When the Euclidean distance of a new point is the same from both centroids k1 and k2 in a k-means clustering algorithm, the point is equidistant from both clusters. In this situation, the algorithm typically has a few possible ways to handle this tie: 1. Assign to a Random Cluster: The point can be assigned to either k1 or k2 randomly. 2. Assign to the Smaller Cluster: If one of the clusters has fewer points, the algorithm might assign the point to that cluster to help balance the sizes of the clusters. 3. Assign Based on Previous Iteration: If the point was part of one of the clusters in a previous iteration, it might be reassigned to the same cluster. The choice among these strategies depends on the specific implementation of the k-means algorithm. In practice, the tie situation is rare, and the impact on the final clustering result is typically minimal.
idk if this algo is correct or not, but the output entirely depends upon the initial procedure . that's why i m little skeptical about the correctness as what if initally points are meant to be in same cluster
@@noorsiddiqui8695 ig he's talking about distance, two different points can have same distance from the centroid, in that case we should assign it to any cluster
Mam please cover the last unit also i.e Advanced concepts (Mining data streams,Text mining,Multimedia data mining,Mining sequance patterns in transactional databaces etc...)
last minute prep before 9hrs for sem... this is better than long lecture video
Exactly
Same here tomorrow semister exam for me 😢
Exactly
Pass ah fail ah cheppu
😮@@pandirisaicharan1206
Example from 5 min engineering..
Tq for more clear exaplantion mam
Having exam at 10am morning and me watching this video at 2.31 am midnight 😁
SAME BRO
Love u madam 😘😍 Tomorrow exam is data mining and big data for us ur videos make our exam easy for tomorrow❤
Control Bhai Control
😂@@darkwarrior6767
When the Euclidean distance of a new point is the same from both centroids k1 and k2 in a k-means clustering algorithm, the point is equidistant from both clusters. In this situation, the algorithm typically has a few possible ways to handle this tie:
1. Assign to a Random Cluster: The point can be assigned to either k1 or k2 randomly.
2. Assign to the Smaller Cluster: If one of the clusters has fewer points, the algorithm might assign the point to that cluster to help balance the sizes of the clusters.
3. Assign Based on Previous Iteration: If the point was part of one of the clusters in a previous iteration, it might be reassigned to the same cluster.
The choice among these strategies depends on the specific implementation of the k-means algorithm. In practice, the tie situation is rare, and the impact on the final clustering result is typically minimal.
idk if this algo is correct or not, but the output entirely depends upon the initial procedure . that's why i m little skeptical about the correctness as what if initally points are meant to be in same cluster
Thankyou for sharing the knowledge !!
The value of X1 is 170 but u have mentioned that 70 plz check it mam
Thank you so much Ma'am😊😊😊😊😊
My exam today at 10 am and I'm seeing this vedio now 😂
Tq sis very use full vedio
Thank you for your teaching mam
1 day to for exam leagend ❤❤
Voice nice to understand ❤
what should we do when k1 and k2 is equal
It can never be, if it does that means it's at the same point from the centroid and two point can't exist at same space
@@noorsiddiqui8695 ig he's talking about distance, two different points can have same distance from the centroid, in that case we should assign it to any cluster
@@EscapeThirty In that case assign as FCFS(first come first serve)
Thank you so much❤
Your content is very helpful..
Could you please create a video on Gaussian mixture models
please re check the video I feel there is some mistake .....you said Y0 but added the value of Yc its a bit confusing
The ans for 4 ; k1 = 7.2 not 6.32
so here each and every time, should we have to divide it in 2 clusters??
someone please answer this question..
Thank you ❤
Nice explanation 👌
Thank You 👍🙏
Please upload videos for k meloids ( PAM algorithm)🙏🙏
Thanku mam ❤❤
Nice voice mam 👏👏
What about pam(K- Medoids(Partitioning Around Medoid(PAM))
Mam for example....oka set ki k1 and k2 find chedtunnappudu same values vaste.... Appudu aa particular set ni eh k1 lo place cheyaala leda k2 lo naaa
Randam ga assign cheyachuu
Is k means partition algorithm or k means algorithm same or different
can the values of height and weight can we take reduced or particulary 12 ?
Super super super super madam
complete the entire problems for better understanding
From dec 27th we have semester exams please mam can you complete data mining syllabus 😢
Mam small request from my side if solving a problem takes too.much time then please explain already solved problem which u have written in ur notes
YEA
nice mam 🤩
Plz mam provide 4 th and 5 th units of data mining and data analysis
Thank you🙏
Madam Manam new centroids Akkadan use chayala
Mam please cover the last unit also i.e Advanced concepts (Mining data streams,Text mining,Multimedia data mining,Mining sequance patterns in transactional databaces etc...)
Thank you mam.
Mam is it k means algorithm
For the data mining lectures no one can replace praksh sir 💥
Channel name??
channel name please?
What is the name of prakash sir channel ??
So why are here😂
Suresh can teach better than him 🏌️
Copied from 5 mins engineering though nice explanation
thank you mam
Iove you mam🥳😊
Mam Can U share Your Notes
Tq madam
tysm
wcss ?
mam, your content is copied from 5 min engineering?!
pure doubt mai hoo
Are gajaa and vishwa sariga chudandi concentrate.... suchitha nuvu kuda
Rey sharath nuv kuda concentrate chey.. focus focus next exam ki
zoom out
😂😢
Thank you mam