Efficient two-dimensional Haar$^+$ synopsis construction for the maximum absolute error measure

Abstract

Several wavelet synopsis construction algorithms were previously proposed for optimal Haar$^+$ synopses. Recently, we proposed the OptExtHP-EB algorithm to find an optimal one-dimensional $\hbox {Haar}^+$ synopsis. By utilizing the novel properties of optimal synopses, OptExtHP-EB represents the set of optimal synopses in a node of a $\hbox {Haar}^+$ tree by a set of extended synopses. While it is much faster than the previous $\hbox {Haar}^+$ synopsis construction algorithms, it can handle only one-dimensional data. In this paper, we propose the OptExtHP-EB2D algorithm for two-dimensional $\hbox {Haar}^+$ synopses by extending OptExtHP-EB. While a one-dimensional $\hbox {Haar}^+$ tree has only two child nodes and three coefficients in a node, a two-dimensional $\hbox {Haar}^+$ tree is much more complex in that it has four child nodes and seven coefficients per node. Thus, for each possible subset of the coefficients selected in a node, we develop the efficient methods to compute a set of optimal synopses denoted by extended synopses. Our experiments confirm the effectiveness of our proposed OptExtHP-EB2D algorithm.

Appendix

Proof of Lemma 5

For a pair of ranges $r\in R$ and $r'\in R'$, $R_{\mathsf {req}}(r\,{\bowtie }_{\delta }r')=\{r \oplus z\cdot \delta \,|\, mat_\mathsf {max}(r,r') \le z\cdot \delta \le mat_\mathsf {min}(r,r'), z\in {\mathbb {Z}}\}$. Let $r_1$ and $r'_1$ be the front ranges of R and $R'$, respectively. Then, for every range $r_{\delta }\in R_{\mathsf {req}}(r\,{\bowtie }_{\delta }r')$ with $r\in R$ and $r'\in R'$, we have the followings.

$$\begin{aligned}&r_{\delta }.{\mathsf {min}}= r.{\mathsf {min}}+ z\cdot \delta&(\text {since }r_{\delta }=r\oplus z\cdot \delta )\\&\quad \ge r.{\mathsf {min}}+ mat_\mathsf {max}(r_{},r')&(\text {due to }z\cdot \delta \ge mat_\mathsf {max}(r_{},r'))\\&\quad = r.{\mathsf {min}}+ (r'.{\mathsf {max}}\mathsf {-} r_{}.{\mathsf {max}})/2&(\text {by the value of } mat_\mathsf {max}(r_{},r'))\\&\quad \ge r_{1}.{\mathsf {min}} + (r'_1.\mathsf {max}\mathsf {-} r_1.\mathsf {max})/2&(\text {since }R\text { and }R'\text { are }\delta \text {-shifted range sets}) \\&\quad = r_{1}.{\mathsf {min}} + mat_\mathsf {max}(r_1,r'_1)&(\text {by the value of } mat_\mathsf {max}(r_{1},r'_1))\\&r_{\delta }.{\mathsf {max}} = r.\mathsf {max}+ z\cdot \delta&(\text {since }r_{\delta }=r\oplus z\cdot \delta )\\&\quad \ge r_{}.{\mathsf {max}}+ mat_\mathsf {max}(r_{},r')&(\text {due to }z\cdot \delta \ge mat_\mathsf {max}(r_{},r')) \\&\quad = (r'.{\mathsf {max}}+ r_{}.{\mathsf {max}})/2&(\text {by }mat_\mathsf {max}(r_{},r')= (r'.{\mathsf {max}}\mathsf {-} r_{}.{\mathsf {max}})/2)\\&\quad \ge (r'_{1}.{\mathsf {max}}+ r_{1}.{\mathsf {max}})/2&(\text {since }R\text { and }R'\text { are }\delta \text {-shifted range set}) \\&\quad = r_{1}.\mathsf {max} + mat_\mathsf {max}(r_1,r'_1)&(\text {by }mat_\mathsf {max}(r_1,r'_1)= (r'_{1}.{\mathsf {max}}\mathsf {-} r_{1}.{\mathsf {max}})/2) \end{aligned}$$

Since $r_{\delta }.\mathsf {min}\ge (R_{\mathsf {F}}.\mathsf {front}).\mathsf {min}$ and $r_{\delta }.\mathsf {max}\ge (R_{\mathsf {F}}. \mathsf {front}).\mathsf {max}$, $R_{\mathsf {req}}(R {\bowtie }_{\delta } R').\mathsf {front}=R_{\mathsf {F}}.\mathsf {front}$. Similarly, we can show that $R_{\mathsf {req}}(R\,{\bowtie }_{\delta }R').\mathsf {rear}=R_{\mathsf {rear}}.\mathsf {rear}$.

Since $R_{\mathsf {req}}(r_{}\,{\bowtie }_{\delta }r')$ is a $\delta$-shifted range set, there always exists a range $r''$ in $R\,{\bowtie }_{\delta }R'$ such that $r''.\mathsf {min}$ is located in $[(R_{\mathsf {req}}.\mathsf {front}).\mathsf {min},(R_{\mathsf {rear}}.\mathsf {rear}).\mathsf {min}]$. Thus, $R_\mathsf {req}(R\,{\bowtie }_{\delta }R')$ is also a $\delta$-shifted range set whose front and rear ranges are $R_{\mathsf {F}}.\mathsf {front}$ and $R_{\mathsf {R}}.\mathsf {rear}$, respectively. $\square$

Proof of Lemma 6

We prove each case as follows:

(a) When every range in R does not contain any range in $R'$: We break the proof into two subcases.

(a-1) When $r_{1}.{\mathsf {min}} <r'_{1}.{\mathsf {min}}$: If $r_{m}.{\mathsf {min}} \ge r'_{1}.{\mathsf {min}}$ holds, since R is a $\delta$-shifted range set, there always exists a range $r_j \in R$ such that $r_{j}.{\mathsf {min}} =r'_{1}.{\mathsf {min}}$. Then, $r_j$ contains $r'_1$, it is a contradiction. Thus, we have $r_{m}.{\mathsf {min}} <r'_{1}.{\mathsf {min}}$. In this case, $r_m$ and $r'_1$ are the pair of the ranges whose minimum values are the closest among all pairs of the ranges in R and $R'$, respectively.

For a pair of ranges $r_j\in R$ and $r'_k\in R'$, let $\mathtt{mbr}(\{r_j, r'_k\})=[e_\mathsf {min}, e_\mathsf {max}]$. Then, we have the property $e_\mathsf {min} \le \min (r_{m}.{\mathsf {min}}, r'_{1}.{\mathsf {min}})$ from the following inequalities.

$$\begin{aligned} e_\mathsf {min}&= \min (r_{j}.{\mathsf {min}},r'_{k}.{\mathsf {min}})&(\text {by Definition}~6) \\&\le \min (r_{m}.{\mathsf {min}},r'_{1}.{\mathsf {min}})&(\text {since }r_j.\mathsf {min}\le r_{m}.{\mathsf {min}}\le r'_{1}.{\mathsf {min}}) \end{aligned}$$

Symmetrically, we can show $\max (r_{m}.{\mathsf {max}},r'_{1}.{\mathsf {max}}) \le e_\mathsf {max}$.

Since $e_\mathsf {min} \le \min (r_{m}.{\mathsf {min}}, r'_{1}.{\mathsf {min}})$, $e_\mathsf {max} \ge \max (r_{m}.{\mathsf {max}}, r'_{1}.{\mathsf {max}})$ and $[\min (r_{m}.{\mathsf {min}},r'_{1}.{\mathsf {min}}), \max (r_{m}.{\mathsf {max}},r'_{1}.{\mathsf {max}})]=\mathtt{mbr}(\{r_m, r'_1\})$, $\mathtt{mbr}(\{r_j, r'_k\})$ always contains $\mathtt{mbr}(\{r_m, r'_1\})$. That is, every range in $R\,{\bowtie }_\mathsf {mbr}R'$ contains $\mathtt{mbr}(\{r_m, r'_1\})$. Thus, by Definition 6, the required range set of $R\,{\bowtie }_\mathsf {mbr}R'$ becomes $\{\mathtt{mbr}(\{r_m, r'_1\})\}$.

(a-2) When $\varvec{r_{1}.{\mathsf {min}} \ge r'_{m}.{\mathsf {min}}}$: We omit the proof since we can show similarly to the case of (a-1).

(b) When there exists a range in R containing a range in $R'$: Let $r'_{k_\mathsf {F}}$ be the range in $R'$ which has the smallest minimum value among all ranges contained by $r_{j_\mathsf {F}}$. Then, we first show that, for every pair of ranges $r_j\in R$ and $r'_k\in R'$ satisfying $j\le j_\mathsf {F}$ and $k\le k_\mathsf {F}$, $\mathtt{mbr}(\{r_j, r'_k\})$ contains $\mathtt{mbr}(\{r_{j_\mathsf {F}}, r'_{k_\mathsf {F}}\})$. It implies that such $\mathtt{mbr}(\{r_j, r'_k\})$s are not included in $R_\mathsf {req}(R\,{\bowtie }_\mathsf {mbr}R')$. We consider two subcases of when (b-1) $j_\mathsf {F}=1$ and (b-2) $j_\mathsf {F}>1$.

(b-1) When $j_\mathsf {F}=1$: Since $\mathtt{mbr}(\{r_{1}, r'_\mathsf {k}\})$ always contains $r_{1}$ and $\mathtt{mbr}(\{r_{1}, r'_{k_\mathsf {F}}\})=r_{1}$, all $\mathtt{mbr}(\{r_{1}, r'_\mathsf {k}\})$s with $k\le k_\mathsf {F}$ contain $\mathtt{mbr}(\{r_{1}, r'_{k_\mathsf {F}}\})$.

(b-2) When $j_\mathsf {F}>1$: If $r_{j_\mathsf {F}}.\mathsf {max}<r'_1.\mathsf {max}$, $r_{j_\mathsf {F}}$ cannot contain any range in $R'$. If $r_{j_\mathsf {F}}.\mathsf {max}>r'_1.\mathsf {max}$, $r_{j_\mathsf {F}-1}$ contains $r'_1$ and it is a contradiction. Thus, we get $r_{j_\mathsf {F}}.\mathsf {max}=r'_1.\mathsf {max}$ and $k_\mathsf {F}=1$. Then, for every $\mathtt{mbr}(\{r_j, r'_1\})$ with $j\le j_\mathsf {F}$, since $\mathtt{mbr}(\{r_j, r'_1\}).\mathsf {max}=r'_1.\mathsf {max}=\mathtt{mbr}(\{r_{j_\mathsf {F}}, r'_1\}).\mathsf {max}$ and $\mathtt{mbr}(\{r_j, r'_1\}).\mathsf {min}=r_j.\mathsf {min}\le \mathtt{mbr}(\{r_{j_\mathsf {F}}, r'_1\}).\mathsf {min}$, $\mathtt{mbr}(\{r_j, r'_1\})$ contains $\mathtt{mbr}(\{r_{j_\mathsf {F}}, r'_{1}\})$.

For every pair of ranges $r_j\in R$ and $r'_k\in R'$ satisfying $j\ge j_\mathsf {R}$ and $k\ge k_\mathsf {R}$, we can symmetrically show that $\mathtt{mbr}(\{r_j, r'_k\})$ contains $\mathtt{mbr}(\{r_{j_\mathsf {R}}, r'_{k_\mathsf {R}}\})$. Thus, we need to consider $\mathtt{mbr}(\{r_j, r'_k\})$s with $j_\mathsf {F}\le j\le j_\mathsf {R}$ and $k_\mathsf {F}\le k\le k_\mathsf {R}$. Since a range $r_j\in R$ contains a range $r'_k\in R'$, $\mathtt{mbr}(\{r_j, r'_k\})=r_j$ and it is contained by $\mathtt{mbr}(\{r_j, r'_{k'}\})$ with every $r'_{k'}\in R'$, the required range set of $\{\mathtt{mbr}(\{r_j, r'_k\})\,|\,j_\mathsf {F}\le j\le j_\mathsf {R},k_\mathsf {F}\le k\le k_\mathsf {R}\}$ becomes $\{r_j\,|\,j_\mathsf {F}\le j\le j_\mathsf {R}\}$. $\square$

Proof of Lemma 7

For a pair of $r_1\in R$ and $r_2\in R$, if $r_1$ contains $r_2$, since $\mathtt{mbr}(\{r_1, r_3\})$ contains $\mathtt{mbr}(\{r_2, r_3\})$ with every range $r_3\in R'$ by Definition 6, $\mathtt{mbr}(\{r_1, r_2\})\not \in R_\mathsf {req}(\{\mathtt{mbr}(\{r, r'\})\,|\,r\in R,r'\in R'\})$ by Definition 7. Thus, $R_\mathsf {req}(\{\mathtt{mbr}(\{r, r'\})\,|\,r\in R,r'\in R'\})=R_\mathsf {req}(\{\mathtt{mbr}(\{r, r'\})\,|\,r\in R_\mathsf {req}(R),r'\in R'\})$. Symmetrically, we can show $R_\mathsf {req}(\{\mathtt{mbr}(\{r, r'\})\,|\,r\in R_\mathsf {req}(R),r'\in R'\})=R_\mathsf {req}(\{\mathtt{mbr}(\{r, r'\})\,|\,r\in R_\mathsf {req}(R),r'\in R_\mathsf {req}(R')\})$. Thus, $R_\mathsf {req}(R\,{\bowtie }_\mathsf {mbr}R')=R_\mathsf {req}(R_\mathsf {req}(R)\,{\bowtie }_\mathsf {mbr}R_\mathsf {req}(R'))$. We can similarly prove $R_\mathsf {req}(R\,{\bowtie }_{\delta }R')=R_\mathsf {req}(R_\mathsf {req}(R)\,{\bowtie }_{\delta } R_\mathsf {req}(R'))$. $\square$