Image reduction method based on the F-transform

Abstract

We present a new method of (color) image reduction based on the F-transform technique with a generalized fuzzy partition. This technique successfully combines approximation (when reduction is performed) and interpolation (when reconstruction is produced). The efficiency of the proposed method is theoretically justified by its linear complexity and by comparison with interpolation, and aggregation-based reductions. We also analyze the measures ($\mathrm{MSE}$, $\mathrm{PEN}$, and $\mathrm{SSIM}$) that are commonly used to estimate the quality of reduced images and show that these measures have better values using the newly proposed method.

Appendix

The proposition below shows that if $\mathrm{MSE}$ measures the quality of reduction, and the reduction (reconstruction) is performed on the basis of the “block-to-pixel” (“pixel-to-block”) scheme, and the blocks are disjoint, then the optimal reduction is the arithmetic mean aggregation over a corresponding block.

Proposition 1

Let $u:N\times M\rightarrow [0,255]$ be an image function, $\bar{u}:n\times m\rightarrow [0,255]$ a reduced representation of u, and $\rho =\frac{NM}{nm}$ a reduction ratio. Assume that $N=M$, $n=m$ and $\frac{N}{n}=\sqrt{\rho }$ is a natural number which we denote by d. Let $B_{1,1},\ldots , B_{n,n}$ be disjoint $d\times d$ blocks that partition the domain of u. Let $\hat{u}:N\times N\rightarrow [0,255]$ be a reconstructed image where the reconstruction follows the scheme “block-to-pixel”, i.e., if $(x,y)\in B_{i,j}$, then $\hat{u}(x,y)= \bar{u}(i,j)$. If $\hat{u}$ minimizes $\mathrm{MSE}$ in (1), then

$$\begin{aligned} \bar{u}(i,j)=\frac{1}{d^2}\sum _{(x,y)\in B_{i,j}}u(x,y),\quad i,j=1,\ldots , n, \end{aligned}$$

(17)

i.e., the value of the reduction $\bar{u}$ at each pixel (i, j) of the reduced domain is the arithmetic mean of values of u over the corresponding block $B_{i,j}$ in the domain of u .

Proof

Let an image function $u:N\times M\rightarrow [0,255]$ be fixed and all assumptions of the proposition be fulfilled. Because the domain of u is partitioned into $n^2$ disjoint blocks $B_{1,1},\ldots , B_{n,n}$, we rewrite the expression for $\mathrm{MSE}$ as follows:

$$\begin{aligned} \mathrm{MSE}(u,\hat{u})\!=\!\frac{\sum _{i=1}^n\sum _{j=1}^n \sum _{(x,y)\in B_{i,j}}(u(x,y)-\hat{u}(x,y))^2}{N^2}. \end{aligned}$$

Furthermore,

$$\begin{aligned} \mathrm{MSE}(u,\hat{u})\!=\!\frac{\sum _{i=1}^n\sum _{j=1}^n \sum _{(x,y)\in B_{i,j}}(u(x,y)-\bar{u}(i,j))^2}{N^2}, \end{aligned}$$

so that $\mathrm{MSE}(u,\hat{u})$ actually depends on $n^2$ unknown values $\bar{u}(i,j),\, i,j=1,\ldots , n$, i.e.,

$$\begin{aligned} \mathrm{MSE}(u,\hat{u})=\mathrm{MSE}(\bar{u}(1,1),\ldots , \bar{u}(n,n)). \end{aligned}$$

By the necessary condition for the existence of a (relative) minimum, all partial derivatives of $\mathrm{MSE}$ by $\bar{u}(i,j),\,i,j=1,\ldots , n$, should be equal to 0. This leads to the following system of $n^2$ linear equations:

$$\begin{aligned} \sum _{(x,y)\in B_{i,j}}(u(x,y)-\bar{u}(i,j))=0,\quad i,j=1,\ldots , n. \end{aligned}$$

Because there are $d^2$ pixels in each block $B_{i,j}$, we easily come to Eq. (17).$\square$

We repeat the formulation of Theorem 1 and give its proof.

Theorem 3

Let f be a continuous function on [a, b]. Then, for any $\varepsilon >0$, there exist $h_{\varepsilon }$ such that for any $h_{\varepsilon }/2<h'\le h_{\varepsilon }$ and any $(h_{\varepsilon }, h')$-uniform fuzzy partition of [a, b], the corresponding inverse F-transform $\hat{f}_{\varepsilon }$ of f fulfills

$$\begin{aligned} |f(x)-\hat{f}_{\varepsilon }(x)|\le \varepsilon ,\, x\in [a,b]. \end{aligned}$$

Proof

Let us choose some $\varepsilon >0$. By assumption, the function f is continuous and thus, uniformly continuous on [a, b]. Therefore, for the chosen $\varepsilon$ we can find $h_\varepsilon >0$ such that for all $x',x''\in [a,b]$, $|x'-x''|<h_\varepsilon$ implies $|f(x')-f(x'')|<\varepsilon /2$. Let us assume that the value $(b-a)/h_{\varepsilon }$ is an integer (otherwise we choose the least natural number $n_{\varepsilon }$ such that $n_{\varepsilon }>2$ and $(b-a)/(n_{\varepsilon }-1)\le h_\varepsilon$) and find the $h_\varepsilon$-equidistant nodes $x_1,\ldots ,x_{n_{\varepsilon }}\in [a,b]$, where $n_{\varepsilon }=(b-a)/h_{\varepsilon }+1$, such that $a=x_1<\cdots < x_{n_{\varepsilon }}=b$. Then we choose $h_{\varepsilon }/2<h'\le h_{\varepsilon }$ and establish a $(h_{\varepsilon }, h')$-uniform fuzzy partition of [a, b] determined by the chosen nodes and constituted by basic functions $A_1,\ldots , A_{n_{\varepsilon }}$.

Let us proof that the inverse F-transform $\hat{f}_{\varepsilon }$ of f fulfills the requested inequality. For this purpose, we choose some $x\in [a,b]$ and find k such that $x\in [x_k,x_{k+1})$.

Let $F_1,\ldots , F_{n_{\varepsilon }}$ be the components of the F-transform of f w.r.t. basic functions $A_1,\ldots , A_{n_{\varepsilon }}$. By the property (d) from the list on the page 7, for the chosen $x\in [x_k,x_{k+1})$, we have

$$\begin{aligned} |f(x)-F_k|\le 2\omega (h_{\varepsilon },f)<\varepsilon , \end{aligned}$$

and analogously,

$$\begin{aligned} |f(x)-F_{k+1}|<\varepsilon , \end{aligned}$$

where we used the fact that $h'\le h_{\varepsilon }$. Therefore, for the chosen x we can write the chain of inequalities:

$$\begin{aligned}&|f(x)-\hat{f}_{\varepsilon }(x)|=\left| f(x)-\frac{\sum _{i=1}^{n_{\varepsilon }} F_iA_i(x)}{\sum _{i=1}^{n_{\varepsilon }} A_i(x)}\right| \\&\quad =\frac{|f(x)\sum _{i=1}^{n_{\varepsilon }} A_i(x)-\sum _{i=1}^{n_{\varepsilon }} F_iA_i(x)|}{\sum _{i=1}^{n_{\varepsilon }} A_i(x)}\\&\quad \le \frac{\sum _{i=1}^{n_{\varepsilon }} A_i(x)|f(x)-F_i|}{\sum _{i=1}^{n_{\varepsilon }} A_i(x)}. \end{aligned}$$

Because $h_{\varepsilon }/2<h'\le h_{\varepsilon }$, there are at most two values $A_k(x)$ and $A_{k+1}(x)$ that may be different from zero. Therefore,

$$\begin{aligned}&|f(x)-\hat{f}_{\varepsilon }(x)|\le \frac{\sum _{i=1}^{n_{\varepsilon }} A_i(x)|f(x)-F_i|}{\sum _{i=1}^{n_{\varepsilon }} A_i(x)}\\&\quad = \frac{\sum _{i=k}^{k+1}A_i(x)|f(x)-F_i|}{\sum _{i=1}^{n_{\varepsilon }} A_i(x)}\\&\quad <\varepsilon \frac{\sum _{i=k}^{k+1}A_i(x)}{\sum _{i=1}^{n_{\varepsilon }} A_i(x)}= \varepsilon \frac{\sum _{i=1}^{n_{\varepsilon }}A_i(x)}{\sum _{i=1}^{n_{\varepsilon }} A_i(x)}=\varepsilon . \end{aligned}$$

Because x was chosen arbitrary inside [a, b], the chain of inequalities proves the claim.$\square$