Convergence of K-means algorithm

[@Dexter4o4]

🎯 Key Takeaways for quick navigation: 02:52 🔄 *The Lloyd's algorithm converges, and the proof involves comparing the distance of points to the mean in the current iteration with the mean in the next iteration.* 05:29 📏 *The point that minimizes the sum of squared distances to a set of points is the mean of those points.* 09:04 🔀 *If the algorithm does not converge at iteration t, it means at least one point found a cluster with a mean closer than its current cluster's mean.* 15:30 🔄 *The objective function after a reassignment strictly reduces, demonstrating progress in convergence.* 25:56 📉 *The algorithm guarantees a reduction in the objective function after each reassignment, showing convergence.* 26:38 🔄 *The objective function strictly reduces after each reassignment, indicating progress in the convergence of the K-means algorithm.* 29:49 🧾 *The finiteness of the number of partitions ensures that the algorithm must converge, as each reassignment eliminates a possible partition, and there are only a finite number of them.* 30:31 🚀 *The monotonic reduction of the objective function implies that the algorithm will eventually reach a partition where every point is happy with its own mean, leading to convergence.* 31:23 ⏰ *The worst-case analysis doesn't imply the algorithm will take k^n iterations to converge, as practical datasets often converge quickly. The argument is about the worst-case scenario.* 32:14 📉 *Convergence is assured, but it doesn't guarantee the algorithm converges to the partition with the smallest objective value; it could reach a local minimum where every point is content.*

[@Dexter4o4]

let's consider a simple numerical example of Lloyd's algorithm, also known as the k-means algorithm. Suppose we have a set of data points in a 1-dimensional space: {1, 2, 3, 5, 6, 8}. We want to cluster these points into two groups (k=2). 1. **Initialization**: We randomly select two points as the initial centroids. Let's say we choose 2 and 6. 2. **Assignment**: We assign each data point to the closest centroid. The groups are now {1, 2, 3} (centroid at 2) and {5, 6, 8} (centroid at 6). 3. **Update**: We calculate the new centroids by finding the mean of the points in each cluster. The new centroids are 2 (mean of {1, 2, 3}) and 6.33 (mean of {5, 6, 8}). 4. **Repeat Assignment and Update**: We repeat the assignment and update steps until the centroids do not change significantly or a maximum number of iterations is reached. In this case, the centroids do not change in the next iteration, so the algorithm has converged. The final clusters are {1, 2, 3} and {5, 6, 8}, with centroids at 2 and 6.33, respectively².

[@akankshaharsh3064]

Explanations made by Sir are really nice.

[@khastakachori123]

This course is pure gold! Like his very much.

[@vatg2001]

Diamond, gold everything this lecture👌

[@Shrikant_Anand]

To see why the objective function at iteration t+1 is less than or equal to the intermediate quantity(summation over i = 1 to n of squared distance b/w x_i and mean of that cluster to which x_i wants to go to at iteration t) see that the objective function at iteration t+1 is equivalent to the squared distance of all data points w.r.t to the cluster mean that it belongs to at iteration (t+1) and this is the lowest sum one could get if we consider the (t+1)th configuration of the points in different clusters. This means the summation of the squared distance of all data points present in clusters 1 to k at iteration (t+1) taken w.r.t some other point apart from their respective cluster's mean would definitely be less than the lowest possible sum for (t+1)th configuration. I hope this helps because I also took time to understand this.

[@rajan_0]

All this could have been better explained with a working example. I see the professor is kind of struggling to explain something trivial, I think most of the ideas being explained here in detail would be easier to grasp if the professor was working on an example as he went along. This is the theme in most of the courses here, the rigid separation between theory and practice only makes things harder to understand.

[@vatg2001]

Kya baat boli hai bhai jai hind

[@shubhamgattani5357]

Agreed. A simple example by taking 4 points x1,x2,x3,x4 and 2 or 3 clusters would have made lecture understandable in one go.

[@abrarmohammed2457]

enjoying every lecture of professor arun sir

[@SakshamKumar-y4p]

my brain is not able to compile this many notations 😢

[@storiesshubham4145]

15:08 Sir is thinking hard to get the correct words 😄

[@vatg2001]

😂😂😂😂😂😂😂😂😂😂😂

[@vishwafuru]

his brain is thinking at a pace faster than his mouth can deliver. nevertheless he nailed it.

[@vatg2001]

🤯🤯🤯🤯🤯🤯

[@AYASHJAIN]

in 7.14 minutes i think by mistake sir has written summation from o to n...it would be i think o to l