Math Problem Statement

For a 3 class problem, following are the discriminant functions; g_{12} = x_{1} = 0 g_{13} = x_{2} = 0 g_{32} = x_{1} + x_{2} = 0

a) Class 1

Using these functions classify the unknown feature vector (6,4).

b) Class 2 c) Class 3

d) Ambiguous point

Solution

For a 3-class problem with three discriminant functions g12,g13,g_{12}, g_{13}, and g32g_{32}, we can determine the class of a feature vector by evaluating these functions. Here's a step-by-step solution:

Given Discriminant Functions:

  1. g12=x1=0g_{12} = x_1 = 0

    • This function tells us the boundary between Class 1 and Class 2 lies along the line x1=0x_1 = 0.
  2. g13=x2=0g_{13} = x_2 = 0

    • This function tells us the boundary between Class 1 and Class 3 lies along the line x2=0x_2 = 0.
  3. g32=x1+x2=0g_{32} = x_1 + x_2 = 0

    • This function tells us the boundary between Class 3 and Class 2 lies along the line x1+x2=0x_1 + x_2 = 0.

To classify a feature vector, we check the position of the vector relative to these boundaries.

Part (a): Classifying the Feature Vector (6,4)(6, 4)

The feature vector (6,4)(6, 4) has:

  • x1=6x_1 = 6 and x2=4x_2 = 4.
  1. Check with g12=x1=0g_{12} = x_1 = 0:

    • Since x1=6>0x_1 = 6 > 0, the vector lies in the region for Class 1 or Class 3.
  2. Check with g13=x2=0g_{13} = x_2 = 0:

    • Since x2=4>0x_2 = 4 > 0, the vector lies in the region for Class 1 or Class 2.
  3. Check with g32=x1+x2=0g_{32} = x_1 + x_2 = 0:

    • x1+x2=6+4=10>0x_1 + x_2 = 6 + 4 = 10 > 0, so the vector lies in the region for Class 1.

Since the vector satisfies all conditions for Class 1, the classification for the feature vector (6,4)(6, 4) is:

Class 1\text{Class 1}

Part (b): Classifying for Class 2

To determine where Class 2 would lie, we need:

  1. x10x_1 \leq 0,
  2. x1+x20x_1 + x_2 \leq 0, and
  3. x2>0x_2 > 0.

Part (c): Classifying for Class 3

To determine where Class 3 would lie, we need:

  1. x20x_2 \leq 0,
  2. x1+x20x_1 + x_2 \geq 0, and
  3. x1>0x_1 > 0.

Part (d): Ambiguous Point

An ambiguous point would satisfy all boundaries, meaning it lies precisely on the boundaries between classes:

  • For example, a point like (0,0)(0, 0) satisfies g12=0g_{12} = 0, g13=0g_{13} = 0, and g32=0g_{32} = 0, thus it lies on the intersection of all three boundaries and is ambiguous.

Summary:

  • (6, 4) belongs to Class 1.
  • Class 2 points satisfy x10x_1 \leq 0, x2>0x_2 > 0, x1+x20x_1 + x_2 \leq 0.
  • Class 3 points satisfy x20x_2 \leq 0, x1>0x_1 > 0, x1+x20x_1 + x_2 \geq 0.
  • Ambiguous point example: (0,0)(0, 0).

Would you like further details or explanations?


  1. How do discriminant functions determine class boundaries?
  2. Can a feature vector lie on a boundary and still belong to a class?
  3. What types of discriminant functions are used for multi-class problems?
  4. How does ambiguity arise in classification boundaries?
  5. What methods can be used to resolve ambiguity in classification?

Tip: When working with multiple discriminant functions, sketching the boundaries helps visualize class regions and intersections.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Classification
Discriminant Functions

Formulas

g_{12} = x_1 = 0
g_{13} = x_2 = 0
g_{32} = x_1 + x_2 = 0

Theorems

Linear Discriminant Analysis

Suitable Grade Level

Undergraduate Level