Abstract
To enhance the efficacy of change detection in remote sensing images, we propose a novel Spatial Complex Fuzzy Inference System (Spatial CFIS). This system incorporates fuzzy clustering to generate complex fuzzy rules and employs a triangular spatial complex fuzzy rule base to predict changes in subsequent images compared to their original versions. The weight set of the rule base is optimized using the ADAM algorithm to boost the overall performance of Spatial CFIS. Our proposed model is evaluated using datasets from the weather image data warehouse of the USA Navy and the PRISMA mission funded by the Italian Space Agency (ASI). We compare the performance of Spatial CFIS against other relevant algorithms, including PFC-PFR, SeriesNet, and Deep Slow Feature Analysis (DSFA). The evaluation metrics include RMSE (Root Mean Squared Error), R2 (R Squared), and Analysis of Variance (ANOVA). The experimental results demonstrate that Spatial CFIS outperforms other models by up to 40% in terms of accuracy. In summary, this paper presents an innovative approach to handling remote sensing images by applying a spatial-oriented fuzzy inference system, offering improved accuracy in change detection.
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Discover the latest articles, news and stories from top researchers in related subjects.Data availability
The data that support the findings of this study are available from the US Navy and Prisma project, but restrictions apply to the availability of these data, which were used under license for the current study, and so are not publicly available. Data are, however, available from the authors upon reasonable request and with permission of the US Navy and Prisma project.
Code availability
Source codes and Dataset of this paper are available at the following address: https://github.com/vietdslab/SpatialCFIS
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Acknowledgements
This research has been funded by the Research Project:22-2024-RD/HĐ-ĐHCN, Hanoi University of Industry; CSCL34.01/22-23, Space Technology Institute - Vietnam Academy of Science and Technology.
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Appendix A Appendix
Appendix A Appendix
1.1 A.1 Numerical examples
In this subsection, we present a numerical example to describe the steps of the Spatial CFIS algorithm in Section 4.2.
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1.
Step 1. Pre-processing data
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(a)
Input Image Data include 3x3 images can be replaced by 1x9 images below: Image 1: [36, 47, 42, 48, 67, 74, 55, 52, 46] Image 2: [36, 42, 43, 58, 59, 84, 55, 54, 41] Image 3: [32, 41, 36, 48, 54, 77, 65, 64, 31] Image 4: [33, 40, 37, 58, 62, 80, 59, 71, 36] Image 5: [34, 42, 27, 55, 52, 72, 58, 66, 39]
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(b)
Step 1.1. Convert satellite color images to gray images As the input image data utilized in the presented example is already in grayscale format, there is no requirement for any additional conversion procedures.
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(c)
Step 1.2. Determine difference matrix (Imaginary part) Imaginary part (difference matrix): determined by subtracting the correspondent pixel in the satellite image by equation
$$ HoD_i = X^{(t)}-X^{(t-1)} $$\(HoD_1\)(Image 2 - Image 1): [0, 5, 1, 10, 8, 10, 0, 2, 5] \(HoD_2\)(Image 3 - Image 2): [4, 1, 7, 10, 5, 7, 10, 10, 10] \(HoD_3\)(Image 4 - Image 3): [1, 1, 1, 10, 8, 3, 6, 7, 5] \(HoD_4\)(Image 5 - Image 4): [1, 2, 10, 3, 10, 8, 1, 5, 3] Our data now are in the form that: \({X}'<{{X}^{(t)}}, HoD>\)
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(a)
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Step 2: Clustering We apply fuzzy clustering for both X(t) and HoD. Inputs can be represented as: \(X_1: [(36, 0), (47, 5), (42,1), (48,10), (67, 8), (74,10),\) \( (55, 0), (52, 2), (46, 5)]\) \(X_2: [(36, 4), (42, 1), (43, 7), (58, 10), (59, 5), (84, 7),\) \( (55,10), (54,10), (41,1)]\) \(X_3: [(32, 1), (41, 1), (36, 1), (48, 10), (54, 8), (77, 3),\) \( (65, 6), (64, 7), (31, 5)]\) \(X_4: [(33, 1), (40, 2), (37, 10), (58, 3), (62, 10), (80, 8),\) \( (59, 1), (71, 5), (36, 3)]\) With the parameters \(C = 2, m = 2, \varepsilon = 0.001\) (threshold of two consecutive output) and \(maxstep = 3\) (number of iterators). Output: - Membership matrix U. - The center of cluster V.
- (a):
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Step 2.1: Recalculate X(t) and HoD values shows a returned value in the range [0,1] Because the values of each pixel of the gray image are in the range of [0, 255], we divide each one by 225. \(X_1\): [(0.1412, 0), (0.1843, 0.0196), (0.1647, 0.0039), (0.1882, 0.0392), (0.2627, 0.0314), (0.2902, 0.0392), (0.2157, 0), (0.2039, 0.0078), (0.1804, 0.0196)]; \(X_2\): [(0.1412, 0.0157), (0.1647, 0.0039), (0.1686, 0.0275), (0.2275, 0.0392), (0.2314, 0.0196), (0.3294, 0.0275), (0.2157, 0.0392), (0.2118, 0.0392), (0.1608, 0.0392)]; \(X_3\): [(0.1255, 0.0039), (0.1608, 0.0039), (0.1412, 0.0039), (0.1882, 0.0392), (0.2118, 0.0314), (0.302, 0.0118), (0.2549, 0.0235), (0.251, 0.0275), (0.1216, 0.0196)]; \(X_4\): [(0.1294, 0.0039), (0.1569, 0.0078), (0.1451, 0.0392), (0.2275, 0.0118), (0.2431, 0.0392), (0.3137, 0.0314), (0.2314, 0.0039), (0.2784, 0.0196), (0.1412, 0.0118)].
- (b):
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Step 2.2: Initiate the centers of cluster matrix V randomly satisfied: – Let \(V_j\) be the center of cluster j, \({{V}_{j1}}\in \left( \min Xi,\dots \right. \) \(\left. \max Xi \right) ;{{V}_{j2}}\in \left( \min HoD_i,...\max HoD_i \right) \) then
$$V_{ }^{(0)}=\,\left[ \begin{array}{*{35}{r}} 0.1416 & 0.1744 \\ 0.0024 & 0.0113 \\ \end{array} \right] $$ - (c):
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Step 2.3: Calculate the degrees of membership matrix U by the cluster center V: Calculate the values of U as a function of the cluster center vector V, using formula (6) as follows:
$$\begin{aligned} {{U}_{11}}= & \frac{1}{{{\left( \frac{\sqrt{{{\left( {{X}_{11}}-{{V}_{11}} \right) }^{2}}+{{\left( {{X}_{12}}-{{V}_{12}} \right) }^{2}}}}{\sqrt{{{\left( {{X}_{11}}-{{V}_{11}} \right) }^{2}}+{{\left( {{X}_{12}}-{{V}_{12}} \right) }^{2}}}} \right) }^{\frac{2}{2-1}}}+{{\left( \frac{\sqrt{{{\left( {{X}_{11}}-{{V}_{11}} \right) }^{2}}+{{\left( {{X}_{12}}-{{V}_{12}} \right) }^{2}}}}{\sqrt{{{\left( {{X}_{11}}-{{V}_{21}} \right) }^{2}}+{{\left( {{X}_{12}}-{{V}_{22}} \right) }^{2}}}} \right) }^{\frac{2}{2-1}}}}\,\\ {{U}_{12}}= & \frac{1}{{{\left( \frac{\sqrt{{{\left( {{X}_{11}}-{{V}_{21}} \right) }^{2}}+{{\left( {{X}_{12}}-{{V}_{22}} \right) }^{2}}}}{\sqrt{{{\left( {{X}_{11}}-{{V}_{11}} \right) }^{2}}+{{\left( {{X}_{12}}-{{V}_{12}} \right) }^{2}}}} \right) }^{\frac{2}{2-1}}}+{{\left( \frac{\sqrt{{{\left( {{X}_{11}}-{{V}_{21}} \right) }^{2}}+{{\left( {{X}_{12}}-{{V}_{22}} \right) }^{2}}}}{\sqrt{{{\left( {{X}_{11}}-{{V}_{21}} \right) }^{2}}+{{\left( {{X}_{12}}-{{V}_{22}} \right) }^{2}}}} \right) }^{\frac{2}{2-1}}}}\\ {{U}^{(0)}}= & \left[ \begin{array}{*{35}{r}} 0.9952 & 0.0048 \\ 0.073 & 0.927 \\ 0.2174 & 0.7826 \\ 0.2156 & 0.7844 \\ 0.3459 & 0.6541 \\ 0.3771 & 0.6229 \\ 0.2501 & 0.7499 \\ 0.1841 & 0.8159 \\ 0.055 & 0.945 \\ \end{array} \right] \end{aligned}$$ - (d):
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Step 2.4: Update center of clusters Using the formulas for cluster center calculation (5), we recalculate the values of the new cluster centers based on the membership degrees U computed in Step 2.3 as follows:
$$\begin{aligned} {{V}_{J1}}= & \frac{\sum \limits _{k=1}^{N}{\mathop {U}_{kj}^{m} \times {{X}_{k}}}}{\sum \limits _{k=1}^{N}{\mathop {U}_{kj}^{m}}};{{V}_{J2}}=\,\frac{\sum \limits _{k=1}^{N}{\mathop {U}_{kj}^{m} \times HoD}_{k}}{\sum \limits _{k=1}^{N}{\mathop {U}_{kj}^{m}}}\\ V_{j1}^{(1)}= & \left[ \begin{array}{*{35}{r}} 0.173 \\ 0.0081 \\ \end{array} \right] ; V_{j2}^{(1)}=\,\left[ \begin{array}{*{35}{r}} 0.2027 \\ 0.0188 \\ \end{array} \right] \\ V_{ }^{(1)}= & \left[ \begin{array}{*{35}{r}} 0.173 & 0.2027 \\ 0.0081 & 0.0188 \\ \end{array} \right] \end{aligned}$$Calculate the difference between \(V^{1}\) and \(V^{0}\) by using Euclidean formula:
$$\begin{aligned} & \left\| {{V}^{(1)}}\,-\,{{V}^{(0)}}\, \right\| =\sqrt{\sum \nolimits _{l=1}^{2}{{{(V_{jl}^{(1)}-V_{jl}^{(0)})}^{2}}}} \\= & \sqrt{{{(V_{j1}^{(1)}-V_{j1}^{(0)})}^{2}}+{{(V_{j2}^{(1)}-V_{j2}^{(0)})}^{2}}}\\= & \sqrt{{{(V_{11}^{(1)}-V_{11}^{(0)})}^{2}}+{{(V_{12}^{(1)}-V_{12}^{(0)})}^{2}}+{{(V_{21}^{(1)}-V_{21}^{(0)})}^{2}}+{{(V_{22}^{(1)}-V_{22}^{(0)})}^{2}}}\\= & 0.2623 \end{aligned}$$ - (e):
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Step 2.5 Repeat steps 2 and 3 until one of the two following conditions are not satisfied: – Condition 1: The number of iterations is less than the max step. – Condition 2: \(\vert V^{(t)}\,-\,V^{(t-1)}\, \vert \, \le \varepsilon =0.0001\) Because Number of iteration is 1 and \(\left\| {{V}^{(1)}}\,-\,{{V}^{(0)}}\, \right\| \) \(= 0.2623 < \varepsilon \) we continue iteration 2.
$$ {{U}^{(1)}}=\,\left[ \begin{array}{*{35}{r}} 0.3003 & 0.6997 \\ 0.4799 & 0.5201 \\ 0.3846 & 0.6154 \\ 0.5492 & 0.4508 \\ 0.6348 & 0.3652 \\ 0.6641 & 0.3359 \\ 0.5031 & 0.4969 \\ 0.4969 & 0.5031 \\ 0.4692 & 0.5308 \\ \end{array} \right] ; V_{ }^{(2)}=\,\left[ \begin{array}{*{35}{r}} 0.2206 & 0.187 \\ 0.0227 & 0.013 \\ \end{array} \right] $$Calculate the difference between \(V^{2}\) and \(V^{1}\) by using Euclidean formula: . Since the number of iterations is 2 and \(\left\| {{V}^{(2)}}\,-\,{{V}^{(1)}}\, \right\| \, = 0.2583 < \varepsilon \). We continue the iteration 3:
$${{U}^{(2)}}=\,\left[ \begin{array}{*{35}{r}} 0.2495 & 0.7505 \\ 0.0369 & 0.9631 \\ 0.1429 & 0.8571 \\ 0.3422 & 0.6578 \\ 0.7666 & 0.2334 \\ 0.689 & 0.311 \\ 0.648 & 0.352 \\ 0.3843 & 0.6157 \\ 0.0509 & 0.9491 \\ \end{array} \right] ; V_{ }^{(3)}=\,\left[ \begin{array}{*{35}{r}} 0.2442 & 0.1808 \\ 0.0234 & 0.0152 \\ \end{array} \right] $$Calculate the difference between \(V^{3}\) và \(V^{2}\) by using Euclidean formula: \(\left\| {{V}^{(2)}}-{{V}^{(1)}} \right\| = 0.0245 < \varepsilon \). Since the number of iterations is 3 and \(\left\| {{V}^{(2)}}-{{V}^{(1)}} \right\| = 0.0245 < \varepsilon \), we stop the algorithm because the number of iterations equals to maxstep. The final result of this example is shown as below:
$$U=\,\left[ \begin{array}{*{35}{r}} 0.2495 & 0.7505 \\ 0.0369 & 0.9631 \\ 0.1429 & 0.8571 \\ 0.3422 & 0.6578 \\ 0.7666 & 0.2334 \\ 0.689 & 0.311 \\ 0.648 & 0.352 \\ 0.3843 & 0.6157 \\ 0.0509 & 0.9491 \\ \end{array} \right] ; V=\,\left[ \begin{array}{*{35}{r}} 0.2442 & 0.1808 \\ 0.0234 & 0.0152 \\ \end{array} \right] $$
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3.
Step 3: Finding rules using Triangular fuzzy method Determining the value of \(a, b, c, a', b', c'\) of input data \(X_1^{'}\): The value of b and \(b'\) can be defined by the center of cluster matrix (\(b_{kj}=V_j\))
$$\begin{aligned} {{b}_{1}}= & {{V}_{11}};{{b}_{2}}=\,{{V}_{21}};{{{b}'}_{1}}=\,{{V}_{12}};{{{b}'}_{2}}=\,{{V}_{22}};\\ {{a}_{kj}}= & \frac{\sum \limits _{i=1,2,\,\,...n\,\,\text {v}\grave{\textrm{a}}\,\,I_{i}^{(k)}\,\le \,{{b}_{kj}}}{{{U}_{\text {i,j}}}\times \,I_{i}^{(k)}}}{\sum \limits _{i=1,2,\,\,...n\,\,\text {v}\grave{\textrm{a}}\,\,I_{i}^{(k)}\,\le \,{{b}_{kj}}}{{{U}_{\text {i,j}}}}}\\ {{c}_{kj}}= & \frac{\sum \limits _{i=1,2,\,\,...n\,\,\text {v}\grave{\textrm{a}}\,\,I_{i}^{(k)}\,\ge \,{{b}_{kj}}}{{{U}_{\text {i,j}}}\times \,I_{i}^{(k)}}}{\sum \limits _{i=1,2,\,\,...n\,\,\text {v}\grave{\textrm{a}}\,\,I_{i}^{(k)}\,\ge \,{{b}_{kj}}}{{{U}_{\text {i,j}}}}} \end{aligned}$$Incorporating the real parts values of \(I_{i}^{(k)}\), we obtain \(a_1=0.1926\), \(a_2=0.1636\), \(c_1=0.2757\), and \(c_2=0.2089\).
$$\begin{aligned} {{{a}'}_{kj}}= & \frac{\sum \limits _{i=1,2,\,\,...n\,\,\text {v}\grave{\textrm{a}}\,\,HoD_{i}^{(k)}\,\le \,{{b}_{kj}}}{{{U}_{\text {i,j}}}\times HoD_{i}^{(k)}}}{\sum \limits _{i=1,2,\,\,...n\,\,\text {v}\grave{\textrm{a}}\,\,HoD_{i}^{(k)}\,\le \,{{b}_{kj}}}{{{U}_{\text {i,j}}}}}\\ {{{c}'}_{kj}}= & \frac{\sum \limits _{i=1,2,\,\,...n\,\,\text {v}\grave{\textrm{a}}\,\,I_{i}^{(k)}\,\ge \,{{b}_{kj}}}{{{U}_{\text {i,j}}}\times HoD_{i}^{(k)}}}{\sum \limits _{i=1,2,\,\,...n\,\,\text {v}\grave{\textrm{a}}\,\,HoD_{i}^{(k)}\,\ge \,{{b}_{kj}}}{{{U}_{\text {i,j}}}}} \end{aligned}$$Using the imaginary parts values \(HoD_{i}^{(k)}\), we obtain \(a'_1=0.0035\), \(a'_2=0.0032\), \(c'_1=0.0359\), and \(c'_2=0.0266\). Similar with the other imaginary parts. We can assume the rule for the first input data below: Rule 1: includes 6 parameters a, b, c and \(a', b', c'\) where a, b and c are corresponding to the first triangle of real part and \(a', b'\) and \(c'\) are corresponding to the first triangle of imaginary part.
$$\begin{aligned} (a,\,b,\,c,\,{a}',\,{b}',\,{c}')=\left[ {{a}_{1}},\,{{b}_{1}},{{c}_{1}},{{{{a}'}}_{1}},{{{{b}'}}_{1}},{{{{c}'}}_{1}} \right] \end{aligned}$$$$\begin{aligned} = \left[ 0.1926,\,\,0.2442,\,0.2757,\,0.0035,\,\,0.0234,\,\,0.0359 \right] \end{aligned}$$The valid values can be limited by polygon \((AA'C'BC)\) where \(A (0, a1, 0); A'(a1', 0, 0); B'(b1', b1, 1); \) \(C(0, c1, 0); C'(c1', 0, 0); B(b1', b1, 0)\) Rule 2: includes 6 parameters a, b, c and \(a', b', c'\) where a, b, c are corresponding to the second triangle of real part and \(a', b', c'\) are corresponding to the second triangle of imaginary part.
$$\begin{aligned} (a,\,b,\,c,\,{a}',\,{b}',\,{c}')=\left[ {{a}_{2}},\,{{b}_{2}},{{c}_{2}},{{{{a}'}}_{2}},{{{{b}'}}_{2}},{{{{c}'}}_{2}} \right] \end{aligned}$$$$\begin{aligned} =\left[ 0.1636,\,\,0.1808,\,0.2089,\,\,0.0032,\,0.0152,\,0.0266 \right] \end{aligned}$$The valid values can be limited by polygon \((AA'C'BC)\) where \(A (0, a2, 0); A'(a2', 0, 0); B'(b2', b2, 1);\) \( C(0, c2, 0); C'(c2', 0, 0); B(b2', b2, 0)\)
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Step 4: Calculate the degree of membership of input data \(X_2'\) Input data \(X_2'\) \(X_2'\):x[(0.1412, 0.0157), (0.1647, 0.0039), (0.1686, 0.0275), (0.2275, 0.0392), (0.2314, 0.0196), (0.3294, 0.0275), (0.2157, 0.0392), (0.2118, 0.0392), (0.1608, 0.0392)]
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Step 4.1: Calculate invalid value (outbound value)
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The polygon \(AA'C'BC\) is formed as in Fig. 14. There are two outbound regions. The first one is formed by polygon \(OAA'\); the other one is formed by the region out of polygon \(OAA'C'BC\).
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The invalid value in out bound region will be updated by the parameter \(\delta \). In which, \(\delta \) is determined step by step as below:
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Step 1: Find invalid values have axis-x or axis-y \(< 0\) and convert satisfied all of the invalid values have axis-x and axis-y \(\ge 0\).
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Step 2: Determine \(\delta \) with invalid value: \(\cdot \) Initial \(a = 1, b = 255\) \(\cdot \) Calculate \(\delta =\frac{a+b}{2}\) \(\cdot \) Divide all invalid values to \(\delta \). If at least one invalid value is outbound, we update \(a=\delta \). Otherwise, if all invalid values are inbound, we update \(b=\delta \). \(\cdot \) Repeat until the difference between two consecutive values of \(\delta \) is less than or equal the threshold \(\varepsilon \).
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Step 3: Update invalid value In the region \(OAA'\), we multiply each invalid value with \(\delta \) In the region out of polygon \(OAA'C'BC\), we divide each invalid value for \(\delta \) or \(-\delta \) satisfied each new value of axis-x and axis-y \(> 0\).
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Step 4.2: Determine the degree of membership - The first input (0.1412, 0.0157) and Rule 1 + Define point D have corresponded \((0.1412, 0.0157, 0)\), D is the correspond of input in polygon \(AA'C'BC\) + E is the intersection of BD and \(A'C'\) + Define point F satisfied \(F \subset (A'B'C')\) and DF is perpendicular with \(AA'C'BC\). So the height DF will be the degree of membership U of input (0.1412, 0.0157). \(\frac{DF}{B{B}'}=\frac{DE}{BE}\) \(\Rightarrow \) \(DF=\frac{B{B}'*DE}{BE}=\frac{1*0.1416}{0.2449}=0.5782\) - The first input (0.1412, 0.0157) and Rule 2 + Define point D have corresponded \((0.1412, 0.0157, 0)\), D is the correspond of input in polygon \(AA'C'BC\) + E is the intersection of BD and \(A'C'\) + Define point F satisfied \(F \subset (A'B'C')\) and DF is perpendicular with \(AA'C'BC\). So the height DF will be the degree of membership U of input (0.1412, 0.0157). \(\frac{DF}{B{B}'}=\frac{DE}{BE}\) \(\Rightarrow \) \(DF=\,\frac{B{B}'\times DE}{BE}=\frac{1 \times 0.1412}{0.1808}=0.7810\) Similiarly with the other point. - The second input (0.1647, 0.0039) + Rule 1: DF = 0.1584 + Rule 2: DF = 0.251 - The third input (0.1686, 0.0275) + Rule 1: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0. 0.2303 + Rule 2: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.311 - The \(four^{th}\) input (0.2275, 0.0392) + Rule 1: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.3106 + Rule 2: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.4192 - The \(fif^{th}\) input (0.2314, 0.0196) + Rule 1: DF = 0.8337 + Rule 2: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.3117 - The \(six^{th}\) input (0.3294, 0.0275) + Rule 1: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.3149 + Rule 2: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.5221 - The \(seven^{th}\) input (0.2157, 0.0392) + Rule 1: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.2944 + Rule 2: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.3977 - The \(eigh^{th}\) input (0.2118, 0.0392) + Rule 1: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.289 + Rule 2: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.3905 - The \(nin^{th}\) input (0.1608, 0.0392) + Rule 1: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.2195 + Rule 2: Because this value is invalid (outbound), we divide it to \(\delta \) = 3; DF = 0.2965
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1.
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Step 5: Determine defuzzification value (DEF) Initial parameter \((h1, h2, h3, h1', h2', h3') = (1, 2, 1, 1, 2, 1)\) and using the defuzzification formulas (24) and (25), we obtain DEF of Rule 1: \(DEF(X_1)= 0.2101;DEF(HoD_1)=0.0083\); DEF of Rule 2: \(DEF(X_1)= 0.0918;DEF(HoD_1)=0.006\); Using Adam Stochastic Optimization algorithm (2) o find optimize value of parameter. After training, we take \((h1, h2, h3, h1', h2', h3') = (1, 0.7, 0.7, 1, 0.7, 0.7)\)
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Step 6: Predict output image
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(a)
Step 6.1: Determine \(O^*_{i.Rel}\) and \(O_{i.Img}^{*}\) In step 5.2 we have MIN of \(U_{A_{_{kj}}}\) \(= MIN (DF \) of rule 1, DF of rule 2) = \(MIN (0.5782, 0.7810) = 0.5782\). Similarly, with the other ones of input data X. We take the degree of membership value U We used two formulae above (22), (23) to determine \(O^*_{i.Rel}\) and \(O_{i.Img}^{*}\).
$$O{{_{i}^{*}}^{ }}=(O_{i.\operatorname {Re}l}^{*},\,O_{i.\operatorname {Im}g}^{*})= \left[ \begin{array}{*{35}{r}} 0.175802 & 0.009461 \\ 0.169767 & 0.009338 \\ 0.175812 & 0.009461 \\ 0.175833 & 0.009462 \\ 0.22323 & 0.010429 \\ 0.168092 & 0.009304 \\ 0.175798 & 0.009461 \\ 0.175789 & 0.009461 \\ 0.175801 & 0.009461 \\ \end{array} \right] $$ -
(b)
Step 6.2: Determine the final prediction result The output pixel values of the predicted image are determined directly from the real part result \(O_{i.Rel}\) (computed in step 4.2), while the imaginary part is computed based on the phase variation ratio of \(O'_{i.Img}\) as given by (27). In this equation, \(X_{i}^{t-1}\) represents the ground truth value at time step \(t-1\).
$$O_{i.\operatorname {Im}g}^{{{*}'}}=\,{{X}_{i}}^{(t-1)}*(1+O_{i.\operatorname {Im}g}^{*})=\left[ \begin{array}{*{35}{r}} 0.142536 \\ 0.186021 \\ 0.166258 \\ 0.189981 \\ 0.26544 \\ 0.2929 \\ 0.217741 \\ 0.205829 \\ 0.182107 \\ \end{array} \right] $$Finally, the final prediction output is calculated by both real and imaginary parts according to formula (28). To get the best result, we used Adam optimization algorithm again to find the optimize parameter \(\gamma \). After training we have \(\gamma = 0.00001\) The predicted image of \(X_3^*\) as follows:
$$X_{3}^{*}=\left[ \begin{array}{*{35}{r}} 0.1425 \\ 0.186 \\ 0.1663 \\ 0.19 \\ 0.2654 \\ 0.2929 \\ 0.2177 \\ 0.2058 \\ 0.1821 \\ \end{array} \right] $$ -
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Step 6.3: Convert to original space and denoising Converting the predicted image of \(X_3^*\) to the original domain and denoising it results in:
$$X_{3}^{*}=\left[ \begin{array}{*{35}{r}} 36 \\ 47 \\ 42 \\ 48 \\ 68 \\ 75 \\ 56 \\ 52 \\ 46 \\ \end{array} \right] $$Prediction output Image (Fig. 17):
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Thang, N.T., Giang, L.T., Son, L.H. et al. A novel spatial complex fuzzy inference system for detection of changes in remote sensing images. Appl Intell 55, 178 (2025). https://doi.org/10.1007/s10489-024-06000-0
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DOI: https://doi.org/10.1007/s10489-024-06000-0