different values, so exact histogram specification is straightforward. Digital images typically have numerous pixels which share the same value. If one imposes the prescribed histogram to a digital image, usually there are numerous ways of assigning the prescribed values to the quantized values of the image. Therefore, exact histogram specification for digital images is an ill-posed problem. In order to guarantee that any prescribed histogram will be satisfied exactly, all pixels of the input digital image must be rearranged in a strictly ordered way. Further, the obtained strict ordering must faithfully account for the specific features of the input digital image. Such a task can be realized if we are able to extract additional representative information (called auxiliary attributes) from the input digital image. This is a real challenge in exact histogram specification for digital images. We propose a new method that efficiently provides a strict and faithful ordering for all pixel values. It is based on a well designed variational approach. Noticing that the input digital image contains quantization noise, we minimize a specialized objective function whose solution is a real-valued image with slightly reduced quantization noise, which remains very close to the input digital image. We show that all the pixels of this real-valued image can be ordered in a strict way with a very high probability. Then transforming the latter image into another digital image satisfying a specified histogram is an easy task. Numerical results show that our method outperforms by far the existing competing methods.

Appendix A

Given a square matrix A, the expression A≻0 means that A is positive definite and A⪰0 that A is positive semi-definite.

1.1 A.1 Proof of Proposition 1

By Hypothesis 1, for any u∈ℝⁿ, the function $\mathcal {J}(\cdot,\mathbf{u})$ in (2) is strictly convex and coercive, hence for any u and β>0, it has a unique minimizer. Each minimizer point $\widehat{\mathbf{f}}$ of $\mathcal{J}(\widehat {\mathbf{f}},\mathbf{u})$ is determined by $D_{1}\mathcal{J}(\widehat{\mathbf{f}},\mathbf{u})=0$. We have

$$ 0=D_1\mathcal{J}(\widehat{\mathbf{f}},\mathbf{u})=D_1\varPsi (\widehat{\mathbf{f}},\mathbf{u})+\beta D_1\varPhi(\widehat {\mathbf{f}}), $$

(19)

where

(20)

Differentiation with respect to $\widehat{\mathbf{f}}$ yet again yields

$$ D_1^2\mathcal{J}(\widehat{\mathbf{f}},\mathbf{u})=D_1^2\varPsi (\widehat{\mathbf{f}},\mathbf{u})+\beta D_1^2\varPhi (\widehat{\mathbf{f}})\in \mathbb{R}^{n\times n}. $$

(21)

Here, $D_{1}^{2}\varPsi(\widehat{\mathbf{f}},\mathbf{u})$ is an n×n diagonal matrix with strictly positive entries according to Hypothesis 1:

$$ D_1^2\varPsi(\widehat{\mathbf{f}},\mathbf{u})[i,i]=\psi''(\widehat {\mathbf{f}}[i]-\mathbf{u}[i]),\quad\,\forall i\in\mathbb{I}_n. $$

(22)

Hence $D_{1}^{2}\varPsi(\mathcal{F}(\mathbf{u}),\mathbf{u})\succ0$. Furthermore,

$$ D_1^2\varPhi(\widehat{\mathbf{f}})=G^T\mathrm{diag}\Big(\phi ''(\mathbf{g}_{1}^T\widehat{\mathbf{f}}),\ldots, \phi''(\mathbf{g}_{r}^T\widehat{\mathbf{f}})\Big)G, $$

(23)

so $D_{1}^{2}\varPhi(\widehat{\mathbf{f}})\succeq0$. It follows that $D_{1}^{2}\mathcal{J}(\widehat{\mathbf{f}},\mathbf{u})\succ0$, for any u∈ℝⁿ.

Consequently, Lemma 1 holds true for any u∈ℝⁿ and any β>0. The same lemma shows that the statement of Proposition 1 holds true for $\mathcal{O}=\mathbb{R}^{n}$.

1.2 A.2 Proof of Lemma 2

Since $\mathcal{F}$ is a local minimizer function,

$$ D_1\mathcal{J}(\mathcal{F}(\mathbf{u}),\mathbf{u})=0,\quad\forall \mathbf{u}\in\mathbb{R}^n. $$

(24)

We can thus differentiate with respect to u on both sides of (24) which yields

$$D_1^2\mathcal{J}(\mathcal{F}(\mathbf{u}),\mathbf {u})D\mathcal{F}(\mathbf{u})+D_2D_1\mathcal{J}(\mathcal{F}(\mathbf {u}),\mathbf{u} )=0,\quad\forall\mathbf{u}\in\mathbb{R}^n. $$

(25)

Note that $D\mathcal{F}(\mathbf{u})$ and $D_{2}D_{1}\mathcal{J}(\mathcal {F}(\mathbf{u}),\mathbf{u})$ are n×n real matrices. The Hessian matrix $H(\mathbf{u})=D_{1}^{2}\mathcal{J}(\mathcal {F}(\mathbf{u}),\mathbf{u})$ can be expanded using (21):

$$H(\mathbf{u})=D_1^2\varPsi(\mathcal{F}(\mathbf{u}),\mathbf {u})+\beta D_1^2\varPhi(\mathcal{F}(\mathbf{u} ))\in \mathbb{R}^{n\times n}. $$

Replacing $\widehat{\mathbf{f}}$ by $\mathcal{F}(\mathbf{u})$ in (22) and (23) yields

as stated in (5). Using Hypothesis 1, it is readily seen that

Then H(u)≻0, hence H(u) is invertible.

Using (19) and (A.1), where we consider $\mathcal {F}(\mathbf{u})$ in place of $\widehat{\mathbf{f}}$, shows that

(26)

Then (25) is equivalent to

(27)

Obviously, $\mathrm{rank}\big(D\mathcal{F}(\mathbf{u})\big)=n$. This result is independent of the value of β>0.

1.3 A.3 Proof of Theorem 1

The proof consists of two parts.

(I) Given $g\in\mathcal{G}$, where $\mathcal{G}$ is given in (8), consider the function f _g :ℝⁿ→ℝ defined by

$$f_g(\mathbf{u})=g^T\mathcal{F}(\mathbf{u}),\quad\,\forall \mathbf {u}\in\mathbb{R}^n. $$

By Lemma 2, $D\mathcal{F}(\mathbf{u})$ is invertible. Hence⁵ $g^{T}D\mathcal {F}(\mathbf{u})\neq0$ and thus

$$\mathrm{rank}\big(f_g(\mathbf{u})\big)=\mathrm{rank}\big (g^TD\mathcal{F}(\mathbf{u})\big)=1,\quad \forall\mathbf{u}\in\mathbb{R}^n. $$

In particular, f _g does not have critical points. Its inverse of the origin K _g reads

$$K_g\stackrel{\mathrm{def}}{=}f_g^{-1}(0)=\{\mathbf {u}\in\mathbb{R}^n\mid g^T\mathcal{F}(\mathbf{u})=0\} . $$

(28)

By an extension⁶ of the constant rank theorem [4], the subset K _g in (28), supposed nonempty, is a $\mathcal{C}^{s-1}$ manifold of ℝⁿ of dimension n−1. Hence $\mathbb{L}^{n}(K_{g})=0$ (see e.g. [20, 31]). Using Proposition 1, f _g is $\mathcal{C}^{s-1}$-continuous, so K _g is closed.

The set $K_{\mathcal{G}}$ in (9) also reads

$$K_{\mathcal{G}}=\bigcup_{g\in\mathcal{G}}K_g. $$

Using that $\mathcal{G}$ is of finite cardinality, it follows that $K_{\mathcal{G}}$ is closed in ℝⁿ and that

$$\mathbb{L}^n\left(K_{\mathcal{G}}\right)=0. $$

The conclusion is clearly independent of the value of β>0. (II) Given $(i,j)\in\mathbb{I}_{n}\times \mathbb{I}_{n}$ (including i=j), define the subset K _i,j ⊂ℝⁿ as

$$K_{i,j}=\mathcal{F}_{i}^{-1}(\mathbf{u}_{j})=\big\{ \mathbf{u}\in\mathbb{R}^n\mid\mathcal{F} _{i}(\mathbf{u})=\mathbf{u}_{j}\big\}. $$

(29)

From Proposition 1, $\mathcal{F}_{i}:\mathbb{R}^{n}\to\mathbb {R}$ is $\mathcal{C}^{s-1}$ continuous, so K _i,j is closed in ℝⁿ. By Lemma 2, all rows of $D\mathcal{F}(\mathbf{u})\in\mathbb{R}^{n\times n}$ are linearly independent, ∀u∈ℝⁿ. Consequently, for any $i\in\mathbb{I}_{n}$,

$$\mathrm{rank}\big(D\mathcal{F}_{i}(\mathbf{u})\big)=1,\quad \forall\mathbf{u}\in\mathbb{R}^n. $$

Using the same arguments as in (I), K _i,j is a $\mathcal{C}^{s-1}$ submanifold of ℝⁿ of dimension n−1, hence

$$\mathbb{L}^n(K_{i,j})=0. $$

Noticing that $K_{\mathcal{I}}$ in (10) is a finite union of (n−1)-dimensional submanifolds in ℝⁿ like K _i,j entails the result. The independence of these results from β>0 is obvious.

1.4 A.4 A Lemma Needed to Prove Proposition 2

Lamma 3

Let (A, B)∈ℝ^n×n×ℝ^n×n satisfy

Consider the n×n matrix $M\stackrel{\mathrm{def}}{=}(A+B)^{-1}A$. Then all eigenvalues of M ^T M belong to (0,1].

Proof

Let λ be an eigenvalue of (A+B)⁻¹ A and v∈ℝⁿ∖{0} a right eigenvector corresponding to λ. Then

(30)

(31)

If λ=0 then (31) yields A v=0 which is impossible because A≻0, A ^T=A and v≠0. Hence

$$\lambda\neq0. $$

(32)

Furthermore, (31) yields

$$\lambda\, \mathbf{v}^TB\mathbf{v}=(1-\lambda)\,\mathbf {v}^TA\mathbf{v}. $$

Using that A≻0 and B⪰0, the last equation shows that

$$ \frac{1}{\lambda}-1=\frac{\mathbf{v}^TB\mathbf{v}}{\mathbf {v}^TA\mathbf{v}}\geq0. $$

(33)

Combining the latter inequality with (32) entails that

$$0<\lambda\leq1. $$

(34)

Hence all eigenvalues of M live in (0,1].

Using that A is diagonal with positive diagonal entries, we can write down

$$A+B=A^{\frac{1}{2}}\left(I+A^{-\frac {1}{2}}BA^{-\frac{1}{2}}\right)A^{\frac{1}{2}}. $$

(35)

Then the definition of M shows that⁷

(36)

Using (35), the expression in (30) is equivalent to

Combining the last result with (36) yields

Consequently, λ ² is an eigenvalue of M ^T M corresponding to an eigenvector given by $(A^{\frac{1}{2}}\mathbf{v})$. Thus all eigenvalues of M ^T M belong to (0,1] as well. □

1.5 A.5 Proof of Proposition 2

Let us denote

as well as

$$M(\mathbf{u})=(A(\mathbf{u})+\beta B(\mathbf{u}))^{-1}A(\mathbf{u}). $$

Then $D\mathcal{F}(\mathbf{u})=M(\mathbf{u})$. We have

Using Lemma 3 and Hypothesis 1, for any v∈ℝⁿ, all eigenvalues of (M(v))^T M(v) are in (0,1]. By the definition of the matrix 2-norm [33] one obtains

$$t\in[0,1]\quad\mbox{and}\quad\boldsymbol{\zeta}\in\mathbb {R}^n\quad\Rightarrow\quad\| M(\mathbf{u}+t\boldsymbol{\zeta} )\|_{2}\leq1. $$

Noticing that M(⋅) is a continuous mapping, the mean value theorem (see e.g. [4, 14]) shows that

$$\|\mathcal{F}(\mathbf{u}+\boldsymbol{\zeta})-\mathcal{F}(\mathbf {u})\|_2\leq \max_{0\leq t\leq1}\|M(\mathbf{u}+t\boldsymbol{\zeta})\|_2\| \boldsymbol{\zeta}\|_2\leq\|\boldsymbol{\zeta}\|_2. $$