Algebraic technique for computationally efficient Hahn moment invariants

Abstract

In image processing applications such as scene analysis problems, moments are used as image descriptors of three dimensional objects. These moments may be sensitive to several transformations such as translation and scaling. Three-dimensional moment invariants are possible solution to such problems. Recently, Tchebichef and Krawtchouk moments and their invariants are proposed. These moment invariants are obtained using indirect method or image normalisation method. Perfect invariance cannot be achieved when moment invariants are not obtained from their corresponding polynomials. In this paper, the translation and scale invariance of Hahn moments for two and three dimensional symmetrical and non symmetrical images are derived directly from discrete orthogonal Hahn polynomials using algebraical method. This also enhances the computational efficiency in terms of processing time as compared to the method based on geometric moment invariants. The performance of proposed descriptor is evaluated using binary characters and 3D images. Hahn moment which is generalization of Tchebichef and Krawtchouk moment, give better results with translation and scale invariance and classification problems under clean and noisy image conditions.

Appendix

Proof of (24) Substituting the value of $H_{n-k}$ from (10) in (23), following equation is obtained

$$\begin{aligned} H_{n}(x+x')=\sum _{k=0}^{n}v_{n(n-k)}(x')\sum _{l=0}^{n-k}B_{(n-k)l}\langle x\rangle _{l} = \sum _{k=0}^{n}\sum _{l=k}^{n}B_{lk}v_{nl}(x')\langle x\rangle _k \end{aligned}$$

(61)

Extracting coefficient of $\langle x\rangle _{n-k}$ from (61),the value obtained is

$$\begin{aligned} coeff. \quad of \quad \langle x\rangle _{n-k}=\sum _{i=0}^{k}B_{(n-i)(n-k)}v_{n(n-i)}(x') \end{aligned}$$

(62)

Comparing (62) with (22), we get

$$\begin{aligned} B_{(n-k)(n-k)}v_{n(n-k)}(x')=\sum _{l=0}^{k}{n-l \atopwithdelims ()k-l}B_{n(n-l)}\langle x'\rangle _{k-l} - \sum _{i=0}^{k-1}B_{(n-i)(n-k)}v_{n(n-i)}(x') \end{aligned}$$

(63)

putting $l=k-l'$ in (63), following expression is obtained

$$\begin{aligned} B_{(n-k)(n-k)}v_{n(n-k)}(x')=\sum _{l=0}^{k}{n-k+l \atopwithdelims ()l}B_{n(n-k+l)}\langle x'\rangle _{l} - \sum _{i=0}^{k-1}B_{(n-i)(n-k)}v_{n(n-i)}(x')\nonumber \\ \end{aligned}$$

(64)

Hence, the expression (24) is obtained.

Proof of (27) For evaluating value of $f_{i}(n,k)$, substitute the value of $B_{nk}$ from (11) in (24) and compare with (26)

$$\begin{aligned} \sum \limits _{i=0}^{k-1}f_{i}(n,k)\langle x'\rangle _{k-i}= & {} \frac{(\alpha +\beta +n-k)!}{(2n+\alpha +\beta -2k)!}\bigg [\sum \limits _{l=0}^{k} \frac{(\beta +n)!(2n+\alpha +\beta -k+l)!}{(\alpha +\beta +n)!(k-l)!(n+\beta -k+l)!}\langle x'\rangle _{l}\nonumber \\- & {} \,\sum \limits _{m=0}^{k-1}\frac{(\beta +n-m)!(2n+\alpha +\beta -k-m)!}{(\beta +n-k)!(n+\alpha +\beta -m)!}\sum \limits _{j=0}^{m-1}f_{j}(n,m)\langle x'\rangle _{m-j}\bigg ]\nonumber \\ \end{aligned}$$

(65)

Extracting the coefficient of $\langle x'\rangle _{k-i}$ from (65), we get

$$\begin{aligned} coeff. \; of \; \langle x'\rangle _{k-i}&= \frac{(\alpha +\beta +n-k)!}{(2n+\alpha +\beta -2k)!}\bigg [\frac{(\beta +n)!(2n+\alpha +\beta -i)!}{(\alpha +\beta +n)!(i)!(n+\beta -i)!} \nonumber \\&-\,\sum _{m=k-i}^{k-1}\frac{(\beta +n-m)!(2n+\alpha +\beta -k-m)!}{(\beta +n-k)!(n+\alpha +\beta -m)!}f_{m+i-k}(n,m)\bigg ]\nonumber \\ \end{aligned}$$

(66)

Putting $m=m'+k-i$ in the last term of (66), we get coefficient of $\langle x'\rangle _{k-i}$

$$\begin{aligned} f_{i}(n,k)&= \frac{(\alpha +\beta +n-k)!}{(2n+\alpha +\beta -2k)!}\bigg [\frac{(\beta +n)!(2n+\alpha +\beta -i)!}{(\alpha +\beta +n)!(i)!(n+\beta -i)!}\nonumber \\&-\,\sum _{m'=0}^{i-1}\frac{(\beta +n-m'-k+i)!(2n+\alpha +\beta -2k-m'+i)!}{(\beta +n-k)!(n+\alpha +\beta -m'-k+i)!}f_{m'}(n,m'+k-i)\bigg ]\nonumber \\ \end{aligned}$$

(67)

As can be seen from (67), that $f_{i}(n,k)$ can be obtained using recursive equation.