# Perceptron update

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/formulas/38

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/formulas/38/

The perceptron uses a threshold to choose a class and an error-based learning rule to change weights. Its threshold convention must be stated, especially at zero. This discrete learning rule is different from sigmoid backpropagation.



$$
\widehat y=\operatorname{step}(w^Tx+b)
$$





$$
w_{new}=w+\eta(y-\widehat y)x
$$





$$
b_{new}=b+\eta(y-\widehat y)
$$



## Perceptrons and the threshold

A perceptron is a simple neuron used for classification with a threshold decision. If a question says output 1 for z ≥ 0 and 0 otherwise, then z = 0 gives 1. If it says output 1 for z > 0 and 0 otherwise, then z = 0 gives 0. That single equality sign can change an entire training table.

![The weighted sum is a real number. The threshold turns that number into a hard decision.](https://deep-learning-study-room.vercel.app/assets/figures/v_threshold.svg)

Diagram: The weighted sum is a real number. The threshold turns that number into a hard decision.. Labels: A perceptron makes a threshold decision | A perceptron makes a threshold decision | Example threshold: return 1 when the weighted sum is greater than 0. | Inputs | x₁ and x₂ | Weighted sum | z = w₁x₁ + w₂x₂ + b | Threshold | z > 0: 1; otherwise: | 0 | Unlike sigmoid, this rule makes a hard 0-or-1 decision.

The weighted sum is a real number. The threshold turns that number into a hard decision. Tap the diagram to enlarge.

**Question.** Implement binary AND with w₁ = w₂ = 0.6, b = −0.8 and output 1 when z ≥ 0.

### Given

w₁ = w₂ = 0.6; bias b = −0.8.

Binary inputs; output 1 when z ≥ 0.

### To find

Calculate the output for 00,01,10,11 and show that the result is AND.

### Formula — which one and why

**Threshold logic.** Compute the weighted sum for every binary input combination and apply the threshold rule exactly. Hidden neurons can provide separate threshold decisions before the output combines them.

step(z) returns 1 at or above zero unless the question specifies a strict inequality; otherwise it returns 0.



$$
z=\sum_iw_ix_i+b
$$





$$
\operatorname{step}(z)=\begin{cases}1&z\ge0\\0&z<0\end{cases}
$$



**Yaad rakhne ka cue:** Har input row ka sum nikalo, phir threshold lagao.

### Solution — step by step

### Input (0, 0)

**Step 1: multiply input 1 by its weight.**



$$
0.6(0)=0
$$



**Step 2: multiply input 2 by its weight.**



$$
0.6(0)=0
$$



**Step 3: add the two products.**



$$
s=0+0=0
$$



**Step 4: add the bias once.**



$$
z=0+(-0.8)=-0.8
$$



**Step 5: apply the stated threshold.**

z = -0.8; output = 0 because output is 1 exactly when z ≥ 0.

### Input (0, 1)

**Step 1: multiply input 1 by its weight.**



$$
0.6(0)=0
$$



**Step 2: multiply input 2 by its weight.**



$$
0.6(1)=0.6
$$



**Step 3: add the two products.**



$$
s=0+0.6=0.6
$$



**Step 4: add the bias once.**



$$
z=0.6+(-0.8)=-0.2
$$



**Step 5: apply the stated threshold.**

z = -0.2; output = 0 because output is 1 exactly when z ≥ 0.

### Input (1, 0)

**Step 1: multiply input 1 by its weight.**



$$
0.6(1)=0.6
$$



**Step 2: multiply input 2 by its weight.**



$$
0.6(0)=0
$$



**Step 3: add the two products.**



$$
s=0.6+0=0.6
$$



**Step 4: add the bias once.**



$$
z=0.6+(-0.8)=-0.2
$$



**Step 5: apply the stated threshold.**

z = -0.2; output = 0 because output is 1 exactly when z ≥ 0.

### Input (1, 1)

**Step 1: multiply input 1 by its weight.**



$$
0.6(1)=0.6
$$



**Step 2: multiply input 2 by its weight.**



$$
0.6(1)=0.6
$$



**Step 3: add the two products.**



$$
s=0.6+0.6=1.2
$$



**Step 4: add the bias once.**



$$
z=1.2+(-0.8)=0.4
$$



**Step 5: apply the stated threshold.**

z = 0.4; output = 1 because output is 1 exactly when z ≥ 0.

**Final answer:** outputs are [0, 0, 0, 1] for inputs 00, 01, 10, 11: exactly AND.

### Summary of the four cases

| Inputs | Full net calculation | Net z | Output | AND target |
| --- | --- | --- | --- | --- |
| 0, 0 | 0.6(0) + 0.6(0) − 0.8 | −0.8 | 0 | 0 |
| 0, 1 | 0.6(0) + 0.6(1) − 0.8 | −0.2 | 0 | 0 |
| 1, 0 | 0.6(1) + 0.6(0) − 0.8 | −0.2 | 0 | 0 |
| 1, 1 | 0.6(1) + 0.6(1) − 0.8 | 0.4 | 1 | 1 |

The neuron fires only when both inputs are 1. No weight update was requested, so this is a forward evaluation, not a training problem.

For binary targets, a common learning rule is w ← w + η(t − a)x and b ← b + η(t − a). For bipolar targets −1 and +1, the class notes use a different rule: on a mismatch, w ← w + η tx and b ← b + ηt; otherwise keep them. The notes give output 0 at zero threshold, +1 above it and −1 below it. Never combine the activation from one convention with an unexplained update from another.

**Training table routine:**  copy current weights → calculate z → apply the stated threshold → compare output with target → compute every weight change → update bias → carry new values to the next row. Reset nothing between rows or epochs unless instructed.

### Binary perceptron: target 1

**Question. A binary perceptron has x = [1,2], weights [-0.2, 0.1], bias 0 and target 1. Use output 1 for net ≥ 0 and learning rate 0.1. Perform one update.**

### Given

x = [1,2]; w = [-0.2, 0.1]; b = 0; target = 1; η = 0.1.

### To find

Net, threshold output, correction and new weights/bias.

### Formula — which one and why

This binary learning rule uses target minus output. Carry the same correction into every input weight and the bias.

t is target; o is threshold output; Δ means the amount added to the current parameter.



$$
z=\sum_iw_ix_i+b
$$





$$
\Delta w_i=\eta(t-o)x_i
$$





$$
\Delta b=\eta(t-o)
$$



**Yaad rakhne ka cue:** Target minus output; correct prediction ho toh correction zero.

### Solution — step by step

**Step 1: calculate the net input.**



$$
z=(-0.2)(1)+(0.1)(2)+0=0
$$



**Step 2: apply the threshold.**

Net = 0; output = 1 under the stated ≥ 0 rule.

**Step 3: calculate the correction.**



$$
t-o=1-1=0
$$



**Step 4: update weight 1.**



$$
w_1^{\mathrm{new}}=-0.2+0.1(0)(1)=-0.2
$$



**Step 5: update weight 2.**



$$
w_2^{\mathrm{new}}=0.1+0.1(0)(2)=0.1
$$



**Step 6: update the bias.**



$$
b_{\mathrm{new}}=0+0.1(0)=0
$$



**Final answer:** New weights = [-0.2,0.1]; new bias = 0.

### Binary perceptron: target 0

**Question. A binary perceptron has x = [1,2], weights [0.2, 0.1], bias 0 and target 0. Use output 1 for net ≥ 0 and learning rate 0.1. Perform one update.**

### Given

x = [1,2]; w = [0.2, 0.1]; b = 0; target = 0; η = 0.1.

### To find

Net, threshold output, correction and new weights/bias.

### Formula — which one and why

This binary learning rule uses target minus output. Carry the same correction into every input weight and the bias.

t is target; o is threshold output; Δ means the amount added to the current parameter.



$$
z=\sum_iw_ix_i+b
$$





$$
\Delta w_i=\eta(t-o)x_i
$$





$$
\Delta b=\eta(t-o)
$$



**Yaad rakhne ka cue:** Target minus output; correct prediction ho toh correction zero.

### Solution — step by step

**Step 1: calculate the net input.**



$$
z=(0.2)(1)+(0.1)(2)+0=0.4
$$



**Step 2: apply the threshold.**

Net = 0.4; output = 1 under the stated ≥ 0 rule.

**Step 3: calculate the correction.**



$$
t-o=0-1=-1
$$



**Step 4: update weight 1.**



$$
w_1^{\mathrm{new}}=0.2+0.1(-1)(1)=0.1
$$



**Step 5: update weight 2.**



$$
w_2^{\mathrm{new}}=0.1+0.1(-1)(2)=-0.1
$$



**Step 6: update the bias.**



$$
b_{\mathrm{new}}=0+0.1(-1)=-0.1
$$



**Final answer:** New weights = [0.1,-0.1]; new bias = -0.1.

## Drill 11 One perceptron learning update

## Question

**Question.** Binary perceptron, w = [0,0], b = 0, x = [1,1], target 1, η = 0.5. Use output 1 if z>0 else 0. Perform one update.

## Answer — step by step

### Given

w = [0,0]; b = 0; x = [1,1]; target t = 1.

Learning rate η = 0.5.

Output 1 only when z > 0; otherwise output 0.

### To find

Perform one update and report new weights and bias.

### Formula — which one and why

**Perceptron learning rule.** Evaluate the stated threshold, compare the predicted class with its target, then correct weights and bias. The next sample uses the newly updated parameters.

t = target; o = thresholded output; η = learning rate; xᵢ = input i. Binary and bipolar output conventions must follow the question.



$$
z=\sum_iw_ix_i+b
$$





$$
w_i^{\mathrm{new}}=w_i+\eta(t-o)x_i
$$





$$
b^{\mathrm{new}}=b+\eta(t-o)
$$



**Yaad rakhne ka cue:** Prediction nikalo; target−prediction se correction nikalo.

### Solution — step by step

### Meaning and formula

A perceptron first computes a weighted sum z and then applies the exact threshold rule in the question. This question uses **strictly greater than zero**, so z = 0 produces output 0.



$$
z=w_1x_1+w_2x_2+b,\qquad o=\begin{cases}1&z>0\\0&z\le0\end{cases}
$$



The perceptron learning rule uses target minus output, written e below. This is a discrete perceptron rule, not the sigmoid backpropagation derivative.



$$
e=t-o,\qquad w_{j,\mathrm{new}}=w_j+\eta e x_j,\qquad b_{\mathrm{new}}=b+\eta e
$$



### Step 1: obtain the old output



$$
z=0\times1+0\times1+0=0,\qquad o=0
$$



### Step 2: compare it with the target



$$
e=1-0=1
$$



### Step 3: update each weight separately



$$
w_{1,\mathrm{new}}=0+0.5\times1\times1=0.5
$$





$$
w_{2,\mathrm{new}}=0+0.5\times1\times1=0.5
$$





$$
b_{\mathrm{new}}=0+0.5\times1=0.5
$$



### Step 4: interpret the updated prediction



$$
z_{\mathrm{new}}=0.5\times1+0.5\times1+0.5=1.5
$$



Since 1.5 > 0, the new output is 1. **Final parameters: w = [0.5, 0.5], b = 0.5.**