By hand. Node 3 is at distances \(1, 1, 1, 2\) from the others: closeness \(1/5\). Node 5 is at \(3, 3, 2, 1\): closeness \(1/9\). All five: \(\left(\tfrac17, \tfrac17, \tfrac15, \tfrac16, \tfrac19\right)\).
NetworkX (and most software) reports the normalised form \((n-1)/\sum_j d(i,j) = 1/\overline{d}(i)\), the reciprocal of the mean distance, so values lie in \((0, 1]\). For node 3 that is \(4/5 = 0.8\).
Interpretation
Large Closeness Centrality:
Indicates that a node is, on average, close to all other nodes in the network.
This means it can quickly interact with or reach other nodes.
Nodes with high closeness centrality are often considered central or influential because they can efficiently spread information or resources throughout the network.
Small Closeness Centrality:
Indicates that a node is, on average, farther away from all other nodes in the network.
These nodes are less central and may take longer to interact with or reach other nodes.
Nodes with low closeness centrality are often on the periphery of the network and are less influential in terms of spreading information or resources.
In this graph, most nodes are close to most other nodes: the normalised values only run from about 0.28 to 0.57.
However we can see that some nodes are slightly more central than others.
Betweenness Centrality
Alternatively we might be interested in the question, is the node on many paths?
Betweenness captures how important a node is to the communication process, or how much information passes through the node.
First, let’s consider the case in which there is only one shortest path between any pair of nodes.
Then, the betweenness centrality of node \(i\) is the number of shortest paths that pass through \(i\).
Mathematically, summing over the unordered pairs \(\{j, k\}\) of nodes other than \(i\),
\[
\text{betweenness}(i) = \sum_{\{j,k\} \subseteq V\setminus\{i\}}
\begin{cases}
1 &\text{if the path from }j\text{ to }k\text{ goes through }i, \\
0 &\text{otherwise}.
\end{cases}
\]
For a graph with \(n\) nodes, we can normalize this to a value between 0 and 1 by dividing by \({n-1 \choose 2} = (n-1)(n-2)/2\), the number of such pairs.
By hand. Node 3 lies on the shortest paths of \(\{1,4\}, \{1,5\}, \{2,4\}, \{2,5\}\): betweenness 4. Node 4 lies on \(\{1,5\}, \{2,5\}, \{3,5\}\): betweenness 3. Nodes 1, 2, 5 lie on none. Normalised by \(\binom42 = 6\): \((0, 0, 0.67, 0.5, 0)\).
General Case: Dependency
In a general graph, there may be multiple shortest paths between \(j\) and \(k\).
To handle this, we define:
\(\sigma(i \mid j,k)\) is the number of shortest paths between \(j\) and \(k\) that pass through \(i\), and
\(\sigma(j,k)\) is the total number of shortest paths between \(j\) and \(k\).
Then we define the dependency of \(i\) on the paths between \(j\) and \(k\) as the ratio of those two quantities:
Reminder: Properties of nonnegative symmetric matrices
A nonnegative symmetric matrix\(A\) has the following properties:
All eigenvalues of \(A\) are real, and its eigenvectors can be chosen orthogonal (symmetric matrices, Lecture 12).
The largest eigenvalue of \(A\) is nonnegative, and no other eigenvalue exceeds it in absolute value. (The other eigenvalues can be negative: a single edge has eigenvalues \(1\) and \(-1\).)
If the graph is connected, the largest eigenvalue has an eigenvector with all positive entries, and it is the only eigenvector with all positive entries.
\(x_i\) is called the eigenvector centrality of node \(i\).
\(\lambda\) is a constant (which we’ll show is an eigenvalue).
Put another way, the importance of node \(i\) is proportional to the sum of the importance of the other nodes \(j\) directly connected to \(i\).
Matrix Form
In matrix form, this can be written as
\[
\mathbf{Ax} = \lambda \mathbf{x},
\]
where \(\mathbf{x}\) is the eigenvector corresponding to the eigenvalue \(\lambda\) of the adjacency matrix \(\mathbf{A}\).
The eigenvector \(\mathbf{x}\) gives the centrality scores for all nodes in the network.
In general, there may be multiple eigenvalues \(\lambda\) for which the above equation holds.
We use the largest eigenvalue: by Perron–Frobenius its eigenvector is the only one whose entries are all positive, so it is the only one that reads as a set of scores.
Computing It: Power Iteration
Start from \(\mathbf{x} = \mathbf{1}\) and repeat \(\mathbf{x} \leftarrow A\mathbf{x}\), rescaling each time. One multiplication = every node replaces its score by the sum of its neighbours’ scores.
node 1
node 2
node 3
node 4
node 5
\(A\mathbf 1\) (= degrees)
2
2
3
2
1
\(A^2\mathbf 1\)
5
5
6
4
2
\(A^3\mathbf 1\)
11
11
14
8
4
\(A^3\mathbf 1\), rescaled so max = 1
0.79
0.79
1
0.57
0.29
converged (NetworkX), max = 1
0.82
0.82
1
0.57
0.26
Nodes 1, 2 and 4 all start at degree 2, but node 4 falls behind: its neighbours are node 3 and the leaf 5, while 1 and 2 are attached to each other and to node 3.
Why it works. Write \(\mathbf 1 = \sum_k c_k \mathbf{v}_k\) in eigenvectors; then \(A^t\mathbf 1 = \sum_k c_k\lambda_k^t\mathbf{v}_k\), and the largest eigenvalue (here \(\lambda = 2.21\)) dominates as \(t\) grows. Perron–Frobenius makes \(c_1 > 0\), since \(\mathbf v_1\) is positive.
Three Measures, One Graph
A = nx.to_numpy_array(T, nodelist=[1, 2, 3, 4, 5])x = np.ones(5)for _ inrange(100): # power iteration x = A @ x x = x / np.linalg.norm(x)ec = nx.eigenvector_centrality(T)table = pd.DataFrame({'degree': dict(T.degree()),'closeness 1/Σd': {v: 1/sum(d.values()) for v, d in nx.shortest_path_length(T)},'betweenness (raw)': nx.betweenness_centrality(T, normalized=False),'eigvec, power iter.': dict(zip(T, x / x.max())),'eigvec, NetworkX': {v: ec[v] /max(ec.values()) for v in T},})print(table.round(3))
Node 3 wins everything. Node 4 beats nodes 1 and 2 on closeness and betweenness (it sits on the way to node 5) but loses on eigenvector centrality (its neighbours are poorer). Each measure encodes a different notion of importance.
Key Points on Eigenvector Centrality
Eigenvector: The vector \(\mathbf{x}\) represents the centrality scores.
Largest Eigenvalue: The centrality scores are derived from the eigenvector corresponding to the largest eigenvalue of the adjacency matrix.
Perron-Frobenius Theorem: For a connected graph, the largest eigenvalue is positive. In addition, there is only 1 corresponding eigenvector with all positive entries.
Power iteration finds it with nothing but matrix–vector products, which is how it is computed on large graphs.
The Perron-Frobenius Theorem is an important theoretical tool when working with adjacency matrices, which are by definition nonnegative.
It appears that we have identified the instructor (node 0) and the club president (node 33) as important nodes based on all three measures. The rest of the ranking is where the measures part ways.
Clustering and Partitioning
Clustering and Partitioning
We now turn to the question of finding important groups of nodes.
Why might we want to cluster graph nodes?
Assigning computations to processors in a parallel computer
Segmenting images (finding boundaries between objects)
Clustering words found together in documents, or documents with similar words
Divide and conquer algorithms
Circuit layout in VLSI
Community detection in social networks
Min \(s\)-\(t\) cut
We’ll start with a problem that is fundamental to many other problems in graph analysis.
It involves finding the smallest set of edges that, if removed, would disconnect a specified source node (s) from a target node (t) in a graph.
The goal of a min \(s\)-\(t\) cut is to partition the graph into two disjoint subsets such that:
\(s\) is in one subset and \(t\) is in the other, and
the total weight (or number) of the edges that need to be removed to achieve this separation is minimized.
Practical Example: Soviet Railway Network
Here is a schematic diagram of the railway network of the Western Soviet Union and Eastern European countries in the Cold War era.
The idea was to determine a minimum cut, which would best halt the flow of materials in the railway network.
Cost of a Cut
Let \(G = (V, E)\) be a weighted graph.
A weighted graph assigns a weight \(w(e)\) to each edge \(e\in E\).
An \(s\)-\(t\) cut \(C\) of \(G\) is a partition of \(V\) into \((U, V-U)\) such that \(s \in U\) and \(t \in V-U\).
The cost of a cut is the total weight of the edges that go between the two parts:
As mentioned, when Wayne Zachary studied the club, a conflict arose between the instructor and the president (nodes 0 and 33, respectively).
Zachary predicted the way the club would split based on an \(s\)-\(t\) min cut between the instructor and the president.
Code
cut_edges = nx.minimum_edge_cut(G, s=0, t=33)Gcopy = G.copy()Gcopy.remove_edges_from(cut_edges)instructor_side = nx.node_connected_component(Gcopy, 0)node_colors = [side_colors[0] if v in instructor_side else side_colors[1] for v in G]fig = plt.figure(figsize=(12, 5))nx.draw_networkx(G, node_color=node_colors, pos=pos, with_labels=True, node_size=1000, font_size=16)plt.axis('off')plt.show()wrong = [v for v in G if (v in instructor_side) != (G.nodes[v]['club'] =='Mr. Hi')]print(f'{len(cut_edges)} cut edges; members on the wrong side of the actual split: {wrong}')
10 cut edges; members on the wrong side of the actual split: [8, 9]
This unweighted cut places every member correctly except two (8 and 9). Zachary’s edges carried interaction counts, which he used as capacities; with those weights the min cut misses only node 8.
Minimum Cuts
In partitioning a graph, we may not have any particular \(s\) and \(t\) in mind.
Rather, we may want to simply find the minimal way to disconnect the graph.
Clearly, we can do this using an \(s\)-\(t\) min cut, by simply trying all \(s\) and \(t\) pairs.
Let’s try this approach of finding the minimum \(s-t\) cut over all possibilities in the karate club graph.
Code
Gcopy = G.copy()Gcopy.remove_edges_from(nx.minimum_edge_cut(G))big_side =max(nx.connected_components(Gcopy), key=len)node_colors = [side_colors[1] if v in big_side else side_colors[0] for v in G]fig = plt.figure(figsize=(12,5))nx.draw_networkx(G, node_color=node_colors, pos=pos, with_labels=True, node_size=1000, font_size=16)plt.axis('off')plt.show()
This is in fact the minimum cut. Node 11 only has one edge to the rest of the graph, so the min cut is 1.
As this example shows, minimum cut is not, in general, a good approach for clustering or partitioning.
To get a more useful partition, we need to define a new goal. The new goal will be to find a balanced cut.
Balanced Cuts
The idea to avoid the problem above is to normalize the cut by the size of the smaller of the two components. The problem above would be avoided because the smaller of the two cuts is just a single node.
where \(E(U, V\setminus U)\) is the number of edges between the two parts.
The idea is that finding \(\alpha(G)\) gives a balanced cut – one that maximizes the number of disconnected nodes per edge removed.
Unfortunately, this is a challenging problem that is not computable in polynomial time.
However, we can make good approximations, which we’ll look at now.
To do so, we’ll introduce spectral graph theory.
Spectral Graph Theory
Spectral graph theory is the use of linear algebra to study the properties of graphs.
To introduce spectral graph theory, we define some terms.
Let \(G\) be an undirected graph with \(n\) nodes. We define the \(n\times n\) matrix \(D\) as a diagonal matrix of node degrees, i.e., \(D = \text{diag}(d_1, d_2, d_3, \dots)\) where \(d_i\) is the degree of node \(i\).
Assuming \(G\) has adjacency matrix \(A\), we define the Laplacian of \(G\) as
\[
L = D - A.
\]
So \(L_{ii} = d_i\) on the diagonal; off the diagonal, \(L_{ij} = -1\) if \((i,j)\) is an edge and \(0\) otherwise. Every row sums to zero.
A Laplacian You Can Check by Hand
The graph from Homework 10, Problem 3: two triangles joined by the bridge \(3\)–\(4\). Degrees \((2,2,3,3,2,2)\).
\[
L = D - A = \begin{bmatrix}
\phantom{-}2&-1&-1&\phantom{-}0&\phantom{-}0&\phantom{-}0\\
-1&\phantom{-}2&-1&\phantom{-}0&\phantom{-}0&\phantom{-}0\\
-1&-1&\phantom{-}3&-1&\phantom{-}0&\phantom{-}0\\
\phantom{-}0&\phantom{-}0&-1&\phantom{-}3&-1&-1\\
\phantom{-}0&\phantom{-}0&\phantom{-}0&-1&\phantom{-}2&-1\\
\phantom{-}0&\phantom{-}0&\phantom{-}0&-1&-1&\phantom{-}2
\end{bmatrix}
\]
Each row: the degree on the diagonal, a \(-1\) for each neighbour, and a row sum of zero.
Laplacian Matrix Interpretation
Now let us think about an \(n\)-component vector \(\mathbf{x} \in \mathbb{R}^n\) as an assignment of values to nodes in the graph \(G\).
For example, \(\mathbf{x}\) could encode node importance or strength or even a more concrete notion like temperature or altitude.
Laplacian Quadratic Form
Here is an amazing fact about the Laplacian of \(G\).
i.e., \(\mathbf{x}^TL\mathbf{x}\) is the sum of squared differences of \(\mathbf{x}\)over the edges in \(G\).
On the two-triangle graph, take \(\mathbf{x} = (1, 1, 1, -1, -1, -1)\). Every edge inside a triangle contributes \(0\); only the bridge contributes, \((1-(-1))^2 = 4\). So \(\mathbf{x}^TL\mathbf{x} = 4\).
We constrain \(\mathbf{x}\) to have a nonzero norm (i.e. \(\Vert \mathbf{x}\Vert = 1\)), otherwise \(\mathbf{x} = \mathbf{0}\) would be a trivial solution.
We can express this in terms of the graph Laplacian:
the smallest eigenvalue of \(L\), attained at the corresponding eigenvector.
This is PCA’s variance argument from Lecture 13 run backwards: there the top eigenvector of the covariance matrix maximised\(\mathbf{w}^TS\mathbf{w}\); here the bottom eigenvector of \(L\)minimises\(\mathbf{x}^TL\mathbf{x}\).
We are connecting functions on the graph \(G\) with eigenvectors of the matrix \(L\). This is quite remarkable.
What do we know about \(L\)?
\(L\) is symmetric. Therefore the eigenvectors of \(L\) are orthogonal and its eigenvalues are real (Lecture 12).
\(L\) is positive semidefinite, i.e. \(\mathbf{x}^T L\mathbf{x} \geq 0\) for all \(\mathbf{x}\).
Therefore the eigenvalues of \(L\) are all nonnegative.
We can order the eigenvalues from largest to smallest \(\lambda_n \geq \dots \geq \lambda_2 \geq \lambda_1 \geq 0.\)
How do we know that \(L\) is positive semidefinite?
Consider \(\sum_{(i,j)\in E} (x_i - x_j)^2.\) This is always a nonnegative quantity.
As a result, \(\mathbf{x}^T L\mathbf{x} \geq 0\) for all \(\mathbf{x}\).
Assume that \(G\) is connected (i.e. there is a path between any two nodes).
Then \(L\) has a single eigenvalue of value \(\lambda_1 = 0\). The corresponding eigenvector is \(\mathbf{w}_1 = \mathbf{1} = [1, 1, 1, \dots]^T\).
Note that \(\mathbf{x} \perp {\mathbf 1}\) means that the sum of the entries in \(\mathbf{x}\) is zero, which implies that \(\mathbf{x}\) is mean-centered.
Fiedler Vector
The corresponding eigenvector to \(\lambda_2\) is called the Fiedler vector.
If we think of \(x_i\) as a 1-D coordinate for node \(i\) in the graph, then choosing \(\mathbf{x} = \mathbf{w}_2\) (the eigenvector corresponding to \(\lambda_2\)) puts each node in a position that minimizes the sum of the squared stretching of each edge.
The Fiedler Vector of the Two-Triangle Graph
L = nx.laplacian_matrix(H, nodelist=[1, 2, 3, 4, 5, 6]).toarray()lam, W = np.linalg.eigh(L) # eigenvalues in increasing orderw2 = W[:, 1] * np.sign(W[0, 1]) # an eigenvector's sign is arbitrary: make node 1 positiveprint('eigenvalues:', np.clip(lam, 0, None).round(3)) # clip round-off below 0print('w2: ', w2.round(3))
\(\lambda_1 = 0\) (connected graph) and the Fiedler value is \(\lambda_2 = 0.438\), small because one edge holds the graph together.
\(\mathbf{w}_2 = (0.465, 0.465, 0.261, -0.261, -0.465, -0.465)\): one triangle positive, the other negative; the bridge endpoints 3 and 4 sit closest to zero.
The entries sum to zero (\(\mathbf{w}_2 \perp \mathbf 1\)). Flipping every sign gives the same eigenvector.
Recall from physics that the energy in a stretched spring is proportional to the square of its stretched length. Placing the nodes at \(\mathbf{w}_2\) is a spring-energy layout in one dimension: springs only, with the constraints \(\Vert\mathbf{x}\Vert = 1\) and \(\mathbf{x}\perp\mathbf 1\) doing the job of repulsion. The spring layout you saw last lecture adds explicit repulsion between nodes, which is why the two pictures differ.
The spectral layout from last lecture uses \(\mathbf{w}_2\) for the first coordinate and \(\mathbf{w}_3\) for the second, where \(\lambda_3 = \min_{\Vert \mathbf{x}\Vert = 1, \;\mathbf{x}\perp \{\mathbf{1}, \mathbf{w}_2\}} \mathbf{x}^TL\mathbf{x}\).
Code
import os, urllib.requestdef data_file(path):"""Course data file: use a local copy (here, or in the repo root above a notebook in jupyter_notebooks/), else fetch it from the Materials repo (Colab)."""for prefix in ('', '../'):if os.path.exists(prefix + path):return prefix + path os.makedirs(os.path.dirname(path) or'.', exist_ok=True) urllib.request.urlretrieve('https://raw.githubusercontent.com/tools4ds/DS701-Materials-FA26/main/'+ path, path)return pathimport reimport matplotlib.patches as mpatchesfrom matplotlib.colors import ListedColormap# data from http://www-personal.umich.edu/~mejn/netdata/football = nx.read_gml(data_file('data/football.gml'))conf = np.array([d['value'] for _, d in football.nodes.data()])conf_name = {}withopen(data_file('data/football.txt'), 'r') as fp:for line in fp: m = re.match(r'\s*(\d+)\s+=\s*([\w\s-]+)\s*\n', line)if m: conf_name[int(m.group(1))] = m.group(2).strip()colours =list(CATEGORICAL_12)colours[5], colours[8] = colours[8], colours[5] # Independents (5) in grey: not a conferenceconf_cmap = ListedColormap(colours)conf_patches = [mpatches.Patch(color=conf_cmap(i), label=conf_name[i]) for i inrange(12)]def draw_football(ax, pos, labels, title, edges=True): nx.draw_networkx_nodes(football, pos=pos, ax=ax, node_size=40, node_color=[conf_cmap(int(c)) for c in labels])if edges: nx.draw_networkx_edges(football, pos=pos, ax=ax, edge_color='gray', alpha=0.3) ax.set_title(title, size=14) ax.axis('off')fig, axes = plt.subplots(1, 2, figsize=(13, 6))draw_football(axes[0], nx.spectral_layout(football), conf, 'Spectral layout $(\\mathbf{w}_2, \\mathbf{w}_3)$')draw_football(axes[1], nx.spring_layout(football, seed=0), conf, 'Spring layout')fig.legend(handles=conf_patches, loc='lower center', ncol=6, fontsize=10)plt.subplots_adjust(bottom=0.15)plt.show()
Spectral Partitioning
This leads to key ideas in node partitioning.
The basic idea is to partition nodes according to the Fiedler vector \(\mathbf{w}_2\).
There are a number of options for how to split based on the Fiedler vector.
If \(\mathbf{w}_2\) is the Fiedler vector, then split nodes according to a value \(s\):
bisection: \(s\) is the median value in \(\mathbf{w}_2\)
sign: separate positive and negative values (\(s = 0\))
gap: separate according to the largest gap in the values of \(\mathbf{w}_2\)
ratio cut (sweep): try every threshold \(s\) and keep the one that minimizes the isoperimetric ratio \(\alpha\). The sweep is the version with a performance guarantee (Cheeger’s inequality).
Bisection (median \(= 0\)): the same split, 3 and 3.
Gap: the largest gap is \(0.261 - (-0.261) = 0.522\), between nodes 3 and 4: the same split.
All three cut only the bridge: \(\alpha = 1/\min(3,3) = 1/3\), which is the best possible here.
When the Rules Disagree
Code
# K3 = {1, 2, 3} and K5 = {4, ..., 8}, joined by the bridge 3-4K = nx.complete_graph([1, 2, 3])K.add_edges_from(nx.complete_graph([4, 5, 6, 7, 8]).edges())K.add_edge(3, 4)nodes =list(range(1, 9))lam_K, W_K = np.linalg.eigh(nx.laplacian_matrix(K, nodelist=nodes).toarray())f = W_K[:, 1] *-np.sign(W_K[0, 1]) # make node 1 negativedef alpha(G, U): cut =sum((a in U) != (b in U) for a, b in G.edges())return cut, cut /min(len(U), G.number_of_nodes() -len(U))splits = {'sign / gap': {v for v, fv inzip(nodes, f) if fv <0},'bisection (median)': {v for v, fv inzip(nodes, f) if fv < np.median(f)}}pos_K = nx.spring_layout(K, seed=2)fig, axes = plt.subplots(1, 2, figsize=(12, 4.5))for ax, (name, U) inzip(axes, splits.items()): cut, a = alpha(K, U) nx.draw_networkx(K, pos=pos_K, ax=ax, node_size=700, font_size=14, node_color=[OKABE_ITO[3] if v in U else OKABE_ITO[5] for v in K]) ax.set_title(f'{name}: {cut} cut edge(s), α = {a:.2f}', size=14) ax.axis('off')plt.show()print('w2 =', f.round(2))
w2 = [-0.51 -0.51 -0.32 0.18 0.29 0.29 0.29 0.29]
\(K_3\) and \(K_5\) joined by one edge. The bridge is still the natural cut, and sign and gap find it (\(\alpha = 1/3\)). Bisection insists on 4 and 4, so it must cut four edges inside the \(K_5\) (\(\alpha = 1\)). Balance costs edges.
Here is a spectral partitioning for the karate club graph, splitting by sign.
Code
f = nx.fiedler_vector(G, weight=None, seed=0) # unweighted, like L = D - A aboveif f[0] >0: f =-f # put the instructor on the negative sideU = {v for v in G if f[v] <0}#fig = plt.figure(figsize=(12,5))node_colors = [side_colors[0] if v in U else side_colors[1] for v in G]nx.draw_networkx(G, pos=pos, node_color=node_colors, with_labels=True, node_size=1000, font_size=16)plt.axis('off')plt.show()cut, a = alpha(G, U)wrong = [v for v in G if (v in U) != (G.nodes[v]['club'] =='Mr. Hi')]print(f'sides {len(U)} / {34-len(U)}, {cut} cut edges, alpha = {a:.2f}; 'f'on the wrong side of the actual split: {wrong}')
sides 15 / 19, 10 cut edges, alpha = 0.67; on the wrong side of the actual split: [2, 8]
With no \(s\) or \(t\) specified, the sign of \(\mathbf{w}_2\) recovers the club’s actual split except for two members, nearly the same partition as the \(s\)-\(t\) min cut between the instructor and the president.
Spectral Clustering
In many cases we would like to move beyond graph partitioning, to allow for clustering nodes into, say, \(k\) clusters.
The idea of spectral clustering takes the observations about the Fiedler vector and extends them to more than one dimension.
The spectral layout has already placed the football teams in a Euclidean space, \((\mathbf{w}_2, \mathbf{w}_3)\), where many conferences form tight groups. So we can run a standard clustering algorithm in that space, and add \(\mathbf{w}_4, \mathbf{w}_5, \dots\) to spread out the conferences that still overlap.
Spectral Clustering: The Algorithm
Given a graph \(G\), an embedding dimension \(d\) and a number of clusters \(k\):
Compute \(L = D - A\), the Laplacian of \(G\).
Compute the eigenvectors of the \(d + 1\) smallest eigenvalues, and drop the first, \(\mathbf{w}_1 = \mathbf{1}\): it gives every node the same coordinate.
Let \(U \in \mathbb{R}^{n\times d}\) have \(\mathbf{w}_2, \mathbf{w}_3, \dots, \mathbf{w}_{d+1}\) as columns.
Let the position of node \(i\) be row \(i\) of \(U\), a point in \(\mathbb{R}^d\).
Cluster the \(n\) points into \(k\) clusters using \(k\)-means (Lecture 3).
Two-way spectral partitioning is the special case \(d = 1\), \(k = 2\).
Spectral Clustering: Example
On the football network we ask for \(k = 11\) clusters: there are 11 real conferences, and the 12th label, the Independents, is not a community.
Spectral clustering with \(d = 3\), drawn in the \((\mathbf{w}_2, \mathbf{w}_3)\) layout. Cluster colours are arbitrary: compare the groupings.
Code
compare_with_conferences(3)
The third eigenvector separates groups that overlap in the two-dimensional picture.
Choosing \(d\) with the Adjusted Rand Index
Code
from sklearn.metrics import adjusted_rand_scoredims =range(1, 15)ari = [adjusted_rand_score(conf, spectral_clusters(d)) for d in dims]plt.figure(figsize=(8, 4.5))plt.plot(dims, ari, 'o-', color=OKABE_ITO[5])plt.xlabel('Embedding dimension $d$ (number of eigenvectors)', size=14)plt.ylabel('Adjusted Rand Index', size=14)plt.title('Spectral clustering of the football network vs. conferences', size=14)plt.ylim(0, 1)plt.xticks(list(dims))plt.grid(alpha=0.3)plt.show()best =int(np.argmax(ari))print(f'ARI: d = 1: {ari[0]:.3f}, d = 2: {ari[1]:.3f}, d = 3: {ari[2]:.3f}, d = 6: {ari[5]:.3f}; 'f'max {ari[best]:.3f} at d = {dims[best]}')
ARI: d = 1: 0.319, d = 2: 0.571, d = 3: 0.804, d = 6: 0.830; max 0.848 at d = 8
The ARI climbs steeply, from 0.32 with one eigenvector to 0.80 with three, then levels off: from \(d = 5\) on it stays between about 0.82 and 0.85 (the maximum, 0.848 at \(d = 8\), is barely above the 0.830 at \(d = 6\)). Extra eigenvectors neither help nor hurt much; the gain is in the first half-dozen.
Summary
Three centralities, three notions of importance: closeness (reach, \(1/\sum_j d(i,j)\)), betweenness (brokerage, shortest paths through \(i\)), eigenvector (status, connected to the well-connected).
Eigenvector centrality solves \(A\mathbf{x} = \lambda\mathbf{x}\) for the largest \(\lambda\); power iteration\(\mathbf{x} \leftarrow A\mathbf{x}/\Vert A\mathbf{x}\Vert\) computes it.
Min cut peels off a leaf; the isoperimetric ratio\(\alpha\) divides by the smaller side and rewards balanced cuts.
Laplacian\(L = D - A\), \(\mathbf{x}^TL\mathbf{x} = \sum_{(i,j)\in E}(x_i - x_j)^2\); the Fiedler vector\(\mathbf{w}_2\) is the smoothest non-constant assignment, and its sign, median or largest gap splits the graph.
Spectral clustering: rows of \([\mathbf{w}_2, \dots, \mathbf{w}_{d+1}]\) as coordinates, then \(k\)-means; choose \(d\) by ARI when labels exist.
Before class, do the knowledge check. In class you will compute these centralities and run power iteration on a real network, build PageRank from scratch, and partition the network with its Fiedler vector.
