Orders on Partial Partitions Based on Block Apportioning

HAL Id: hal-02882868

https://hal.archives-ouvertes.fr/hal-02882868

Submitted on 6 Nov 2020

HAL is a multi-disciplinary open access

archive for the deposit and dissemination of

sci-entific research documents, whether they are

pub-lished or not. The documents may come from

teaching and research institutions in France or

abroad, or from public or private research centers.

L’archive ouverte pluridisciplinaire HAL, est

destinée au dépôt et à la diffusion de documents

scientifiques de niveau recherche, publiés ou non,

émanant des établissements d’enseignement et de

recherche français ou étrangers, des laboratoires

publics ou privés.

Orders on Partial Partitions Based on Block

Apportioning

Christian Ronse

To cite this version:

Christian Ronse. Orders on Partial Partitions Based on Block Apportioning. Acta Applicandae

Mathematicae, Springer Verlag, 2016, 141 (1), pp.69-105. �10.1007/s10440-014-0004-z�. �hal-02882868�

(will be inserted by the editor)

Orders on partial partitions based on block apportioning

Received: date / Revised: date

Abstract Some problems in image analysis require considering an order on space partitions: in image filtering the partition of image zones must grow, while in image segmentation one maximizes the partition into image objects.

In the classical refinement order, the growth of a partition is achieved through the merging of blocks. We generalize this operation into block apportioning: some blocks in a partial partition may be split, and their parts are merged with some remaining blocks; this includes the possibility of a block being merged with an-other without being split. We study this operation in detail; we obtain then the

apportioning order, which extends the refinement order.

For partial partitions, the standard order combines the merging of blocks with the growth of existing blocks and the creation of new blocks. Combining these operations with block apportioning leads to theapportioning-inflating andextended

orders; the latter contains the standard order. Our analysis rests on a study of Serra’s building order on partial partitions, which intervenes in the above new orders.

Except the building order, all these orders are graded, with a compound grad-ing for the apportiongrad-ing-inflatgrad-ing and extended orders; this gives the number of elementary operations involved in a growth of a partial partition.

Keywords partial partition·partial order·block apportioning·image analysis

Mathematics Subject Classification (2010) 03E02·06A99·68U10

1 Introduction

This paper is a sequel to [16], which studied some new orders on partial partitions; it also elaborates on some parts of [15].

This work received funding from the Agence Nationale de la Recherche, contract ANR-2010-BLAN-0205-01.

Christian Ronse

ICube, Universit´e de Strasbourg, CNRS, 300 Boulevard S´ebastien Brant, CS 10413, 67412 IL-LKIRCH CEDEX, FRANCE — E-mail: [email protected], URL: http://icube-miv.unistra.fr/

Given a setE, apartitionofEis a family of non-void mutually disjoint subsets of E, called blocks [11], whose union is E. Apartial partition ofE is a family of non-void mutually disjoint blocks, but their union is not necessarily equal to E; the union of all blocks of a partial partitionπis thesupport ofπ,supp(π); thusπis a partition ofsupp(π); let the background ofπ be the complement of the support,

back(π) =E \ supp(π). We writeΠ(E) for the set of all partitions ofE, andΠ∗(E) for the set of all partial partitions ofE.

The customary order onΠ(E) isrefinement[11]: forπ1, π2∈ Π(E), we say that π1 isfiner thanπ2, or thatπ2 iscoarser thanπ1, and writeπ1≤ π2 (orπ2≥ π1),

if and only if every block of π1 is included in a block of π2, equivalently, every

block ofπ2is a union of blocks ofπ1. Then≤is a partial order relation onΠ(E),

for whichΠ(E) is a complete lattice, whose least (finest) and greatest (coarsest) elements are the identity partition 0_E (whose blocks are all singletons in E) and theuniversal partition 1E (withEas single block) [11]. See [14] for more details.

Partial partitions of a setEwere first studied by Dra˘skovi˘cov´a [4, 5], who called them “partitionsin E”. She orderedΠ∗(E) similarly to the refinement order: for

π1, π2 ∈ Π∗(E), we write π1 ≤ π2 (or π2 ≥ π1), if and only if every block of π1 is included in a block of π2. This order ≤constitutes Π∗(E) into a complete

lattice whose least and greatest elements are theempty partial partition Øand the universal partition 1E; it was further studied in [13, 14, 16]. Note however that

for π1, π2∈ Π∗(E) with π1 ≤ π2, a block ofπ2 will not necessarily be a union of

blocks ofπ1(as it was the case with partitions): blocks ofπ2can be constructed by

merging several blocks ofπ1, inflating individual blocks ofπ1, or both combined,

or else by creating new blocks outsidesupp(π1). Hence [15, 16] we do not call this

order onΠ∗(E) refinement, but thestandard order. In fact the refinement order is the restriction to partitions of the standard order on partial partitions.

Partitions and the refinement order are relevant to image segmentation. A grey-level or colour image can be represented as a function F :E → T, where E

is the space of pixels andT is the set of image values (grey-levels or colours). A

segmentation of the image F is a partition of E into visually significant regions representing objects or the surrounding background. As explained in [15, 16], the conventional requirements for image segmentation outlined in [25] mean that the segmentation process selects, among all partitions with connected blocks on which the image is homogeneous, one that is coarsest possible. More precisely:

1. To each imageF on E, the segmentation method associates the family CF of all subsets of E that are “connected” (according to a choice of connectivity) and on which the imageF is “homogeneous” (according to some criterion). 2. Let Π(E, CF) = Π(E)∩ P CF \ {∅}
be the set of all partitions of E whose

blocks belong to CF \ {∅}. Then the segmentation of F must be a maximal element, for the refinement order, ofΠ(E, CF).

Serra [21, 17] introduced the paradigm ofconnective segmentation. Here the set

CF must be aconnection, i.e.,CF comprises all singletons, and a union of overlap-ping elements ofCF will belong toCF; equivalently,0E∈ Π(E, CF) andΠ(E, CF)

must be closed under the supremum operation of the complete latticeΠ(E). Then the segmentation of F will be the greatest element of Π(E, CF). A well-known examples is whenCF _{consists of all connected subsets ofEon whichF is constant,}

and then the segmentation will be the partition of allflat zonesofF, i.e., maximal connected subsets ofE on whichF has a constant value.

Fig. 1 (From [17], following [6].) (a) Initial image of the silhouette to be segmented. (b) Seg-mentation of the face (by a colour criterion). (c) Marker for the bust. (d) Final segSeg-mentation.

Connective segmentation underlies the connected filtering of images [21, 19, 17, 18, 24]. To every image F : E → T corresponds a partition π(F) of E obtained by a connective segmentation. Given an operatorψ:TE → TE for transforming images, we say thatψisconnected if it coarsens the segmentation partition corre-sponding to the image:π(F)≤ π(ψ(F)). Initially, this notion was considered for the segmentation into flat zones, then ψ is connected if it coarsens the partition into flat zones; since visible edges (in particular object contours) lie at the frontier between distinct flat zones, it follows that a connected operator can either preserve or eliminate such edges, but never displace or deform them.

Working exclusively with partitions is limiting, since the known operations have been mostly merging and splitting blocks, in other words coarsening and refining a partition, which were used in alternation in the split-and-merge approaches to image segmentation [12]. The use of partial partitions in image segmentation was justified in [13, 16]. Let us just mention two facts:

– Segmentations are often built by iteratively growing partial partitions; for ex-ample: inregion growing, blocks are inflated fromseeds; nowcompound segmen-tation [20, 17, 15] proceeds by constructing an initial partial partition according to a first criterion, then a partial partition of the background according to a second criterion, and the final segmentation is the union of the two, see for example Figure 1.

– Some segmentation algorithms produce a partial partition, whose background consists of frontiers separating the blocks; this frontier can be thick, see for example Figure 2. Some segmentation algorithms can miss whole regions that remain thus in the background.

In [13, 17], the framework of connective segmentation was extended to partial partitions. LetΠ∗(E, CF) =Π∗(E)∩ P CF\ {∅}
be the set of all partial partitions ofE whose blocks belong toCF\ {∅}. HereCF must be apartial connection, i.e., a union of overlapping elements ofCF will belong toCF; equivalently,Ø ∈ Π∗(E, CF) andΠ∗(E, CF) must be closed under the supremum operation of the complete lat-ticeΠ∗(E). Then the segmentation ofF will be the greatest element ofΠ∗(E, CF). An example is given by the regional Lipschitz segmentation [21] (also called

smoothsegmentation [17]). It depends on two numerical parameters, a radiusr >0 and a slopes >0, and has two versions, “by erosion” and “by opening”. Given a function F :E → T, letLE_Fr,s be the set of allp ∈ E such that the restriction of

Fig. 2 (From [21].) Left: original grey-level image. Middle and right: regional Lipschitz seg-mentation, respectively “by erosion” and “by opening”, with radius r = 7 and slope s = 6; the blocks are the white connected components, they correspond to regions, and the background is in black, it represents the frontiers.

of slopes; letLOr,s_F be the union of all ballsBr(p) forp ∈ LE_Fr,s, in other words

the dilation of LEr,s_F by the ball Br. The segmentation “by erosion” (resp., “by

opening”) consists of the partial partition of connected components ofLE_Fr,s(resp.,

LOr,s_F ). We illustrate both versions of this segmentation in Figure 2, where we see that they produce thick frontiers between regions.

When a segmentation produces a partial partition with a background made of thick frontiers, the latter can be thinned by applying a region growing algorithm to the blocks, this time relying on another criterion than the one used in the segmentation. Region growing, whether in the segmentation algorithm or in the post-segmentation processing, follows a partial order relation onΠ∗(E):

– The inflating order ≤i (written E in [15, 16]): for π1, π2 ∈ Π∗(E),π1

i ≤ π2 iff

every block ofπ1 is included in a (unique) block ofπ2, and every block ofπ2

contains exactly one block of π1; in other words π2 is obtained from π1 by

inflating its blocks.

On the other hand, the compound segmentation paradigm [20, 17, 15] follows an-other order onΠ∗(E):

– The inclusion order ⊆: for π1, π2 ∈ Π∗(E), π1 ⊆ π2 simply means that each

block ofπ1 is a block of π2, in other wordsπ2 is obtained from π1 by adding

new blocks to it.

A third order onΠ∗(E) is involved in the merging of blocks:

– The merging order ≤m (written ⊑ in [15, 16] and called pure refinement order in [15]): for π1, π2 ∈ Π∗(E), π1

m

≤ π2 iff π1 and π2 share the same support

and π1≤ π2 according to the refinement order for partitions of that support;

equivalently, every block ofπ1 is included in a (unique) block ofπ2, and every

block of π2 is a union of blocks ofπ1.

These 3 orders are thebasicorders onΠ∗(E) [16], they are included in the standard order≤. Combining any two of them, one generates 3compound orders:

– Combining inclusion ⊆ and inflating ≤i, one gets the inclusion-inflating order

i

⊆ (written⊆E in [15, 16]): forπ1, π2 ∈ Π∗(E), π1

i

(a) (b) (c) (d) (e) Fig. 3 From [10]: (a) Original image. (b,c) Segmentation: we show (b) the blocks in pseudo-colour, and (c) the contours of the blocks. Following [22, 23]: (d) The blocks of size less than 20 pixels are removed (they are shown in black), and (e) the remaining blocks of size > 20 are inflated to cover the removed ones, we show the contours of the blocks.

included in a (unique) block ofπ2 and every block ofπ2 contains at most one

block of π1; equivalently, π2 is obtained from π1 by inflating existing blocks

and creating new blocks.

– Combining merging ≤m and inflating≤i, one gets themerging-inflating order mi≤

(written⊑E in [15, 16] and calledrefinement-inflating order in [15]): forπ1, π2∈ Π∗(E),π1

mi

≤ π2 iff every block ofπ1 is included in a (unique) block ofπ2 and

every block ofπ2contains at least one block ofπ1; equivalently,π2is obtained

fromπ1by inflating and merging its blocks.

Finally combining inclusion⊆and merging≤m, one obtains the standard order≤, which contains the 5 others. The 3 basic and 3 compound orders were studied in [16], where we gave examples of their possible applications in image analysis and segmentation.

Beside thick frontiers separating regions, or incomplete segmentations, there is another problem in image segmentation, which was addressed by Serra [22, 23]. Many image segmentation algorithms will produce, on a gradual transition between two zones, small “parasitic” blocks in the partial partition, See Figure 3 (b,c). Serra proposed to first remove “parasitic” blocks, next to inflate the remaining blocks until the removed blocks are completely covered (for instance, one can take the Voronoi diagram), see Figure 3 (d,e).

Now we consider this compound process, with two successive steps, as a single operation that we callapportioning. Some blocks are removed, and their contents is apportioned between one or several remaining blocks. In other words, a disap-pearing block either is merged with one remaining block, or it is split into several pieces, and each piece is merged with some remaining block. Block merging is thus a special case of apportioning, when the disappearing blocks are not split. See Fig-ure 4. This operation leads to a new partial order on partitions, which we call the

apportioning order and write≤a; it contains the merging order:∀ π1, π2 ∈ Π∗(E), π1

m

≤ π2 ⇒ π1

a

≤ π2. This order was briefly suggested in Serra’s work [22, 23], but

0000 0000 0000 0000 1111 1111 1111 1111 00 00 00 00 00 00 00 11 11 11 11 11 11 11 00000000000000 00000000000000 00000000000000 00000000000000 00000000000000 00000000000000 00000000000000 00000000000000 11111111111111 11111111111111 11111111111111 11111111111111 11111111111111 11111111111111 11111111111111 11111111111111 000000000000000 000000000000000 000000000000000 000000000000000 000000000000000 000000000000000 000000000000000 000000000000000 111111111111111 111111111111111 111111111111111 111111111111111 111111111111111 111111111111111 111111111111111 111111111111111 00000000000000 00000000000000 00000000000000 00000000000000 00000000000000 00000000000000 00000000000000 00000000000000 11111111111111 11111111111111 11111111111111 11111111111111 11111111111111 11111111111111 11111111111111 11111111111111 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111 000 000 000 000 000 111 111 111 111 111 00000 00000 00000 00000 00000 11111 11111 11111 11111 11111 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 000 000 000 111 111 111 00000000 00000000 00000000 00000000 11111111 11111111 11111111 11111111 00000000 00000000 00000000 00000000 11111111 11111111 11111111 11111111 000000000 000000000 000000000 000000000 111111111 111111111 111111111 111111111 000000000 000000000 111111111 111111111 00000 1111100000000 0000 1111 1111 1111 π1 π π2

Fig. 4 Apportioning. Left: π1, the blocks to be removed are in grey, and the preserved ones

are in colour with a black hatching. Middle: partial partition π with blocks removed. Right: one removed block is split into three parts, another one into two parts, and two remain in one part; then each part is merged with a preserved block, resulting in π2. The removed blocks of

π1have been apportioned. We have π1⊇ π i

≤ π2and supp(π1) = supp(π2). We write π1 a

≤ π2.

This problem with “parasitic” zones on gradual transitions between objects, and the solution provided by apportioning, arise also in image filtering. Suppose that one wants to sharpen the edge between two homogeneous zones separated by a gradual transition. We show in Figure 5 (a) the grey-level profile across the transition. By connected filtering, we can merge flat zones; for instance in (b,c) all singleton zones have been merged with a neighbouring non-singleton zone, getting the latter’s grey-level; in (d,e) the merging goes further, all small flat zones have been merged with a large flat zone. However the resulting edge will not be located at the middle of the original transition, since the latter lies inside a flat zone. In (f) we show a sharpening where the edge lies at the middle of the transition; it is obtained by eliminating all small zones, then inflating equally the two large zones (through a Voronoi diagram); in other words, small flat zones have been apportioned to the large ones, using the method proposed by Serra. Such an edge sharpening that breaks image zones cannot be achieved by connected filtering; it is similar to the behaviour ofshock filters(see [26] for a recent survey and extension). An alternative view of this type of filtering, where transition zones can be eliminated, is that we associate to an image F defined on E a partial partition

π∗(F) of zones corresponding to significant homogeneous regions, excluding those corresponding to noise or transitions. Then an image processing operatorψcan be characterized in terms of its behaviour on π∗(F). For instance, if in Figure 5 (a) we discard the singleton zones, working on the partial partition of non-singleton flat zones, we can obtain the results (b,c,d,e); now if we discard all small flat zones and work on the partial partition of large flat zones, we can obtain the result (f). Apportioning intervenes implicitly in some region growing methods such as watershed segmentation. Algorithms for constructing an image watershed often face the situation where a crest line emerges through a narrow passage into a wide plateau surrounded by higher crests, see for instance the left part of Figure 6 (from of [1]); for some algorithms, the plateau and the surrounding crests will be incorporated into the watershed (it becomes thus a “thick boundary”), while for others the plateau will be discarded, belonging neither to a basin nor to the watershed; for a few algorithms, this plateau will be included into one basin. Now the algorithm in [1] nicely apportions this plateau between the watershed line and the two basins separated by it, as seen in Figure 6, right.

(b) E T (a) E T (c) E T (d) E T _(e) E T _(f) E T

Fig. 5 (a) The graph of a one-dimensional signal across a gradual step edge; below we show its segmentation into flat zones, with light grey rectangles for non-singleton zones, and vertically hatched ones for groups of singleton zones. (b,c) The edge is partially sharpened by merging singleton zones with a neighbouring non-singleton zone. (d,e) We can merge both singleton and small non-singleton zones with the closest large zone, but for the middle zone there are two choices. (f) The edge is sharpened by eliminating all small zones, then inflating equally the two large ones; the edge between them lies inside the previous middle zone.

3 3 3 5 5 5 10 10 10 10 15 20 20 3 3 3 5 5 5 10 10 10 10 15 20 20 3 3 3 5 5 30 30 30 10 15 15 20 20 3 3 3 5 30 20 20 20 30 15 15 20 20 40 40 40 40 40 20 20 20 40 40 40 40 40 10 10 10 10 40 20 20 20 40 10 10 10 10 5 5 5 5 10 40 20 40 10 10 5 5 5 1 1 3 5 10 15 20 15 10 5 1 0 0 1 1 3 5 10 15 20 15 10 5 1 0 0 1 1 3 5 10 15 20 15 10 5 1 0 0 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 30 30 30 3 3 3 3 3 3 3 3 3 30 1 20 0 30 3 3 3 3 30 30 30 30 1 1 20 0 0 30 30 30 30 1 1 1 1 1 1 20 0 0 0 0 0 0 1 1 1 1 1 1 20 0 0 0 0 0 0 1 1 1 1 1 1 20 0 0 0 0 0 0 1 1 1 1 1 1 20 0 0 0 0 0 0 1 1 1 1 1 1 20 0 0 0 0 0 0

Fig. 6 (From [1].) Each square represents a pixel, with the number inside giving its grey-level. Left: two regional minima originating from the bottom left and right corners have their catchment basins meeting at a crest line of level 20 (bottom middle), and the latter prolongs itself into a square plateau (center) surrounded by crests with levels 30 and 40. Right: in the segmentation, the plateau is apportioned between the two basins and the watershed line.

Instead of the apportioning order, Serra [22, 23] studied a more general partial order on Π∗(E) that he called the building order. He wrote it , but we write it ⋐, since we use the symbol ≺for covering relations [15, 16]. Forπ1, π2 ∈ Π∗(E), π1⋐π2 if and only if every block ofπ2contains at least one block ofπ1. Then the

two successive operations of removing some blocks and then inflating remaining blocks, are both growing for the order ⋐:π1⊇ π

i

≤ π2 ⇒ π1⋐π⋐π2. Conversely,

we will see that when π1⋐π2, there existsπ ∈ Π∗(E) such that we obtainπ by

removing some blocks ofπ1, then π2by inflating blocks ofπ:π1⊇ π

i ≤ π2.

Now the apportioning order is the restriction of the building order to partial partitions sharing the same support: for π1, π2 ∈ Π∗(E), π1

a

≤ π2 if and only if π1 ⋐ π2 and supp(π1) = supp(π2). Following the generalization of merging into

apportioning, we will generalize the merging-inflating and standard orders into the following two orders:

– The apportioning-inflating order ≤ai, where π1

ai

≤ π2 ifπ2 is obtained fromπ1 by

apportioning and inflating its blocks, see Figure 11 ; equivalently,π1⋐π2and supp(π1)⊆ supp(π2).

– Theextended order≤e, whereπ1

e

≤ π2ifπ2is obtained fromπ1by creating then

apportioning blocks, see Figure 13 ; equivalently, π1 ⋐ π2∩ P(supp(π1)) and supp(π1)⊆ supp(π2).

We saw above that thick boundaries between regions can be thinned by inflat-ing blocks of the partial partition, while small “parasitic” segmentation classes can be eliminated by apportioning blocks. Thus when both problems arise together, the solution will combine inflating and apportioning blocks, thus the partial partition will grow according to the apportioning-inflating order.

As explained before, algorithms for image segmentation often construct the partial partition by inflating and creating blocks from initial seeds. Thus if we consider the compound process of first creating the segmentation, then improving it, we will combine inflating, creating and apportioning blocks, thus the partial partition will grow according to the extended order. Block apportioning could also be applied in the initial construction and growth of the segmentation, for instance if a region ceases to satisfy a criterion; this idea is implicitly contained in the split-and-merge strategy for segmentation [12].

Image segmentation has been integrated into the framework of hierarchies, where coarser segmentation partitions are built by merging some segmentation classes from finer ones [8]; in the case of watersheds, the merging can be guided by the saliency of watershed lines [9]. Combining this with the growth of zones, the merging-inflating order underlies such a hierarchical segmentation. Note also that one can also remove markers, and the unmarked basins will no more grow into regions, but will be apportioned between neighbouring regions. Thus the apportioning-inflating order could be used for modeling the growth and evolution of regions in hierarchical segmentation.

This paper gives a detailed mathematical study of these 3 new orders on partial partitions based on the operation of apportioning blocks. The apportioning order is basic (as the inflating, inclusion and merging orders), while the apportioning-inflating and extended orders are compound (as the inclusion-apportioning-inflating, inflating and standard orders). In fact, the extension of the merging, merging-inflating and standard orders into the apportioning, apportioning-merging-inflating and extended orders respectively translates several properties of the former into similar properties of the latter.

A significant aspect of orders onΠ∗(E) is theJordan-Dedekind chain condition, namely that all maximal chains in an interval have the same length. This is im-portant in practice (whenE is finite), since it means that when one constructs a partial partition by a succession of elementary growth operations, the number of steps always remains constant, it will be equal to theheight of that partial parti-tion. Our 3 new orders satisfy that property (as do the 6 orders studied in [16]).

In fact, the apportioning-inflating and extended orders, being compound, use two elementary growth operations, and the height function is the sum of two functions counting the number of times that each operation is performed.

On the other hand the building order does not satisfy the chain condition, and its maximal chains are counter-intuitive. Probably this is linked to the fact that the growth in this order involves two operations, inflating and removing blocks, one increasing the support and the other decreasing it. In our opinion, the building order is not specific enough to be meaningful in practice.

The 6 (3 basic and 3 compound) orders on Π∗(E) studied in [16], the 3 new ones introduced here and the identity, are ordered by inclusion and constitute a lattice whose Hasse diagram is illustrated in Figure 14. Note that the standard order is the only one among the 10 that forms a lattice on Π∗(E); however the merging order is the disjoint union of the lattices Π(A), A ∈ P(E), ordered by refinement.

Paper organization

Section 2 gives our notation and summarizes the main concepts and results of [16], about multiple grading and height functions, then the merging, inflating and inclusion orders and their combinations. Section 3 studies the apportioning order and its combinations with basic orders: the apportioning-inflating and extended orders. Section 4 concludes, summarizing our results, discussing their relevance and putting them into perspective. The Appendix analyses the “bad” properties of the building order.

2 Mathematical preliminaries

Subsection 2.1 recalls general notions about orders, covering relations, maximal chains, grading and height functions, then the multiple grading introduced in [16]. Subsection 2.2 summarizes the mathematical definition and main properties of the 6 orders studied in [16].

But we first give some general notation. In mathematical formulas, we will write “&” for the logical “and”. For a setA, we will writeP(A) for the set of parts ofA, and|A|for the cardinal ofA. Given two subsetsAand B of a setE, we say thatAandB overlap, and writeA≬B, if A ∩ B 6=∅.

Every binary relation R is identified with the set of ordered pairs (a, b) such thata R b; thus the inclusionR ⊆ S means that (a R b) ⇒ (a S b), and the union

R ∪ S is given bya(R ∪ S)b ⇔ (a R b) or (a S b). Thecomposition R · S of two relations is defined by a R · S b ⇔ ∃ c, (a R c) & (c S b). In particular, we will express each compound order onΠ∗(E) as a composition of two basic orders.

2.1 Orders, grading and height

Our terminology on orders and lattices follows [2, 3, 7]. See also [16]. We will con-sider several distinct (partial) order relations, they will be identified by a super-script on the usual order symbol, say≤x for “less than or equal to”,<x for “strictly

less than”, ≥x for “greater than or equal to”, >x for “strictly greater than”. The only exception will be for the building order on partial partitions, where we use ⋐ for “less than or equal to” and ⋑ for “greater than or equal to”, without any specific symbol for “strictly less than” and “strictly greater than”.

Aquasi-order is a reflexive and transitive binary relation. A non-empty inter-section of quasi-orders is a quasi-order, and the interinter-section of an order and a quasi-order is an order.

The set of all partial order relations on a setP is closed under non-void inter-section, but it is not a lattice; however, given a fixed partial order ≤on P, the setO(≤) of all partial order relations onP that are included in≤ is a complete lattice, since it is closed under intersection and has a greatest element.

Let≤be a partial order relation on a setP. We callisolated anyx ∈ P that is incomparable to any other element ofP:∀ y ∈ P, neithery < x nor x < yholds. For x, y ∈ P we say that y covers x (or x is covered by y), and write x ≺ y or

y ≻ x, if x < ybut there is no z ∈ P such thatx < z < y; when we consider the covering relation for distinct orders, we can distinguish them with superscripts like

x

≺. Givenx, y ∈ P withx ≤ y, thelength of theinterval [x, y] ={z ∈ P | x ≤ z ≤ y}

is the supremum of all integersn withx=z0 < · · · < zn =y; when this length is

finite (for instance whenP is finite), it is the greatest such n, and the sequence takes the formx=z0≺ · · · ≺ zn=y, we call it acovering chain between x and y.

WhenP has a least element 0, theheightofx ∈ P is the length of the interval [0, x]. WhenP has no least element, but for everyx ∈ P there exists a minimal element

msuch thatm ≤ x, we call theheight of x w.r.t. mthe length of the interval [m, x]. Given a mapg:P → Z, we say that P is graded by g [2, 7] if for any x, y ∈ P,

x < y ⇒ g(x)< g(y) and x ≺ y ⇒ g(y) =g(x) + 1. Let us now describe the generalization of grading introduced in [16]:

Proposition 1 In a poset (P, ≤), let the covering relation ≺ be partitioned into t

mutually disjoint non-void relations≺, . . . ,1 ≺, and consider t maps gt 1, . . . , gt:P → Z.

Suppose that:

1. For all x, y ∈ P and i= 1, . . . , t we have x≺ yi =⇒

(

gi(y) =gi(x) + 1 & gj(y) =gj(x)for j 6=i .

Then the following two statements are equivalent: 2. Every interval in P has finite length.

3. For all x, y ∈ P ,

x < y =⇒

(

∀ i ∈ {1, . . . , t}, gi(y)≥ gi(x) , & ∃ i ∈ {1, . . . , t}, gi(y)> gi(x) .

When these conditions are met, we obtain the following: 4. For all x, y ∈ P and i= 1, . . . , t we have

x≺ y ⇐⇒ x ≤ yi & (

gi(y) =gi(x) + 1 & gj(y) =gj(x)for j 6=i .

5. In a covering chain z0 ≺ · · · ≺ zn in P , among the n coverings zℓ−1 ≺ zℓ (ℓ =

1, . . . , n), there are gi(zn)− gi(z0)occurrences of zℓ−1

i

≺ zℓ for i= 1, . . . , t.

6. P is graded byPt

i=1gi.

When conditions 1 and 2 (equivalently, 1 and 3) are satisfied, we will say that

P is graded by(g1, . . . , gt)for( 1

≺, . . . ,≺t). A particular case is whent= 1, that is,

P is graded bygfor≺, here conditions 1 and 3 mean thatP is graded byg; then item 4 gives x ≺ y ⇐⇒ [x ≤ y & g(y) = g(x) + 1], and item 5 means that for

x < y, all covering chains betweenxand yhave the same lengthg(y)− g(x): this is theJordan-Dedekind chain condition. Fort >1, item 5 is a stronger version of it. There is an alternative expression of Proposition 1 in terms of vectors. For

x ∈ P, write G(x) = g1(x). . . , gt(x)
; fori= 1, . . . , t, let Ei ∈ Zt be the vector

withi-th coordinate equal to 1 and all other coordinates equal to 0. We take the componentwise order on Zt, namely (a1, . . . , at) ≤ (b1, . . . , bt) iff for i = 1, . . . , t

we have ai ≤ bi, and (a1, . . . , at) < (b1, . . . , bt) iff (a1, . . . , at) ≤ (b1, . . . , bt) and

(a1, . . . , at)6= (b1, . . . , bt). Then items 1, 3 and 4 can be written as:

1. For allx, y ∈ P andi= 1, . . . , twe havex≺ yi =⇒ G(y) =G(x) +Ei.

3. For allx, y ∈ P,x < y =⇒ G(x)< G(y).

4. For allx, y ∈ P andi= 1, . . . , twe havex≺ y ⇐⇒i [x ≤ y&G(y) =G(x)+Ei].

2.2 Partial partitions, basic and compound orders

Our general notation follows [13, 14, 16]. A partial partition of E is made of mu-tually disjoint non-void subsets of E called blocks. We write Π(E) for the set of all partitions of E, and Π∗(E) for the set of all partial partitions ofE; thus

Π∗(E) =S

A∈P(E)Π(A). WriteØfor the empty partial partition (with no block);

then Π(∅) = Π∗(∅) ={Ø}. Set 1∅ = 0∅ = Ø, while for anyA ∈ P(E)\ {∅}, let 1A = {A} (the partition of A into a single block) and 0A =



{p} | p ∈ A (the partition of Ainto its singletons); following [11], we call 0A the identity partition

ofA, and1Atheuniversal partitionofA. Forπ ∈ Π∗(E), thesupport ofπ, written supp(π), is the union of its blocks:supp(π) =S

π; the complement of the support is the background of π, back(π) =E \ supp(π). For π ∈ Π∗(E), a transversal ofπ

is a subset of E made by choosing one point in each block of π, in other words a set A ⊆ supp(π) such that |A ∩ B| = 1 for any B ∈ π; a crossing ofπ is a set

A ⊆ supp(π) such that A ∩ B 6=∅for any B ∈ π; in other words, it is a subset of

supp(π) containing a transversal ofπ.

Let us now recall the order relations onΠ∗(E) considered in [16]. Thestandard order ≤is given by:∀ π1, π2∈ Π∗(E),π1≤ π2 ⇐⇒ ∀ B ∈ π1, ∃ C ∈ π2, B ⊆ C.

Next, thebuilding order ⋐ [22, 23] is defined by:∀ π1, π2∈ Π∗(E),π1⋐π2 ⇐⇒ ∀ C ∈ π2, ∃ B ∈ π1, B ⊆ C.

Note that a block of π1 is included in at most one block of π2; on the other

hand a block ofπ2can contain several blocks ofπ1. We define thus thesingularity

relation ⇚ [16]: forπ1, π2∈ Π∗(E), writeπ1⇚π2 (orπ2⇛π1) if every block of π2 contains at most one block ofπ1:

π1⇚π2 ⇐⇒  ∀ B, B ′ ∈ π1, ∀ C ∈ π2,  B ⊆ C &B′⊆ C =⇒ B=B′ .

In [16] we considered theinclusion order ⊆and defined the following 4 orders, obtained by intersecting the standard order ≤ with the building order ⋐, the singularity relation ⇚ or the support equality relation:

– Themerging order m≤:∀ π1, π2∈ Π∗(E), π1

m

≤ π2 ⇐⇒ π1≤ π2 &supp(π1) =supp(π2) . (1)

It is thus the union of the refinement orders on all Π(A),A ∈ P(E).

– Theinflating order ≤i:∀ π1, π2∈ Π∗(E), π1

i

≤ π2 ⇐⇒ π1≤ π2&π1⋐π2& π1⇚π2 . (2) – Themerging-inflating order mi≤:∀ π1, π2∈ Π∗(E),

π1

mi

≤ π2 ⇐⇒ π1≤ π2& π1⋐π2 . (3) – Theinclusion-inflating order ⊆i:∀ π1, π2∈ Π∗(E),

π1

i

⊆ π2 ⇐⇒ π1≤ π2 &π1⇚π2 .

We next consider the various covering relations on Π∗(E) associated to the above orders [16]. Givenπ1, π2∈ Π∗(E), let us write:

– π1

m

≺ π2 ifπ2 is obtained by merging two blocks ofπ1: π1 m ≺ π2 ⇐⇒  |π1| ≥2, ∃ C1, C2∈ π1, C16=C2, π2= π1\ {C1, C2}
∪ {C1∪ C2}  ; (4) – π1 s

≺ π2 ifπ2 is obtained by adding a singleton block toπ1: π1

s

≺ π2 ⇐⇒ supp(π1)⊂ E, ∃ p ∈ back(π1), π2=π1∪{p}  ; (5) – π1

c

≺ π2 ifπ2 is obtained by adding a block toπ1: π1

c

≺ π2 ⇐⇒ supp(π1)⊂ E, ∃ B ⊆ back(π1), B 6=∅, π2=π1∪ {B} ; (6) – π1

i

≺ π2 ifπ2 is obtained by inflating one block of π1 by exactly one point: π1 i ≺ π2 ⇐⇒  supp(π1)⊂ E, π16=Ø, ∃ p ∈ back(π1), ∃ B ∈ π1, π2= π1\ {B}
∪B ∪ {p}  . (7) Whenπ1 x

≺ π2 (x=m, s, c, i), we say thatπ2 x-covers π1, and we call the relation

x

≺thex-covering.

Assume now thatE is finite. For anyπ ∈ Π∗(E), we define itsm-height hm(π), s-height hs(π), andc-height hc(π), as follows:

Table 1 Properties of the 3 basic orders.

Designation Maximal Minimal Cover Grading Height of π

m

≤ merging 1A, A ∈ P(E) 0A, A ∈ P(E) m

≺ hm w.r.t. 0_supp(π)

⊆ inclusion partitions Ø(least) ≺c hc i

≤ inflating Ø ; partitions 0A, A ∈ P(E) i

≺ hm w.r.t. 0A, A transversal of π

These three heights are non-negative integers. We have

hm(π) =hs(π)− hc(π) , hs(π) =hm(π) +hc(π) , hc(π) =hs(π)− hm(π) .

Now theheight ofπin the standard order [14] is the sum:

h(π) =hm(π) +hs(π) = 2|supp(π)| − |π| . (9)

We summarize into Tables 1 and 2 the main properties of the 6 orders studied in [16]. The Maximal and Minimal columns give the maximal and minimal ele-ments; when it is the least or the greatest element, this is indicated by(least)and

(greatest) respectively. The 3 compound orders are obtained as a composition of two basic orders, given in the Composition column. The Cover column gives the corresponding covering relation. WhenE is finite, the Grading column gives the grading function. Basic orders have a simple grading, which gives the height func-tion; however, when there is no least element, we show in theHeight of πcolumn w.r.t. to which minimal element the height of πis given. Compound orders have a double grading, so we indicate also there the function giving the height ofπ, it is in fact the sum of the two grading functions.

Table 1 describes the 3 basic orders: merging, inclusion and inflating. Note that for the merging and inflating orders,Ø=1∅ =0∅ is isolated. Table 2 (split into

two parts) describes the 3 compound orders: merging-inflating, inclusion-inflating and standard. Note that for the merging-inflating order,Ø is isolated, and1E is

the greatest element ofΠ∗(E)\ {Ø}. Also, for the merging-inflating and inclusion-inflating orders, the expression as composition of two basic orders commutes.

For the refinement order onΠ(E), we restrict the merging order to the support

E, so the greatest and least elements are1Eand0E, the covering relation is m ≺, and whenE is finite,Π(E) is graded byhm, and the latter gives the height function.

3 Apportioning and derived orders

We introduce three new orders: theapportioning, apportioning-inflatingandextended

orders. Since they are based on the building order, we will first briefly study the latter; a more detailed analysis will be given in the Appendix, where we will show that its covering chains have variable length and a strange form, see Figure 15; even by “normalizing” them, their length can still vary, see Figure 16. In Sub-section 3.1 we study the first two new orders, both restrict the building order: in the apportioning order, the support remains constant, while in the apportioning-inflating order, the support grows. Subsection 3.2 studies the extended order; it

Table 2 Properties of the 3 compound orders.

Designation Maximal Minimal Composition ≤ standard 1E (greatest) Ø (least) ⊆ ·

m ≤ mi ≤ merging-inflating Ø; 1E 0A, A ∈ P(E) m ≤ ·≤ ;i ≤ ·i ≤m i

⊆ inclusion-inflating partitions Ø(least) ⊆ ·≤ ;i ≤ · ⊆i Order Cover Grading Height of π

≤ ≺ ∪m ≺ (hs m, hs) for ( m ≺,≺) h(π)s mi ≤ ≺ ∪m ≺ (−hi c, hs) for ( m ≺,≺) hi m(π) w.r.t. 0A, A crossing of π i ⊆ ≺ ∪s ≺ (hi c, hm) for ( s ≺,≺) hi s(π)

also involves a growth of the support, but it is not included in the building or-der. All three orders are graded, with a simple grading for apportioning, and a compound grading for the other two. We will see that they have many similarities with themerging, merging-inflating andstandard orders.

Note that the building, apportioning, apportioning-inflating and extended or-ders donotconstitute a lattice onΠ∗(E): the supremum or infimum of two partial partitions do not necessarily exist. Of all orders considered here and in [16], only the standard order on Π∗(E) and the refinement order on Π(E) are complete lattices.

Recall the definition of the building order ⋐: π1 ⋐ π2 iff every block of π2

contains at least one block ofπ1. The following properties are known [22, 23, 16]: Proposition 2 The building order⋐is a partial order relation on Π∗(E). It contains the merging ≤, inflatingm ≤, merging-inflatingi mi≤ and inverse inclusion ⊇ orders: for

any π1, π2∈ Π∗(E), each of π1 m ≤ π2, π1 i ≤ π2, π1 mi ≤ π2 and π1⊇ π2implies π1⋐π2; we have also π1≤ π2 =⇒ π1⋐π2\ P back(π1)
⋐π2∩ P(supp(π1)) . (10) Furthermore, ∀ A ∈ P(E), π1⋐π2 =⇒ π1∩ P(A) ⋐π2∩ P(A) . (11)

For the analysis of our 3 new orders, we will require the following:

Proposition 3 The building order is generated by inverse inclusion followed by

inflat-ing: for any π1, π2∈ Π∗(E),

π1⋐π2 ⇐⇒ ∃ π ∈ Π∗(E), π1⊇ π

i ≤ π2 .

The least element is 0E, the greatest element is Ø; in Π∗(E)\ {Ø} (hence in Π(E)),

the greatest element is 1E. For any π0, π1, π2∈ Π∗(E),

 π0⋐π1⋐π2& π0 i ≤ π2  =⇒ π0 i ≤ π1 i ≤ π2 . (12)

0000 0000 0000 0000 1111 1111 1111 1111 000 000 000 000 000 000 000 000 000 111 111 111 111 111 111 111 111 111 00000000 00000000 11111111 11111111 0000 00 00 00 11 11 11 11 11 0000 0000 0000 0000 1111 1111 1111 1111 000 000 000 000 000 111 111 111 111 111 π1 π2 remove π inflate

Fig. 7 Illustration of Proposition 3: from π1⋐ π2 we get π1⊇ π i

≤ π2.

Proof If π1 ⊇ π

i

≤ π2, then by Proposition 2 we get π1 ⋐π ⋐π2, thus π1 ⋐π2.

Conversely, letπ1⋐π2. For eachC ∈ π2, choose oneB ∈ π1such thatB ⊆ C, and

setf(C) =B. Letπ={f(C)| C ∈ π2}; thenπ1⊇ π; as f(C)⊆ C for allC ∈ π2, π ≤ π2 and π⋐π2 by construction; now for any otherD ∈ π2, as f(D)⊆ Dand D∩C=∅, we havef(D)6⊆ C, henceπ⇚π2. Thusπ

i

≤ π2by (2). This construction π1⊇ π

i

≤ π2 is illustrated in Figure 7.

In a partial partition π, every block of π contains at least one singleton {p}; hence0_E⋐π. AsØ has no block,π⋐Ø. Whenπ ∈ Π∗(E)\ {Ø},πhas at least one blockB, included in E, soπ⋐1E, with1E∈ Π(E).

Let π0, π1, π2∈ Π∗(E) such thatπ0 ⋐π1⋐π2 andπ0

i

≤ π2; thusπ0≤ π2 and π0 ⇚ π2. We first show that for any A ∈ π0, there exist BA ∈ π1 and CA ∈ π2

such thatA ⊆ BA⊆ CA. Indeed, asπ0≤ π2, there isCA∈ π2 such thatA ⊆ CA;

as π1 ⋐ π2, there is BA ∈ π1 such that BA ⊆ CA; as π0 ⋐ π1, there is A′ ∈ π0

such that A′ ⊆ BA; nowA, A′ ∈ π0 and A, A′ ⊆ CA; asπ0 ⇚π2, we get A=A′,

henceA ⊆ BA⊆ CA. It follows thatπ0≤ π1. We next show that for any B ∈ π1

there existsC ∈ π2 such thatB ⊆ C. Indeed, asπ0⋐π1, there isA ∈ π0such that A ⊆ B; by the above, there are BA ∈ π1 and CA ∈ π2 such that A ⊆ BA ⊆ CA;

butB, BA ∈ π1 with ∅ 6=A ⊆ B, BA, so BA =B; hence B ⊆ CA. It follows that π1≤ π2. Since π0≤ π1≤ π2 andπ0⋐π1⋐π2, by (3) we haveπ0

mi ≤ π1 mi ≤ π2; now π0 i

≤ π2, so Equation (26) in Theorem 12 of [16] givesπ0

i ≤ π1

i

≤ π2. ⊓⊔

We see that for the building order, the growth of a partial partition is achieved by first removing some blocks, then inflating some of the remaining blocks. This is exactly what is done in Serra’s method [22, 23] for eliminating “small parasitic” segmentation classes, which we described in the Introduction. Since the merging order is included in the building order, a particular case is that merging blocks can be achieved by removing some blocks then inflating the remaining blocks, an elementary fact proven in item 2 of Proposition 10 of [16]. However, when

π1⊇ π

i

≤ π2, we have indeedπ1⋐π⋐π2, but alsoπ1≥ π ≤ π2; it is thus somewhat

counterintuitive to have two successive growths of a partial partition where the support first decreases fromπ1toπ, next increases fromπtoπ2. In the Appendix

we will exhibit some bizarre properties of the building order, in particular its strange covering chains, see Figure 15. In our opinion, this is probably related to the fact that the building order involves no constraint on the support of partial partitions. This will not happen with our three new orders: the apportioning order requires support equalitysupp(π1) =supp(π2), while the merging-apportioning and

0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 1111 1111 000000000000000 000000000000000 000000000000000 000000000000000 111111111111111 111111111111111 111111111111111 111111111111111 00000 00000 00000 00000 00000 00000 00000 11111 11111 11111 11111 11111 11111 11111 00000 00000 00000 00000 00000 00000 00000 11111 11111 11111 11111 11111 11111 11111 0000000000000000 0000000000000000 0000000000000000 0000000000000000 1111111111111111 1111111111111111 1111111111111111 1111111111111111 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 1111 1111 000000000000000 000000000000000 000000000000000 000000000000000 111111111111111 111111111111111 111111111111111 111111111111111

π

π

β

β

α

α

α

α

Fig. 8 From left to right: first π1, second π2; here π1 a

≤ π2, the blocks of π1 included in a

block of π2are hatched. Third and fourth, two possible choices for β, and the resulting α, in

order to apportion π1 into π2.

3.1 Apportioning as a generalization of block merging

We will study apportioning from two points of view: first as an operation that generalizes block merging, as it splits some blocks and merges their parts with other blocks; next as an order relation on Π∗(E) built from existing orders and quasi-orders. This double point of view, in terms of operations on blocks and of order relations, will also apply to theapportioning-inflating order.

Let π1 and π2 be two non-empty partial partitions having the same support,

and letπ2 be obtained from π1 by apportioning some of its blocks. We will

de-compose π1 into a set πa of blocks to be apportioned, and the non-empty set πr=π1\ πa of remaining blocks. A block ofπ1 that is split into parts will belong

toπa, and each part will be merged with a block ofπr; when several blocks ofπ1

are merged, one chooses one of them to be inπr, and the others will be inπa. Thus

each block ofπ2 will be made of one block ofπr, to which we add entire blocks, or

parts of blocks, ofπa. Hence a block ofπ2 contains at least one block of π1, but

exactly one of them belongs to πr; when this block of π2 contains several blocks B1, . . . , Bm ofπ1, we must choose oneBi to be remaining,Bi∈ πr, and the other

ones will be apportioned,Bj∈ πa forj 6=i. Thus the decomposition ofπ1 intoπa

andπr is not necessarily unique. So we postulate a mapβ:π2→ π1 that chooses

for each blockC ∈ π2, among all blocksB ∈ π1such thatB ⊆ C, the one that will

belong toπr; thusβ(C)⊆ C. Since a block ofπ1 cannot belong to several blocks

ofπ2,βwill be injective. Thenπr={β(C)| C ∈ π2}=β(π2) andπa =π1\ β(π2).

In order to describe the apportioning, we must indicate for every p ∈ supp(πa)

to which block of πr it will be added; this block will be written α(p). We define

thus a mapα:supp(πa)→ πrimplementing this apportioning, namely each block B ∈ πr will be enlarged by adding to itα−1(B) ={p ∈ supp(πa)| α(p) =B}. As πrand πa are given in terms of the mapβ, we haveα:supp π1\ β(π2)
→ β(π2),

and each block B ∈ β(π2) gives rise to a block B ∪ α−1(B)∈ π2. Conversely, for

everyC ∈ π2,C=β(C)∪ α−1(β(C)), whereβ(C) andα−1(β(C)) are disjoint. See

Figure 8. Thus π2= {B ∪ α−1(B)| B ∈ β(π2)}= {β(C)∪ α−1(β(C)) | C ∈ π2}.

Note that forC ∈ π2 andB ∈ π1\ β(π2), the portion ofB apportioned toβ(C) is B ∩ C=B ∩ α−1(β(C)).

In fact, only the injectionβ:π2→ π1 really matters, sinceαcan be uniquely

determined fromπ1,π2 andβ:

Lemma 4 Let π1, π2 ∈ Π∗(E)\ {Ø} and β : π2 → π1. Then the following two

conditions are equivalent:

1. supp(π1) =supp(π2)and for every C ∈ π2, β(C)⊆ C.

2. There is a map α:supp π1\ β(π2)
→ β(π2)such that for every C ∈ π2, C = β(C)∪ α−1_(β(C)).

Furthermore, β is an injection π2→ π1, and for every C ∈ π2 we have: C ∩ supp(β(π2)) =β(C) , C ∩ supp π1\ β(π2)
=C \ supp(β(π2)) =C \ β(C) .

(13)

Finally, the map α of item 2 is unique, it is given by:

∀ p ∈ supp π1\ β(π2)
, ∀ C ∈ π2, α(p) =β(C) ⇐⇒ p ∈ C . (14)

Proof First, suppose condition 1:supp(π1) =supp(π2) andβ(C)⊆ Cfor allC ∈ π2.

For two distinct C, D ∈ π2 we have β(C) ⊆ C, β(D) ⊆ D and C ∩ D = ∅, so β(C)6=β(D); henceβis injective. FixC ∈ π2; for any blockD ∈ π2 distinct from C,β(D)∩ C ⊆ D ∩ C=∅; thus C ∩ supp(β(π2)) =C ∩  _[ D∈π2 β(D)= [ D∈π2 C ∩ β(D)
=C ∩ β(C) =β(C) .

ThenC \ supp(β(π2)) =C \ C ∩ supp(β(π2))) =C \ β(C). Now

supp π1\ β(π2)
=supp(π1)\ supp(β(π2)) =supp(π2)\ supp(β(π2)) ,

withC ⊆ supp(π2). Hence (13) holds.

Forp ∈ supp π1\β(π2)
,p ∈ supp(π1) =supp(π2), sopbelongs to a unique block

ofπ2, and we can define α:supp π1\ β(π2)
→ β(π2) by (14). TakeC ∈ π2. For p ∈ α−1(β(C)), we have α(p) =β(C), sop ∈ C by (14); thusα−1(β(C))⊆ C. For

p ∈ C \ β(C),p ∈ supp π1\ β(π2)
by (13), soα(p) is defined; asp ∈ C, (14) gives α(p) =β(C), sop ∈ α−1(β(C)); thusC \ β(C)⊆ α−1_{(β(C)). From the inequalities} β(C)⊆ C andC \ β(C)⊆ α−1(β(C))⊆ C, we deduce thatC=β(C)∪ α−1(β(C)). Therefore condition 2 holds.

Next, suppose condition 2: we have α:supp π1\ β(π2)
→ β(π2) such that for

allC ∈ π2,C=β(C)∪ α−1(β(C)); thenβ(C)⊆ β(C)∪ α−1(β(C)) =C. Sinceαis supp π1\ β(π2)
→ β(π2), we have

[

C∈π2

α−1(β(C)) =α−1(β(π2)) =supp π1\ β(π2)
=supp(π1)\ supp(β(π2)) ;

thus supp(π2) = [ C∈π2  β(C)∪ α−1(β(C)) =h [ C∈π2 β(C)i∪h [ C∈π2 α−1(β(C))i =supp(β(π2))∪ supp(π1)\ supp(β(π2))
=supp(π1) .

Let us now show that αsatisfies (14). LetC ∈ π2 and p ∈ supp π1\ β(π2)
. If α(p) = β(C), thenp ∈ α−1(β(C)); but C =β(C)∪ α−1(β(C)), thus p ∈ C. Now ifα(p)6=β(C), then there is some D ∈ π2 such that α(p) = β(D), and we have D 6=C; by the argument in the previous sentence,p ∈ D, and as C ∩ D=∅, we getp /∈ C. Thereforeα(p) =β(C) ⇐⇒ p ∈ Cand (14) holds. ⊓⊔

A direct consequence of (13) is thatsupp π1\β(π2)
= S_C∈π2 C \β(C)
. When

the conditions of Lemma 4 are satisfied, we will say thatβ apportions π1 into π2.

We now define theapportioning order ≤a mathematically as the intersection of the building order and the support equality relation:

∀ π1, π2∈ Π∗(E), π1

a

≤ π2 ⇐⇒ π1⋐π2 &supp(π1) =supp(π2) . (15)

Note that here we do not requireπ1andπ2to be non-empty; in fact whenπ1

a ≤ π2, π1=Ø ⇔ π2=Ø. The two versions of apportioning are equivalent:

Proposition 5 Given π1, π2 ∈ Π∗(E)\ {Ø}, we have π1

a

≤ π2 if and only if there

exists β:π2→ π1 such that β apportions π1 into π2.

Proof Ifβ:π2→ π1apportionsπ1 intoπ2, thensupp(π1) =supp(π2) and for every C ∈ π2,β(C)⊆ C; asβ(C)∈ π1, this means thatπ1⋐π2. Hence π1

a ≤ π2.

Conversely, ifπ1

a

≤ π2, thensupp(π1) =supp(π2) and π1⋐π2. Thus for every C ∈ π2 there exists at least oneB ∈ π1 such that B ⊆ C; we choose one suchB

and set β(C) =B. Thenβ is a mapπ2→ π1 andβ(C)⊆ C for allC ∈ π2. Hence

βapportionsπ1 intoπ2. ⊓⊔

We haveπ1⊇ β(π2)

i

≤ π2 (cf. Proposition 3). As explained above, the mapβ: π2→ π1is not necessarily unique, see Figure 8. However the number of apportioned

blocks, that is, the size ofπ1\ β(π2), remains constant:

Lemma 6 Let π1, π2∈ Π∗(E)\ {Ø}, and β:π2→ π1such that β apportions π1into π2. Then:

1. π1=π2 ⇐⇒ β(π2) =π1 ⇐⇒ β(π2) =π2 .

2. For any other β′:π2→ π1 such that β′ apportions π1 into π2, there is a bijection

between π1\ β′(π2)and π1\ β(π2).

Proof 1. We have

β(π2) =π1 ⇐⇒ π1\ β(π2) =Ø ⇐⇒ supp π1\ β(π2)
=∅ .

Let α : supp π1\ β(π2)
→ β(π2) be the map associated to β in Lemma 4. If β(π2) = π1, then for any C ∈ π2 we have α−1(β(C))⊆ supp π1\ β(π2)
= ∅, so C=β(C)∪ α−1(β(C)) =β(C), and we getβ(π2) =π2. Henceπ1=β(π2) =π2.

Ifβ(π2)6=π1, thensupp π1\ β(π2)
6=∅, so there exists p ∈ supp π1\ β(π2)
,

and we getα(p) =β(C) for someC ∈ π2; thusα−1(β(C))6=∅, soβ(C)⊂ β(C)∪ α−1(β(C)) =C; this means thatβ(π2)6=π2, and asβ(C)∈ π1,π16=π2.

2. For any C ∈ π2, if β(C) = β′(C), then β(C) = β′(C) ∈ β(π2)∩ β′(π2).

β′(D); butβ(C)⊆ C andβ′(D)⊆ D, soB ⊆ C ∩ D, henceC=DandB=β(C) =

β′(C). Thus forC ∈ π2, ifβ(C)6=β′(C), thenβ(C), β′(C)∈ β/ (π2)∩ β′(π2), hence β(C)∈ β(π2)\ β′(π2) and β′(C)∈ β′(π2)\ β(π2). Therefore we have a bijection

betweenβ(π2)\ β′(π2) andβ′(π2)\ β(π2), associatingβ(C) toβ′(C) for allC ∈ π2

such thatβ(C)6=β′(C). Combining it with the identity on π1\ β(π2)∪ β′(π2)
,

we get a bijection between

π1\ β′(π2) = β(π2)\ β′(π2)
∪hπ1\ β(π2)∪ β′(π2)
i and π1\ β(π2) = β′(π2)\ β(π2)
∪ h π1\ β(π2)∪ β′(π2)
i . ⊓ ⊔

Let π1, π2 ∈ Π∗(E)\ {Ø}. By item 2, when π1

a

≤ π2, for any β : π2 → π1

that apportions π1 into π2, the cardinal |π1\ β(π2)|, which gives the number of

blocks that are apportioned to other blocks, will remain constant. We call that constant cardinal the apportioning index ofπ1 into π2, and write it I(π1, π2). We

also trivially defineI(Ø, Ø) = 0.

By item 1, if π1 = π2 we have π1 = β(π2) = π2, while if π1

a < π2 we have π1⊃ β(π2) i < π2. Thus π1 a

< π2 ⇐⇒ supp(π1) =supp(π2) &∃ π ∈ Π∗(E), π1⊃ π

i

< π2 . (16)

AlsoI(π1, π2) = 0 ⇐⇒ π1=π2(this remains true forπ1=π2=Ø.)

Now in a composition of apportionings, the apportioning indexes add up:

Proposition 7 Let π0, π1, π2 ∈ Π∗(E)\ {Ø}. Suppose that β1 apportions π0 into π1 and β2 apportions π1 into π2. Then β1β2 apportions π0 into π2, and I(π0, π2) = I(π0, π1) +I(π1, π2).

Let α1:supp π0\ β1(π1)
→ β(π2), α2 :supp π1\ β2(π2)
→ β(π2) and α12: supp π0\β1β2(π2)
→ β1β2(π2)be the maps associated to β1, β2and β1β2respectively,

according to (14). The domain of definition of α12is the union of those of α1and α2: supp π0\ β1β2(π2)
=supp π0\ β1(π1)
∪ supp π1\ β2(π2)
. (17)

For p ∈ supp π0\ β1β2(π2)
, α12(p)depends on the last apportioning to which the p

is subjected (see Figure 9):

1. If p belongs to a block of π0that is apportioned and to a block of π1that is enlarged,

i.e., if p ∈ supp π0\ β1(π1)
and p /∈ supp π1\ β2(π2)
, then α12(p) =α1(p).

2. If p belongs to a block of π1 that is apportioned, i.e., if p ∈ supp π1\ β2(π2)
, then α12(p) =β1α2(p).

Proof Clearly β1β2 is an injection π2 → π0 and for every C ∈ π2, β1β2(C) ⊆ β2(C)⊆ C. Also supp(π0) =supp(π1) =supp(π2). Thusβ1β2 apportions π0 into π2.

Since β1 is injective, its restriction toπ1\ β2(π2) is a bijectionπ1\ β2(π2)→ β1(π1)\ β1β2(π2), so|π1\ β2(π2)|=|β1(π1)\ β1β2(π2)|. Nowπ0\ β1β2(π2) is the

disjoint union ofπ0\ β1(π1) andβ1(π1)\ β1β2(π2). Thus

I(π0, π2) =|π0\ β1β2(π2)|=|π0\ β1(π1)|+|β1(π1)\ β1β2(π2)|

π

π

π

π

β

β

β

α

α

α

Fig. 9 From left to right: π0, then π1, with β1 apportioning π0 into π1, next π2, with β2

apportioning π1 into π2. Finally we show the apportioning of π0 into π2, with the map α12

associated to β1β2, given by α12(p) = α1(p) for p ∈ supp(β2(π2)) \ supp(β1(π1)), and α12(p) =

β1α2(p) for p ∈ supp(π0) \ supp(β2(π2)).

Letp ∈ supp(π0). Ifp ∈ supp π0\β1(π1)
, then we havep /∈ β1(B) for allB ∈ π1,

in particular for any C ∈ π2, β2(C) ∈ π1, so p /∈ β1β2(C), hencep ∈ supp π0\ β1β2(π2)
; thus (a)supp π0\β1(π1)
⊆ supp π0\β1β2(π2)
. Ifp ∈ supp π1\β2(π2)
,

then we havep /∈ β2(C) for allC ∈ π2, and asβ1β2(C)⊆ β2(C),p /∈ β1β2(C), hence p ∈ supp π0\ β1β2(π2)
; thus (b) supp π1\ β2(π2)
⊆ supp π0\ β1β2(π2)
. Now

suppose thatp ∈ supp π0\β1β2(π2)
andp /∈ supp π0\β1(π1)
, sop /∈ β1β2(C) for all C ∈ π2, but there existsB ∈ π1such thatp ∈ β1(B); thenB 6=β2(C) for allC ∈ π2;

asβ1(B)⊆ B, we getp ∈ B, sop /∈ β2(C) for allC ∈ π2, hencep ∈ supp π1\β2(π2)
;

thus (c)supp π0\ β1β2(π2)
⊆ supp π0\ β1(π1)
∪ supp π1\ β2(π2)
. From the three

inclusions (a,b,c), the equality (17) follows.

Letp ∈ supp π0\ β1β2(π2)
. Ifp ∈ supp π0\ β1(π1)
andp /∈ supp π1\ β2(π2)
,

there is someC ∈ π2such thatp ∈ β2(C) butp /∈ β1β2(C); then (14) applied toβ1

givesα1(p) =β1β2(C); nowp ∈ C, so (14) applied toβ1β2givesα12(p) =β1β2(C).

Thus α12(p) = α1(p) and item 1 holds. If p ∈ supp π1\ β2(π2)
, there is some C ∈ π2 such that p ∈ C but p /∈ β2(C), so p /∈ β1β2(C); then (14) applied to β1β2 givesα12(p) =β1β2(C), while (14) applied toβ2 givesα2(p) =β2(C), hence β1α2(p) =β1β2(C). Thusα12(p) =β1α2(p) and item 2 holds. ⊓⊔

Note that in case 2, the result does not depend on whether p belongs to a block of π0 that is apportioned or one that is enlarged, i.e., whether p ∈ or p /∈ supp π0\ β1(π1)
, see Figure 9.

Now we complement Proposition 7 by showing that when the apportioning in-dex is>1, the apportioning can be decomposed into two successive apportionings:

Lemma 8 Let π0, π2 ∈ Π∗(E)\ {Ø}, and let c, d be two integers >0. Suppose that β apportions π0 into π2, with I(π0, π2) =c+d. Then there exists π1∈ Π∗(E)\ {Ø}

and two maps βc:π1→ π0 and βd:π2→ π1 such that βc apportions π0 into π1, βd

apportions π1 into π2, β=βcβd, I(π0, π1) =c and I(π1, π2) =d.

Proof Since β(π2) ⊆ π0 and |π0\ β(π2)| = I(π0, π2) = c+d, we can partition π0\ β(π2) into two partial partitionsπc and πd with |πc|= cand |πd|=d; thus π0=β(π2)∪πc∪πd. LetV =supp(β(π2)) andW =supp β(π2)∪πc
=supp(β(π2))∪

2

π

π

π

π

π

π

π

π

β

β

β

Fig. 10 From left to right: π0, then π2, with β apportioning π0into π2and π0\β(π2) = πc∪πd

(here c = d = 1); next π1, with βdapportioning π₁ into π₂ and π₁\ βd(π₂) = πd, and finally

π0, with βcapportioning π₀into π₁ and π₀\ βc(π₁) = πc. We have β = βcβd.

supp(πc) = V ∪ supp(πc); thus V ⊆ W, W ∩ supp(πd) = ∅ and W ∪ supp(πd) = supp(β(π2))∪ supp(πc)∪ supp(πd) =supp(π0).

By (13), for anyC ∈ π2 we haveβ(C) =C ∩ V; thus ∅ ⊂ β(C)⊆ C ∩ W ⊆ C.

Letπ∗={C ∩ W | C ∈ π2}, then π∗ is a partial partition and supp(π∗) = [

C∈π2

(C ∩ W) = [

C∈π2

C∩ W =supp(π2)∩ W =W .

Let π1 = π∗∪ πd; as W ∩ supp(πd) = ∅, π1 ∈ Π∗(E) and supp(π1) = W ∪ supp(πd) =supp(π0). Defineβd:π2→ π1:C 7→ C ∩ W; thenβd apportionsπ1 into π2 and π1\ βd(π2) = π1\ π∗ = πd, so I(π1, π2) = |πd|= d. Define βc : π1 → π0

by βc(B) = B for B ∈ πd, and βc(B) = B ∩ V for B ∈ π∗; in other words, for C ∈ π2we haveβc(C ∩ W) =C ∩ V =β(C). Thusβc(πd) =πd andβc(π∗) =β(π2),

so βc(π1) = βc(π∗)∪ βc(πd) = β(π2)∪ πd. Then βc apportions π0 into π1 and π0\ βc(π1) =π0\ β(π2)∪ πd
=πc, henceI(π0, π1) =|πc|=c. Now for allC ∈ π2, βc(βd(C)) =βc(C ∩ W) =C ∩ V =β(C), henceβ=βcβd. See Figure 10. ⊓⊔

We now define the covering relation corresponding to the apportioning order

a

≤. Forπ1, π2 ∈ Π∗(E)\ {Ø}, we writeπ1

a

≺ π2 and say that π2 a-covers π1, ifπ2

is obtained by apportioning a single block ofπ1among remaining blocks, that is, π1

a

< π2 andI(π1, π2) = 1. In other words: π1

a

≺ π2 ⇐⇒

"

supp(π1) =supp(π2) &|π1| ≥2

&∃ A ∈ π1, π1\ {A}

i < π2

#

. (18)

We give the form of (18) with the mapαof Lemma 4:

π1

a

≺ π2 ⇐⇒



|π1| ≥2 &∃ A ∈ π1 &∃ α:A → π1\ {A}, π2=B ∪ α−1(B)| B ∈ π1\ {A}



. (19)

We can now give the main properties of the apportioning order:

Theorem 9 Apportioning≤ is a partial order relation on Πa ∗(E); it is included in the building order and it contains the merging order: for any π1, π2∈ Π∗(E), π1

m ≤ π2 ⇒ π1

a

The poset (Π∗(E),≤a) is the disjoint union of the posets (Π(A),⋐)for all A ∈

P(E), where for distinct A, A′ ∈ P(E), elements of Π(A) and Π(A′) are mutually incomparable. The maximal and minimal elements are all 1Aand 0A respectively, for A ∈ P(E); every π ∈ Π∗(E) majorates a unique minimal element, namely 0supp(π).

The covering relation is≺.a

Let E be finite. Then(Π∗(E),≤a)is graded by hm, that is, for any π1, π2∈ Π∗(E)

we have π1 a < π2 =⇒ hm(π1)< hm(π2) , π1 a ≺ π2 =⇒ hm(π2) =hm(π1) + 1 .

For π ∈ Π∗(E), the height of π w.r.t. 0supp(π)is hm(π). For π1, π2∈ Π∗(E)such that π1

a

≤ π2, we have I(π1, π2) =hm(π2)− hm(π1).

Proof By (15), ≤a is the intersection of the building order ⋐ (a partial order by Proposition 2) and of the quasi-order given by support equality; it is thus a partial order relation included in ⋐. The merging order≤mis included in the building order (by Proposition 2) and in the support equality, cf. (1); hence≤mis included in the intersection of the two, namely the apportioning order≤a.

ObviouslyΠ∗(E) is the disjoint union of theΠ(A) for allA ∈ P(E), andπ1

a ≤ π2

means that for A =supp(π1) =supp(π2), we have A ∈ P(E) withπ1, π2 ∈ Π(A)

andπ1⋐π2. ForA ∈ P(E) andπ ∈ Π(A), we have0A⋐π⋐1Awith supp(0A) = supp(π) =supp(1A), thus0A and1A are the least and greatest elements inΠ(A),

so they are respectively minimal and maximal elements inΠ∗(E), and all minimal and maximal elements in Π∗(E) are of this form. For π ∈ Π∗(E), the unique minimal element majorated byπis0A forA=supp(π).

Letπ0, π2∈ Π∗(E) such thatπ0

a

< π2. Thenπ0, π2∈ Π∗(E)\{Ø}. By Lemma 6, I(π0, π2) ≥ 1. If π2 does not cover π0, then there is π1 ∈ Π∗(E)\ {Ø} such

that π0

a < π1

a

< π2; by Lemma 6, I(π0, π1) ≥ 1 and I(π1, π2) ≥ 1, and then

Proposition 7 givesI(π0, π2) =I(π0, π1)+I(π1, π2)≥2. Conversely, ifI(π0, π2)≥2,

by Lemma 8 there isπ1∈ Π∗(E)\ {Ø} such thatπ0

a ≤ π1 a ≤ π2,I(π0, π1) = 1 and I(π1, π2) =I(π0, π2)−1≥1; by Lemma 6, π0 a < π1 a

< π2, thusπ2 does not cover π1. We have thus shown thatπ2 does not cover π1 if and only if I(π0, π2) ≥2,

thereforeπ2 coversπ1 if and only ifI(π0, π2) = 1, that is, π1

a ≺ π2.

LetE be finite andπ1, π2∈ Π∗(E) withπ1

a

≤ π2. Since|supp(π1)|=|supp(π2)|,

by (8) we get

hm(π2)− hm(π1) = |supp(π2)| − |π2|
− |supp(π1)| − |π1|
=|π1| − |π2| .

If π1 = π2 = Ø, then I(π1, π2) = I(Ø, Ø) = 0 = hm(Ø)− hm(Ø); otherwise, π1, π2∈ Π∗(E)\ {∅}, and givenβ that apportionsπ1 intoπ2,

I(π1, π2) =|π1\ β(π2)|=|π1| − |β(π2)|=|π1| − |π2|=hm(π2)− hm(π1) . If π1 a < π2, then I(π1, π2) > 0, hence hm(π2)− hm(π1) > 0. If π1 a ≺ π2, then I(π1, π2) = 1, hence hm(π2)− hm(π1) = 1. Therefore (Π∗(E), a ≤) is graded byhm.

For π ∈ Π∗(E), hm(0supp(π)) = 0, cf. Table 1, so the height ofπ w.r.t.0supp(π) is

We note that the statement of Theorem 9 is similar to the corresponding one for the merging order (cf. Table 1, see also Theorem 7 of [16] for more details), where we only replace≤m and≺m by≤a and≺a.

The apportioning order does not constitute a lattice. Indeed, let{A, B, C, D} ∈ Π∗(E), i.e.,A, B, C, Dare mutually disjoint non-empty subsets ofE. Then each of the 3 partial partitions obtained by mergingA, B, C, Dby pairs will cover each of the 6 partial partitions obtained by merging just two ofA, B, C, D, that is:

{A ∪ B, C, D} , {C ∪ D, A, B} , {A ∪ C, B, D} , {B ∪ D, A, C} , {A ∪ D, B, C} , {B ∪ C, A, D}    a ≺    {A ∪ B, C ∪ D} , {A ∪ C, B ∪ D} , {A ∪ D, B ∪ C} .

This is impossible in a lattice, because each of the 3 greater ones is a minimal upper bound of the 6 lower ones, and each of the 6 lower ones is a maximal lower bound of the 3 greater ones.

Now consider the following 4 partial partitions built fromA, B, C, D:

π0={A, B, C, D} , π1={A ∪ B, C, D} , π2={A ∪ C, B ∪ D} , π3={A ∪ B ∪ C ∪ D} .

Then π0 ≺ πm 1 _{≺ π}a 2 _{≺ π}m 3_{, π}0 _{< π}m 2_{, π}1 _{< π}m 3 _{and π}0 m_{< π}3_{. This shows that we}

cannot separate apportioning from merging in that order.

In Property 5 of [16] we gave an example showing that (when |E| ≥ 5) the intersection of the apportioning order and of the singularity relation is not transi-tive. We recall it here. Let{J, K, L, M, N } ∈ Π∗(E), i.e.,J, K, L, M, Nare mutually disjoint non-empty subsets ofE, and let

π0={J, K, L ∪ M, N } , π1={J, K ∪ L, M ∪ N } , π2={J ∪ K, L ∪ M ∪ N } . Then: π0 a ≺ π1 a ≺ π2, π0 6 m ≺ π1 6 m ≺ π2, π0 ⇚ π1 ⇚ π2, π0 m < π2 and π0 6⇚ π2. In

other words,π1(resp.,π2) is obtained by apportioning one block ofπ0 (resp.,π1)

to two other blocks, and not by merging blocks, thus every block ofπ1 (resp.,π2)

contains exactly one block ofπ0(resp.,π1); however, each of the two blocks ofπ2

is obtained by merging two blocks ofπ0.

The apportioning order was initially defined by the succession of two oper-ations: first removing some block, next inflating the remaining blocks until the initial support of the partial partition is covered. One can admit the possibility of continuing to inflate these remaining blocks beyond the recovery of the initial sup-port; this leads to theapportioning-inflating order ≤ai that we define mathematically as the intersection of the building order and the support inclusion relation:

∀ π1, π2∈ Π∗(E), π1

ai

≤ π2 ⇐⇒ π1⋐π2 &supp(π1)⊆ supp(π2) . (20)

Before giving an interpretation of it in terms of a map β as in Lemma 4, we first give its main order-theoretic properties:

Theorem 10 Apportioning-inflating is a partial order relation on Π∗(E); it is in-cluded in the building order and it contains the apportioning and merging-inflating orders: for any π1, π2 ∈ Π∗(E), π1

a ≤ π2 ⇒ π1 ai ≤ π2, π1 mi ≤ π2 ⇒ π1 ai ≤ π2 and