K-Means Clustering

K- Mean is used to solves the clustering problem:

• one of the most widely used clustering algorithms.
• K-Means is an Iterative Algorithm

It is simple, fast, and widely used in practice.

Scikit Learn k-mean example:

from sklearn.cluster import KMeans
import numpy as np

X = np.array([[1, 2], [1, 4], [1, 0],
              [10, 2], [10, 4], [10, 0]])
              
kmeans = KMeans(
    n_clusters=2,    # Number of Cluster
    random_state=0,  # Dont Starts with random centroid locations
    n_init="auto")   # automatically chooses the number of runs.
    .fit(X)          # Train Model

# print Cluster centers
kmeans.cluster_centers_

# print groups label for data items
kmeans.labels_

Clustering

Given an *unlabeled dataset automatically group the data into coherent subsets, called clusters.

Cluster analysis is a powerful unsupervised learning technique for discovering groups in data.

When K-Means Works Well

K-Means works best when:

• Clusters are roughly spherical.
• Clusters have similar sizes.
• Features are properly scaled.

Feature scaling is important:

$$x_j := \frac{x_j - \mu_j}{\sigma_j}$$

Without scaling, features with larger magnitude dominate distance calculations.

When K-Means Struggles

K-Means performs poorly when:

• Clusters are non-convex.
• Clusters have very different densities.
• Data contains significant outliers.
• Clusters are not well separated.

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K-Mean Algorithm

K-Means works as follows:

Takes $K$ and unlabeled data as input.

Given:

Training set
$\{x^{(1)}, x^{(2)}, ..., x^{(m)}\}$

where $x^{(i)} \in \mathbb{R}^n$.

And number of clusters $K$

Repeats: {

1. Assignment Step

   Assign points to nearest centroid.

2. Centroid update step

   Move centroids $\mu$ to mean of assigned points.

   Minimizes Cost function (Distortion):

   $$J = \frac{1}{m} \sum_{i=1}^{m} \left\| x^{(i)} - \mu_{c^{(i)}} \right\|^2$$

   }

Stops when assignments stabilize.

Output is cluster assignments for each example and cluster centroids.

Suppose we want to group data into $K = 2$ clusters.

Algorithm

Repeat for $t = 1, \dots, T$:

where T = number of runs

1. Randomly initialize centroids
2. Run K-Means to convergence
3. Compute distortion:

$$J^{(t)} = \frac{1}{m} \sum_{i=1}^{m} \left\| x^{(i)} - \mu_{c^{(i)}} \right\|^2$$

Finally:

$$\text{Choose clustering with smallest } J^{(t)}$$

Input

• $K$ (number of clusters) = 2 in this case
• Dataset $\{x^{(1)}, x^{(2)}, ..., x^{(m)}\}$, where
  • $x^{(i)} \in \mathbb{R}^n$
  • $m$ = number of examples

Example: 🍓🍡🍬🍭

Candies scatter on table with $K=2$ bowls to collect them.

1. Initialize Cluster centroids($\mu$) 🥣

Randmly put 2 Bowls on the table to start collecting candies.

Randomly initialize $K$ cluster centroids

Total number of clusters

$K$

We have $K$ centroids:

• $k \in \{1, 2, ..., K\}$
• $\mu_k$ = centroid of cluster $k$

$\mu_1, \mu_2, \mu_3 \dots \mu_k$

where $\mu_k \in \mathbb{R}^n$

• $\mu_k$ is a point in the same space as the data (i.e., $\mathbb{R}^n$)

Choosing the Number of Clusters (K)

There is no universal rule for choosing $K$.

Proper Random Initialization

When running K-Means:

$$K < m$$

Where:

• $K$ = number of clusters
• $m$ = number of training examples

It does not make sense to choose $K \ge m$.

Recommended Initialization Method

1. Randomly select $K$ distinct training examples.
2. Set initial centroids equal to those examples:

$$\mu_1 = x^{(i_1)}, \quad \mu_2 = x^{(i_2)}, \quad \dots, \quad \mu_K = x^{(i_K)}$$

where $i_1, \dots, i_K$ are randomly chosen distinct indices.

This ensures centroids start within the data distribution.

Elbow Method

When we increase $K$, distortion $J$ decreases because we have more centroids to fit the data.

Idea is to find point where adding more clusters does not significantly reduce distortion.

1. Run K-Means for different values of $K$.
2. Compute distortion $J$ for each.
3. Plot $J$ vs $K$.
4. Look for an “elbow” where improvement slows down.

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Repeatedly Steps 2-3 until convergence {

2. 📏 Cluster Assignment Step

For each training example($x^{(i)}$) where $i = 1, \dots, m$:

Assign $x^{(i)}$ to the cluster with the closest centroid.

In Mathematical terms:

Index of Cluster $c^{(i)}$

Index of cluster is Centroid closest to $x^{(i)}$ in $\{1, \dots, K\}$

Examples:

• If $x^{(i)}$ is closest to $\mu_1$, then $c^{(i)} = 1$
• If $x^{(i)}$ is closest to $\mu_2$, then $c^{(i)} = 2$

$\mu_c^{(i)}$ = CLuster centroid of cluster

Value of centroid of cluster assigned to $x^{(i)}$ is $\mu_{c^{(i)}}$

• If $c^{(i)} = 1$, then $\mu_{c^{(i)}} = \mu_1$
• If $c^{(i)} = 2$, then $\mu_{c^{(i)}} = \mu_2$

Which can be expressed as:

$$c^{(i)} := \arg\min_{k \in \{1, \dots, K\}} \left\| x^{(i)} - \mu_k \right\|^2$$

Formally:

$$c^{(i)} = \arg\min_k \left\| x^{(i)} - \mu_k \right\|^2$$

Where:

• $\mu_k$ = centroid of cluster $k$
• $c^{(i)}$ = cluster assigned to example $i$

We typically use squared Euclidean distance.

Example

Given

Centroids:

$$\mu_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$ $$\mu_2 = \begin{bmatrix} -3 \\ 0 \end{bmatrix}$$ $$\mu_3 = \begin{bmatrix} 4 \\ 2 \end{bmatrix}$$

Training example:

$$x^{(i)} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$$

Distance to $\mu_1$

$$\| x^{(i)} - \mu_1 \|^2 = (-1 - 1)^2 + (2 - 2)^2$$ $$= (-2)^2 + 0^2 = 4$$

Distance to $\mu_2$

$$\| x^{(i)} - \mu_2 \|^2 = (-1 + 3)^2 + (2 - 0)^2$$ $$= (2)^2 + (2)^2 = 4 + 4 = 8$$

Distance to $\mu_3$

$$\| x^{(i)} - \mu_3 \|^2 = (-1 - 4)^2 + (2 - 2)^2$$ $$= (-5)^2 + 0^2 = 25$$

Assigning Cluster

• Distance to $\mu_1$: 4
• Distance to $\mu_2$: 8
• Distance to $\mu_3$: 25

The smallest distance is to $\mu_1$ so $c^{(i)} = 1$

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3. 🧿 Move Centroid Step

For each cluster $k$ where $k = 1, \dots, K$:

• $K$ = total number of clusters
• $k \in \{1, 2, ..., K\}$

Update centroid $\mu_k$ to be the mean of all points assigned to it.

$$\mu_k = \frac{1}{|C_k|} \sum_{i : c^{(i)} = k} x^{(i)}$$

Where:

• $C_k$ = set of points assigned to cluster $k$
• $|C_k|$ = number of points in cluster $k$

This moves the centroid to the center of its assigned points.

• Each centroid becomes the mean of the points assigned to it.

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⛔ Stop Condition

The algorithm converges when:

• Cluster assignments $c^{(i)}$ no longer change, or
• Centroids $\mu_k$ stop moving.

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What If a Cluster Gets No Points?

If some centroid has zero assigned points:

1. Eliminate the cluster (resulting in $K - 1$ clusters), or
2. Reinitialize the centroid randomly

In practice, empty clusters are uncommon with reasonable initialization.

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💰 Cost Function (Distortion)

Minimize th Distance between points and their assigned centroids.

$$J(c^{(1)}, \dots, c^{(m)}, \mu_1, \dots, \mu_K) = \frac{1}{m} \sum_{i=1}^{m} \left\| x^{(i)} - \mu_{c^{(i)}} \right\|^2$$

Or equivalently:

$$J(\mu_1, \dots, \mu_K) = \frac{1}{m} \sum_{i=1}^{m} \left\| x^{(i)} - \mu_{c^{(i)}} \right\|^2$$

Where:

• $J$ = cost function (distortion)
• $m$ = number of training examples
• $x^{(i)}$ = $i$-th training example
• $\mu_{c^{(i)}}$ = centroid of the cluster assigned to $x^{(i)}$
• $c^{(i)}$ = index of cluster assigned to $x^{(i)}$

Each iteration consists of:

1. Assignment step
   Minimizes $J$ with respect to $c^{(i)}$ (cluster assignments).

2. Centroid update step
   Minimizes $J$ with respect to $\mu_k$ (centroid locations).

Since each step does not increase $J$, the algorithm is guaranteed to converge.

Each iteration of K-Means guarantees that $J$ does not increase.

However, it may converge to a local minimum, not necessarily the global minimum.

A good solution:

• Each natural cluster is captured by one centroid.

A bad solution:

• Merge two true clusters into one
• Split one true cluster into multiple parts
• Assign very few points to some clusters

All of these correspond to higher distortion:

$$J_{\text{bad}} > J_{\text{good}}$$

To reduce the risk of poor local minima: