Contrast preserving decolorization based on the weighted normalized L1 norm

Abstract

Image decolorization is to transform a color image into a grayscale image with the preserved contrast and consistent details. It is an important tool in image processing and realistic applications, such as monochrome printing and e-ink display. In this paper, we propose a novel contrast preserving method for image decolorization. Our main contribution is threefold: Firstly, we define a new contrast feature for a color image which combine the correlated information among R, G and B channels with the color contrast in each channel. Secondly, we propose to use the weighted normalized L1 norm to measure the distance between the grayscale image and the color image contrast features, and formulate an constrained optimization problem. Finally, we utilize a discrete searching solver to solve the optimization problem efficiently. The proposed decolorization method is good at preserving low contrast as well as high contrast structures in the color image. The objective and subjective evaluation on three benchmark datasets demonstrates that our decolorization method is effective and competitive with some state-of-the-art decolorization methods.

Appendix

1.1 A.1 Proof of Proposition 1

Proof

It is equivalent to prove: $\forall x,y,z \in {\mathscr{A}}$,

$$ |x-y|(y+z)(x+z)+|y-z|(x+y)(x+z)-|x-z|(x+y)(y+z)\leq 0. $$

• (1) x ≤ y ≤ z or x ≥ y ≥ z, we need to proof

  $$ |y-z|x^{2}-|y^{2}-z^{2}|x+yz|y-z|\geq 0. $$

  (32)

  As δ = (y − z)⁴ ≥ 0, (32) holds.

• (2) x ≤ z ≤ y or x ≥ z ≥ y, we need to proof

  $$ |y-z|x^{2}+3|y^{2}-z^{2}|x+yz|y-z|\geq 0. $$

  (33)

  As δ = (y − z)²(9z² + 9y² + 14yz) ≥ 0, (33) holds.

• (3) y ≤ x ≤ z or y ≥ x ≥ z, we need to proof

  $$ |y-x|z^{2}+3|y^{2}-x^{2}|x+xy|y-x|\geq 0. $$

  (34)

  As δ = (x − y)²(9x² + 9y² + 14xy) ≥ 0, (34) holds.

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1.2 A.2 Proof of Proposition 2

Proof

For k ≠ 0,

$$ f(kx,ky) = \frac{\|kx-ky\|}{\|kx\|+\|ky\|} = \frac{|k|\|x-y\|}{|k|(\|x\|+\|y\|)} = f(x,y). $$

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1.3 A.3 Proof of Proposition 3

Proof

$$ f(x,y) = \frac{\|x-y\|}{\|x\|+\|y\|} = \frac{\|y-x\|}{\|y\|+\|x\|} = f(y,x). $$

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Proof

(of Proposition4) We set c ≥ 0 of the sublevel set A_c because q ≥ 0 on the domain.

$$ x=(x_{i}) \in A_{c} \Longleftrightarrow \sum\limits_{i=1}^{n} |x_{i}-y_{i}| \leq \sum\limits_{i=1}^{n} c(|x_{i}|+|y_{i}|). $$

∀x, z ∈ A_c, ∀t ∈ (0, 1),

$$ \begin{array}{@{}rcl@{}} && \sum\limits_{i=1}^{n} |tx_{i}+(1-t)z_{i}-y_{i}| \end{array} $$

(35)

$$ \begin{array}{@{}rcl@{}} &\leq & \sum\limits_{i=1}^{n} |tx_{i}-ty_{i}|+|(1-t)z_{i}-(1-t)y_{i}| \end{array} $$

(36)

$$ \begin{array}{@{}rcl@{}} &\leq & \sum\limits_{i=1}^{n} ct(|x_{i}|+|y_{i}|)+c(1-t)(|z_{i}|+|y_{i}|) \end{array} $$

(37)

$$ \begin{array}{@{}rcl@{}} &= & \sum\limits_{i=1}^{n} c(|tx_{i}+(1-t)z_{i}|+|y_{i}|). \end{array} $$

(38)

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