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Journal of Applied Mathematics
Volume 2013 (2013), Article ID 973501, 10 pages
A Discrete Dynamical Model of Signed Partitions
Dipartimento di Matematica, Università della Calabria, Via Pietro Bucci, Cubo 30B, 87036 Arcavacata di Rende, Italy
Received 28 November 2012; Accepted 19 February 2013
Academic Editor: Luigi Muglia
Copyright © 2013 G. Chiaselotti et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We use a discrete dynamical model with three evolution rules in order to analyze the structure of a partially ordered set of signed integer partitions whose main properties are actually not known. This model is related to the study of some extremal combinatorial sum problems.
A dynamical system is by definition a system whose state changes with time . We have a discrete dynamical system when is an integer or a natural number, and the elements of the system can be obtained in the form , where is some global function which describes the evolution rule of the system (see ). In a series of very recent works, the theory of discrete dynamical systems has been applied in several contexts. In , the theory of discrete dynamical systems is applied in order to analyze some models of concurrent computing systems. In , a dynamical model of parallel computation on bi-infinite time scale with an approach similar to two-sided symbolic dynamics is constructed. In , the authors analyze the orbit structure of parallel discrete dynamical systems over directed dependency graphs, with Boolean functions as global functions. In , the authors extend the manner of defining the evolution update of discrete dynamical systems on Boolean functions, without limiting the local functions to being dependent restrictions of a global one. Finally, in  is given a complete characterization of the orbit structure of parallel discrete dynamical systems with maxterm and minterm Boolean functions as global functions. In , the authors have introduced the poset of all the signed integer partitions , where the positive parts and the negative parts are all distinct between them and the number of nonzero parts is exactly . In this case, the partial order is that on the components. The concept of signed integer partition has been introduced in  and studied in  from an arithmetical point of view. In [7, 10], it is shown that the structure of the lattice is strictly related to the study of some combinatorial extremal sum problems. In this paper, we study some particular type of signed partitions (specifically, of the signed partitions of ) by adopting the interesting point of view of the discrete dynamical models. In particular, we analyze the lattice as a discrete dynamical model with three local evolution rules whose dynamics is studied in a sequential mode. The way to study a lattice of classical partitions as a discrete dynamic model having some particular evolution rules is implicit in , where Brylawski proposed a dynamical approach to study the lattice of all the partitions of the fixed positive integer with the dominance ordering. In such a context, a configuration of the system is represented by an ordered partition of an integer , that is, a decreasing sequence having sum , whose blocks of the corresponding Young diagram are considered as mobile sand grain, and the movement of a sand grain, respects the following rules.
Rule 1 (vertical rule). One grain can move from a column to the next column if the difference of height of these two columns is greater than or equal to 2.
Rule 2 (horizontal rule). If a column containing grains is followed by a sequence of columns containing grains and then one column containing grains, then one grain of the first column can slip to the last column.
In this paper, we prove that the covering relation in the lattice is uniquely determined by three evolution rules of our discrete dynamical model. We articulate the proof of this result by contradiction by using only the first-order logic. This method of proof is very laborious; however, its advantage is that it can be easily implemented on a proof assistant checker based on the first-order logic (e.g., Mizar). The paper is articulated as follows. In Section 2, we recall some basic definitions and preliminary results, for example, the definition of and some of its properties. In Section 3, we explain how to see the signed partitions of as configurations of our discrete dynamical model and we also describe its evolution rules. In Section 4, we assume by contradiction that there is a generic element of that covers another generic element but is not generated by with none of our evolution rules. From this assumption we deduce several propositions and conditions that we use in order to prove our principal result, that is, Theorem 4.
2. Definitions and Preliminary Results
If is a poset and , we write () if covers . Now we briefly recall the definition of the lattice that we have introduced in  in a more formal context. Let and be two nonnegative integers such that .
We call -string a -tuplet of integers such that(i);(ii);(iii);(iv)the unique element in (1) which can be repeated is .If is a -string, we call parts of the integers , nonnegative parts of the integers , and nonpositive parts of the integers . We set and . In some cases, we do not distinguish between nonnegative and nonpositive parts of , and we write more simply instead of . We also denote by the number of parts of that are strictly positive, with the number of parts of that are strictly negative, and we set . is the set of all the -strings. If and are two -strings, we set if for all if for all , and if and . On , we consider the partial order on the components that we denote by . To simplify the notations, in all the numerical examples the integers on the right of the vertical bar will be written without minus sign. Since is a finite distributive lattice, it is also graded, with minimal element and maximal element .
We recall now the concept of involution poset (see [12, 13] for some recent studies on such class of posets). An involution poset (IP) is a poset with a unary operation , such that (I1), for all ; (I2) if and if , then .The map is called complementation of and the complement of . Let us observe that if is an involution poset, by it follows that is bijective and by and it holds that if are such that , then . If is an involution poset and if , we will set . We note that if is an involution poset, then is a self-dual poset because from and it follows that if , we have that , if and only if , and this is equivalent to say that the complementation is an isomorphism between and its dual poset . In , it has been shown that is an involution poset and its complementation map is the following: where is the usual complement of in , and is the usual complement of in (e.g., in , we have that ). If is an integer such that , we set now . Its easy to see that is a sublattice of and obviously .
We call signed Young diagrams (briefly SYD) an ordered couple , where is a Young diagram of an integer partition built with decreasing columns instead with decreasing rows, and is a Young diagram of an integer partition built with increasing columns. In the sequel, we will call pile a column of a Young diagram and grain a square of a pile. For example, (3) is determined by the partitions 4331 and 113. If , we denote by the signed Young diagram , where is the Young diagram of the partition and is the Young diagram of the partition . Now, if is a SYD, where is the decreasing partition having as Young diagram and is the increasing partition having as Young diagram, we set and we call the signed partition of . We say that a signed Young diagram is a -SYD if it results that and we denote with the set of all the -SYD. It is clear that the map is a bijection between and , whose inverse function is the map ; therefore, in the sequel, we will identify the elements of with the -SYDs of . If and are two usual Young diagrams, we set if is a subdiagram of . If and are two SYDs, we set Therefore, it is clear that we can identify with . Now we set . Obviously we can identify with . If is a statement, in the sequel we denote by the negation of .
3. Description of the Evolution Rules
In our dynamic model, the set of all configurations is exactly . Our goal is to define some rules of evolution that starting from the minimum of allow us to reconstruct the Hasse diagram of (and therefore to determine the covering relations in ).
Let and . If , we call -plus pile of , and we denote it by , the pile of that corresponds to part of and, if , we call -minus pile of , and we denote it by , the pile of that corresponds to part of . We call the pile plus singleton pile if and minus singleton pile if . Let us note that if there exists a plus singleton pile, then it is necessarily unique, analogously for a minus singleton pile. If , we set and we call the plus height difference of in . If , we set and we call the minus height difference of in . If , we say that has a plus cliff at if . If , we say that has a minus cliff at if .
Remark 1. When we apply the following rules to one element , we impose that there is an “invisible” extra pile in the imaginary place of having exactly grains. This is only a formal trick for decreas the number of rules necessary for our model; therefore, the pile must be not considered as a part of .
3.1. Evolution Rules
Rule 1. If the -plus pile has at least one grain and if has a plus cliff at , then one grain must be added on the -plus pile: (5)
Rule 2. If there is not a plus singleton pile and there is a minus singleton pile, then the latter must be shifted to the side of the lowest non empty plus pile: (6)
Rule 3. One grain must be deleted from the -minus pile if has a minus cliff at : (7)
Remark 2. (i) Under the hypothesis in Rule 3, the plus must have at least 2 grains.
(ii) In Rule 2, the lowest non empty plus pile can also be the invisible column in the place . In this case, all the plus piles are empty and a possible minus singleton pile must be shifted in the place .
4. Main Result
In the sequel, we write (or ) to denote that is an -tuplet of integers obtained from applying the Rule , for , and that if the condition is false. We also set
Proposition 3. If , then .
Proof. Let , (the invisible pile in the place ), and . We distinguish the three possible cases related to the previous rules.
Case 1. Let us assume that , , and that has a plus cliff at . If , then . It is clear that because . Since there is a plus cliff at , we have ; hence, , and this implies that . We must show now that covers in . Since and differ between them only in the place for and , respectively, it is clear that there does not exist an element such that . Hence, .
Case 2. Let us assume that in there is not a plus singleton pile and that there is a minus singleton pile (we say , for some ). Since , we can assume that , for some . This means that has the following form: , where (otherwise has a plus singleton pile). Applying Rule 2 to , we obtain , where . It is clear then that and since is obtained from with only a shift of the pile-1 to the left in the place . Let us note that the only elements such that and are and , but and ; hence, are not elements of . This implies that covers in .
Case 3. If and has a minus cliff at , we apply Rule 3 to on the pile and we obtain , where . Since has a minus cliff at , we have ; therefore, because and since implies . As in Case 1, we note that covers in because they differ between them only for a grain in the place .
The main result of this paper is to prove that if , then covers in if and only if for some . Obviously this result can be reformulated as follows.
Theorem 4. If , then .
By Proposition 3, to prove Theorem 4, it suffices to show that . To show this inclusion, we proceed by contradiction as follows. We start with the following hypotheses: and we also assume that Then the proof of Theorem 4 will consist into show that the hypotheses (9) and the conditions (10) always lead to contradictory cases, so that it must be necessarily verified the inclusion .
We will articulate the proof of Theorem 4 in several cases. Each possible case will be placed in the form of a proposition and we will obtain a contradiction in all cases, showing, hence, the thesis of the theorem. Before proceeding, let us note that in some cases we also use the alternative notations and .
The description of Rules 1, 2, and 3 in terms of parts of and is the following: (1) is equivalent to AND ( AND ()); (2) is equivalent to () AND ( such that ) AND ( such that , ) AND (, , ); (3) is equivalent to AND ( AND ()).
Negation of Rules 1, 2, and 3. (1) [(1) OR (1B)], where (1B) is (), (1) is [ AND ( () OR ( OR ))]. (2) [(2) OR (2B) OR (2C) OR (2D)], where (2) is () AND ( such that ) AND ( such that ) AND (( OR ) OR OR ); (2B) is ( such that ); (2C) is (); (2D) is ( OR ). (3) [(3) OR (3B)], where is (); () is [ AND ( () OR ( OR ))]. we set
Proposition 5. Consider (1B) ((1.1b) OR (1.2b)).
Proof. The condition means that for all , or (i.e., or ). We have then the following possibilities: (j); (jj) (and therefore since ); (jjj). If for all then (12) holds. Let us assume therefore that is the smallest index such that ; hence, and . This means that for all indexes holds necessarily (jjj). In particular, for we have ; hence, . Now, if , , that is, . Iterating (13) follows.
Proposition 6. Consider (2D) (2.1d).
Proof. By (2D), if we take , we have or ; therefore, it is necessarily because by hypothesis, . Thus, still by (2D), for , we have or ; therefore, it must be . By iterating, we can deduce that ,, ; hence, , ,…, because is an element of , that is, exactly (14). On the other side, if , then it is immediate to note that (2D) is verified.
Proposition 7. Consider (3B) ((3.1b) OR (3.2b)).
Proof. The condition means that for all , (i.e., or ). We have then the following possibilities:(i) (and therefore since );(ii).
If (i) holds for all , then (15) holds. We can assume therefore that is the smallest index such that (ii) holds. Let us observe then that if in a place , (ii) holds; then also in the places , (ii) holds. In fact, if and by absurd , then which is a contradiction because . Therefore, if (ii) holds in , then it also holds in , and therefore, by iteration, it holds for . Now we assume at first that . In this case, (ii) holds for all indexes in ; therefore, or and , . If , then and if , then ; thus in both cases (ii) holds. Let now . Then (because in the place (i) holds) and hence . Moreover, since (i) holds in the places , we have ; therefore, , , where . This completes the proof.
Proposition 8. Consider (3.1b) ().
Proof. Since , it must be , where for all because (15) implies . Therefore, for all ; that is, .
Proposition 9. Consider(i)(1.1b) AND (2B) absurd,(ii)(1.1b) AND (2.1d) absurd,(iii)(2C) AND (3.2b) absurd,(iv)(1.1b) AND (3.1b) absurd,(v)(1.2b) AND (3.1b) absurd.
Proposition 10. Consider (1.2b) AND (2B) (2.1d).
Proof. By (2B), it follows that in (13) it must be ; therefore, .
Proposition 11. Consider (2.1d) AND (3.2b) absurd.
Proposition 12. Consider () absurd.
Proof. We set(F) [() AND ( such that ) AND ( such that )]; (G) ( OR );(H) ( OR ).
Then () is equivalent to ((F) AND (G)) OR ((F) AND (H)). We show at first that ((F) AND (H)) leads to a contradiction. Next we show that ((F) AND (G) AND (H)) also leads to an absurd. This will prove the thesis;
((F) AND (H)) absurd.
By (F), it follows that satisfies the hypothesis of Rule 2; therefore, by Proposition 3 we can take an element such that . As in the proof of the Case 2 in Proposition 3, it results that and must have the following form: where and . We distinguish now several cases.
(I) . Then , and also because ; hence .
We note now that because . Moreover, since , it follows that because . Now, since , from the form of it results that . Moreover, since also , we have . Therefore, since in (19), there are exactly inequalities and in (20) there are exactly inequalities; it follows that
(II) . Since , by (17) we have two possible cases: has the form (21) or with and . If (22) holds, then , which is absurd because ; hence it must hold necessarily (21).
(III) . Since and , by (17) it follows that Then from the hypothesis and by (18), we have , which is a contradiction.
Hence the unique possibility is that has the form (21). Since , we have for and for . Moreover, at least one inequality of the type or the type must hold. We distinguish now the several cases.
() We assume that a unique inequality of the type holds for some . Therefore, we have for all and for all . Then, since (let us note that the previous inequalities also hold if because ), we have that and . Hence satisfies the hypotheses of Rule 1 in the place and so , where with . Now, if , it results that ; therefore, , which is a contradiction with the hypothesis . Otherwise, if , then we have , which is absurd because .
() If a unique inequality of the type holds for some , we deduce a contradiction as in the proof of the previous case by using Rule 3 instead of Rule 1.
() We assume now that at least two strict inequalities among the and hold, and let us also assume that the first (by left) of such strict inequalities is for some . We consider then the following -string: Since is the first strict inequality by left, we have (also if ); hence because ; this implies that ; therefore because . Moreover, it is clear that . Let us observe now that ; in fact, by hypothesis, must have at least another place where or and differs by only in the place . This shows that , with , which is absurd because .
() If at least two strict inequalities among the and hold and the first (by left) of such strict inequalities is for some , then we deduce a contradiction in the same way as in the previous case.
This concludes the proof that ((F) AND (H)) absurd. To complete the proof we show now that ((F) AND (G) AND (H)) absurd.
Since (H) is false, we have and ; therefore, has the form (17) and . We take as in (18), so that and differ between them only in the places , and . Moreover, since , by (G), (18), and (17) it follows that . Therefore, we obtain that is in contrast with the hypothesis .
We set now
Proposition 13. Consider(i)(1) (1A),(ii)(3) (3A).
Proof. (i) The implication is obvious. We assume now that (1) holds and there exists such that and also that (i.e., there exists such that ) or (i.e., there exists such that ). Since , we can apply Rule 1 to and we can take . By Proposition 3, we know that and . Moreover, since () or , we also have ; therefore, , which is in contrast with the hypothesis . Hence this proves that the condition must be verified for all , which is exactly (26).
(ii) The proof is similar to (i) applying Rule 3 to instead of Rule 1.
Proposition 14. Consider(i)(1A) (1a),(ii)(3A) (3a).
Proof. (i) By (1A), we have , and by the definition of this is equivalent to say that satisfies the hypotheses of Rule 1 for some . Therefore, by Proposition 3, there exists an element such that , where and obviously . Since , we distinguish two cases.(a). In this case , that is in contrast with the hypothesis .(b). In this case ; hence because from the hypothesis .
(ii) The proof of this case is similar to (i).
Lemma 15. Assume that (1a) holds. If , , and , then .
Proof. If , then the thesis holds by the definition of (1a) in (28). We assume now that . Then or , and since , we necessarily have . Now, if , we have since , which is in contradiction with . Hence, it must be ; that is, . Moreover, because and . Therefore, since , we have , which implies because . By subtracting from each term of the previous inequalities, we obtain ; therefore, it must be .
Proposition 16. Consider (1a) (1.1a).
Proof. By (28), there exists at least one element . In this place we have and moreover . Since , it must be or . If , then , which is in contrast with . Hence and therefore there exists such that and . We take ; that is, . Since and , by Lemma 15 it follows that ; that is, . Let now . If , then we have the thesis otherwise ; that is, ; therefore because . Then since , we can apply Lemma 15 and we deduce ; that is, . By iteration the thesis follows.
Lemma 17. Assume that (3a) holds. If , , then .
Proof. If , then the thesis holds by definition of (3a) in (29). We assume now that . Now, if , we have since , which is in contradiction with the hypothesis . Hence it must be ; that is, ; therefore since and . The last inequalities provide . By subtracting from each term of the previous inequalities, we obtain ; therefore it must be .
Proposition 18. If (3a) holds, then has the following form: , with and . Moreover, if, we also have .
Proof. By (3a) in (29), we know that . Moreover, if , then and because . Therefore, we can take the minimum such that ; hence and this means that . Assume now that . Since , we have ; therefore . By (3a), follows then that . Now, if , the thesis is true, otherwise, if , by Lemma 17 we have . Again, if , the thesis is true; otherwise, if , by Lemma 17 we have . Proceeding repeatedly in this way we obtain , . Finally, since and , we also have . This proves that .
Proposition 19. Consider ((2C) AND (3a)) .
Proof. (2C) implies that has the form , with . Since (3a) holds, by Proposition 18, the thesis follows.
Proposition 20. Consider ((1.1a) AND (2C) AND (3a)) absurd.
Proposition 21. Consider ((1.1b) AND (2C) AND (3a)) absurd.
Proposition 22. Consider ((1.1a) AND (2B)) .
Proof. By (1.1a) in (30) and (2B), it follows that and . Now, since and , we have that ; that is, . Therefore .
Proposition 23. Consider ((1.1a) AND (2B) AND (3.1b)) absurd.
Proposition 24. Consider ((1.1a) AND (3.1b)) absurd.
Proposition 25. Consider ((1.1a) AND (2B) AND (3a)) absurd.
Proof. By Proposition 18, we have with and . By Proposition 22,. Now, if , by Proposition 18 we have and therefore , which is absurd. So we can assume that . Since , and , it follows that cannot be ; otherwise, it is . This implies that it must be . We distinguish now two cases.
Case 1. If , since , by Lemma 17 (with ) we obtain . Now, if , then , and as before this provides a contradiction. If , we can still apply Lemma 17 (with ) and therefore have . Proceeding in this way in all cases we always obtain and hence is absurd.
Case 2. If , then , and since , , and , it must also be ; that is, , absurd.
Proposition 26. Consider ((1.1a) AND (2B) AND (3.2b)) absurd.
Proposition 27. Consider ((1.2b) AND (2C) AND (3a)) absurd.
Proposition 28. Consider ((2.1d) AND (3a)) absurd.
We complete now the proof of our main result; that is, we provide three syntactic trees where each possible path from the roots to the leaves leads to a contradiction. In the following we draw each tree and next we also write all the implication chains for each path from the roots to the leafs.
We recall at first some useful implications and equivalences already proved in the previous section:() (1) [by Proposition 13 (i)] (1A) [by Proposition 14] (1a) [by Proposition 16] (1.1a)() (2) [by Proposition 12] absurd() (2D) [by Proposition 6] (2.1d)() (3) [by Proposition 13 (ii)] (3A) [by Proposition 14] (3a) [see statement of Proposition 18] Proposition 18.
The first tree that we consider is that having the condition (1) as root (we denote such tree with TREE(1)).
TREE (1): (33)
We observe now that by virtue of Proposition 5 when we start from the condition (1B), we must consider two subtrees, the first with root (1.1b) (which we denote by TREE(1.1b)) that is
TREE (1.1b): (34)
and the second (which we denote by TREE(1.2b)) that is
TREE (1.2b): (35)
Let us note that the ramification (36)
in each of the previous trees follows by Proposition 7.
We have then exactly possible paths from the roots to the leafs in all the previous trees (we exclude the paths that lead to (2) because we already know that they lead to a contradiction by ). In the following we write all these paths and, for each, the related proposition that leads to a contradiction together with some implication among the , , and . In the sequel, the node (3B) is omitted for brevity because it is only an intermediate node and in the paths that involve (3B) we use only the leafs (3.1b) and (3.2b) to deduce the related contradictions: (1) AND (2B) AND (3) absurd by , , and Proposition 25 (1) AND (2B) AND (3.1b) absurd by and Proposition 23 (1) AND (2B) AND (3.2b) absurd by and Proposition 26 (1) AND (2C) AND (3) absurd by , , and Proposition 20 (1) AND (2C) AND (3.1b) absurd by and Proposition 24 (1) AND (2C) AND (3.2b) absurd by and Proposition 9 (iii) (1) AND (2D) AND (3) absurd by , , and Proposition 28 (1) AND (2D) AND (3.1b) absurd by and Proposition 24 (1) AND (2D) AND (3.2b) absurd by and Proposition 11
(1.1b) AND (2B) AND (3) absurd by Proposition 9 (i) (1.1b) AND (2B) AND (3.1b) absurd by Proposition 9 (i) (1.1b) AND (2B) AND (3.2b) absurd by Proposition 9 (i) (1.1b) AND (2C) AND (3) absurd by and by Proposition 21 (1.1b) AND (2C) AND (3.1b) absurd by Proposition 9 (iv) (1.1b) AND (2C) AND (3.2b) absurd by Proposition 9 (iii) (1.1b) AND (2D) AND (3) absurd by and by Proposition 9 (ii) (1.1b) AND (2D) AND (3.1b) absurd by and by Proposition 9 (ii) (1.1b) AND (2D) AND (3.2b) absurd by and by Proposition 9 (ii)
(1.2b) AND (2B) AND (3) (1.2b) AND (2D) AND (3) (that we examine below) by and by Proposition 10 (1.2b) AND (2B) AND (3.1b) (1.2b) AND (2D) AND (3.1b) (that we examine below) by and by Proposition 10 (1.2b) AND (2B) AND (3.2b) (1.2b) AND (2D) AND (3.2b) (that we examine below) by and by Proposition 10 (1.2b) AND (2C) AND (3) absurd by and by Proposition 27 (1.2b) AND (2C) AND (3.1b) absurd by Proposition 9 (v) (1.2b) AND (2C) AND (3.2b) absurd by Proposition 9 (iii) (1.2b) AND (2D) AND (3) absurd by , , and Proposition 28 (1.2b) AND (2D) AND (3.1b) absurd by Proposition 9 (v) (1.2b) AND (2D) AND (3.2b) absurd by and by Proposition 11
This concludes the proof of Theorem 4. Below we draw the Hasse diagram of the lattice by using the evolution Rules 1, 2, and 3 starting with the minimum element of this lattice, which is . We label a generic edge of the next diagram with the symbol if it leads to a production that uses the Rule , for : (37)
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