• Entities have transitivity: if $a$ and $c$ are the same things, $a$ and $d$ are not the same things, then $a$ and $c$ can not be the same things.
• We already have probabilities $p$ for each pairs telling they are matching or not, e.g. $p(r_a, r_b) = 0.46$

Only six pairs is needed for querying to Oracle:

$(r_a, r_b)$, $(r_a, r_c)$, $(r_d, r_e)$, $(r_a, r_d)$, $(r_a, r_f )$, $(r_d , r_f )$

Previous:

$(r_a,r_b)$, $(r_a,r_c)$, $(r_d,r_e)$, $(r_a,r_d)$, $(r_a,r_f)$, $(r_d,r_f)$.

Another ordering:

$(r_a, r_d)$, $(r_a, r_f )$, $(r_d, r_f )$, $(r_d, r_e)$, $(r_a,r_b)$, $(r_a,r_c)$.

Considering pairs with probabilities:

Matching:

$p(r_d, r_e) = 0.80$, $p(r_b, r_c) = 0.60$, $p(r_a, r_c) = 0.54$, $p(r_a, r_b) = 0.46$.

Non-matching: $p(r_a, r_d) = 0.84$, $p(r_d, r_f) = 0.81$, $p(r_c, r_e) = 0.72$, $p(r_a, r_e)=0.65$, $p(r_c, r_d)=0.59$, $p(r_e, r_f) = 0.59$, $p(r_a, r_f)=0.55$, $p(r_b, r_d) = 0.51$, $p(r_b, r_e) = 0.46$, $p(r_c, r_f) = 0.45$, $p(r_b, r_f) = 0.29$

A total of 7 pairs will be queried $(r_a,r_d)$, $(r_d,r_f)$, $(r_d,r_e)$, $(r_c,r_e)$, $(r_b,r_c)$, $(r_a,r_f)$, $(r_a,r_c)$, if using minimizing # queries strategy

In an entity graph,

with $E^+$ as the ground truth of positive edges,

and $E_T^+$ as the positive edges we can get with $t = |T|$ queries and $T$ is the set of the query.

Then the progressive recall will be defined as

$precall(t) =\sum_{t'=1}^t recall(t')$, and $precall^+(t)=\sum_{t'=1}^t recall^+(t')$.

A $recall$ of $T$ can be defined as $recall(t) = \frac{|E^+_T|}{|E^+|}$, and $recall^+(t) = \frac{|E^+_{T^+}|}{|E^+|}$

For matching edges, with probability $(1- \frac{\alpha}{n} )$, $\alpha \gt 0$ the score is above 0.7, and the remaining probability mass is distributed uniformly in $[0, 0.7)$.

For non-matching edges, with probability $(1-\frac{\beta}{n})$ , $\beta \gt 0$, the score is below 0.1 and the remaining n probability mass is distributed uniformly in $(0.1,1]$.

$O(log^2n)-approximation$ under edge noise model

$\Omega(n)-approximation$ under progressive recall

A total of 7 pairs will be queried $(r_a,r_d)$, $(r_d,r_f)$, $(r_d,r_e)$, $(r_c,r_e)$, $(r_b,r_c)$, $(r_a,r_f)$, $(r_a,r_c)$.

$O(log^2n)-approximation$ under edge noise model

$\Sigma(\sqrt n)$ under progressive recall

A total of 7 pairs will be queried $(r_e,r_d)$, $(r_a,r_d)$, $(r_c, r_e)$, $(r_c, r_a)$, $(r_f, r_d)$. $(r_f, r_a)$, $(r_b, r_c)$.

A maximized PRecall strategy $S^*$ is:

1. Presume we already know the ground truth
2. For each cluster $C_k$ in the graph (which is a same real-world entity), we ask $|C_k|$ questions to the Oracle.
3. Questions are asked in the sequence of $C_1, C_2, ..., C_k$, where $|C_k|$ has a decreasing order.
4. Finally asking $\binom k2$ non-matching questions.
5. Ideally, $s^*$ will ask $n - k + \binom k2$ questions

$b_e$ is the benefit of a edge between $u$ and $v$, calculated as $|c_T(u) | * | c_T(v) |* p(u,v)$

$b_v$ is the benefit of a node when adding this node to the $P$ set.

$b_{vc}(v,c) = p_v(v,c) * |c|$

$b_v(v,P) = \max b_{vc}(v,c)$

A total of 6 pairs will be queried $(r_a,r_d)$, $(r_d,r_f)$, $(r_d,r_e)$, $(r_c,r_e)$, $(r_b,r_c)$, $(r_a,r_c)$.

A total of 6 pairs will be queried $(r_a,r_d)$, $(r_d,r_f)$, $(r_d,r_e)$, $(r_c,r_e)$, $(r_b,r_c)$, $(r_a,r_c)$.

$b_e(r_a,r_c)$ = 2 * 1 * 0.54 = 1.08

$b_e(r_a, r_c)$ = 1 * 1 * 0.55 = 0.55

if $w = 1$:

if $\tau = n$ and $\theta = 0$ then $s_{hybrid} = s_{vesd}$

if $\tau = 0$ or $\theta = n$, then $s_{hybrid} = s_{wang}$

n: num of records

k: num of entities (clusters)

k': num of non-singleton entities (clusters)

$|c_1|$: largest clusters

$t^*_r$: size of the spanning forest

ER: dirty or clean-clean

origin: real or synthetic of data

R: # record pairs

$n'$: $\binom n2 = \lfloor R \rfloor$

$s_{edge}$

$s_{hybrid}$

$s_{hybrid}$ is better than $s_{edge}$ with same parameter, but...