VIPSolutionsConsider the following set of data points: input desired Xo ox label 10 1 o 1 1 1 0 0 1 1 0 o 1 o As the above table shows, the data points are categorized (labeled) into two groups specified by the labels “1” and “0”. (A) Use the perceptron learning rule to train a single-neuron perceptron on the data points given above. Initialize the weights to (WO =-0.7, wl = -0.2, w2 = 0.95 ), assume the learning rate is 1, and use the hard-limiter activation function (If output 1; otherwise output 0), i.e.: _(L ifvz0 ow) = 6 ifv
Question:
Consider the following set of data points:
input desired
Xo ox label
10 1
o 1 1
1 0 0
1 1 0
o 1 o
As the above table shows, the data points are categorized (labeled) into two groups specified
by the labels “1” and “0”.
(A) Use the perceptron learning rule to train a single-neuron perceptron on the data points
given above. Initialize the weights to (WO =-0.7, wl = -0.2, w2 = 0.95 ), assume the learning
rate is 1, and use the hard-limiter activation function (If output 1; otherwise output 0), i.e.:
_(L ifvz0
ow) = 6 ifv
This is AI generated Response.
Asked by: 55
Created at: 2025-06-03 06:05:04
Asked by: 55
Created at: 2025-06-03 06:05:04
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Let's solve the perceptron update steps and predictions step by step.
### **Given:**
- Initial weights: \( w_0 = -0.7 \), \( w_1 = -0.2 \), \( w_2 = 0.95 \)
- Learning rate (\( \eta \)) = 1
- Activation: \( \varphi(v) = 1 \) if \( v \geq 0 \), else 0
- Inputs: Add bias input \( x_0 = 1 \) for all patterns
### **Training Data:**
| x₀ | x₁ | x₂ | d |
|----|----|----|---|
| 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | -1 | 0 | 0 |
| 1 | -1 | 1 | 0 |
| 1 | 0 | -1 | 0 |
---
## **(A) Perceptron Learning Rule**
### **First iteration (using all data points in sequence):**
#### **Pattern 1:** (1, 1, 0), d = 1
\( v = -0.7 + 1 \times (-0.2) + 0 \times 0.95 = -0.7 - 0.2 = -0.9 \)
\( y = 0 \) (since \( v < 0 \)), error = \( 1 - 0 = 1 \)
Update:
- \( w_0 = -0.7 + 1 \times 1 \times 1 = 0.3 \)
- \( w_1 = -0.2 + 1 \times 1 \times 1 = 0.8 \)
- \( w_2 = 0.95 + 1 \times 1 \times 0 = 0.95 \)
#### **Pattern 2:** (1, 0, 1), d = 1
\( v = 0.3 + 0 \times 0.8 + 1 \times 0.95 = 0.3 + 0 + 0.95 = 1.25 \)
\( y = 1 \), error = \( 1 - 1 = 0 \)
No update.
#### **Pattern 3:** (1, -1, 0), d = 0
\( v = 0.3 + (-1) \times 0.8 + 0 \times 0.95 = 0.3 - 0.8 = -0.5 \)
\( y = 0 \), error = \( 0 - 0 = 0 \)
No update.
#### **Pattern 4:** (1, -1, 1), d = 0
\( v = 0.3 + (-1) \times 0.8 + 1 \times 0.95 = 0.3 - 0.8 + 0.95 = 0.45 \)
\( y = 1 \), error = \( 0 - 1 = -1 \)
Update:
- \( w_0 = 0.3 + 1 \times (-1) \times 1 = -0.7 \)
- \( w_1 = 0.8 + 1 \times (-1) \times (-1) = 1.8 \)
- \( w_2 = 0.95 + 1 \times (-1) \times 1 = -0.05 \)
#### **Pattern 5:** (1, 0, -1), d = 0
\( v = -0.7 + 0 \times 1.8 + (-1) \times -0.05 = -0.7 + 0 + 0.05 = -0.65 \)
\( y = 0 \), error = \( 0 - 0 = 0 \)
No update.
### **Weights after one iteration:**
- \( w_0 = -0.7 \)
- \( w_1 = 1.8 \)
- \( w_2 = -0.05 \)
---
### **Second iteration (repeat with updated weights):**
#### **Pattern 1:** (1, 1, 0), d = 1
\( v = -0.7 + 1 \times 1.8 + 0 \times -0.05 = -0.7 + 1.8 = 1.1 \)
\( y = 1 \), error = \( 1 - 1 = 0 \)
No update.
#### **Pattern 2:** (1, 0, 1), d = 1
\( v = -0.7 + 0 \times 1.8 + 1 \times -0.05 = -0.7 - 0.05 = -0.75 \)
\( y = 0 \), error = \( 1 - 0 = 1 \)
Update:
- \( w_0 = -0.7 + 1 \times 1 \times 1 = 0.3 \)
- \( w_1 = 1.8 + 1 \times 1 \times 0 = 1.8 \)
- \( w_2 = -0.05 + 1 \times 1 \times 1 = 0.95 \)
#### **Pattern 3:** (1, -1, 0), d = 0
\( v = 0.3 + (-1) \times 1.8 + 0 \times 0.95 = 0.3 - 1.8 = -1.5 \)
\( y = 0 \), error = \( 0 - 0 = 0 \)
No update.
#### **Pattern 4:** (1, -1, 1), d = 0
\( v = 0.3 + (-1) \times 1.8 + 1 \times 0.95 = 0.3 - 1.8 + 0.95 = -0.55 \)
\( y = 0 \), error = \( 0 - 0 = 0 \)
No update.
#### **Pattern 5:** (1, 0, -1), d = 0
\( v = 0.3 + 0 \times 1.8 + (-1) \times 0.95 = 0.3 - 0.95 = -0.65 \)
\( y = 0 \), error = \( 0 - 0 = 0 \)
No update.
### **Weights after two iterations:**
- \( w_0 = 0.3 \)
- \( w_1 = 1.8 \)
- \( w_2 = 0.95 \)
---
## **(A) Fillable Answers**
**After one iteration:**
- w0 = **-0.7**
- w1 = **1.8**
- w2 = **-0.05**
**After two iterations:**
- w0 = **0.3**
- w1 = **1.8**
- w2 = **0.95**
---
## **(B) Predictions on Test Data**
**Use weights:** \( w_0 = 0.3, w_1 = 1.8, w_2 = 0.95 \)
| x₀ | x₁ | x₂ | Local field \( v \) | Predicted label |
|----|----|----|------------------------------|-----------------|
| 1 | 2 | 2 | \( 0.3+2 \times 1.8+2 \times 0.95 = 0.3+3.6+1.9=5.8 \) | 1 |
| 1 | -2 | -2 | \( 0.3+(-2)\times 1.8+(-2)\times 0.95=0.3-3.6-1.9=-5.2 \)| 0 |
| 1 | 0 | 0 | \( 0.3+0+0=0.3 \) | 1 |
| 1 | -2 | 0 | \( 0.3+(-2)\times 1.8+0=0.3-3.6=-3.3 \) | 0 |
---
### **Fill in the table:**
| x₁ | x₂ | Local field | Predicted label |
|----|----|-------------|-----------------|
| 2 | 2 | 5.8 | 1 |
| -2 | -2 | -5.2 | 0 |
| 0 | 0 | 0.3 | 1 |
| -2 | 0 | -3.3 | 0 |
---
## **Final Answers**
### **(A) Weights after:**
- **One iteration:**
w0 = **-0.7**, w1 = **1.8**, w2 = **-0.05**
- **Two iterations:**
w0 = **0.3**, w1 = **1.8**, w2 = **0.95**
### **(B) Test Set**
| Local field | Predicted label |
|-------------|-----------------|
| 5.8 | 1 |
| -5.2 | 0 |
| 0.3 | 1 |
| -3.3 | 0 |
Question:
i need 1 line pure theory
This is AI generated Response.
Asked by: 55
Created at: 2025-06-03 06:12:05
Asked by: 55
Created at: 2025-06-03 06:12:05
Final Answer:
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Using the perceptron learning rule, weights are updated based on classification errors to find a linear boundary separating the data points into classes 0 and 1.
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