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Basic Classification Which is that?.

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Presentation on theme: "Basic Classification Which is that?."— Presentation transcript:

1 Basic Classification Which is that?

2 The Classification Problem
On the basis of some examples, determine in which class a previously unobserved instance belongs Can be analogous to learning Supervised: a teacher prescribes class composition Unsupervised: class memberships are formed autonomously

3 Common Classification Methods
Template Matching Correlation Bayesian Classifier Neural Networks Fuzzy Clustering Support Vector Machines Principle Component Analysis Independent Component Analysis

4 Template Matching Identify or create class templates
For a given entity x Find the distances from x to each of the class templates Associate x with the class whose template is minimally distant Optionally, update the class template

5 Example 1 x2 m2 m1 x1

6 Example 1: Create Class Templates
Class exemplars are ordered pairs <x1, x2>, which may be written as vectors xT = <x1, x2> The mean vectors mi are obtained by averaging the component values of the class exemplars for each class i

7 Example 1: Find Minimum Distance
Distance from a vector x to each class mean mi, Distancei(x) = ||x-mi|| = [(x-mi)T(x-mi)]½ Note: [(x-mi)T(x-mi)] = xTx-xTmi-miTx+miTmi = ||x||2-2xTmi+||mi||2 = ||x||2 – 2 (xTmi – ½ ||mi||2) xTx =||x||2 is fixed for all i Thus, Distancei(x) is minimized when the quantity (xTmi – ½ ||mi||2) is maximized

8 Example 1: The Decision Boundary
The decision boundary d with respect to classes i and j, dij(x) = Distancei(x)-Distancej(x) = 0 → (||x||2 – 2 (xTmi – ½ ||mi||2)) – (||x||2 – 2 (xTmj – ½ ||mj||2)) = 0 → (xTmi – ½ ||mi||2) - (xTmj – ½ ||mj||2) = 0 → dij(x) = xT (mi-mj) - ½ (mi-mj)T (mi+mj) = 0 Note: This is not the same as Eq

9 Example 2: Details Class 1 exemplars Class 2 Exemplars x1 x2 1 2 4 5.2
1.7 1.8 3.8 4.2 2.1 2.3 4.5 5.9 1.5 5 2.2 4.9 5.3 1.2 2.5 3.7 5.4 1.15 5.8 1.4 5.7 m1 m2 5.1

10 Example 2: Decision Boundary
dij(x) = xT (mi-mj) - ½ (mi-mj)T (mi+mj) = 0 (m1-m2)T = <1.5, 1.8>-<4.5, 5.1> = <-3, -3.3> (m1+m2)T = <1.5, 1.8>+<4.5, 5.1> = <6, 6.9> -½(m1-m2)T (m1+m2)= d12(x) = <x1, x2><-3, -3.3>T = = -3x x = = 3x x = 0

11 Correlation Commonly used to locate similar patterns in 1- or 2-dimensional domain Identify pattern x to which to correlate For x Find the correlation of x to samples Associate x with the samples whose correlation to x are largest Report location of highly correlated samples

12 Example 3: Finding Eyes x

13 Computational Matters
Normalized correlation is typically computed using Pearson’s r

14 Notation and Interpretation
The number n of pairs of values x and y for which the degree of correlation is to be determined |r| ≤ 1 r = 0, if x and y are uncorrelated r > 0, if y increases (decreases) as x increases (decreases), i.e., x and y are positively correlated (to some degree) r < 0, if y decreases (increases) as x increases (decreases), i.e., x and y are negatively correlated (to some degree) To assess the relative strengths of two values r1 and r2, compare their squares. If r1= 0.2 and r2=0.4, r2 indicates 4 times as strong a correlation.

15 Example 4: 5x5 Grid Patterns
1 1 1 r = 0.343 r = 0.514 1 r = -1.0

16 Bayesian Classifier Optimal for Gaussian data

17 Fuzzy Classifiers Jang, Sun, and Mizutani, Neuro-Fuzzy and Soft Computing Fuzzy C-Means (FCM)

18 Neural Networks Feedforward networks and the backpropagation training algorithm Adaptive resonance theory Kohonen netowrks


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