GENETIC ALGORITHMS FOR THE UNSUPERVISED CLASSIFICATION OF SATELLITE IMAGES Ankush Khandelwal(200601011) Vaibhav Kedia(200602022)

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Presentation transcript:

GENETIC ALGORITHMS FOR THE UNSUPERVISED CLASSIFICATION OF SATELLITE IMAGES Ankush Khandelwal( ) Vaibhav Kedia( )

Motivation behind Genetic algorithms Problems in classification of scenes with dynamic objects.  Temporal variation of cluster centroids.  Temporal variation in number of classes.

Outline Feature Space Basic Algorithm Problems in statistical tools used in the paper Solutions to the problems faced.

Feature Space In multispectral images, the bands form the axis of the feature space Here we have used a three dimensional feature space consisting of three bands of a LANDSAT image  520nm-600nm(green band)  630nm-690nm(red band)  760nm-900nm(Near InfraRed band)

BASES OF GENETIC ALGORITHM(GA) In GA applications, the unknown parameters are encoded in the form of strings, so-called chromosomes. Each unit represents a combination of brightness values, one for each band, and thus a potential cluster centroid.

Chromosome Representation The length of the chromosome, K, is equivalent to the number of clusters in the classification problem.

Chromosome initialization At the beginning, for each chromosome i (i =1, 2,…,.P, where P is the size of population) all values are chosen randomly from the data space. One (arbitrary) chromosomes of the parent generation is given here: -1 (110, 88, 246) (150, 78, 226) -1 (11, 104, 8) (50,100, 114) -1 (227, 250, 192)

Crossover The purpose of the crossover operation is to create two new individual chromosomes from two existing chromosomes selected randomly from the current population.

Crossover Example Parent1 : -1 (110, 88, 246) (150, 78, 226) -1 (11, 104, 8) (50, 100, 114) -1 (227, 250, 192) Parent2 : (210, 188, 127) (110, 88, 246) (122, 98, 45) -1 (98, 174, 222) (125, 101, 233) Child1 : -1 (110, 88, 246) (150, 78, 226) -1 (122, 98, 45) -1 (98, 174, 222) (125, 101, 233) Child2 : (210, 188, 127) (110, 88, 246) (11, 104, 8) (50, 100, 114) -1 (227, 250, 192)

Mutation During mutation, all the chromosomes in the population are checked unit by unit and according to a pre-defined probability all values of a specific unit may be randomly changed.

Mutation Example Old string: (210, 188, 127) (110, 88, 246) (122, 98, 45) -1 (98, 174, 222) (125, 101, 233) New string: (210, 188, 127) (97, 22, 143) (122, 98, 45) -1 (98, 174, 222) (125, 101, 233)

Indices identification Based on crossover and mutation the chromosomes, once initialized, iteratively evolve from one generation to the next. In order to be able to stop this iterative process, a so-called fitness function needs to be defined to measure the fitness or adaptability of each chromosome in the population. The value of fitness is also called index. Here, the DBI was adopted.

Basic Algorithm

The Davies-Bouldin's Index

Problems in Statistical Tools used in the Paper The random selection of the chromosome from the huge universal data set makes this algorithm not an efficient way of classifying the image. Sometimes the index favored wrong chromosome for classification because of the favoritism towards high interclass distance rather than the low sum of the standard deviations.

Solutions to the problems Using local maximas of histograms to decrease the size of the alphabet producing chromosome units. We took multiplication of count and standard deviation as our maximizing factor.

Results from Given Method

Results from Our Method

References Genetic Algorithms for the unsupervised classification of satellite images, Y.F Yang, P. Lohmann, C. Heipke Genetic Algorithms in search optimization and machine learning by David E Goldberg Genetic clustering for automatic evolution of clusters and application to image classification, Bandyopadhyay, Maulik