Out-of-core Algorithms for Binary Partition Hierarchies

Abstract

Binary partition hierarchies (BPH) and minimum spanning trees are essential data structures for hierarchical analysis, such as quasi-flat zones and watershed segmentation. Traditional BPH construction algorithms are limited by their requirement to load the data entirely into memory, making them impractical for processing large images whose processing exceeds the capacity of the computer’s main memory. To overcome this limitation, an algebraic framework was introduced, enabling the out-of-core computation of BPH leveraging three key operations: select, join, and insert. In this publication, we present two distinct calculi based on these operations: one designed for general spatial partitions and another optimized for causal partitioning. The second calculus is specifically tailored to meet out-of-core constraints, ensuring efficient processing of large-scale data. We provide detailed algorithms, including pseudo-code and complexity analysis, and conduct experimental comparisons between the two approaches.

Appendices

Appendix A Proof of Property 7

Before proving Property 7, let us first introduce some lemmas.

Lemma 12

Let $A$ and $B$ be two disjoint subsets of $V$ and let $A^\prime$ and $B^\prime$ be two subsets of $A$ and $B$, respectively, then we have:

$$\begin{aligned}&\text{ select }\left( {A^\prime \cup B^\prime },{{\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B}\right) \\&= \text{ select }\left( {A^\prime },{{\mathcal {B}}^\prec _A}\right) \sqcup \text{ select }\left( {B^\prime },{{\mathcal {B}}^\prec _B}\right) \end{aligned}$$

Lemma 13

Let $A$ and $B$ be two disjoint subsets of $V$ and $Z$ be a set such as $Z \cap A \ne \emptyset$ and $Z \cap B \ne \emptyset$. Then we have: $\text{ select }\left( {Z},{\text{ join }\left( {{\mathcal {B}}^\prec _A},~{{\mathcal {B}}^\prec _B} \right) }\right) = {\text {j}oin} \left( {\text{ select }\left( {Z\cap A},{{\mathcal {B}}^\prec _A}\right) }\right. \left. {\text{ select }\left( {Z\cap B},{{\mathcal {B}}^\prec _B}\right) }\right)$

Proof

(of Lemma 13) Let ${\mathcal {H}} = \text{ select }\left( {Z},{\text{ join }\left( {{\mathcal {B}}^\prec _A},~{{\mathcal {B}}^\prec _B} \right) }\right)$. We are going to prove that ${\mathcal {H}} = {\text {j}oin} \left( { \text{ select }\left( {Z\cap A},{{\mathcal {B}}^\prec _A}\right) }\right. \left. {\text{ select }\left( {Z\cap B},{{\mathcal {B}}^\prec _B}\right) }\right)$. We have

$$\begin{aligned} {\mathcal {H}}&= \text {select}(Z, {\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B \boxplus \gamma (A,B))\\&\qquad \text {by definition of join}\\&= \text {select}(Z, \text {insert}( \text {select}(Z, {\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B \boxplus \gamma (A,B)), \\ &\qquad {\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B ))\\&\qquad \text {by Lemma 13 of} ~[26]\\&= \text {insert}( \text {select}(Z, {\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B \boxplus \gamma (A,B)), \\ &\qquad \text {select}(Z, {\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B) )\\&\qquad \text {by Lemma 12 of } [26]\\&= \text {select}(Z, {\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B) \boxplus \gamma (A,B) \\ &\qquad \sqcup \text {select}(Z, {\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B) \end{aligned}$$

From Property 7.3 of [26], we have

$$\begin{aligned} {\mathcal {H}}&= \text{ select }\left( {Z},{{\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B}\right) \sqcup \text{ select }\left( {Z},{{\mathcal {B}}^\prec _A \sqcup {\mathcal {B}}^\prec _B}\right) \\ &\qquad \boxplus \gamma (A,B) \end{aligned}$$

Since $Z\cap A \subseteq A$ and $Z\cap B \subseteq B$, by Lemma 12 we deduce that:

$$\begin{aligned} {\mathcal {H}}&= \text{ select }\left( {Z\cap A},{{\mathcal {B}}^\prec _A}\right) \sqcup \text{ select }\left( {Z\cap B},{{\mathcal {B}}^\prec _B}\right) \\ &\qquad \boxplus \gamma (A,B)\\&= \text{ join }\left( {\text{ select }\left( {Z\cap A},{{\mathcal {B}}^\prec _A}\right) },~{\text{ select }\left( {Z\cap B},{{\mathcal {B}}^\prec _B}\right) } \right) \\&\qquad \text {by definition of join} \end{aligned}$$

$\square$

At initialization step of Algorithm 1, we have ${\mathcal {M}}_0 = \text{ select }\left( { \gamma ^\bullet _{{V {\setminus } S_{0} }}({S_0})},{{\mathcal {B}}^\prec _{S_{0}}}\right)$. Then at the first iteration we have

$$\begin{aligned} {\mathcal {M}}_1&= \text {join}\left( \text{ select }\left( { \gamma ^\bullet _{{V \setminus S_{1} }}({S_1})},{{\mathcal {B}}^\prec _{S_{1}}}\right) \right) \\ &\qquad \left( \text{ select }\left( { \gamma ^\bullet _{{V \setminus S_{0} }}({S_0})},{{\mathcal {B}}^\prec _{S_{0}}}\right) \right) \\&=\text {select}( \gamma ^\bullet _{{V \setminus S_{0}}}({S_{0}}) \cup \gamma ^\bullet _{{V \setminus S_{1}}}({S_{1}}) ,\\&\qquad \text{ join }\left( {{\mathcal {B}}^\prec _{S_0}},~{{\mathcal {B}}^\prec _{S_1}} \right) )\\&\qquad \text {by Lemma}~\\&=\text{ select }\left( { \gamma ^\bullet _{{V \setminus S_{0}}}({S_{0}}) \cup \gamma ^\bullet _{{V \setminus S_{1}}}({S_{1}}) },{ {\mathcal {B}}^\prec _{S_0\cup S_1} }\right) \\&\qquad \text {by Property 9 of } [26] \end{aligned}$$

By recurrence, we have

$$\begin{aligned} {\mathcal {M}}_i = \text{ select }\left( {F},{{\mathcal {B}}^\prec _{\cup \{S_j \; | \;j \in \{0, \ldots i\}\}}}\right) \\ \text {with} \quad F = \bigcup _{j=0}^i \left\{ \gamma ^\bullet _{{V \setminus S_j}}({S_j}) \right\} \end{aligned}$$

Then, if we consider a partition into $k+1$ regions, when $i=k$ we have

$$\begin{aligned} {\mathcal {M}}_k = \text{ select }\left( {F},{{\mathcal {B}}^\prec _{V}}\right) \\ \text {with} \quad F = \bigcup _{j=0}^k \left\{ \gamma ^\bullet _{{V \setminus S_j}}({S_j}) \right\} \end{aligned}$$

This latter rewriting fits Definition 6. $\square$

Appendix B Proof of Lemma 8

Proof

(of statement 1) At each step of the first loop of Algorithm 3, we have ${\mathcal {M}}^\uparrow = {\text {j}oin} \left( { \text{ select }\left( {\gamma ^\bullet _{{i}}({S_{i-1}})},{B^\uparrow _{i-1}}\right) }\right. \left. { \text{ select }\left( {\gamma ^\bullet _{{{i-1}}}({S_i})},{B^\prec _{S_i}}\right) }\right)$. Theorem 17 of [26] states that $B^\uparrow _{i-1} = \text{ select }\left( {S_i},{{\mathcal {B}}^\prec _{\cup \{S_j \; | \;j \in \{0, \ldots i\}\}}}\right)$. Then, since $\delta ^\bullet (\gamma (S_{i\text {-}1},S_{i}))\cap \gamma ^\bullet _{{i}}({S_{i-1}}) \subseteq \gamma ^\bullet _{{i}}({S_{i-1}})$ and $\delta ^\bullet (\gamma (S_{i\text {-}1},S_{i}))\cap \gamma ^\bullet _{{{i-1}}}({S_i}) \subseteq \gamma ^\bullet _{{{i-1}}}({S_i})$, by Lemma 13 we can deduce that ${\mathcal {M}}^\uparrow = \text{ select }\left( {\delta ^\bullet (\gamma (S_{i\text {-}1},S_{i}))},{{\mathcal {B}}^\prec _{\cup \{S_j \; | \;j \in \{0, \ldots i\}\}}}\right)$ $\square$

Proof

(of statement 2) Let $A$, $B$ and $C$ be three pairwise disjoint subsets of $V$ such that $\gamma (A,C)=\emptyset$. Let ${\mathcal {M}}^\uparrow = \text{ join }\left( { \text{ select }\left( { \gamma ^\bullet _{{B}}({A}) },{ {\mathcal {B}}^\prec _{A} }\right) },~{ \text{ select }\left( { \gamma ^\bullet _{{B}}({A}) },{ {\mathcal {B}}^\prec _{A} }\right) } \right)$ and ${\mathcal {M}}^\downarrow = \text{ insert }({ \text{ select }\left( { \gamma ^\bullet _{{B}}({A}) },{ {\mathcal {B}}^\prec _{A\cup B \cup C} }\right) },{ {\mathcal {M}}^\uparrow })$. Then we have

$$\begin{aligned} {\mathcal {M}}^\downarrow&= \text {join}( \text{ select }\left( { \gamma ^\bullet _{{B}}({A}) },{ {\mathcal {B}}^\prec _{A} }\right) ,\\&\qquad \text{ select }\left( { \gamma ^\bullet _{{A}}({B}) },{ {\mathcal {B}}^\prec _{B \cup C} }\right) )\\&\qquad \text {by Lemma 15 of} [26]\\&=\text {select}\left( \gamma ^\bullet _{{B}}({A}) \cup \gamma ^\bullet _{{A}}({B}), \text{ join }\left( {{\mathcal {B}}^\prec _{A}},~{{\mathcal {B}}^\prec _{B \cup C}} \right) \right) \\&\qquad \text {by Lemma}~13\\&=\text {select}\left( \gamma ^\bullet _{{B}}({A}) \cup \gamma ^\bullet _{{A}}({B}), {\mathcal {B}}^\prec _{A \cup B \cup C} \right) \\&\qquad \text {by definition of join} \end{aligned}$$

Since $\gamma ^\bullet _{{B}}({A}) \cup \gamma ^\bullet _{{A}}({B}) = \delta ^\bullet (\gamma (A,B))$, we have ${\mathcal {M}}^\downarrow =\text {select}\left( \delta ^\bullet (\gamma (A,B)), {\mathcal {B}}^\prec _{A \cup B \cup C} \right)$. $\square$

Appendix C Proof of Property 9

Proof

In order to establish Property 9, we first introduce the following lemma:

Lemma 14

Let ${\mathcal {H}}$ be a BPH on $V$. Let $A \subseteq V$ and $B \subseteq V$ such that $A \subseteq B$. The we have $\text{ select }\left( {A},{{\mathcal {H}}}\right) \sqsubseteq \text{ select }\left( {B},{{\mathcal {H}}}\right)$.

Then, we proceed in two steps:

• (1) ${\mathcal {M}}^\uparrow _i \sqsubseteq {\mathcal {M}}_i$; and

• (2) ${\mathcal {M}}^\downarrow _i \sqsubseteq {\mathcal {M}}$.

(1) As stated in Sect. 4.3, ${\mathcal {M}}_i$ built on ${\textbf{P}}$ is defined as follows

$$\begin{aligned} \begin{aligned} {\mathcal {M}}_i = \text{ select }\left( {F},{{\mathcal {B}}^\prec _{\cup \{S_j \; | \;j \in \{0, \ldots i\}\}}}\right) \end{aligned} \end{aligned}$$

(C1)

                                                            where

$$\begin{aligned} F = \bigcup _{j=0}^i \left\{ \gamma ^\bullet _{{V \setminus S_j}}({S_j})\right\} \end{aligned}$$

(C2)

From the definition of $\delta ^\bullet$, $\gamma$ and $\gamma ^\bullet$, it can be seen that

$$\begin{aligned} F&= \bigcup _{j=0}^i \left\{ \gamma ^\bullet _{{V \setminus S_j}}({S_j}) \right\} \end{aligned}$$

(C3)

$$\begin{aligned}&= \bigcup _{j=1}^i \left\{ \delta ^\bullet (\gamma (S_{j\text {-}1},S_{j})) \right\} \end{aligned}$$

(C4)

As stated in Property 8, we have

$$\begin{aligned} \begin{aligned} {\mathcal {M}}^\uparrow _i = \text{ select }\left( {C},{{\mathcal {B}}^\prec _{\cup \{S_j \; | \;j \in \{0, \ldots i\}\}}}\right) \end{aligned} \end{aligned}$$

(C5)

                                                            where

$$\begin{aligned} C = \delta ^\bullet (\gamma (S_{i\text {-}1},S_{i})) \end{aligned}$$

(C6)

We can deduce that for any $i \in \{1, \ldots , k\}$ we have $C \subseteq F$. Therefore, given Lemma 14 we have ${\mathcal {M}}^\uparrow _i \sqsubseteq {\mathcal {M}}_i$.

(2) For this second statement, we can express ${\mathcal {M}}$ and ${\mathcal {M}}_i$ with $C$ from Equation (C6) and $F$ from Equation (C4).

$$\begin{aligned} {\mathcal {M}} = \text{ select }\left( {F},{{\mathcal {B}}^\prec _{V}}\right) \end{aligned}$$

(C7)

$$\begin{aligned} {\mathcal {M}}^\downarrow _i = \text{ select }\left( {C},{{\mathcal {B}}^\prec _{V}}\right) \end{aligned}$$

(C8)

We have $C \subseteq F$, and with Lemma 14 we have ${\mathcal {M}}^\downarrow \sqsubseteq {\mathcal {M}}$. $\square$

Appendix D Proof of Property 11

Proof

Given $n_i = |\text{ select }\left( {S_{i -1}},{{\mathcal {H}}}\right) |\ + |\text{ select }\left( {S_{i}},{{\mathcal {H}}}\right) |$, we will prove statements of Property 11 in the following order:

• (1) $|{\mathcal {M}}^\downarrow _i |\le n_i$; and

• (2) $|{\mathcal {M}}^\uparrow _i |\le n_i$.

(1) From Property 8, we have ${\mathcal {M}}^\downarrow _i = \text{ select }\left( {F},{{\mathcal {H}}}\right)$ where $F = \delta ^\bullet (\gamma (S_{i\text {-}1},S_{i}))$. We can see that $F \subseteq (S_{i-1} \cup S_i)$ which gives us Equation (D9). Each region $R$ of $\text{ select }\left( {S_{i-1} \cup S_i},{ {\mathcal {H}}}\right)$ either intersects an element of $S_{i -1}$ or $S_i$ or both. As a result, $R$ belongs either to $\text{ select }\left( {S_{i -1}},{{\mathcal {H}}}\right)$ or $\text{ select }\left( {S_{i}},{{\mathcal {H}}}\right)$ or both. As a result, we have Equation (D10).

$$\begin{aligned} |{\mathcal {M}}^\downarrow _i |&\le |\text{ select }\left( {S_{i-1} \cup S_i},{ {\mathcal {H}}}\right) | \end{aligned}$$

(D9)

$$\begin{aligned}&\le |\text{ select }\left( {S_{i -1}},{{\mathcal {H}}}\right) |\ + |\text{ select }\left( {S_{i}},{{\mathcal {H}}}\right) | \end{aligned}$$

(D10)

(2) Given $C=\delta ^\bullet (\gamma (S_{i-1}, S_i))$, we can write the two structures as follows:

$$\begin{aligned}&{\mathcal {M}}^\downarrow _i = \text{ select }\left( {C},{{\mathcal {B}}^\prec _V}\right) \end{aligned}$$

(D11)

$$\begin{aligned}&{\mathcal {M}}^\uparrow = \text{ select }\left( {C},{{\mathcal {B}}^\prec _{\cup \{S_j \; | \;j \in \{0, \ldots k\}\}}}\right) \end{aligned}$$

(D12)

Considering Lemma 15, we have $|{\mathcal {M}}^\uparrow _i |\le |{\mathcal {M}}^\downarrow _i |$. Thus we have $|{\mathcal {M}}^\uparrow _i |\le |{\mathcal {M}}^\downarrow _i |\le n_i$

Lemma 15

Let $(S_0, \ldots , S_k)$ be a causal partition of V, $A \subseteq V$ and $u$ be a pair of distinct elements of $V$. For any $i \in \{0,\ldots ,k\}$, the following inequalities holds true:

$$\begin{aligned}&|\text{ select }\left( {A},{{\mathcal {H}}}\right) |\le |\text{ select }\left( {A},{{\mathcal {H}} \boxplus \{u\}}\right) |\end{aligned}$$

(D13)

$$\begin{aligned}&|\text{ select }\left( {X},{{\mathcal {B}}^\prec _{{\cup \{S_j \; | \;j \in \{0, \ldots i\}\}}}}\right) |\le |\text{ select }\left( {X},{{\mathcal {B}}^\prec _V}\right) |\end{aligned}$$

(D14)

$\square$