Consistent updating of databases with marked nulls

Abstract

This paper revisits the problem of consistency maintenance when insertions or deletions are performed on a valid database containing marked nulls. This problem comes back to light in real-world linked data or RDF databases when blank nodes are associated with null values. This paper proposes solutions for the main problems one has to face when dealing with updates and constraints, namely update determinism, minimal change and leanness of an RDF graph instance. The update semantics is formally introduced and the notion of core is used to ensure a database as small as possible (i.e.   the RDF graph leanness). Our algorithms allow the use of constraints such as tuple-generating dependencies, offering a way for solving many practical problems.

Appendices

Proof of Proposition 6.1

Proposition 6.1

Let $\Delta = ({\mathfrak {D}},{\mathbb {C}})$ be a database and ${\textsf {iRequest}}$ a finite set of facts. If Algorithm 1 returns $\Delta '=({\mathfrak {D}}', {\mathbb {C}})$ where ${\mathfrak {D}}' \ne {\mathfrak {D}}$, then ${\mathfrak {D}}'$ satisfies the following statements:

• 1. Effectiveness: (a) for every $\varphi$ in ${\textsf {iRequest}}$, ${\mathfrak {D}}'$ contains an instance of $\varphi$, (b) $R\_core({\mathfrak {D}}')={\mathfrak {D}}'$ and (c) ${\mathfrak {D}}' \models {\mathbb {C}}$.

• 2. Monotonicity:${\mathfrak {D}}\sqsubseteq {\mathfrak {D}}'$.

• 3. Minimal change: For every $\varphi$ in ${\mathfrak {D}}'$ not in ${\mathfrak {D}}$ and not isomorphic to an atom in ${\textsf {iRequest}}$, ${\mathfrak {D}}' {\setminus } \{\varphi \} \not \models {\mathbb {C}}$.

Proof

1(a) Let $\varphi$ in ${\textsf {iRequest}}$. Then, $\varphi$ is in $T^*_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})={\mathfrak {D}}_1$ (line 3 of Algorithm 1). Thus, if $\varphi$ is not in ${\mathfrak {D}}'$, then $\varphi$ has been instantiated due to the computation of the restricted core on line 4 of Algorithm 1. Therefore, in this case $R\_core({\mathfrak {D}}_1)={\mathfrak {D}}'$ contains an instance of $\varphi$.

1(b) The fact that $R\_core({\mathfrak {D}}')={\mathfrak {D}}'$ is an obvious consequence of Algorithm 1.

1(c) If ${\mathfrak {D}}_1$ contains no null N such that $\delta (N) \ge k$, then ${\mathfrak {D}}' \models {\mathbb {C}}$ holds because, by Proposition 4.1, ${\mathfrak {D}}_1$ satisfies ${\mathbb {C}}$ and thus, by Lemma 3.1, so does its core which is equal to its restricted core due to Lemma 4.2(1).

Now, assume that ${\mathfrak {D}}_1$ contains at least one null N with $\delta (N) \ge \delta _{max}$ and that ${\mathfrak {D}}' \not \models {\mathbb {C}}$. This means that there exists $c:B(\mathbf{X },\mathbf{Y }) \Rightarrow L(\mathbf{X },\mathbf{Z })$ in ${\mathbb {C}}$ such that ${\mathfrak {D}}' \not \models c$. Thus, there exists a homomorphism h such that $h(body(c)) \subseteq {\mathfrak {D}}'$ and, for any extension $h'$ of h to the variables occurring in $\mathbf{Z }$, $h'(head(c))$ is not in ${\mathfrak {D}}'$. Denoting h(body(c)) by $B(\alpha , \beta )$, this means that $B(\alpha , \beta ) \subseteq {\mathfrak {D}}'$ and that ${\mathfrak {D}}'$ contains no atom of the form $L(\alpha , \_)$. Since ${\mathfrak {D}}'\subseteq {\mathfrak {D}}_1$, we have $h(body(c)) \subseteq {\mathfrak {D}}_1$. As for every null N in h(body(c)), $\delta (N) < \delta _{max}$, ${\mathfrak {D}}_1$ contains an atom of the form $L(\alpha , \nu )$ where $\nu$ is made of nulls whose degree is $\delta _{max}$. Therefore, $L(\alpha , \nu )$ has been removed during the core processing, meaning that ${\mathfrak {D}}_1$ also contains an instance $L(\alpha ', \gamma )$ of $L(\alpha , \nu )$. Denoting by h the homomorphism computing the restricted core ${\mathfrak {D}}'$, ${\mathfrak {D}}_1$ contains $h(L(\alpha , \mathbf{N }))=L(\alpha ', \gamma )$ along with all atoms in $B(\alpha , \beta )$ or in $h(B(\alpha , \beta ))$. The case $h(\alpha ) \ne \alpha$ is not possible because this implies that $B(\alpha ,\beta ) \subseteq {\mathfrak {D}}'$ does not hold. Hence, $h(\alpha ) = \alpha$, and so, $h(L(\alpha ,\nu ))=L(\alpha , \gamma )$ is in ${\mathfrak {D}}'$, a contradiction to the fact that ${\mathfrak {D}}'$ contains no atom of the form $L(\alpha , \_)$. Therefore, this part of the proof is complete.

2. In Algorithm 1, on line 3, we have ${\mathfrak {D}}\subseteq T^*_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})$, thus ${\mathfrak {D}}\sqsubseteq T^*_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})$ holds. As earlier noticed, $T^*_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}}) \sqsubseteq R\_core(T^*_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}}))$, which implies by transitivity that on line 4, ${\mathfrak {D}}\sqsubseteq {\mathfrak {D}}_1$ holds. Therefore, the proof of this part is complete.

3. Let $\varphi$ in ${\mathfrak {D}}'$, not in ${\mathfrak {D}}$ and not isomorphic to an atom in ${\textsf {iRequest}}$, and assume that ${\mathfrak {D}}' {\setminus } \{\varphi \} \models {\mathbb {C}}$. Then, $\varphi$ has been generated during the computation of ${\mathfrak {D}}_1=T^*_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})$. Hence, there exists $c:B(\mathbf{X },\mathbf{Y }) \Rightarrow L(\mathbf{X },\mathbf{Z })$ in ${\mathbb {C}}$ and a homomorphism h such that $h(c) \subseteq {\mathfrak {D}}_1$ and $h(head(c))=\varphi$.

Since ${\mathfrak {D}}' {\setminus } \{\varphi \} \models c$, c cannot be full, meaning that c has existentially quantified variables and that ${\mathfrak {D}}' {\setminus } \{\varphi \}$ contains an atom of the form $L(\alpha , \gamma )$ where $\alpha =h(\mathbf{X })$ and $\gamma \ne h(\mathbf{Z })$. Let p be the least integer such that $h(body(c)) \subseteq T^p_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})$ and $\varphi \not \in T^p_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})$. Since $\varphi$ is in ${\mathfrak {D}}'$, for every null N in $\varphi$, $\delta (N) < \delta _{\max }$. Thus, when computing $T^{p+1}_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})$, all nulls occurring in the instantiated body of the corresponding constraint have a degree strictly less than $\delta _{\max }-1$, meaning that $\varphi$ is generated as the atom $L(\alpha , \mathbf{N })$ where $\mathbf{N }$ is a vector of fresh nulls whose degrees are strictly less than $\delta _{max}$. As a consequence, by definition of $T_\textsf {fw}$, $L(\alpha , \gamma )$ is not in $T^p_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})$ (otherwise $L(\alpha , \mathbf{N })$ would not be generated), meaning that $L(\alpha , \gamma )$ is generated at a subsequent computation step q (i.e.   $q>p+1$).

Now, let $h_0$ be the homomorphism defined by $h_0(\mathbf{N })=\gamma$ and $h_0(N)=N$ for every N not occurring in $\mathbf{N }$. We first notice that $h_0$ is restricted because, since $L(\alpha , \gamma )$ is in ${\mathfrak {D}}'$ (this so because $L(\alpha , \gamma ) \in {\mathfrak {D}}'{\setminus } \{\varphi \}$), all nulls in $\gamma$ have a degree strictly less than $\delta _{\max }$. Denoting by $Atoms(\mathbf{N })$ the set of atoms in ${\mathfrak {D}}_1$ in which at least one null in $\mathbf{N }$ occurs, we show that for every A in $Atoms(\mathbf{N })$, $h_0(A)$ is also in ${\mathfrak {D}}_1$. Consider now the step m where the first atom A in $Atoms(\mathbf{N })$ different from $\varphi$ appears in $T^m_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})$. During the computation, a constraint $c_A$ is instantiated into $h_A(c_A)$ so as to contain $\varphi$ in its body (the only atom so far in which $\mathbf{N }$ occurs), in order to generate $A=h_A(head(c_A))$. Hence, as soon as $h_0(\varphi )$ appears in $T^q_\textsf {fw}({\mathfrak {D}}\cup {\textsf {iRequest}})$, $h_0(h_A(c_A))$ applies and generates $h_0(A)$, showing that $h_0(A)$ is in ${\mathfrak {D}}_1$. Since this reasoning applies step by step for every A in $Atoms(\mathbf{N })$, this implies that for every A in $Atoms(\mathbf{N })$, $h_0(A)$ is in ${\mathfrak {D}}_1$.

Hence, in the computation of $R\_core({\mathfrak {D}}_1)$, $\mathbf{N }$ can be instantiated into $\gamma$ using $h_0$ (or one of its extensions), which implies that $\varphi$ is not in ${\mathfrak {D}}'$, a contradiction to our hypothesis that $\varphi$ is in ${\mathfrak {D}}'$. Therefore, the proof is complete. $\square$

Proof of Proposition 6.2

Proposition 6.2

Let $\Delta = ({\mathfrak {D}},{\mathbb {C}})$ be a database and ${\textsf {dRequest}}$ a finite set of instantiated atoms where every null occurs at most once. Algorithm 2 returns $\Delta '=({\mathfrak {D}}', {\mathbb {C}})$ where ${\mathfrak {D}}'$ satisfies the following statements:

• 1. Effectiveness (a) for every $\varphi$ in ${\textsf {dRequest}}$, ${\mathfrak {D}}'$ contains no atom isomorphic to $\varphi$, (b) $R\_core({\mathfrak {D}}')={\mathfrak {D}}'$, and (c) ${\mathfrak {D}}' \models {\mathbb {C}}$.

• 2. Monotonicity${\mathfrak {D}}' \sqsubseteq {\mathfrak {D}}$.

Proof

1(a) Let $\varphi$ in ${\textsf {dRequest}}$ and suppose an atom $\varphi _1$ isomorphic to $\varphi$ that belongs to ${\mathfrak {D}}'$, when running Algorithm 2. If $\varphi _1$ is in ToDel on line 1, then for $\varphi _1$ to belong to ${\mathfrak {D}}'$, it must be that on line 5, $\varphi _1$ is in ${\mathfrak {D}}_1$ and thus that, on line 7, $\varphi _1$ is in Back. If $\varphi _1$ is not in ToDel, then $\varphi _1$ is not in ${\mathfrak {D}}$ showing that this atom appears in ${\mathfrak {D}}_1$ on line 4, the only step in Algorithm 2 where atoms are added to the instance. In this case, we again have that $\varphi _1$ is in Back. But then, on line 11, $\varphi _1$ belongs to $T^*_\textsf {bw}(Disallowed \cup Back, {\mathfrak {D}}_1)$ showing that $\varphi _1$ cannot belong to ${\mathfrak {D}}_2$. Since ${\mathfrak {D}}'$ is a subset of ${\mathfrak {D}}_2$, it is not possible that $\varphi _1$ belongs to ${\mathfrak {D}}'$, a contradiction that completes this part of the proof.

1(b) $R\_core({\mathfrak {D}}')={\mathfrak {D}}'$ is an obvious consequence of the statements on lines 5 and 12 of Algorithm 2.

1(c) If ${\mathfrak {D}}'$ is output by Algorithm 2 on line 9, then Disallowed is empty, meaning that every null N in ${\mathfrak {D}}_1$ is such that $\delta (N) < \delta _{max}$. Thus, by Proposition 4.1, we have $T^*_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)\models {\mathbb {C}}$, implying that, by Lemma 3.1, ${\mathfrak {D}}' \models {\mathbb {C}}$. Otherwise, ${\mathfrak {D}}'$ is output on line 13, in which case Proposition 4.2 shows that ${\mathfrak {D}}_2 \models {\mathbb {C}}$, and thus, by Lemma 3.1, we have ${\mathfrak {D}}' \models {\mathbb {C}}$, which completes this part of the proof.

2. To prove that ${\mathfrak {D}}' \sqsubseteq {\mathfrak {D}}$, we first show that for ${\mathfrak {D}}_1=R\_core(T^*_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel))$ as computed on line 5 of Algorithm 2, we have ${\mathfrak {D}}_1 \sqsubseteq {\mathfrak {D}}$. To this end, let us first prove by induction that $T^*_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel) \sqsubseteq {\mathfrak {D}}$. The base case of the induction, that is, ${\mathfrak {D}}{\setminus } ToDel \sqsubseteq {\mathfrak {D}}$, holds because ${\mathfrak {D}}{\setminus } ToDel \subseteq {\mathfrak {D}}$. Then, for a given integer p, assuming that $T^k_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel) \sqsubseteq {\mathfrak {D}}$ for every $0 \le k \le p$, we show that $T^{p+1}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel) \sqsubseteq {\mathfrak {D}}$. To see this, let $h^p$ be the homomorphism such that $h^p(T^p_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)) \subseteq {\mathfrak {D}}$ and consider the two cases as follows:

• Assume first that $T^{p+1}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)= T^{p}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)$. This means that the least fixed point has been reached, in which case $h^{k+1}$ is equal to $h^k$.

• When $T^{p+1}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)\ne T^{p}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)$, for every null N in $T^{p}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)$ let $h^{p+1}(N)= h^p(N)$. Then, considering a null N in $T^{p+1}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)$ but not in $T^{p}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)$ implies that $\delta (N)\le \delta _{\max }$ and that there exists c in ${\mathbb {C}}$ and a homomorphism h such that $h(body(c)) \subseteq T^{p}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)$ and N occurs in $\varphi =h(head(c))$. In this case, we have $h^p(h(body(c)) \subseteq {\mathfrak {D}}$ because, by our induction hypothesis, $h^p(T^{p}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel))\subseteq {\mathfrak {D}}$. Since ${\mathfrak {D}}\models {\mathbb {C}}$, c applies in ${\mathfrak {D}}$, meaning that ${\mathfrak {D}}$ contains an atom $A=v(\varphi )$ where v is defined over the nulls $\nu$ in $\varphi$ by $v(\nu )=t$ where t is the corresponding term in A. Notice that v is well defined because any null $\nu$ in $\varphi$ either occurs in $T^{p}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)$, in which case $v(\nu )=\nu$, or $\nu$ is a fresh null such N, in which case $v(\nu )$ is uniquely defined. In this case, let $h^{p+1}(N)$ be v(N). We thus obtain a homomorphism $h^{p+1}$ such that $h^{p+1}(T^{p+1}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)) \subseteq {\mathfrak {D}}$, showing that $T^{p+1}_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel) \sqsubseteq {\mathfrak {D}}$.

Therefore, $T^*_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel) \sqsubseteq {\mathfrak {D}}$ holds. Moreover, since ${\mathfrak {D}}_1=R\_core(T^*_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel))$, ${\mathfrak {D}}_1\sqsubseteq T^*_\textsf {fw}({\mathfrak {D}}{\setminus } ToDel)$ holds, implying ${\mathfrak {D}}_1 \sqsubseteq {\mathfrak {D}}$. Thus, if the test on line 9 succeeds, we trivially have ${\mathfrak {D}}' \sqsubseteq {\mathfrak {D}}$. Otherwise, since ${\mathfrak {D}}' \subseteq {\mathfrak {D}}_2 \subseteq {\mathfrak {D}}_1$, ${\mathfrak {D}}' \sqsubseteq {\mathfrak {D}}$ holds as well, and the proof is complete. $\square$