Cleaning uncertain graphs via noisy crowdsourcing

Abstract

Uncertain graph is an important data model for many real-world applications. To answer the query on the uncertain graphs, the edges in these graphs are associated with existential probabilities that represent the likelihood of the existence of the edge. Almost all works on this area focus on how to promote the efficiency of the query processing. However, another issue should be notable, that is, the query results from the uncertain graphs are sometimes uninformative due to the edge uncertainty. We adopt a crowdsourcing-based approach to make the query results more informative. To save the monetary and time cost of crowdsourcing, we should select the optimal edges to clean to maximize the quality improvement. However, the noise of the crowdsourcing results will make the problem more complex. We prove that the problem is #P-hard and propose an efficient algorithm to derive the optimal edge. Our experimental results show that our proposed algorithm outperforms random-selection up to 22 times in quality improvement and each-edge-comparison way up to 5 times fast in elapsed time, which proves this algorithm is both effective and efficient.

Appendix

1.1 New Reachability

The reachability when crowd returns ‘yes’ is

$$ R_{G,q}^{y}=R_{G,q}+P_{q,e}^{*}({p_{e}^{y}}-p_{e}) $$

(24)

And the reachability when crowd returns ‘no’ is

$$ R_{G,q}^{n}=R_{G,q}+P_{q,e}^{*}({p_{e}^{n}}-p_{e}) $$

(25)

where ${p_{e}^{y}}$ is new edge probability if the crowd’s answer is ‘yes’, similar form for ${p_{e}^{n}}$.

Proof

First, we divide the whole G into two parts: graphs containing e (G_e) and graphs without e ($G_{\overline {e}}$). Assume for every $pg \in PG_{G_{e}}$, there is a corresponding $pg^{\prime } \in PG_{G_{\overline {e}}}$ such that all edges $E_{pg} \in E$ and $E_{pg^{\prime }} \in E$ are the same except that $E_{pg^{\prime }}$ doesn’t have e. Then, we have

$$ Pr(pg^{\prime})=\frac{(1-p_{e})Pr(pg)}{p_{e}} $$

(26)

By (2), $R_{G,q}^{0}$ can also be represented as:

$$ R_{G,q}^{0}=\sum\limits_{pg \in PG_{G_{e}}} Pr(pg)r_{q}^{pg} + Pr(pg^{\prime})r_{q}^{pg^{\prime}} $$

(27)

Furthermore, $G_{e}$ can be divided into $G(P_{e},P_{\overline e})$ and $G_{e}-G(P_{e},P_{\overline e})$. Obviously, for $pg \in PG_{G(P_{e}, P_{\overline e})}$, $r_{q}^{pg}= 1$ and $r_{q}^{pg^{\prime }}= 0$. For $pg \in PG_{G_{e}-G(P_{e}, P_{\overline e})}$, since whether e exists or not does not influence $r_{q}$, we have $r_{q}^{pg}=r_{q}^{pg^{\prime }}$.

Then, for $pg \in PG_{G_{e}}$, the new graph probability $Pr_{e \rightarrow y}(pg) = \frac {{p_{e}^{y}} Pr(pg)}{p_{e}}$ and $Pr_{e \rightarrow n}(pg) = \frac {{p_{e}^{n}} Pr(pg)}{p_{e}}$, where $e \rightarrow y$ ($e \rightarrow n$) represents edge e is cleaned to existence (nonexistence). For $pg \in PG_{G_{\overline e}}$, $Pr_{e \rightarrow y}(pg^{\prime }) = \frac {(1-{p_{e}^{y}}) Pr(pg)}{p_{e}}$ and $Pr_{e \rightarrow n}(pg^{\prime }) = \frac {(1-{p_{e}^{n}}) Pr(pg)}{p_{e}}$. Hence, for $pg \in PG_{G_{e}}$,

$$\begin{array}{@{}rcl@{}} Pr(pg) + Pr(pg^{\prime}) &=& Pr_{e \rightarrow y}(pg) + Pr_{e \rightarrow y}(pg^{\prime})\\ &=& Pr_{e \rightarrow n}(pg) + Pr_{e \rightarrow n}(pg^{\prime}) \end{array} $$

From above analysis, for $pg \in PG_{G_{e}-G(P_{e}, P_{\overline e})}$ we have

$$\begin{array}{@{}rcl@{}} Pr(pg)r_{q}^{pg} + Pr(pg^{\prime})r_{q}{pg^{\prime}}\\ = Pr_{e \rightarrow y}(pg)r_{q}(pg) + Pr_{e \rightarrow y}(pg^{\prime})r_{q}(pg^{\prime})\\ = Pr_{e \rightarrow n}(pg)r_{q}(pg) + Pr_{e \rightarrow n}(pg^{\prime})r_{q}(pg^{\prime}) \end{array} $$

(28)

For $pg \in PG_{G(P_{e}, P_{\overline e})}$, we have

$$ Pr(pg)r_{q}^{pg} + Pr(pg^{\prime})r_{q}{pg^{\prime}}=Pr(pg) $$

(29)

$$ Pr_{e \rightarrow y}(pg)r_{q}^{pg} + Pr_{e \rightarrow y}(pg^{\prime})r_{q}^{pg^{\prime}} = \frac{{p_{e}^{y}} Pr(pg)}{p_{e}} $$

(30)

$$ Pr_{e \rightarrow n}(pg)r_{q}^{pg} + Pr_{e \rightarrow n}(pg^{\prime})r_{q}^{pg^{\prime}} = \frac{{p_{e}^{n}} Pr(pg)}{p_{e}} $$

(31)

Then, $R_{G,q}^{0}$ is equal to sum of each element: (29) + the first line of (28); ${R_{q}^{y}}$ is equal to sum of each element: (30) + the second line of (28) and ${R_{q}^{n}}$ is equal to sum of each element: (31) + the third line of (28). Therefore,

$$\begin{array}{@{}rcl@{}} R_{G,q}^{y}-R_{G,q}^{0} &= &\sum\limits_{pg \in PG_{G(P_{e},P_{\overline e})}} \frac{({p_{e}^{y}} -p_{e})Pr(pg)}{p_{e}}\\ &=& ({p_{e}^{y}}-p_{e})P_{q,e}^{*}\\ R_{G,q}^{n}-R_{G,q}^{0} &= &\sum\limits_{pg \in PG_{G(P_{e},P_{\overline e})}} \frac{({p_{e}^{n}} -p_{e})Pr(pg)}{p_{e}}\\ &= &({p_{e}^{n}}-p_{e})P_{q,e}^{*} \end{array} $$

□

1.2 NECP is # P-hard

Proof

To prove NECP is #P-hard, we reduce ECP in [17] to NECP as Xin Lin [17] has proven ECP (17) is #P-hard. ECP corresponds to situation where a crowd is exactly accurate.

In detail, for B to-clean edges, there are totally $2^{B}$ cleaning results: $CS={000...000,000...001,000...011,......,111...111}$, each element of which is a B-bits sequence where each bit represents the cleaning result of the corresponding edge. For simplicity, we assume all edges’ existential probabilities of an uncertain graph G are the same value p.

For ECP, the expected query result quality after cleaning is

$$ Q=(1-p)^{B} Q_{1} + (1-p)^{B-1} p Q_{2} + (1-p)^{B-2} p^{2} Q_{3} + ... + p^{B} Q_{2^{B}} $$

(32)

where $Q_{1}$, $Q_{2}$, $Q_{3}$, ..., $Q_{2^{B}}$ are the corresponding query result quality of CS.

For NECP, the expected query result quality after cleaning is

$$\begin{array}{@{}rcl@{}} Q^{N} & =&(1-Pr(C_{r}))^{B} Q_{1} + (1-Pr(C_{r}))^{B-1} Pr(C_{r}) Q_{2}\\ &&+ (1-Pr(C_{r}))^{B-2} (Pr(C_{r}))^{2} Q_{3} + ... + (Pr(C_{r}))^{B} Q_{2^{B}} \end{array} $$

(33)

Simplifying (5), we have

$$\begin{array}{@{}rcl@{}} Pr(C_{r})&=&(2P_{c} - 1)p + 1-P_{c}\\ 1-Pr(C_{r})&=&-(2P_{c} - 1)p + P_{c} \end{array} $$

We can see $Pr(C_{r})$ is the linear expression with respect to p. Therefore, the fact computing $Q^{N}$ implies that we can compute Q shows solving NECP is $\#$P-hard. □

1.3 Calculating $P_{q,e}^{*}$

Proposition 2: Calculating $P_{q,e}^{*}$ can be reduced to calculating reachability.

Proof

Similar to calculating reachability $R_{q}$ in (3) which in theory needs to enumerate all $sg \in SG_{G,q}$, calculating $P_{q,e}^{*}$ accordingly needs to enumerate all $sg \in SG_{G(P_{e},P_{\overline e}),q}$. Also, it is impossible to identify $SG_{G(P_{e},P_{\overline e})}$, even more troublesome than enumerating $SG_{G,q}$.

In Example 2, we have mentioned Monte-carlo method can approximate the result of $Pr(p_{1} \vee p_{2} \vee p_{3} \vee ... \vee p_{n})$ (assume $n=|AP|$). Similarly, Monte-carlo method is also applicative to calculating $P_{q,e}^{*}$. First, we denote all paths passing through (not passing though) edge e by $AP_{e}$ ($AP_{\overline e}$). Then, we have

$$ \sum\limits_{sg \in SG_{G(P_{e},P_{\overline e})}} Pr(sg)=Pr(\bigvee_{p \in AP_{e}})(1-Pr(\bigvee_{p \in AP_{\overline e}})) $$

(34)

By (34), we just need to respectively compute two parts: $Pr(\bigvee _{p \in AP_{e}})$ and $Pr(\bigvee _{p \in AP_{\overline e}})$, calculating each of which is equivalent to calculating reachability. □