a.

Both A and C have well defined clusters in their similarity matricies. A is the best suited to hierarchical clustering because there appears to be clusters within clusters. Even B has visible clusters, but not as well defined as the other cases.

b.

C is the most well suited to kmeans clustering because kmeans is very sensitive to noise. C has almost none.

c.

Density based clustering is most well suited to remove noise, and it can also determine the number of clusters. B has a lot of noise, so it is well suited for density based clustering.

d.

If the above matricies were not ordered, it would be impossible to tell where the clustering was, so I could not make the above judgments based on the matrix alone. However, if I had access to the data, I could make judgments based on similarity.

2.

a.

b.

c.

• f₀₀ means (not in C1 and not in C2)
• f₀₁ means (not in C1 and in C2)
• f₁₀ means (in C1 and not in C2)
• f₁₁ means (in C1 and in C2)

from itertools import combinations
# represent letters as indexes for conveniance
D = (0, 1, 2, 3, 4)

C1 = [[1, 1, 1, 0, 0],
      [1, 1, 1, 0, 0],
      [1, 1, 1, 0, 0],
      [0, 0, 0, 1, 1],
      [0, 0, 0, 1, 1]]
C2 = [[1, 1, 1, 1, 0],
      [1, 1, 1, 1, 0],
      [1, 1, 1, 1, 0],
      [1, 1, 1, 1, 0],
      [0, 0, 0, 0, 1]]

# get pairs of letters
pairs = list(combinations(D, 2))
f00 = 0
f01 = 0
f10 = 0
f11 = 0

for [p1, p2] in pairs:
    if C1[p1][p2] and C2[p1][p2]:
        f11 +=1
    if C1[p1][p2] and not C2[p1][p2]:
        f10 +=1
    if not C1[p1][p2] and not C2[p1][p2]:
        f00 +=1
    if not C1[p1][p2] and  C2[p1][p2]:
        f01 +=1
return list(zip(["f00", "f01", "f10", "f11"], [f00, f01, f10, f11]))

d.

\(R = \frac{f_{11} + f_{00}}{f_{01} + f_{10} + f_{00} + f_{11} }= \frac{3 + 3}{10 } = 0.6\)

e.

\(J(A, B) = \frac{|A \cap B |}{|A \cup B|} = \frac{f_{11}}{f_{10} + f_{01} + f_{11}} = \frac{3}{7} \)

3.

• C4 matches with M1 because it is the furthest away from the other clusters, making it have a high level of separation
• C5 goes to M3 because it has a smaller number of points than C2
• C2 goes to M2 because there are less points (height is different)
• C1 goes to M4 because it is closer to the other clusters
• C3 goes to M5 because it is the only one left and it is either C1 or C3 because they are close together, and have few points.

4.

a.

b.

A cophenetic correltion coefficient is computed by taking the original pairwise distances and comparing them to the dendrogrammatic distances (shown in the matrix above) using the following formula:

• x(i, j) = euclidean distance between point i and point j (\(\overline x\)) is the average
• t(i, j) = the dendrogrammtic distance between point and point j \(\overline t\) is the average
• \(c = \frac{\sum_{i<j}[x(i, j) - \overline x][t(i,j) - \overline t]}{\sqrt{\sum_{i<j}[x(i, j) - \overline x]^2\sum_{i<j}[t(i, j) - \overline t]^2}}\)

c.

The complete link clustering does not preserve the pairwise distances as well in the dendrogram as the single link clustering. A higher cophenetic correlation coefficient implies that the preservation of the pairwise distances is better.