AdaSim: A Recursive Similarity Measure in Graphs

This repository provides the python implementation of AdaSim as well as the proof of AdaSim scores properties.

Installation and usage

AdaSim is a recursive link-based similarity measure based on the classic Adamic/Adar philosophy. AdaSim is applicable to both directed and undirected graphs. In the case of directed graphs, similarity scores can be computed based on any of in-links or out-links. In order to use AdaSim, the following packages are required:

Python       >= 3.8
networkx     =2.6.*
numpy        =1.21.*
scipy        =1.7.*
scikit-learn =1.0.*

Graph file format: A graph must be represented as a text file under the edge list format in which, each line corresponds to an edge in the graph, tab is used as the separator of the two nodes, and the node index is started from 0.

Proof of AdaSim Scores Properties

AdaSim scores are symmetric, bounded, monotonic, unique, and always existent.

(1) Symmetry: according to Equation (10), if $a\!=\!b$, $S_k(a,b)\!=\!S_k(b,a)\!=\!1$; if $a\!\neq\!b$, $S_k(a,b)\!=\! S_k(b,a)\! =\! \widehat{S}(a,b)$ for all $k \! \ge\! 0$.

(2) Bounding: for all $k$, $0 \le S_k(a,b) \le 1$.

According to Equation (10):

1- If $a\!=\!b$, then $S_0(a,b)\!=\!S_1(a,b)\!=\!\cdots=1$; therefore, $0\! \le\! S_k(a,b)\! \le\! 1$ for all values of $k$.

2- If $a\!\neq \!b$ and $I_a\!=\!\emptyset$ or $I_b\!=\!\emptyset$, then $S_0(a,b)\!=\!S_1(a,b)\!=\!\cdots\!=\!0$; therefore, $0\! \le \!S_k(a,b)\! \le\! 1$ for all values of $k$.

3- If $a\neq b$, $I_a \neq \emptyset$, and $I_b \neq \emptyset$, then $S_0(a,b) = \widehat{S}_0(a,b) = 0$; therefore, $0 \le S_0(a,b) \le 1$, which means the property holds for $k=0$. Now, we assume that the property holds for $k$, which means $0 \le S_k(a,b) = \widehat{S}_k(a,b) \le 1$, According to the assumption $\widehat{S}_k(a,b) \ge 0$, thus

$$\widehat{S}_{k+1}(a,b) = C \cdot \Bigg( \frac{\alpha}{m} \sum_{i \in I_a \cap I_b} w_i + (1-\alpha) \bigg( \frac{1}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t } \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot \widehat{S}_k(i,j) \cdot w_j \bigg) \Bigg)$$ $$\ge C \cdot \Bigg( \frac{\alpha}{m} \sum_{i \in I_a \cap I_b} w_i + (1-\alpha) \bigg( \frac{1}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t } \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot 0 \cdot w_j \bigg) \Bigg)$$ $$\ge C \cdot \frac{\alpha}{m} \sum_{i \in I_a \cap I_b} w_i$$

$m$ is the maximum Ada score in the dataset, thereby leading to the fact that $0 \le \frac{1}{m} \sum_{i \in I_a \cap I_b} w_i \le 1$; also, $0 < C < 1$ and $0 < \alpha \le 1$, which means $\widehat{S}_{k+1}(a,b) = S_{k+1}(a,b) \ge 0$.

According to the assumption $\widehat{S}_k(a,b) \le 1$, thus

$$\widehat{S}_{k+1}(a,b) = C \cdot \Bigg( \frac{\alpha}{m} \sum_{i \in I_a \cap I_b} w_i + (1-\alpha) \bigg( \frac{1}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t } \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot \widehat{S}_k(i,j) \cdot w_j \bigg) \Bigg)$$ $$\le C \cdot \Bigg( \frac{\alpha}{m} \sum_{i \in I_a \cap I_b} w_i + (1-\alpha) \bigg( \frac{1}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t } \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot 1 \cdot w_j \bigg) \Bigg)$$

since $\sum_{r \in I_a} \!w_r\! \cdot \!\sum_{t \in I_b} w_t$ = $\sum_{r \in I_a} \sum_{t \in I_b} w_r \!\cdot\! w_t$ = $\sum_{i \in I_a} \sum_{j \in I_b}\! w_i\! \cdot\! 1 \!\cdot w_j$, then $\widehat{S}_{k+1}(a,b) \le C \cdot \big( \frac{\alpha}{m} \sum_{i \in I_a \cap I_b} w_i + (1-\alpha) \cdot 1 \big)$; we also know that $\frac{1}{m} \sum_{i \in I_a \cap I_b} w_i \le 1$, which means $\widehat{S}_{k+1}(a,b) \le C \cdot \alpha + C \cdot (1-\alpha) $ = $C$; since $0 < C < 1$, then also $\widehat{S}_{k+1}(a,b) = S_{k+1}(a,b) \le 1$.

(3) Monotonicity: for every node-pair $(a,b)$, the sequence $\{S_k(a,b)\}$ is non-decreasing as $k$ increases.

If $a=b$, $S_0(a,b) = S_1(a,b) = \cdots = 1$; thus, the property holds. If $a \neq b$ and $I_a=\emptyset$ or $I_b=\emptyset$, $S_0(a,b) = S_1(a,b) = \cdots = 0$; thus, the property holds. If $a \neq b$, $I_a \neq \emptyset$, and $I_b \neq \emptyset$, according to Equation (10), $S_0(a,b)=0$ and by the bounding property, $0 \le S_1(a,b) \le 1$; therefore, $S_0(a,b) \le S_1(a,b)$, which means for $k=0$, the property holds. We assume that the property holds for all $k$ where $S_{k-1}(a,b) = \widehat{S}_{k-1}(a,b) \le S_k(a,b) = \widehat{S}_k(a,b)$, which means $\widehat{S}_k(a,b) - \widehat{S}_{k-1}(a,b) \ge 0$. Now, we show the property holds for $k+1$ as follows:

$$\widehat{S}_{k+1}(a,b) - \widehat{S}_k(a,b) = C \cdot \Bigg( \frac{\alpha}{m} \sum_{i \in I_a \cap I_b } w_i + (1-\alpha) \bigg( \frac{1}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t} \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot \widehat{S}_k(i,j) \cdot w_j \bigg) \Bigg)$$ $$- \; C \cdot \Bigg( \frac{\alpha}{m} \sum_{i \in I_a \cap I_b } w_i + (1-\alpha) \bigg( \frac{1}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t} \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot \widehat{S}_{k-1}(i,j) \cdot w_j \bigg) \Bigg)$$ $$= \; \frac{C \cdot (1-\alpha)}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t} \Bigg( \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot \widehat{S}_k(i,j) \cdot w_j - \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot \widehat{S}_{k-1}(i,j) \cdot w_j \Bigg)$$ $$= \; \frac{C \cdot (1-\alpha)}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t} \Bigg( \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot w_j \cdot \big(\widehat{S}_k(i,j) - \widehat{S}_{k-1}(i,j)\big) \Bigg)\\$$

according to the assumptions, $\widehat{S}_k(a,b) - \widehat{S}_{k-1}(a,b) \ge 0$ and we already know that $C \cdot (1-\alpha) \ge 0$ and $\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t \ge 0$; therefore, $\widehat{S}_{k+1}(a,b) - \widehat{S}_k(a,b) \ge 0$, which means $\widehat{S}_{k+1}(a,b) = S_{k+1}(a,b) \ge \widehat{S}_k(a,b) = S_k(a,b)$.

(4) Existence: the fixed points $S(*,*)$ of the AdaSim equation always exists.

By the bounding and monotonicity properties, for any node-pairs $(a,b)$, $\widehat{S}_k(a,b)$ is bounded and non-decreasing as $k$ increases. A sequence $\widehat{S}_k(a,b)$ converges to a $\lim \widehat{S}(a,b) = S(a,b)$ in $[0,1]$, according to the Completeness Axiom of calculus. $\displaystyle\lim_{k\to\infty} \widehat{S}_{k+1}(a,b) = \displaystyle\lim_{k\to\infty} \widehat{S}_k(a,b) = \widehat{S}(a,b)$ and the limit of a sum is identical to the sum of the limits, therefore

$$\widehat{S}(a,b) = \displaystyle \lim_{k\to\infty} \widehat{S}_{k+1} = C \cdot \Bigg( \frac{\alpha}{m} \sum_{i \in I_a \cap I_b} w_i + \frac{(1-\alpha)}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t} \cdot \displaystyle \lim_{k\to\infty} \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot \widehat{S}_k(i,j) \cdot w_j \Bigg)$$ $$= \; C \cdot \Bigg( \frac{\alpha}{m} \sum_{i \in I_a \cap I_b} w_i + \frac{(1-\alpha)}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t} \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot w_j \cdot \displaystyle \lim_{k\to\infty} \widehat{S}_k(i,j) \Bigg)$$ $$= \; C \cdot \Bigg( \frac{\alpha}{m}\sum_{i \in I_a \cap I_b} w_i + \frac{(1-\alpha)}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t} \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot w_j \cdot \widehat{S}(i,j) \Bigg) = S(a,b)\\$$

(5) Uniqueness: the solution for the fixed-point $S(*,*)$ is always unique.

Suppose that $S(*,*)$ and $S^\prime(*,*)$ are two solutions for the AdaSim equation. Also, for all node-pairs $(a,b)$, let $\delta(a,b) = S(a,b) - S^\prime(a,b)$ be the difference between these two solutions. Let $M\!=\!\max\limits_{(a,b)} |\delta(a,b)|$ be the maximum absolute value of all differences observed for some nod-pairs $(a,b)$ (i.e., $|\delta(a,b)| = M)$. We need to prove that $M=0$. If $a=b$, $M=0$ since $S(a,b) = S^\prime(a,b) = 1$. If $a \neq b$ and $I_a = \emptyset$ or $I_b = \emptyset$, $M=0$ since $S(a,b) = S^\prime(a,b) = 0$. Otherwise, $S(a,b) = \widehat{S}(a,b)$ and $S^\prime(a,b) = \widehat{S}^\prime(a,b)$ are computed by AdaSim. When $\alpha=1$, $M=0$ since $\widehat{S}(a,b) = \widehat{S}^\prime(a,b) = \frac{C}{m} \sum_{i \in I_a \cap I_b} w_i$. When $0 < \alpha < 1$, we have

$$\delta(a,b) = \widehat{S}(a,b) - \widehat{S}^\prime(a,b)$$ $$= \frac{C \cdot (1 - \alpha)}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t} \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot w_j \cdot (\widehat{S}(i,j) - \widehat{S}^\prime(i,j))$$ $$= \frac{C \cdot (1-\alpha)}{\sum_{r \in I_a} w_r \cdot \sum_{t \in I_b} w_t} \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot w_j \cdot \delta(i,j)$$

thus,

$$M = |\delta(a,b)| = \Bigg|\frac{ C \cdot (1 - \alpha)}{\sum_{r \in I_a}w_r \cdot \sum_{t \in I_b} w_t} \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot w_j \cdot \delta(i,j) \Bigg|$$ $$\le \frac{ C \cdot (1 - \alpha)}{\sum_{r \in I_a}w_r \cdot \sum_{t \in I_b} w_t} \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot w_j \cdot |\delta(i,j)|$$ $$\le \frac{ C \cdot (1 - \alpha)}{\sum_{r \in I_a}w_r \cdot \sum_{t \in I_b} w_t} \cdot \sum_{i \in I_a} \sum_{j \in I_b} w_i \cdot w_j \cdot M = C \cdot (1 - \alpha) \cdot M$$

Since $0 < \alpha < 1$ and $0 < C < 1$, then $0 < C \cdot (1-\alpha) < 1$, which means $M=0$.

Citation:

Masoud Reyhani Hamedani and Sang-Wook Kim. 2021. AdaSim: A Recursive Similarity Measure in Graphs. In Proceedings of the 30th ACM International Conference on Information and Knowledge Management, October 2021, Pages 1528–1537. https://doi.org/10.1145/3459637.3482316