00:01
Here it is given that in terms of a and b, 1, 2, 3, 4 and 5, that is a value it is 16, 3, 8, 9 and 5, whereas b value it is 15, 19, 12, 8 and 4.
00:23
Starting point is observation 2 and 4.
00:28
So observation 2 and 4 are the starting points of the concept.
00:35
So a and b, 2 and 4, it is nothing but 3, 19, 9, 18, 8.
00:46
The remaining terms are 1, 3 and 5.
00:49
So it will be in terms of 16, 15, 15, 8, 12, 5 and 4.
00:58
Now, the cluster will be in terms of cluster in terms of a and b.
01:08
It will be, cluster will be 24.
01:11
That is 3 plus 9 divided by 2 which is equal to 6 and 19 plus 8 divided by 2 which is equal to 6.
01:21
And next one it will be 135 which is nothing but 16 plus 8 plus 5 divided by 3 which gives you 9 .67 and 15 12 14 which gives you 10 .3p.
01:42
So the euclidean distance can be calculated in terms of euclidean distance can be calculated in terms of euclidean distance can be calculated in terms of.
01:53
Of d square of 1 comma 2 and 4 which is nothing but 3 minus 6 the whole square plus 19 minus 13 .5 the whole square which gives you 39 .25.
02:12
3 .25 similarly in terms of 2 .24 you will be getting 39 .25 then in terms of 2 .24 you will be getting 39 .25 then in terms of of 3 comma 24 it will be 1 or 2 .25 and this process continues for both the term and we will be obtaining the concept in terms of cluster 24 24 and for 135 now in terms of 1 .35 now in terms of 1 2 3 4 and 5 the solutions will be 39 .25 and 5.
02:53
The solutions will be 39 .25.
02:57
39 .25, 102 .25, 6 .25 and 91 .25.
03:08
Similarly, it will be 61 .88, 119 .66, 5 .58, again 5 .58, then 61 .83.
03:23
So here the minimum is the fourth one.
03:32
The total will be 101 .23, 158 .91, 107 .83, 13 .13 .13...