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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 847958, 15 pages
A Minimum Problem for Finite Sets of Real Numbers with Nonnegative Sum
1Dipartimento di Matematica, Universitá della Calabria, Via Pietro Bucci, Cubo 30B, 87036 Arcavacata di Rende, Italy
2Dipartimento di Matematica, Universitá della Calabria, Via Pietro Bucci, Cubo 30C, 87036 Arcavacata di Rende, Italy
Received 6 February 2012; Accepted 2 March 2012
Academic Editor: Yonghong Yao
Copyright © 2012 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.
Let and be two integers such that ; we denote by the minimum [maximum] number of the nonnegative partial sums of a sum , where are real numbers arbitrarily chosen in such a way that of them are nonnegative and the remaining are negative. We study the following two problems: which are the values of and for each and , ? if is an integer such that , can we find real numbers , such that of them are nonnegative and the remaining are negative with , such that the number of the nonnegative sums formed from these numbers is exactly ?
In  Manickam and Miklós raised several interesting extremal combinatorial sum problems, two of which will be described below. Let and be two integers such that ; we denote by the minimum [maximum] number of the nonnegative partial sums of a sum , when are real numbers arbitrarily chosen in such a way that of them are nonnegative and the remaining are negative. Put . In  the authors answered the following question.
Which is the value of ?
In [1, Theorem 1] they found that . On the other side, from the proof of Theorem 1 of  also it follows that for each and that ; therefore (since ). It is natural to set then the following problem which is a refinement of .
Which are the values of and for each and with ?
In the first part of this paper we solve the problem , and we prove (see Theorem 4.1) that and for each positive integer .
A further question that the authors raised in  is the following.
“We do not know what is the range of the possible numbers of the nonnegative partial sums of a nonnegative -element sum. The minimum is as it was proven and the maximum is obviously but we do not know which are the numbers between them for which we can find reals with such that the number of the nonnegative sums formed from these numbers is equal to that number.’’
The following problem is a natural refinement of (Q2).
If is an integer such that , can we find real numbers , such that of them are nonnegative and the remaining are negative with , such that the number of the nonnegative sums formed from these numbers is exactly ?
In the latter part of this paper (see Theorem 4.5) we give a partial solution to the problem .
To be more precise in the formulation of the problems that we study and to better underline the links with some interesting problems raised in , it will be convenient-identify a finite set of real-numbers with an appropriate real-valued function. Let then and be two fixed integers such that and let (we call the index set). We denote by , the set of all the functions such that and . If , we set . It is easy to observe that and . We can reformulate the problem in an equivalent way using the functions terminology instead of the sets terminology.
If is an integer such that , does there exist a function with the property that ?
To solve the problem and (partially) , we use some abstract results on a particular class of lattices introduced in [2, 3]. In this paper we substantially continue the research project started in , which is the attempt to solve some extremal sum problems as started in  and further studied in [3, 5–11].
2. A Partial Order on the Subsets of
Of course if we take two functions such that , then . This implies that if we define on the equivalence relation iff and we denote by the equivalence class of a function , then the definition is well placed. It is also clear that it holds and . Now, when we take an equivalence class /~, there is a unique such that We can then identify the quotient set /~ with the subset of all the functions that satisfy the condition (2.1). This simple remark conducts us to rename the indexes of as follows: instead of instead of 1, instead of instead of . Therefore, if we set , we can identify the quotient set with the set of all the functions that satisfy the following two conditions: Now, if a generic function that satisfies (2.2) is given, we are interested to find all the subsets such that . This goal becomes then easier if we can have an appropriate partial order on the power set “compatible” with the total order of the partial sums inducted by , that is, a partial order that satisfies the following monotonicity property: if , then whenever . To have such a partial order on that has the monotonicity property we must introduce a new formal symbol that we denote by. We add this new symbol to the index set , so we set . We introduce on the following total order:If , then we write: for or . We denote by the set of all the formal expressions (hereafter called strings) that satisfy the following properties:(i), (ii), (iii), (iv)the unique element which can be repeated is .
In the sequel we often use the lowercase letters to denote a generic string in . Moreover to make smoother reading, in the numerical examples the formal symbols which appear in a string will be written without ~ − and §; in such way the vertical bar will indicate that the symbols on the left of are in and the symbols on the right of are elements in . For example, if and , then and .
Note that there is a natural bijective setcorrespondence between and defined as follows: if , then is the subset of made with the elements and such that and . For example, if , then . In particular, if , then .
Now, if and are two strings in , we define: iff .
It is easily seen that (1) is a finite distributive (hence also graded) lattice with minimum element and maximum element ; (2)has the following unary complementary operation : where is the usual complement of in and is the usual complement of in (e.g., in , we have that ).
Since we have the formal necessity to consider functions defined on the extended set instead of on the indexes set , then we will put . Precisely we can identify the quotient set with the set , defined by
We call an element of a -weight function, and if , we will continue to set . Therefore, with these last notations we have that , and the question becomes equivalent to the following.
If is an integer such that , does there exist a function with the property that ?
3. Boolean Maps Induct by Weight Functions
We denote by the boolean lattice composed of a chain with 2 elements that we denote (the minimum element) and (the maximum element). A Boolean map (briefly BM) on is a map ; in particular if ,we also say that is a Boolean total map (briefly BTM) on . If is BM on , we set .
If , the sum function induced by on is the function that associates to the real number , and therefore we can associate to the map setting:
Let us note that , and so .
Our goal is now to underline that some properties of such maps simplify the study of our problems. It is easy to observe that the map has the following properties.(i) is orderpreserving.(ii)If is such that , then ,(iii), , and .
Example 3.1. Let be the following -weight function:
We represent the map defined on by using the Hasse diagram of this lattice, on which we color green the nodes where the map assumes value and red the nodes where it assumes value (See Figure 1).
Note that if we have a generic Boolean total map which has the properties (i), (ii), and (iii) of , that is, the following:(BM1) is orderpreserving,(BM2)if is such that , then ,(BM3), , and,in general there does not exist a function such that (see  for a counterexample).
We denote by the set of all the maps which satisfy (BM1) and (BM2) and by the subset of all the maps in which satisfy also (BM3). We also set and . Let us observe that . A natural question raises at this point.
:if is an integer such that , does there exist a map with the property that ?
The question is the analogue of expressed in terms of Boolean total maps on instead of -weight functions, and if we are able to respond to , we provide also a partial answer to . In Section 4 we give an affirmative answer to the question , and also we give a constructive method to build the map .
4. Main Results
In the sequel of this paper we adopt the classical terminology and notations usually used in the context of the partially ordered sets (see [12–14] for the general aspects on this subject). If , we will set . In particular, if , we will set , . is called a downset of if for and with , then . is called an upset of if for and with , then . is the smallest down-set of which contains , and is a downset of if and only if . Similarly, is the smallest up-set of which contains , and is an upset in if and only if .
Theorem 4.1. If and are two integers such that , then and .
Proof. Assume that . We denote by the sublattice of of all the strings of the form , with and by the sublattice of of all the strings of the form , with . It is clear that and . We consider now the following further sublattices of :
It occurs immediately that , for and is a distributive sublattice of with elements, for .
Now we consider the following -weight function : Then it follows that is such that It means that the boolean map is such that This shows that .
In [1, Theorem 1] it has been proved that and . Since , using a technique similar to that used in the proof Theorem 1 of , it easily follows that . As shown above, it results that . Hence , that is, . This proves; the first part of theorem; it remains to prove the latter part.
We consider now the following -positive weight function : It results then that the sum is such that if ; that is, the boolean map is such that if . This shows that = .
On the other hand, it is clear that for any it results for each and for each .The number is the biggest number of values that a boolean map can assume. Hence, since , we have ; that is be; . This concludes the proof of Theorem 4.1.
To better visualize the previous result, we give a numerical example on a specific Hasse diagram. Let and and let be as given in previous theorem; that is In Figure 2 we have shown the Hasse diagram of the lattice (6,2), where is black, is violet, is red, is blue, is brown, is green. Therefore assume the following values.(i)The blue, black, and brown nodes correspond to values of ,(ii)The violet, red, and green nodes correspond to values of .
First to give the proof of Theorem 4.5 we need to introduce some useful results and the concept of basis in . In the following first lemma we show some properties of the sublattices of .
Lemma 4.2. Here the following properties hold, where and . (i).(ii).(iii).(iv).(v).
Proof. (i) If , then ; that is, it has the form , where ; therefore . If , it has the form , where ; if , it has the form , where . In both cases it results that , that is, .(ii)It is analogue to (i).(iii)The minimum of the sublattice is ; since , it is sufficient to show that . The inclusion follows by the definition of and of . On the other side, if , it follows that ; that is, , with and . Therefore , and this proves the other inclusion.(iv)Let us consider the maximum of the sublattice ; that is . Since , it is sufficient to show that ; this proof is similar to (iii).(v)If ; it has the form , with and ; therefore its complement has the form , with and ; hence . This shows that .
Now, if , we write in the form ; then ; therefore . This shows that .
At this point let us recall the definition of basis for , as given in  in a more general context. In the same way as an antichain uniquely determines a Boolean order-preserving map, a basis uniquely determines a Boolean map that has the properties (BM1) and (BM2) (see  for details). Hence the concept of basis will be fundamental in the sequel of this proof.
Definition 4.3. A basis for is an ordered couple , where and are two disjoint antichains of such that(B1),
(B3). In the proof of Theorem 4.5 we will construct explicitly such a basis.
We will also use the following result that was proved in .
Lemma 4.4. Let be a basis for . Then the map is such that .
Theorem 4.5. If is a fixed integer with , then there exists a boolean map such that .
Proof. Let be a fixed integer such that . We determine a specific Boolean total map such that . We proceed as follows.
The case is proved in the previous Theorem 4.1. Let us assume then . Since is a finite distributive sublattice of graded lattice , also is a graded lattice. We denote by the rank of and with its rank function. Note that the bottom of is and the top is .
We write in the form with . We will build a map such that .
If , we take , with as in Theorem 4.1, and hence we have .
If , we take , with as in Theorem 4.1, and we have .
Therefore we can assume that .
If , we denote by the set of elements such that , and we also set to simplify the notation. We write each in the following form: . If , we set .
If , we set and . We can then write in the form , for some and some . Depending on the previous number we partition into the following two disjoint subsets: , where the first subset is considered empty if .
In the sequel, to avoid an overload of notations, we write simply instead of , for .
Let us note that .
We define now the map : Let us observe that = = = = .
Therefore, if we show that , the theorem is proved.
We write in the following way: ,where = and = . Analogously = , where = and = .
We can see a picture of this partition of the sublattice in Figure 3.
Depending on and , we build now a particular basis for .
To such aim, let us consider the minimum of and the subsets and . Let us distinguish two cases:,. We define two different couples of subsets as follows:
In the case we set ;in the case we set and .
Step 1. is a couple of two disjoint antichains of .
In both cases (a1) and (a2) it is obvious that .
The elements are not comparable between them because they have all rank and analogously for the elements that have all rank . Let now and let ; then because , and because by construction. For the elements in different from , we can proceed as for . On the other side, we can observe that is not comparable with none of the elements since these are all in while . Thus is an antichain. Case (a2)
In this case we only must show that is not comparable with none of the elements . At first from the fact that it follows for each , and for each . Moreover, the elements and are all in ; hence they have the form , with , while , so and for each and each .Step 2. is a basis for .
We must see that (B1), (B2), and (B3) hold in both cases (a1) and (a2).
(B1)Let us begin to observe that . Since , we have then .On the other hand, since , by Lemma 4.2(iv) we have also that . Hence . This proves (B1).(B2)We show at first that . Since , we have that . By Lemma 4.2(i) and (iii) we have then . On the other side, since , by Lemma 4.2(iv) it follows that . By Lemma 4.2(ii), we have . Hence it holds . This proves that . We show now that also ; we proceed by contradiction. Let us suppose that there exists an element , then there are two elements and such that , hence . We will distinguish the following five cases, and in each of them we will find a contradiction.(a) and . In this case and are two distinct elements having both rank and such that ; it is not possible.(b) and . In this case has rank while has rank , and this contradicts the condition .(c) and This case is similar to the previous because has rank while has rank.(d) and This is similar to the previous because has rank while has rank .(e) and . In this case the condition implies that ; since this is not possible since .(B3)Since is the minimum of we have that , moreover, ; therefore . Since , it follows that . By Lemma 4.2(i) we have then that . By Lemma 4.2(ii) we also have that .
To complete the proof of (B3) let us observe that , where , .
Moreover, since ,with and , then . This shows that . Since the six sublattices , and for are a partition of , the property (B3) is proved.Case (a2)
(B1)We begin to show that . In fact, it holds that ; suppose that , and this shows that we obtain a contradiction. By it follows . But is the top of , so , since is antichain. This means that and this is not possible, of course. Since , we have , and, moreover, as in the proof of (B1) in the case (a1), we also have that ; therefore, to prove (B1) it is sufficient to show that ; As in the proof of (B1) in the case (a1), we have . Since is the minimum of and , it follows that the unique element of that can belong to is , but we have shown before that this is not true.(B2)As in the previous case we have ; moreover . As in the case (a1) we have ; therefore, to prove (B2) also in the case (a2), it is sufficient to show that . As in the proof of (B2) in the case (a1) it results that , and, by definition of , it is easy to observe that . Hence .(B3)It is identical to that of case (a1).Step 3. The map defined in (4.8) is such that .
Since we have proved that is a basis for , by Lemma 4.4 it follows that the map if the two following identities hold: We prove at first (4.10). By definition of and it easy to observe that , and, moreover, by Lemma 4.2(ii) we also have that ; hence, since , it results that . On the other hand, by Lemma 4.2(iv) we have that because is a subset of . At this point let us note that the elements of that are also in must belong necessarily to the subset . This proves the other inclusion and hence (4.10).
To prove now (4.9) we must distinguish the cases (a1) and (a2). We set . Let us first examine the case (a1). Since is the minimum of , we have . Moreover, since , it follows that by lemma 3.2(i). Finally, since , the inclusion in (4.9) is proved. To prove the other inclusion we begin to observe that ; therefore the elements of that are not in must be necessarily in , and such elements, by definition of , must be necessarily in This proves that , For , we have , where by Lemma 4.2(i), and, since , also , by Lemma 4.2(v). Therefore by Lemma 4.2(iii). This shows that , hence, the inclusion . The proof of (4.9) in the case (a1) is therefore complete.
Finally, to prove (4.9) in the case (a2), it easy to observe that the only difference with respect to case (a1) is when we must show that . In fact, in the case (a2) it results that and , while is the same in both cases (a1) and (a2). Therefore, in the case (a2), the elements of that are not in must be in or in . As in the case (a1) we have , and, since is the minimum of , it results that by Lemma 4.2(iii); hence . Therefore, also in the case (a2), the elements of that are not in must be necessarily in . The other parts of the proof are same as in case (a1). Hence we have proved the identities (4.9) and (4.10). By Lemma 4.4 it follows then that the map . Finally, by definition of , we have obviously , , and . This shows that . The proof is complete.
To conclude we emphasize the elegant symmetry of the induced partitions on from the boolean total maps ’s constructed in the proof of the Theorem 4.5.
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