Hierarchical Classification of Average Linkage Euclidean Distance

The classification hierarchy will use Euclidean distance.
The data we will use is as follows:

The initial stage is to calculate data X and data Y.
At this stage, the calculation method is different from the others.
The calculation stages are as follows:

• D1,D1 means sqrt((X1-X1) ^ 2 + (Y1-Y1) ^ 2) (using euclidean distance)
• D1,D2 means sqrt((X1-X2) ^ 2 + (Y1-Y2) ^ 2) (using euclidean distance)
• D1,D3 means sqrt((X1-X3) ^ 2 + (Y1-Y3) ^ 2) (using euclidean distance)
• D1,D4 means sqrt((X1-X4) ^ 2 + (Y1-Y4) ^ 2) (using euclidean distance)
• D1,D5 means sqrt((X1-X5) ^ 2 + (Y1-Y5) ^ 2) (using euclidean distance)
• D2,D2 means sqrt((X2-X2) ^ 2 + (Y2-Y2) ^ 2) (using euclidean distance)
• D2,D3 means sqrt((X2-X3) ^ 2 + (Y2-Y3) ^ 2) (using euclidean distance)
• D2,D4 means sqrt((X2-X4) ^ 2 + (Y2-Y4) ^ 2) (using euclidean distance)
• D2,D5 means sqrt((X2-X5) ^ 2 + (Y2-Y5) ^ 2) (using euclidean distance)
• D3,D3 means sqrt((X3-X3) ^ 2 + (Y3-Y3) ^ 2) (using euclidean distance)
• D3,D4 means sqrt((X4-X4) ^ 2 + (Y4-Y4) ^ 2) (using euclidean distance)
• D3,D5 means sqrt((X5-X5) ^ 2 + (Y5-Y5) ^ 2) (using euclidean distance)
• D4,D4 means sqrt((X4-X4) ^ 2 + (Y4-Y4) ^ 2) (using euclidean distance)
• D4,D5 means sqrt((X4-X5) ^ 2 + (Y4-Y5) ^ 2) (using euclidean distance)
• D5,D5 means sqrt((X5-X5) ^ 2 + (Y5-Y5) ^ 2) (using euclidean distance)

The results are as follows:

Next is to create a matrix diagram :

From the matrix above, we will choose the smallest result except 0 and it turns out that the smallest value is 1.414213562 which belongs to data 3 to 5 and 5 to 3.
We will combine the data to form the following:

• D(35)1
• D(35)2
• D(35)4

Next, the average value will be calculated from the data, as follows:

• D(35)1 = average(5.830951895 , 6.32455532) = 6.077753608
• D(35)2 = average(2.828427125 , 3.16227766) = 2.995352392
• D(35)4 = average(5.385164807 , 4.123105626) = 4.754135216

The results obtained are as follows:

After obtaining the data, the smallest value is selected to form a matrix as follows:

After that, the smallest data from the data above is selected again, so that what is obtained is as follows:

• D(352)1
• D(352)4

The results obtained are as follows:

After obtaining the data, the smallest value is selected to form a matrix as follows:

After that, the smallest data from the data above is selected again, so that what is obtained is as follows:

• D(3524)1

The results obtained are as follows:

After obtaining the data, the smallest value is selected to form a matrix as follows:

From the results above, we can create a diagram, so that the diagram is formed as follows: