Generalizing unweighted network measures to capture the focus in interactions

Abstract

Unweighted network measures are commonly used to analyze real-world networks due to their simplicity and intuitiveness. This motivated the search for generalizations of unweighted network measures that take weights into account. We propose a new generalization methodology that captures how focused are the interactions over edges. The less focused the interaction (more uniform over edges) the closer is our generalization to the original unweighted measure. None of the previously developed generalizations capture this aspect of weighted networks. We analyze several real-world networks using our generalizations of the degree and the clustering coefficient. The analysis shows that our generalizations reveal interesting observations.

Appendix: Proofs of effective cardinality properties

Theorem 1

The effective cardinality satisfies the three properties described above: the maximum cardinality, the minimum cardinality, and the consistent partial ordering.

Proof

The proof follows from the following three lemmas. □

Lemma 1

The effective cardinality satisfies the maximum cardinality property.

Proof

When all the weights are equal to a constant C we have

$$ \forall e \in E^{\prime}: {\frac{w(e)}{\sum_{o\in E^{\prime}}w(o)}}={\frac{C}{C|E^{\prime}|}}={\frac{1}{|E^{\prime}|}} $$

We then have

$$ \begin{aligned} c(E^{\prime}) &=2^{\sum_{e\in E'}{\frac{1}{|E'|}}\log_2(|E^{\prime}|)}\\ &=2^{\log_2(|E^{\prime}|)} \\ &=|E^{\prime}| \end{aligned} $$

In other words, both the cardinality and the effective cardinality of a weighted set of edges become equivalent when the weights are uniform. The effective cardinality is also maximum in this case, because the exponent is the entropy of the weight probability distribution, which is maximum when weights are uniform over edges. □

Lemma 2

The effective cardinality satisfies the minimum cardinality property.

Proof

When the set of edges is empty, then the effective cardinality is zero by definition. When all weights are zero except only one weight that is greater than zero, then weight probability distribution is deterministic and the entropy is zero; therefore, the effective cardinality will be 1. □

Lemma 3

The effective cardinality satisfies the consistent partial order property.

Proof

Let $E^{\prime}_1$ and $E^{\prime}_2$ be two (edge) sets such that $|E^{\prime}_1|=|E^{\prime}_2|$ (both have the same cardinality). Let W ₁ and W ₂ be the corresponding sets of weights, where $\sum_{e1 \in E^{\prime}_1} w(e1)=\sum_{e2 \in E^{\prime}_2} w(e2) = S$ (the total weights are equal). Furthermore, let $|W_1 \bigcap W_2|=n-2,${$w_{11},w_{12}$}$= W_1 - W_2,${$w_{21},w_{22}$}$= W_2 - W_1$, where the ‘−’ operator is the “set difference" operator (the two sets share the same weights except for two elements in each set), and $|w_{11}-w_{12}| < |w_{21}-w_{22}|$ (the weights of W ₁ are more uniform than the weights of W ₂). To prove that the effective cardinality satisfies the consistent partial ordering property, we need to prove that $c(E^{\prime}_1)>c(E^{\prime}_2)$. □

Without loss of generality, we can assume that $w_{11} \ge w_{12}$ and $w_{21} \ge w_{22};$ therefore, $w_{11}-w_{12} < w_{21}-w_{22}$. We then have

$$ w_{11}+w_{12}=S - \sum_{w \in W_1 \bigcap W_2}w = w_{21}+w_{22} $$

or

$$ {\frac{w_{11}+w_{12}}{S}}=1 - \sum_{w \in W_1 \bigcap W_2}{\frac{w}{S}} = {\frac{w_{21}+w_{22}}{S}} = L $$

therefore

$$ L \ge {\frac{w_{21}}{S}} > {\frac{w_{11}}{S}} \ge {\frac{L}{2}} \ge L-{\frac{w_{11}}{S}} > L-{\frac{w_{21}}{S}} $$

where $\frac{w_{12}}{S} = L-\frac{w_{11}}{S}$ and $\frac{w_{22}}{S} = L-\frac{w_{21}}{S}$. Then from Lemma 6 we have $h(L,\frac{w_{11}}{S}) > h(L,\frac{w_{21}}{S})$, or

$$ -\frac{w_{11}}{S}lg\left(\frac{w_{11}}{S}\right) - \left(L-\frac{w_{11}}{S}\right)lg\left(c-\frac{w_{11}}{S}\right) > -\frac{w_{21}}{S}lg\left(\frac{w_{21}}{S}\right) - \left(L-\frac{w_{21}} {S}\right)lg\left(c-\frac{w_{21}}{S}\right) $$

Therefore $H(E^{\prime}_1) > H(E^{\prime}_2)$, because the rest of the entropy terms (corresponding to $W_1 \bigcap W_2$) are equal, and consequently $c(E^{\prime}_1)>c(E^{\prime}_2)$.

Lemma 4

The quantityh(C, x) =  − x lg(x) − (C − x)  lg(C − x) is symmetric around and maximized at$x={\frac{C}{2}}$for$C \ge x \ge 0$.

Proof

$$ h\left(C,{\frac{C}{2}}+\delta\right)=-\left({\frac{C}{2}}+\delta\right)\lg\left({\frac{C} {2}}+\delta\right) - \left({\frac{C}{2}}-\delta\right) \lg \left({\frac{C}{2}}-\delta\right) = h\left(C,{\frac{C}{2}}-\delta\right) $$

Therefore h(C, x) is symmetric around c/2. Furthermore, h(C, x) is maximized when

$$ {\frac{\partial h(C,x)}{\partial x}} = 0 = -1 -\lg x + 1 + \lg(C-x) $$

or

$$ \lg x = \lg(C-x) $$

Therefore h(C, x) is maximized at $x=C-x = {\frac{C}{2}}$. □