The results of the paper are shown in this section. First, the equivalence relation and the partial order relation “faster” on the equivalence classes of neighborhood sequences are shown, and then some algebraic properties of this partial order relation are presented. Finally, the “componentwise dominate” relation will be investigated with details.
4.2. A Partial Order Defined by B-Distances
Since there are several, there are actually infinitely many distance functions that are defined by neighborhood sequences; thus, the question arises of how they are related to each other. The next definition is recalled from [
9].
Definition 8. For any , we define the relation “faster” denoted by in the following way: for all points .
The relation can be understood on the set of equivalence classes of neighborhood sequences as well; however, for simplicity, we use the above notations, where and represent their equivalence sets, respectively. When we are talking about periodic neighborhood sequences, then only the elements of are counted.
Notice that equality is allowed by the relation ; thus, the term “is faster than or has the same speed” would match better to it; however, for traditional reasons, we use the short term “faster”.
Similarly, the relation
is defined on the sets of
S and
, and it was studied in [
8,
13] as it also plays an important role in the metricity condition.
Proposition 2 (Fazekas et. al. [
8]).
On the square grid forms a complete distributive lattice. On the other hand, is not a lattice. The equivalence classes for the neighborhood sequences on the square grid are singleton sets, i.e., each of them has exactly one element, and therefore this element can be used to represent this class or used instead of the class itself.
The next examples demonstrate the non-triviality of our partial order for the case of periodic neighborhood sequences.
Example 2. Let such that , . Further, let the points , , and . Then, , but . Therefore, the functions -distance and -distance are not comparable, in general.
We should note here that similar examples can also be constructed on the square and other grids using B-distances. Our next example, however, does not work on point lattices but does work on the triangular grid.
Example 3. Let , and the points , and . Then, one may see that but .
This example shows that, although we have the same sum of the values in the sequences, (i.e.,
), their order makes a difference on the distance: the steps of a shortest path may not commute [
33]. We note here that, on the square- and
nD grids the
B-distance of the point-pair
depends only on the multiset of used elements of the neighborhood sequence and does not depend on the order of the first
elements of the neighborhood sequence [
13,
14].
Now, after these introductory examples and thoughts, let us see how and which neighborhood sequences are comparable on the triangular grid.
Theorem 2. The relation is a partial order on the sets of equivalence classes of T and .
Proof. By definition, the relation is reflexive and transitive. In defining it for the equivalence sets, it is clear that, if and , then , and thus ; therefore, they represent the same class. □
As we have previously seen in Example 3, the neighborhood sequences
and
are not comparable under the relation
. In terms of digital disks, this means that
and
; see also
Figure 2, where digital disks with radii up to 3 are presented with the (periodic) neighborhood sequences
,
and
.
Based on Lemma 2, and by observing the digital disks obtained by neighborhood sequences (see [
34,
35,
40] for details), we may consider the following important observations.
Proposition 3. In the triangular grid, from any point
- (a)
By a 2-step, exactly the same region (i.e., set of points) is reached as by two consecutive 1-steps.
- (b)
By two consecutive 2-steps, a larger region can be reached than by using a 1-step and a 3-step in any order.
We can transform part (b) of the above proposition into the next lemma about the partial order .
Lemma 4. The following statements hold for the specific periodic neighborhood sequences mentioned below. However, and .
Proof. The first and the second statements follow from the formula of Lemma 2.
The first part of the last statement of the lemma is proven, for instance, by looking the points that can be reached in one step: in one step, a larger set of points is reached by than by , i.e., . To prove the other part, consider the B-distances of the points and . By the formula of Lemma 2, and , proving the incomparability of the neighborhood sequences and under on the triangular grid. □
Now, we establish our results on the structure of
T and
under
. Some of the results may seem simple or easy to prove; however, they give important information about the sets
T and
of neighborhood sequences. They also underline the differences between the distances defined on the square and on the triangular grids.
Figure 3 shows the partial order
on the sets of periodic neighborhood sequences with periods 1, 2 and 3 for both of the square and triangular grids. As one can observe, the diagram of
with period 2 already appears more complicated than the diagram of
with period 3 (the diagrams on the top are recalled from [
7]). The Hasse diagrams shown here compress a number of comparisons of
B-distances.
One can observe on
Figure 3, bottom-right, that, for instance, the neighborhood sequence
is faster than
,
,
,
and all the neighborhood sequences below them, i.e.,
,
,
,
,
,
,
,
and
. By, e.g., the formula of Lemma 2, one can see that there is no point that can be reached in one step by any of the listed neighborhood sequences
B but not with the neighborhood sequence
. That could be the base of the induction. The inheritance can be proven for each particular case (applying also, e.g., Proposition 3) by checking the sums of the prefix of the (minimal equivalent) neighborhood sequence
B and comparing the same length prefix of
, i.e., to the number that is twice the length of the prefix.
On the other hand, , , , , , and are faster than . All those relations can be seen since any path defined by is also a path with any of the listed neighborhood sequences; thus, the distance determined by cannot be smaller than the B-distance defined by another neighborhood sequence in this list. (Trivially, is faster than itself. We mention this only for the sake of completeness).
Further, there are neighborhood sequences such that they are incomparable with under the relation faster. Each of those neighborhood sequences B are listed below with two points such that the B-distance of exactly one of the points, p, is smaller from the origin than its -distance, while, for the second point, q, its B-distance is larger than its -distance from the origin.
, and ;
, and ;
, and ;
, and ;
, same points as above;
, and .
The relations among other neighborhood sequences can be shown in a similar manner.
Now, as the first of our main theoretical results about , we show the following.
Theorem 3. The set of periodic neighborhood sequences is not a lattice.
Proof. Let
and
be periodic neighborhood sequences. Notice that
, while
The least upper bound of these sequences in can be written as
, which belongs to the equivalence set
Notice that contains only three values of 2. Clearly, there is no element of that is in . Now, let us suppose, on the contrary, that there is a neighborhood sequence in and for some value . Then, in T but ; therefore, there must be such that holds.
may contain an element larger than 2 among its first three elements. (Clearly, it cannot contain values smaller than 2 at these positions without contradicting the fact that .) However, in this case, for the periodic sequence , we have (and ), but (as one can easily check) and . That leads to a contradiction.
In the other case, it means that there is a value (for a value ) such that ; however, . Without loss of generality, suppose that, for all : holds. Then, defining for and as a periodic neighborhood sequence with period , clearly and . However, in this case, , which leads to a contradiction.
Therefore, cannot be in . □
By the previous proof, we demonstrated the following fact.
Corollary 1. Even if the supremum of two periodic neighborhood sequences exists, it may not be periodic.
By the following theorem, we show the case with the set of general neighborhood sequences.
Theorem 4. The set of neighborhood sequences on the triangular grid is not a lattice.
Proof. Let
and
(The last two values are repeated for each of these sequences up to infinity). Observe that
and
form a singleton equivalence set under ∼. Clearly,
and
. Let
such that
and
. Then, one can easily infer that
. Further, the first two elements of
can be at most 2. Let us assume that the first difference of
and
occurs at their
st position, i.e.,
for all
; however,
. Then,
would imply that
. It can only be if
and
, which contradicts
.
Clearly, but (since the order of elements 1 and 3 matters). Therefore, and have no greatest lower bound, completing the proof that T is not a lattice. □
In the previous proof, we used non-periodic sequences. Thus, it seems that none of T and form a lattice under the relation . This also highlights an important difference between the square and the triangular grid. In the next subsection, another relation will be investigated on T and .
4.3. Lattices
The results of the previous subsection recommend that, instead of the relation
, we may use another relation to find some structure in
T and in
. Now, we adopt another partial order relation that was investigated in [
8] for
S and
and is called “componentwise dominates” in [
5] for some subsets of
. We have also seen that it plays a natural role when we defined and used minimal equivalent neighborhood sequences in
T.
Definition 9. For any two neighborhood sequences , , the relation ⊒ is defined in the following way: Let us have a more detailed view of the structure that this relation gives to the sets
S and
on the square grid. It should be noted that
is a proper refinement of ⊒ in
S and
; more formally, we can state the following results (based, e.g., on the formula to compute
B-distances on the square grid [
14] or on the properties of a faster relation from [
13]).
Proposition 4. Let . If , then . Furthermore, if and , then .
Proof. It is a corollary of the formula to compute B-distances on the square grid; see Proposition 1 and Definition 9. □
Moreover, we state the following.
Proposition 5. Let . If and , then and cannot be at the same time, i.e., at most, one of them may hold.
Proof. Generally, for , there could be three cases, either both and hold, or maybe exactly one of them, or maybe none of them.
If the condition and holds:
We may have that and are incomparable under . Let us consider, e.g., and . Then, considering the points and , their -distance is 2; however, their -distance is 3. On the other hand, by considering and , their -distance is 4, while their -distance is 5.
However, it may also happen that, e.g., , and we check the case, e.g., with and . (By changing the role of and , the opposite relation can be fulfilled.)
Now, to complete the proof, let us assume that and for . It follows that . Thus, there is a position i where the first mismatch occurs in these two sequences, i.e., for all but . Then, considering the points and with , i.e., k is the number of 2s among the first elements of (and also of since their long prefix sequence are identical). Then, one of the - and -distances of these points is i, the other is , and hence it cannot be that both and hold. □
Moreover, the following results are known.
Proposition 6 (Fazekas et al. [
8]).
is a complete distributive lattice with the greatest lower bound and least upper bound . is a distributive lattice with the greatest lower bound and least upper bound ; however, it is not a complete lattice.
Our first result in this subsection is a kind of strengthening of (the first part of) the previous result on the square grid.
Theorem 5. is a Boolean lattice (Boolean algebra).
Proof. A Boolean algebra, by definition, is a complemented distributive lattice. By Proposition 6, is a complete distributive lattice, and we need to show only that it is also complemented. Having exactly two types of neighborhood, the complementary relation can easily be defined on them: with 1 and 2 as complements of each other. This complementarity relation can also be extended to the neighborhood sequences: the complement of the neighborhood sequence , contains the opposite element in every of its positions. In this way, both the axioms of complements: and are satisfied for every neighborhood sequence B on the square grid. Thus, the statement is proven. □
Now, we turn to analyze the componentwise dominate relation on neighborhood sequences of the triangular grid. In the following part, the sets T and are analyzed under ⊒.
Theorem 6. is a complete distributive lattice with the greatest lower bound and least upper bound .
Proof. By Definition 9, ⊒ is a partial order on T. Moreover, for any , , we can define as and as . Furthermore, for any set P of neighborhood sequences , their greatest lower bound
and least upper bound
is determined; thus, the completeness of the lattice is shown. This implies also that and . Finally, the distributivity of the lattice is straightforward. □
However, there is a remarkable difference between the T and S under ⊒.
Proposition 7. is not a Boolean lattice with and where and .
Proof. In the proof of the previous theorem, we previously proved that is a distributive lattice with the given operators. We need to show that the complement operation cannot be introduced on T fulfilling the required axioms. Let , if there is its complement , then both and would be fulfilled. However, then should be to make true; however, would be required to fulfill . This contradiction proves the result. □
Now, we continue with a result on .
Theorem 7. is a distributive lattice with the greatest lower bound and least upper bound .
Proof. The proof is similar to the proof of the previous theorem except the completeness by noting that the constructions mentioned there provide periodic neighborhood sequences with period of the least common multiplier of the periods of the given neighborhood sequences. □
In the next part of the paper, we establish connections on the relations studied in this paper. Let us see how the ordering ⊒ and the sets T and can be connected to the B-distances. First, we use the relation ⊒ on specific subsets of T.
Theorem 8. Let B be any neighborhood sequence in T. The relation ⊒ defines a complete distributive lattice on .
Proof. The reflexivity, the antisymmetry and the transitivity follows from the the definition of ⊒. Therefore, is a partially ordered set for any .
It is clear that, for every , and exist in T and can be defined as , and .
It can be checked (e.g., by using Lemma 1) that both and are in . Thus, is a lattice.
Let be a subset of with some index set K. Let now, . Clearly, is the least upper bound of the subset X, and thus is a complete lattice.
The statement that this lattice is distributive is trivial. Moreover, , the minimal equivalent neighborhood sequence for B, and
Thus, the proof of the theorem is completed. □
The neighborhood sequence defined by the last formula of the previous proof can be called the maximal equivalent neighborhood sequence of B. Analogously to Lemma 1, it can be shown that it is uniquely defined for every . The case of periodic neighborhood sequences is more interesting due to the fact that there are periodic neighborhood sequences such that their minimal/maximal equivalent neighborhood sequences are not periodic and vice versa.
Example 4. is periodic. It is the same as its maximal equivalent neighborhood sequence. However, its minimal equivalent neighborhood sequence is not periodic, it is (having only 2’s after the first 3). The next non-periodic neighborhood sequence belongs to the same equivalence class: having 3 at every place i where for some .
Example 5. Let , and then its minimal equivalent neighborhood sequence is the same; however, its maximal equivalent neighborhood sequence,
, is not periodic (having exactly one value 2).
Therefore, it is interesting to analyze what we can say about subsets of T containing only periodic neighborhood sequences.
Theorem 9. Let B be any neighborhood sequence in . The relation ⊒ defines a distributive lattice on that may not be complete.
Proof. Since ⊒ is a partial order on any set (see (the proof of) Theorem 8), it is a partial order on the subset . Now, we show that, for any two representatives and of the class have and elements in . Let and be the periods of and , respectively. For simplicity, let . (The least common multiplier of and suffices.) Then, let us construct with period l: . Similarly, the construction .
Therefore, form a lattice for any . By the constructions above, it is clear that this lattice is distributive.
Now, we show an example for that proves that this lattice is not complete.
Let . Then, contains every neighborhood sequence that have the following properties: , and for every .
Now, a monotonously increasing and a monotonously decreasing sequence in will be shown, such that they have a common “limit sequence” but is not in . In this way, we show that is not complete. For technical purposes, we introduce the following notation. Any periodic neighborhood sequence can also be written in the form having period : , and this form is denoted by .
Now, for any neighborhood sequence that has period l, i.e., , and for any with let defined with the elements for and . In this way, only the i-th (modulo l) element of B is changed from 2 to 3 or vice versa, each other element remains the same.
Further, for any , we say that k is even-square-bounded (ESB) if there is an even such that , else k is odd-square-bounded (OSB). Clearly, the even-square-bounded and the odd-square-bounded numbers form longer and longer intervals in as they are growing. Let , , and now we define the sequences and in as follows. For an ESB , use and , else (in case k is OSB) let and .
In this way, to describe and , we give the long periods of and after and , respectively, such that exactly one of the -th long periods is modified at the -th position; where the modified period depends on the ESBness of k. For the sake of a better view, we show some of the first constructed values of these sequences:
,
,
,
and
,
,
,
,
.
As one may observe, and as can also be seen from the construction, every element of both sequences for is in , i.e., and for all non-negative values of k. It is also clear that and for every non-negative k, moreover, in both sequences, there are strict inequalities infinitely often. However, by comparing the elements of the two sequences with the same index k, from the construction, we have the following. They both have long periods, i.e., and .
Moreover, their elements for every and for every (the mismatch is at the last position of the long period). Then, for each , let us choose a value with and let and, further, . By the previously analyzed property of the sequences and , the sequence is well-defined. It is clear that holds. Now, we prove that is not periodic.
The definitions of and imply that, for each non-negative j with , and , but . However, there is no periodic sequence with this property.
Finally, let . It is clear that . Now, suppose that in for some . However, in this case, in (such as in T) but ; therefore, there is such that . However, this would yield in , if .
However, by , for each , which contradicts in . Thus, there is no least upper bound for X in ; therefore, is not complete. □
Now, we are ready to complement the results of Theorem 7.
Theorem 10. is not a complete lattice.
Proof. Observe that, in the previous proof, we demonstrated that is not complete as all the neighborhood sequences constructed were in ; however, as , the mentioned property is also inherited to itself. □
Thus, the sets T and have some similar properties under the partial order ⊒ as S and with the differences mentioned.
We are closing in on results that connect the orderings and ⊒ on the triangular grid. These results, with respect to Propositions 4 and 5, show that these relations give a more complex (theoretical) structure on the triangular grid compared with on the square grid.
Theorem 11. Let . The following statements hold:
- (a)
If , then .
- (b)
Moreover, and both may hold, even if and .
- (c)
Further, if none of and hold, all the following possibilities may occur:
both and hold, i.e., and have the same “speed” according to the faster relation;
exactly one of and holds, i.e., one of them is (really) faster (and not equally fast) compared with the other; or
none of and hold: and are also incomparable under the faster relation.
Proof. From Part (a) from , it follows that every -path is also a -path, and thus is immediately proven.
If the class contains more than one element, then one can easily pick two elements from there, e.g., the minimal and the maximal equivalent neighborhood sequences and , respectively. Then, clearly , but . On the other hand, since these neighborhood sequences are equivalent (; they provide the same distance function), both and hold.
Finally, the proof of part (c) is by examples.
Considering and , it is clear that none of and hold; however, they provide the same distance function with their minimal equivalent neighbor sequence with exactly one value of 3 and all other values are 2. Thus, both and hold in this case.
Now, consider and . Clearly, and . On the other hand, the minimal equivalent neighborhood sequence of is contains exactly one value of 1 and exactly one value of 3 and then only 2s. The minimal equivalent neighborhood sequence of is itself. These facts imply that is faster than but not vice versa, i.e., and .
To show the last possibility, let
and
. One may observe, e.g., on the bottom right of
Figure 3 that these two neighborhood sequences are incomparable under the faster relation, i.e.,
and
. □
In the next result, the equivalence relation ∼ is also involved.
Theorem 12. Let every class be represented by the minimal (maximal) equivalent neighborhood sequence (, respectively) of the class. Then, the relation ⊒ provides a proper refinement of defined on the sets of these classes of T, i.e., for any two neighborhood sequences that are minimal (resp. maximal) equivalent neighborhood sequences of themselves implies that . However, it may occur that holds without .
Proof. Let and be two neighborhood sequences such that both of them are equivalent to their minimal (maximal, resp.) neighborhood sequences. Assume that holds. Let p and q be any two points of the grid. Then, let be determined by a short/the shortest -path . Observe that, by the definition of , the -path is also a path from p to q, but then clearly holds.
It is also clear from the definitions that and if and only if , and then both and trivially hold.
The refinement is proper, since there are neighborhood sequences and , such that but . Let and . Notice that each of these sequences defines a singleton equivalence class, therefore and , and thus and . □
We need to note that the previous proof works for any representatives of the classes. On the other hand, this last result shows also one of the good applicability features of our result, and this will lead us to the next section.