lecture 22 

from math import sqrt
import numpy as np

D = {"A":[sqrt(2), sqrt(3), sqrt(5), 0, 0, 0],
     "B":[sqrt(2), 0, sqrt(5), 0, sqrt(6), 0],
     "C":[sqrt(2), 0, sqrt(5), 1, sqrt(6), 0],
     "D":[0, 0, 0, 0, sqrt(6), 2],
     "E":[0, 0, 0, 1, 0, 2]}
affinity = np.zeros((5, 5))
for xi in range(len(D)):
    for xj in range(len(D)):
        xivec = D[list(D.keys())[xi]]
        xjvec = D[list(D.keys())[xj]]
        affinity[xi, xj] = round(np.dot(np.array(xivec), np.array(xjvec)))
affinity.tolist()
10.07.07.00.00.0
7.013.013.06.00.0
7.013.014.06.01.0
0.06.06.010.04.0
0.00.01.04.05.0
import networkx as nx
import matplotlib.pyplot as plt

G = nx.Graph()
G.add_nodes_from(list(D.keys()))

for i in range(len(D.keys())):
    for j in range(i+1, len(D.keys())):
        i_node = list(D.keys())[i]
        j_node = list(D.keys())[j]
        G.add_edge(i_node, j_node, weight=affinity[i, j])
ax_1 = plt.subplot(111)
pos=nx.spring_layout(G, seed=0)
nx.draw_networkx(G, pos, with_labels=True)
for edge in G.edges(data="weight"):
    nx.draw_networkx_edges(G, pos, edgelist=[edge], width=edge[2])
plt.show()
![](./.ob-jupyter/b9ae460d0829f20394bc4a381df69a095b43ed4b.png)❌

It looks like ABCD are in one cluster and E is in another cluster.

2.

a.

  • compute affinity matrix A
  • Set L = A
  • compute k(number of clusters) largest eigenvalues/eigenvectors of L
  • compute matrix Y from eigenvalue matrix U using \(y_i = \frac{1}{\sqrt{\sum^k_{j=1}u^2_{n-j+1, i}}}(\vec{u_i}^T)\)
  • find clusters using k-means on Y

b. 

import scipy
A = affinity
k = 2
L = A
eigenvalues, eigenvectors = scipy.linalg.eigh(L)
U = eigenvectors[:,[-k, -1]]
row_sums = np.sqrt(np.square(U).sum(axis=1, keepdims=True))

Y = U / row_sums
Centroids, euclidian_mean = scipy.cluster.vq.kmeans(Y, k)
Clusters, distances = scipy.cluster.vq.vq(Y, Centroids)
list(zip(Y.tolist(),Clusters))
Eigen Pointcluster
(0.8252524486793774 -0.5647640179302249)0
(0.12086906435542212 -0.9926684588934237)0
(0.06222575627649513 -0.9980620998995093)0
(-0.9131796757110512 -0.40755721054627325)1
(-0.9887063042057109 -0.1498660869706164)1

c.

  • compute affinity matrix
  • Compute Degree matrix
    • diagonal matrix
    • each entry is the sum of weights of all edges incident with node
  • Compute L
    • L = D - W
  • compute k(number of clusters) smallest eigenvalues/eigenvectors of L
  • compute matrix Y from eigenvalue matrix U using \(y_i = \frac{1}{\sqrt{\sum^k_{j=1}u^2_{n-j+1, i}}}(\vec{u_i}^T)\)
  • find clusters using k-means on Y

d. 

import scipy
A = affinity
k = 2
# network x does weighted degrees for us!
D = np.diag(list(dict(G.degree(weight="weight")).values()))
L = D-A

eigenvalues, eigenvectors = scipy.linalg.eigh(L)
U = eigenvectors[:,:2]
row_sums = np.sqrt(np.square(U).sum(axis=1, keepdims=True))

Y = U / row_sums
Centroids, euclidian_mean = scipy.cluster.vq.kmeans(Y, k)
Clusters, distances = scipy.cluster.vq.vq(Y, Centroids)
list(zip(Y.tolist(),Clusters))
Eigen Pointcluster
(-0.8264994349350939 -0.5629375489803203)1
(-0.9670898580131041 -0.25443507330593085)1
(-0.9806921960094918 -0.19555770679285542)1
(-0.8201574486531152 0.5721378849707581)0
(-0.21393030744777544 0.9768489256560097)0

e.

The only difference is that L is computed either like La = D-1 L or like Ls = D^-.5 L D^-.5. We'll do both. 

import scipy
A = affinity
k = 2
# network x does weighted degrees for us!
D = np.diag(list(dict(G.degree(weight="weight")).values()))
L = D-A
L_a = np.matmul(np.diag((1/D.diagonal())), L)
L_s = np.matmul(np.matmul(np.diag(1/np.sqrt(D.diagonal())), L), np.diag(1/np.sqrt(D.diagonal())))
def normalized_cut(L):
    eigenvalues, eigenvectors = scipy.linalg.eigh(L)
    U = eigenvectors[:,:2]
    row_sums = np.sqrt(np.square(U).sum(axis=1, keepdims=True))

    Y = U / row_sums
    Centroids, euclidian_mean = scipy.cluster.vq.kmeans(Y, k)
    Clusters, distances = scipy.cluster.vq.vq(Y, Centroids)
    return list(zip(Y.tolist(),Clusters))
[ (a[0], a[1], b[0], b[1]) for  a, b, in zip(normalized_cut(L_a), normalized_cut(L_s))]
LaLs
/<
Eigen pointclustereigen pointcluster
(-0.2574691955687816 -0.9662865068566178)1(-0.6468914151117793 -0.7625821247935725)0
(-0.5595893949251859 -0.8287699976997625)1(-0.8375946839401586 -0.5462921795478918)0
(-0.7106527764955659 -0.7035429135874678)1(-0.8893139185105294 -0.4572972275702614)0
(-0.9658324732672688 0.2591671923342739)0(-0.8346244494862799 0.5508194153438352)1
(-0.7742723335582469 0.6328525527216169)0(-0.4879301656055612 0.8728826688004119)1

It looks like I was almost right, ABC and DE are clustered.