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Journal of Applied Mathematics
VolumeΒ 2012Β (2012), Article IDΒ 847958, 15 pages
doi:10.1155/2012/847958
Research Article

A Minimum Problem for Finite Sets of Real Numbers with Nonnegative Sum

G. Chiaselotti,1Β G. Marino,2Β and C. Nardi2

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: YonghongΒ Yao

Copyright Β© 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.

Abstract

Let 𝑛 and π‘Ÿ be two integers such that 0 < π‘Ÿ ≀ 𝑛 ; we denote by 𝛾 ( 𝑛 , π‘Ÿ ) [ πœ‚ ( 𝑛 , π‘Ÿ ) ] the minimum [maximum] number of the nonnegative partial sums of a sum βˆ‘ 𝑛 1 = 1 π‘Ž 𝑖 β‰₯ 0 , where π‘Ž 1 , … , π‘Ž 𝑛 are 𝑛 real numbers arbitrarily chosen in such a way that π‘Ÿ of them are nonnegative and the remaining 𝑛 βˆ’ π‘Ÿ are negative. We study the following two problems: ( 𝑃 1 ) which are the values of 𝛾 ( 𝑛 , π‘Ÿ ) and πœ‚ ( 𝑛 , π‘Ÿ ) for each 𝑛 and π‘Ÿ , 0 < π‘Ÿ ≀ 𝑛 ? ( 𝑃 2 ) if π‘ž is an integer such that 𝛾 ( 𝑛 , π‘Ÿ ) ≀ π‘ž ≀ πœ‚ ( 𝑛 , π‘Ÿ ) , can we find 𝑛 real numbers π‘Ž 1 , … , π‘Ž 𝑛 , such that π‘Ÿ of them are nonnegative and the remaining 𝑛 βˆ’ π‘Ÿ are negative with βˆ‘ 𝑛 1 = 1 π‘Ž 𝑖 β‰₯ 0 , such that the number of the nonnegative sums formed from these numbers is exactly π‘ž ?

1. Introduction

In [1] Manickam and Miklós raised several interesting extremal combinatorial sum problems, two of which will be described below. Let 𝑛 and π‘Ÿ be two integers such that 0 < π‘Ÿ ≀ 𝑛 ; we denote by 𝛾 ( 𝑛 , π‘Ÿ ) [ πœ‚ ( 𝑛 , π‘Ÿ ) ] the minimum [maximum] number of the nonnegative partial sums of a sum βˆ‘ 𝑛 1 = 1 π‘Ž 𝑖 β‰₯ 0 , when π‘Ž 1 , … , π‘Ž 𝑛 are 𝑛 real numbers arbitrarily chosen in such a way that π‘Ÿ of them are nonnegative and the remaining 𝑛 βˆ’ π‘Ÿ are negative. Put 𝐴 ( 𝑛 ) = m i n { 𝛾 ( 𝑛 , π‘Ÿ ) ∢ 0 < π‘Ÿ ≀ 𝑛 } . In [1] the authors answered the following question.

( 𝑄 1 ) Which is the value of 𝐴 ( 𝑛 ) ?

In [1, Theorem 1] they found that 𝐴 ( 𝑛 ) = 2 𝑛 βˆ’ 1 . On the other side, from the proof of Theorem 1 of [1] also it follows that 𝛾 ( 𝑛 , π‘Ÿ ) β‰₯ 2 𝑛 βˆ’ 1 for each π‘Ÿ and that 𝛾 ( 𝑛 , 1 ) ≀ 2 𝑛 βˆ’ 1 ; therefore 𝛾 ( 𝑛 , 1 ) = 2 𝑛 βˆ’ 1 (since 2 𝑛 βˆ’ 1 = 𝐴 ( 𝑛 ) ≀ 𝛾 ( 𝑛 , 1 ) ≀ 2 𝑛 βˆ’ 1 ). It is natural to set then the following problem which is a refinement of ( 𝑄 1 ) .

( 𝑃 1 ) Which are the values of 𝛾 ( 𝑛 , π‘Ÿ ) and πœ‚ ( 𝑛 , π‘Ÿ ) for each 𝑛 and π‘Ÿ with 0 < π‘Ÿ ≀ 𝑛 ?

In the first part of this paper we solve the problem ( 𝑃 1 ) , and we prove (see Theorem 4.1) that 𝛾 ( 𝑛 , π‘Ÿ ) = 2 𝑛 βˆ’ 1 and πœ‚ ( 𝑛 , π‘Ÿ ) = 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ for each positive integer π‘Ÿ ≀ 𝑛 .

A further question that the authors raised in [1] is the following.

( 𝑄 2 )   “We do not know what is the range of the possible numbers of the nonnegative partial sums of a nonnegative 𝑛 -element sum. The minimum is 2 𝑛 βˆ’ 1 as it was proven and the maximum is obviously 2 𝑛 βˆ’ 1 but we do not know which are the numbers between them for which we can find reals π‘Ž 1 , … , π‘Ž 𝑛 with βˆ‘ 𝑛 1 = 1 π‘Ž 𝑖 β‰₯ 0 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).

( 𝑃 2 )   If π‘ž is an integer such that 𝛾 ( 𝑛 , π‘Ÿ ) ≀ π‘ž ≀ πœ‚ ( 𝑛 , π‘Ÿ ) , can we find 𝑛 real numbers π‘Ž 1 , … , π‘Ž 𝑛 , such that π‘Ÿ of them are nonnegative and the remaining 𝑛 βˆ’ π‘Ÿ are negative with βˆ‘ 𝑛 1 = 1 π‘Ž 𝑖 β‰₯ 0 , such that the number of the nonnegative sums formed from these numbers is exactly π‘ž ?

In the latter part of this paper (see Theorem 4.5) we give a partial solution to the problem ( 𝑃 2 ) .

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 [1], it will be convenient-identify a finite set of real-numbers with an appropriate real-valued function. Let then 𝑛 and π‘Ÿ be two fixed integers such that 0 < π‘Ÿ ≀ 𝑛 and let 𝐼 𝑛 = { 1 , 2 , … , 𝑛 } (we call 𝐼 𝑛 the index set). We denote by π‘Š ( 𝑛 , π‘Ÿ ) , the set of all the functions 𝑓 ∢ 𝐼 𝑛 β†’ ℝ such that βˆ‘ π‘₯ ∈ 𝐼 𝑛 𝑓 ( π‘₯ ) β‰₯ 0 and | { π‘₯ ∈ 𝐼 𝑛 ∢ 𝑓 ( π‘₯ ) β‰₯ 0 } | = π‘Ÿ . If 𝑓 ∈ π‘Š ( 𝑛 , π‘Ÿ ) , we set 𝛼 ( 𝑓 ) = | { π‘Œ βŠ† 𝐼 𝑛 ∢ βˆ‘ 𝑦 ∈ π‘Œ 𝑓 ( 𝑦 ) β‰₯ 0 } | . It is easy to observe that 𝛾 ( 𝑛 , π‘Ÿ ) = m i n { 𝛼 ( 𝑓 ) ∢ 𝑓 ∈ π‘Š ( 𝑛 , π‘Ÿ ) } and πœ‚ ( 𝑛 , π‘Ÿ ) = m a x { 𝛼 ( 𝑓 ) ∢ 𝑓 ∈ π‘Š ( 𝑛 , π‘Ÿ ) } . We can reformulate the problem ( 𝑃 2 ) in an equivalent way using the functions terminology instead of the sets terminology.

( 𝑃 2 )   If π‘ž is an integer such that 𝛾 ( 𝑛 , π‘Ÿ ) ≀ π‘ž ≀ πœ‚ ( 𝑛 , π‘Ÿ ) , does there exist a function 𝑓 ∈ π‘Š ( 𝑛 , π‘Ÿ ) with the property that 𝛼 ( 𝑓 ) = π‘ž ?

To solve the problem ( 𝑃 1 ) and (partially) ( 𝑃 2 ) , 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 [4], which is the attempt to solve some extremal sum problems as started in [1] and further studied in [3, 5–11].

2. A Partial Order on the Subsets of 𝐼 𝑛

Of course if we take two functions 𝑓 , 𝑔 ∈ π‘Š ( 𝑛 , π‘Ÿ ) such that 𝑓 ( 𝐼 𝑛 ) = 𝑔 ( 𝐼 𝑛 ) , then 𝛼 ( 𝑓 ) = 𝛼 ( 𝑔 ) . This implies that if we define on π‘Š ( 𝑛 , π‘Ÿ ) the equivalence relation 𝑓 ∼ 𝑔 iff 𝑓 ( 𝐼 𝑛 ) = 𝑔 ( 𝐼 𝑛 ) and we denote by [ 𝑓 ] the equivalence class of a function 𝑓 ∈ π‘Š ( 𝑛 , π‘Ÿ ) , then the definition 𝛽 ( [ 𝑓 ] ) = 𝛼 ( 𝑓 ) is well placed. It is also clear that it holds 𝛾 ( 𝑛 , π‘Ÿ ) = m i n { 𝛽 ( [ 𝑓 ] ) ∢ [ 𝑓 ] ∈ π‘Š ( 𝑛 , π‘Ÿ ) / ∼ } and πœ‚ ( 𝑛 , π‘Ÿ ) = m a x { 𝛽 ( [ 𝑓 ] ) ∢ [ 𝑓 ] ∈ π‘Š ( 𝑛 , π‘Ÿ ) / ∼ } . Now, when we take an equivalence class [ 𝑓 ] ∈ π‘Š ( 𝑛 , π‘Ÿ ) /~, there is a unique 𝑓 βˆ— ∈ [ 𝑓 ] such that 𝑓 βˆ— ( π‘Ÿ ) β‰₯ β‹― β‰₯ 𝑓 βˆ— ( 1 ) β‰₯ 0 > 𝑓 βˆ— ( π‘Ÿ + 1 ) β‰₯ β‹― β‰₯ 𝑓 βˆ— ( 𝑛 ) . ( 2 . 1 ) We can then identify the quotient set π‘Š ( 𝑛 , π‘Ÿ ) /~  with the subset of all the functions 𝑓 βˆ— ∈ π‘Š ( 𝑛 , π‘Ÿ ) that satisfy the condition (2.1). This simple remark conducts us to rename the indexes of 𝐼 𝑛 as follows: Μƒ π‘Ÿ instead of Μƒ 1 π‘Ÿ , … , instead of 1, 1 instead of π‘Ÿ + 1 , … , 𝑛 βˆ’ π‘Ÿ instead of 𝑛 . Therefore, if we set Μƒ 𝐼 ( 𝑛 , π‘Ÿ ) = { 1 , … , Μƒ π‘Ÿ , 1 , … , 𝑛 βˆ’ π‘Ÿ } , we can identify the quotient set π‘Š ( 𝑛 , π‘Ÿ ) / ∼   with the set of all the functions 𝑓 ∢ 𝐼 ( 𝑛 , π‘Ÿ ) β†’ ℝ that satisfy the following two conditions: ξ€· Μƒ 1 ξ€Έ ξ‚€ 𝑓 ( Μƒ π‘Ÿ ) + β‹― + 𝑓 + 𝑓 1  ξ€· + β‹― + 𝑓 ξ€Έ ξ€· Μƒ 1 ξ€Έ ξ‚€ 𝑛 βˆ’ π‘Ÿ β‰₯ 0 , 𝑓 ( Μƒ π‘Ÿ ) β‰₯ β‹― β‰₯ 𝑓 β‰₯ 0 > 𝑓 1  ξ€· β‰₯ β‹― β‰₯ 𝑓 ξ€Έ . 𝑛 βˆ’ π‘Ÿ ( 2 . 2 ) Now, if a generic function 𝑓 ∢ 𝐼 ( 𝑛 , π‘Ÿ ) β†’ ℝ that satisfies (2.2) is given, we are interested to find all the subsets π‘Œ βŠ† 𝐼 ( 𝑛 , π‘Ÿ ) such that βˆ‘ 𝑦 ∈ π‘Œ 𝑓 ( 𝑦 ) β‰₯ 0 . This goal becomes then easier if we can have an appropriate partial order βŠ‘ on the power set 𝒫 ( 𝐼 ( 𝑛 , π‘Ÿ ) ) “compatible” with the total order of the partial sums inducted by 𝑓 , that is, a partial order βŠ‘ that satisfies the following monotonicity property: if π‘Œ , 𝑍 ∈ 𝒫 ( 𝐼 ( 𝑛 , π‘Ÿ ) ) , then βˆ‘ 𝑧 ∈ 𝑍 βˆ‘ 𝑓 ( π‘₯ ) ≀ 𝑦 ∈ π‘Œ 𝑓 ( 𝑦 ) whenever 𝑍 βŠ‘ π‘Œ . To have such a partial order βŠ‘ on 𝒫 ( 𝐼 ( 𝑛 , π‘Ÿ ) ) that has the monotonicity property we must introduce a new formal symbol that we denote by 0 Β§ . We add this new symbol to the index set 𝐼 ( 𝑛 , π‘Ÿ ) , so we set 𝐴 ( 𝑛 , π‘Ÿ ) = 𝐼 ( 𝑛 , π‘Ÿ ) βˆͺ { 0 Β§ } . We introduce on 𝐴 ( 𝑛 , π‘Ÿ ) the following total order: 𝑛 βˆ’ π‘Ÿ β‰Ί β‹― β‰Ί 2 β‰Ί 1 β‰Ί 0 Β§ β‰Ί Μƒ Μƒ 1 β‰Ί 2 β‰Ί β‹― β‰Ί Μƒ π‘Ÿ . ( 2 . 3 ) If 𝑖 , 𝑗 ∈ 𝐴 ( 𝑛 , π‘Ÿ ) , then we write: 𝑖 βͺ― 𝑗 for 𝑖 = 𝑗 or 𝑖 β‰Ί 𝑗 . We denote by 𝑆 ( 𝑛 , π‘Ÿ ) the set of all the formal expressions 𝑖 1 β‹― 𝑖 π‘Ÿ ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ (hereafter called strings) that satisfy the following properties:(i) 𝑖 1 , … , 𝑖 π‘Ÿ Μƒ ∈ { 1 , … , Μƒ π‘Ÿ , 0 Β§ } , (ii) 𝑗 1 , … , 𝑗 𝑛 βˆ’ π‘Ÿ ∈ { 1 , … , 𝑛 βˆ’ π‘Ÿ , 0 Β§ } , (iii) 𝑖 1 ≽ β‹― ≽ 𝑖 π‘Ÿ ≽ 0 Β§ ≽ 𝑗 1 ≽ β‹― ≽ 𝑗 𝑛 βˆ’ π‘Ÿ , (iv)the unique element which can be repeated is 0 Β§ .

In the sequel we often use the lowercase letters 𝑒 , 𝑀 , 𝑧 , … to denote a generic string in 𝑆 ( 𝑛 , π‘Ÿ ) . 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 { Μƒ 1 , … , Μƒ π‘Ÿ , 0 Β§ } and the symbols on the right of ∣ are elements in { 0 Β§ , 1 , … , 𝑛 βˆ’ π‘Ÿ } . For example, if 𝑛 = 3 and π‘Ÿ = 2 , then Μƒ Μƒ 𝐴 ( 3 , 2 ) = { 2 ≻ 1 ≻ 0 Β§ ≻ 1 } and 𝑆 ( 3 , 2 ) = { 2 1 | 0 , 2 1 | 1 , 1 0 | 0 , 2 0 | 0 , 1 0 | 1 , 2 0 | 1 , 0 0 | 1 , 0 0 | 0 } .

Note that there is a natural bijective setcorrespondence βˆ— ∢ 𝑀 ∈ 𝑆 ( 𝑛 , π‘Ÿ ) ↦ 𝑀 βˆ— ∈ 𝒫 ( 𝐼 ( 𝑛 , π‘Ÿ ) ) between 𝑆 ( 𝑛 , π‘Ÿ ) and 𝒫 ( 𝐼 ( 𝑛 , π‘Ÿ ) ) defined as follows: if 𝑀 = 𝑖 1 β‹― 𝑖 π‘Ÿ ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ ∈ 𝑆 ( 𝑛 , π‘Ÿ ) , then 𝑀 βˆ— is the subset of 𝐼 ( 𝑛 , π‘Ÿ ) made with the elements 𝑖 π‘˜ and 𝑗 𝑙 such that 𝑖 π‘˜ β‰  0 Β§ and 𝑗 𝑙 β‰  0 Β§ . For example, if 𝑀 = 4 3 1 0 ∣ 0 1 3 ∈ 𝑆 ( 7 , 4 ) , then 𝑀 βˆ— Μƒ Μƒ Μƒ = { 1 , 3 , 4 , 1 , 3 } . In particular, if 𝑀 = 0 β‹― 0 ∣ 0 β‹― 0 , then 𝑀 βˆ— = βˆ… .

Now, if 𝑣 = 𝑖 1 β‹― 𝑖 π‘Ÿ ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ and 𝑀 = 𝑖 ξ…ž 1 β‹― 𝑖 ξ…ž π‘Ÿ ∣ 𝑗 ξ…ž 1 β‹― 𝑗 ξ…ž 𝑛 βˆ’ π‘Ÿ are two strings in 𝑆 ( 𝑛 , π‘Ÿ ) , we define: 𝑣 βŠ‘ 𝑀 iff 𝑖 1 βͺ― 𝑖 ξ…ž 1 , … , 𝑖 π‘Ÿ βͺ― 𝑖 ξ…ž π‘Ÿ , 𝑗 1 βͺ― 𝑗 ξ…ž π‘Ÿ , … , 𝑗 𝑛 βˆ’ π‘Ÿ βͺ― 𝑗 ξ…ž 𝑛 βˆ’ π‘Ÿ .

It is easily seen that (1) ( 𝑆 ( 𝑛 , π‘Ÿ ) , βŠ‘ ) is a finite distributive (hence also graded) lattice with minimum element 0 β‹― 0 ∣ 1 2 β‹― ( 𝑛 βˆ’ π‘Ÿ ) and maximum element π‘Ÿ ( π‘Ÿ βˆ’ 1 ) β‹― 2 1 ∣ 0 β‹― 0 ; (2) ( 𝑆 ( 𝑛 , π‘Ÿ ) , βŠ‘ ) has the following unary complementary operation 𝑐 : ξ€· 𝑝 1 β‹― 𝑝 π‘˜ 0 β‹― 0 ∣ 0 β‹― 0 π‘ž 1 β‹― π‘ž 𝑙 ξ€Έ 𝑐 = 𝑝 ξ…ž 1 β‹― 𝑝 ξ…ž π‘Ÿ βˆ’ π‘˜ 0 β‹― 0 ∣ 0 β‹― 0 π‘ž ξ…ž 1 β‹― π‘ž ξ…ž 𝑛 βˆ’ π‘Ÿ βˆ’ 𝑙 , ( 2 . 4 ) where { 𝑝 ξ…ž 1 , … , 𝑝 ξ…ž π‘Ÿ βˆ’ π‘˜ } is the usual complement of { 𝑝 1 , … , 𝑝 π‘˜ } in { Μƒ 1 , … , Μƒ π‘Ÿ } and { π‘ž ξ…ž 1 , … , π‘ž ξ…ž 𝑛 βˆ’ π‘Ÿ βˆ’ 𝑙 } is the usual complement of { π‘ž 1 , … , π‘ž 𝑙 } in { 1 , … , 𝑛 βˆ’ π‘Ÿ } (e.g., in 𝑆 ( 7 , 4 ) , we have that ( 4 3 1 0 ∣ 0 0 1 ) 𝑐 = 2 0 0 0 ∣ 0 2 3 ).

Since we have the formal necessity to consider functions 𝑓 defined on the extended set 𝐴 ( 𝑛 , π‘Ÿ ) instead of on the indexes set 𝐼 ( 𝑛 , π‘Ÿ ) , then we will put 𝑓 ( 0 Β§ ) = 0 . Precisely we can identify the quotient set π‘Š ( 𝑛 , π‘Ÿ ) / ∼ with the set π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) , defined by ξ€½ ξ€· Μƒ 1 ξ€Έ ξ€· 0 π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) = 𝑓 ∢ 𝐴 ( 𝑛 , π‘Ÿ ) β†’ ℝ ∢ 𝑓 ( Μƒ π‘Ÿ ) β‰₯ β‹― β‰₯ 𝑓 β‰₯ 𝑓 Β§ ξ€Έ ξ‚€ = 0 > 𝑓 1  ξ€· β‰₯ β‹― β‰₯ 𝑓 ξ€Έ ξ€· Μƒ 1 ξ€Έ ξ‚€ 𝑛 βˆ’ π‘Ÿ , 𝑓 ( Μƒ π‘Ÿ ) + β‹― + 𝑓 + 𝑓 1  ξ€· + β‹― + 𝑓 ξ€Έ ξ‚Ό . 𝑛 βˆ’ π‘Ÿ β‰₯ 0 ( 2 . 5 )

We call an element of π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) a ( 𝑛 , π‘Ÿ ) -weight function, and if 𝑓 ∈ π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) , we will continue to set βˆ‘ 𝛼 ( 𝑓 ) ∢ = | { π‘Œ βŠ† 𝐼 ( 𝑛 , π‘Ÿ ) ∢ 𝑦 ∈ π‘Œ 𝑓 ( 𝑦 ) β‰₯ 0 } | . Therefore, with these last notations we have that 𝛾 ( 𝑛 , π‘Ÿ ) = m i n { 𝛼 ( 𝑓 ) ∢ 𝑓 ∈ π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) } , πœ‚ ( 𝑛 , π‘Ÿ ) = m a x { 𝛼 ( 𝑓 ) ∢ 𝑓 ∈ π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) } and the question ( 𝑃 2 ) becomes equivalent to the following.

( 𝑃 2 )   If π‘ž is an integer such that 𝛾 ( 𝑛 , π‘Ÿ ) ≀ π‘ž ≀ πœ‚ ( 𝑛 , π‘Ÿ ) , does there exist a function 𝑓 ∈ π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) with the property that 𝛼 ( 𝑓 ) = π‘ž ?

3. Boolean Maps Induct by Weight Functions

We denote by 𝟐 the boolean lattice composed of a chain with 2 elements that we denote 𝑁 (the minimum element) and 𝑃 (the maximum element). A Boolean map (briefly BM) on 𝑆 ( 𝑛 , π‘Ÿ ) is a map 𝐴 ∢ d o m ( 𝐴 ) βŠ† 𝑆 ( 𝑛 , π‘Ÿ ) β†’ 𝟐 ; in particular if d o m ( 𝐴 ) = 𝑆 ( 𝑛 , π‘Ÿ ) ,we also say that 𝐴 is a Boolean total map (briefly BTM) on 𝑆 ( 𝑛 , π‘Ÿ ) . If 𝐴 is BM on 𝑆 ( 𝑛 , π‘Ÿ ) , we set 𝑆 + 𝐴 ( 𝑛 , π‘Ÿ ) = { 𝑀 ∈ d o m ( 𝐴 ) ∢ 𝐴 ( 𝑀 ) = 𝑃 } .

If 𝑓 ∈ π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) , the sum function Ξ£ 𝑓 ∢ 𝑆 ( 𝑛 , π‘Ÿ ) β†’ ℝ induced by 𝑓 on 𝑆 ( 𝑛 , π‘Ÿ ) is the function that associates to 𝑀 = 𝑖 π‘Ÿ β‹― 𝑖 1 ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ ∈ 𝑆 ( 𝑛 , π‘Ÿ ) the real number Ξ£ 𝑓 ( 𝑀 ) = 𝑓 ( 𝑖 1 ) + β‹― + 𝑓 ( 𝑖 π‘Ÿ ) + 𝑓 ( 𝑗 1 ) + β‹― + 𝑓 ( 𝑗 𝑛 βˆ’ π‘Ÿ ) , and therefore we can associate to 𝑓 ∈ π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) the map 𝐴 𝑓 ∢ 𝑆 ( 𝑛 , π‘Ÿ ) β†’ 𝟐 setting: 𝐴 𝑓 ⎧ βŽͺ ⎨ βŽͺ ⎩ ( 𝑀 ) = 𝑃 i f Ξ£ 𝑓 ( 𝑀 ) β‰₯ 0 , 𝑀 β‰  0 β‹― 0 ∣ 0 β‹― 0 , 𝑁 i f 𝑀 = 0 β‹― 0 ∣ 0 β‹― 0 , 𝑁 i f Ξ£ 𝑓 ( 𝑀 ) < 0 . ( 3 . 1 )

Let us note that | 𝑆 + 𝐴 𝑓 ( 𝑛 , π‘Ÿ ) | = | { 𝑀 ∈ 𝑆 ( 𝑛 , π‘Ÿ ) ∢ 𝐴 𝑓 ( 𝑀 ) = 𝑃 } | = 𝛼 ( 𝑓 ) , and so 𝛾 ( 𝑛 , π‘Ÿ ) = m i n { | 𝑆 + 𝐴 𝑓 ( 𝑛 , π‘Ÿ ) | ∢ 𝑓 ∈ π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) } , πœ‚ ( 𝑛 , π‘Ÿ ) = m a x { | 𝑆 + 𝐴 𝑓 ( 𝑛 , π‘Ÿ ) | ∢ 𝑓 ∈ π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) } .

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 𝐴 𝑓 has the following properties.(i) 𝐴 𝑓 is orderpreserving.(ii)If 𝑀 ∈ 𝑆 ( 𝑛 , π‘Ÿ ) is such that 𝐴 𝑓 ( 𝑀 ) = 𝑁 , then 𝐴 𝑓 ( 𝑀 𝑐 ) = 𝑃 ,(iii) 𝐴 𝑓 ( 1 0 β‹― 0 ∣ 0 β‹― 0 ) = 𝑃 , 𝐴 ( 0 β‹― 0 ∣ 0 β‹― 0 ) = 𝑁 , and 𝐴 𝑓 ( π‘Ÿ β‹― 2 1 ∣ 1 2 β‹― ( 𝑛 βˆ’ π‘Ÿ ) ) = 𝑃 .

Example 3.1. Let 𝑓 be the following ( 5 , 3 ) -weight function: Μƒ 3 Μƒ 2 Μƒ 1 𝑓 ∢ 1 2 ↓ ↓ ↓ ↓ ↓ 1 1 0 . 9 βˆ’ 0 . 8 βˆ’ 2 . 1 ( 3 . 2 ) We represent the map 𝐴 𝑓 defined on 𝑆 ( 5 , 3 ) by using the Hasse diagram of this lattice, on which we color green the nodes where the map 𝐴 𝑓 assumes value 𝑃 and red the nodes where it assumes value 𝑁 (See Figure 1).
Note that if we have a generic Boolean total map 𝐴 ∢ 𝑆 ( 𝑛 , π‘Ÿ ) β†’ 𝟐 which has the properties (i), (ii), and (iii) of 𝐴 𝑓 , that is, the following:(BM1) 𝐴 is orderpreserving,(BM2)if 𝑀 ∈ 𝑆 ( 𝑛 , π‘Ÿ ) is such that 𝐴 ( 𝑀 ) = 𝑁 , then 𝐴 ( 𝑀 𝑐 ) = 𝑃 ,(BM3) 𝐴 ( 1 0 β‹― 0 ∣ 0 β‹― 0 ) = 𝑃 , 𝐴 ( 0 β‹― 0 ∣ 0 β‹― 0 ) = 𝑁 , and 𝐴 ( π‘Ÿ β‹― 2 1 ∣ 1 2 β‹― ( 𝑛 βˆ’ π‘Ÿ ) ) = 𝑃 ,in general there does not exist a function 𝑓 ∈ π‘Š 𝐹 ( 𝑛 , π‘Ÿ ) such that 𝐴 𝑓 = 𝐴 (see [3] for a counterexample).
We denote by 𝒲 + ( 𝑆 ( 𝑛 , π‘Ÿ ) , 𝟐 ) the set of all the maps 𝐴 ∢ 𝑆 ( 𝑛 , π‘Ÿ ) β†’ 𝟐 which satisfy (BM1) and (BM2) and by 𝒲 + ( 𝑛 , π‘Ÿ ) the subset of all the maps in 𝒲 + ( 𝑆 ( 𝑛 , π‘Ÿ ) , 𝟐 ) which satisfy also (BM3). We also set 𝛾 βˆ— ( 𝑛 , π‘Ÿ ) ∢ = m i n { | 𝑆 + 𝐴 ( 𝑛 , π‘Ÿ ) ∢ 𝐴 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) } and πœ‚ βˆ— ( 𝑛 , π‘Ÿ ) ∢ = m a x { | 𝑆 + 𝐴 ( 𝑛 , π‘Ÿ ) | ∢ 𝐴 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) } . Let us observe that 𝛾 βˆ— ( 𝑛 , π‘Ÿ ) ≀ 𝛾 ( 𝑛 , π‘Ÿ ) ≀ πœ‚ ( 𝑛 , π‘Ÿ ) ≀ πœ‚ βˆ— ( 𝑛 , π‘Ÿ ) . A natural question raises at this point.
( 𝑄 ) : if π‘ž is an integer such that 𝛾 βˆ— ( 𝑛 , π‘Ÿ ) ≀ π‘ž ≀ πœ‚ βˆ— ( 𝑛 , π‘Ÿ ) , does there exist a map 𝐴 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) with the property that | 𝑆 + 𝐴 ( 𝑛 , π‘Ÿ ) | = π‘ž ?
The question ( 𝑄 ) is the analogue of ( 𝑃 2 ) expressed in terms of Boolean total maps on 𝑆 ( 𝑛 , π‘Ÿ ) instead of ( 𝑛 , π‘Ÿ ) -weight functions, and if we are able to respond to ( 𝑄 ) , we provide also a partial answer to ( 𝑃 2 ) . In Section 4 we give an affirmative answer to the question ( 𝑄 ) , and also we give a constructive method to build the map 𝐴 .

847958.fig.001
Figure 1

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 𝑍 βŠ† 𝑆 ( 𝑛 , π‘Ÿ ) , we will set ↓ 𝑍 = { π‘₯ ∈ 𝑆 ( 𝑛 , π‘Ÿ ) ∢ βˆƒ 𝑧 ∈ 𝑍 s u c h t h a t π‘₯ βŠ‘ 𝑧 } , ↑ 𝑍 = { π‘₯ ∈ 𝑆 ( 𝑛 , π‘Ÿ ) ∢ βˆƒ 𝑧 ∈ 𝑍 s u c h t h a t 𝑧 βŠ‘ π‘₯ } . In particular, if 𝑧 ∈ 𝑆 ( 𝑛 , π‘Ÿ ) , we will set ↓ 𝑧 = ↓ { 𝑧 } = { π‘₯ ∈ 𝑆 ( 𝑛 , π‘Ÿ ) ∢ 𝑧 βŠ’ π‘₯ } , ↑ 𝑧 = ↑ { 𝑧 } = { π‘₯ ∈ 𝑆 ( 𝑛 , π‘Ÿ ) ∢ 𝑧 βŠ‘ π‘₯ } . 𝑍 is called a downset of 𝑆 ( 𝑛 , π‘Ÿ ) if for 𝑧 ∈ 𝑍 and π‘₯ ∈ 𝑆 ( 𝑛 , π‘Ÿ ) with 𝑧 βŠ’ π‘₯ , then π‘₯ ∈ 𝑍 . 𝑍 is called an upset of 𝑆 ( 𝑛 , π‘Ÿ ) if for 𝑧 ∈ 𝑍 and π‘₯ ∈ 𝑆 ( 𝑛 , π‘Ÿ ) with 𝑧 βŠ‘ π‘₯ , then π‘₯ ∈ 𝑍 . ↓ 𝑍 is the smallest down-set of 𝑆 ( 𝑛 , π‘Ÿ ) which contains 𝑍 , and 𝑍 is a downset of 𝑆 ( 𝑛 , π‘Ÿ ) if and only if 𝑍 = ↓ 𝑍 . Similarly, ↑ 𝑍 is the smallest up-set of 𝑆 ( 𝑛 , π‘Ÿ ) which contains 𝑍 , and 𝑍 is an upset in 𝑆 ( 𝑛 , π‘Ÿ ) if and only if 𝑍 = ↑ 𝑍 .

Theorem 4.1. If 𝑛 and π‘Ÿ are two integers such that 0 < π‘Ÿ < 𝑛 , then 𝛾 ( 𝑛 , π‘Ÿ ) = 𝛾 βˆ— ( 𝑛 , π‘Ÿ ) = 2 𝑛 βˆ’ 1 and πœ‚ ( 𝑛 , π‘Ÿ ) = πœ‚ βˆ— ( 𝑛 , π‘Ÿ ) = 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ .

Proof. Assume that 0 < π‘Ÿ < 𝑛 . We denote by 𝑆 1 ( 𝑛 , π‘Ÿ ) the sublattice of 𝑆 ( 𝑛 , π‘Ÿ ) of all the strings 𝑀 of the form 𝑀 = 𝑖 1 β‹― 𝑖 π‘Ÿ ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ( 𝑛 βˆ’ π‘Ÿ ) , with 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ∈ { 0 Β§ , 1 , … , 𝑛 βˆ’ π‘Ÿ βˆ’ 1 } and by 𝑆 2 ( 𝑛 , π‘Ÿ ) the sublattice of 𝑆 ( 𝑛 , π‘Ÿ ) of all the strings 𝑀 of the form 𝑀 = 𝑖 1 β‹― 𝑖 π‘Ÿ ∣ 0 𝑗 2 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ , with 𝑗 2 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ ∈ { 0 Β§ , 1 , … , 𝑛 βˆ’ π‘Ÿ βˆ’ 1 } . It is clear that 𝑆 ( 𝑛 , π‘Ÿ ) = 𝑆 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) 2 ( 𝑛 , π‘Ÿ ) and 𝑆 1 ( 𝑛 , π‘Ÿ ) β‰… 𝑆 2 ( 𝑛 , π‘Ÿ ) β‰… 𝑆 ( 𝑛 βˆ’ 1 , π‘Ÿ ) . We consider now the following further sublattices of 𝑆 ( 𝑛 , π‘Ÿ ) : 𝑆 + 1 ξ€½ ( 𝑛 , π‘Ÿ ) ∢ = 𝑀 ∈ 𝑆 1 ( 𝑛 , π‘Ÿ ) ∢ 𝑀 = π‘Ÿ ( π‘Ÿ βˆ’ 1 ) β‹― 2 1 ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ξ€Ύ , 𝑆 ( 𝑛 βˆ’ π‘Ÿ ) Β± 1 ξ€½ ( 𝑛 , π‘Ÿ ) ∢ = 𝑀 ∈ 𝑆 1 ( 𝑛 , π‘Ÿ ) ∢ 𝑀 = 𝑖 1 β‹― 𝑖 π‘Ÿ βˆ’ 1 0 ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ( 𝑛 βˆ’ π‘Ÿ ) , 𝑖 1 ≻ 0 Β§ ξ€Ύ , 𝑆 βˆ’ 1 ξ€½ ( 𝑛 , π‘Ÿ ) ∢ = 𝑀 ∈ 𝑆 1 ( 𝑛 , π‘Ÿ ) ∢ 𝑀 = 0 β‹― 0 ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ξ€Ύ , 𝑆 ( 𝑛 βˆ’ π‘Ÿ ) + 2 ξ€½ ( 𝑛 , π‘Ÿ ) ∢ = 𝑀 ∈ 𝑆 2 ( 𝑛 , π‘Ÿ ) ∢ 𝑀 = π‘Ÿ ( π‘Ÿ βˆ’ 1 ) β‹― 2 1 ∣ 0 𝑗 2 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ ξ€Ύ , 𝑆 Β± 2 ( ξ€½ 𝑛 , π‘Ÿ ) ∢ = 𝑀 ∈ 𝑆 2 ( 𝑛 , π‘Ÿ ) ∢ 𝑀 = 𝑖 1 β‹― 𝑖 π‘Ÿ βˆ’ 1 0 ∣ 0 𝑗 2 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ , 𝑖 1 ≻ 0 Β§ ξ€Ύ , 𝑆 βˆ’ 2 ξ€½ ( 𝑛 , π‘Ÿ ) ∢ = 𝑀 ∈ 𝑆 2 ( 𝑛 , π‘Ÿ ) ∢ 𝑀 = 0 β‹― 0 ∣ 0 𝑗 2 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ ξ€Ύ . ( 4 . 1 ) It occurs immediately that 𝑆 𝑖 ( 𝑛 , π‘Ÿ ) = 𝑆 + 𝑖 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) Β± 𝑖 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 𝑖 ( 𝑛 , π‘Ÿ ) , for 𝑖 = 1 , 2 and 𝑆 Β± 𝑖 ( 𝑛 , π‘Ÿ ) is a distributive sublattice of 𝑆 𝑖 ( 𝑛 , π‘Ÿ ) with 2 𝑛 βˆ’ 1 βˆ’ 2 β‹… 2 𝑛 βˆ’ π‘Ÿ βˆ’ 1 = 2 𝑛 βˆ’ 1 βˆ’ 2 𝑛 βˆ’ π‘Ÿ elements, for 𝑖 = 1 , 2 .
Now we consider the following ( 𝑛 , π‘Ÿ ) -weight function 𝑓 ∢ 𝐴 ( 𝑛 , π‘Ÿ ) β†’ ℝ : Μƒ 𝑓 ∢ Μƒ π‘Ÿ β‹― 1 0 Β§ 1 β‹― ( 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ) ( 𝑛 βˆ’ π‘Ÿ ) ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ( 𝑛 βˆ’ π‘Ÿ ) β‹― ( 𝑛 βˆ’ π‘Ÿ ) 0 βˆ’ 1 β‹― βˆ’ 1 ( 𝑛 βˆ’ π‘Ÿ ) ( 1 βˆ’ π‘Ÿ ) βˆ’ 1 ( 4 . 2 ) Then it follows that Ξ£ 𝑓 ∢ 𝑆 ( 𝑛 , π‘Ÿ ) β†’ ℝ is such that Ξ£ 𝑓 ξ‚» ( 𝑀 ) β‰₯ 0 i f 𝑀 ∈ 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) , < 0 i f 𝑀 ∈ 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) . ( 4 . 3 ) It means that the boolean map 𝐴 𝑓 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) is such that 𝐴 𝑓 ξ‚» ( 𝑀 ) 𝑃 i f 𝑀 ∈ 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) , 𝑁 i f 𝑀 ∈ 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) . ( 4 . 4 ) This shows that | 𝑆 + 𝐴 𝑓 ( 𝑛 , π‘Ÿ ) | = | 𝑆 + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) Β± 2 ( 𝑛 , π‘Ÿ ) | = 2 𝑛 βˆ’ π‘Ÿ βˆ’ 1 + 2 𝑛 βˆ’ π‘Ÿ βˆ’ 1 + 2 𝑛 βˆ’ 1 βˆ’ 2 β‹… 2 𝑛 βˆ’ π‘Ÿ βˆ’ 1 = 2 𝑛 βˆ’ 1 .
In [1, Theorem 1] it has been proved that 𝛾 ( 𝑛 , 1 ) = 2 𝑛 βˆ’ 1 and 𝛾 ( 𝑛 , π‘Ÿ ) β‰₯ 2 𝑛 βˆ’ 1 . Since 𝛾 ( 𝑛 , π‘Ÿ ) β‰₯ 𝛾 βˆ— ( 𝑛 , π‘Ÿ ) , using a technique similar to that used in the proof Theorem 1 of [1], it easily follows that 𝛾 βˆ— ( 𝑛 , π‘Ÿ ) β‰₯ 2 𝑛 βˆ’ 1 . As shown above, it results that | 𝑆 + 𝐴 𝑓 ( 𝑛 , π‘Ÿ ) | = 2 𝑛 βˆ’ 1 . Hence 2 𝑛 βˆ’ 1 β‰₯ 𝛾 ( 𝑛 , π‘Ÿ ) β‰₯ 𝛾 βˆ— ( 𝑛 , π‘Ÿ ) β‰₯ 2 𝑛 βˆ’ 1 , that is, 𝛾 ( 𝑛 , π‘Ÿ ) = 𝛾 βˆ— ( 𝑛 , π‘Ÿ ) = 2 𝑛 βˆ’ 1 . This proves; the first part of theorem; it remains to prove the latter part.
We consider now the following ( 𝑛 , π‘Ÿ ) -positive weight function 𝑔 ∢ 𝐴 ( 𝑛 , π‘Ÿ ) β†’ ℝ : Μƒ Μƒ π‘Ÿ β‹― 1 0 Β§ 1 β‹― ( 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ) ( 𝑛 βˆ’ π‘Ÿ ) 𝑔 ∢ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 1 β‹― 1 0 βˆ’ 1 β‹― 𝑛 βˆ’ π‘Ÿ βˆ’ 1 𝑛 βˆ’ π‘Ÿ βˆ’ 1 𝑛 βˆ’ π‘Ÿ ( 4 . 5 ) It results then that the sum Ξ£ 𝑔 ∢ 𝑆 ( 𝑛 , π‘Ÿ ) β†’ ℝ is such that Ξ£ 𝑔 ( 𝑀 ) β‰₯ 0 if 𝑀 ∈ 𝑆 Β± 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) Β± 2 ( 𝑛 , π‘Ÿ ) ; that is, the boolean map 𝐴 𝑔 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) is such that 𝐴 𝑔 ( 𝑀 ) = 𝑃 if 𝑀 ∈ 𝑆 Β± 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) Β± 2 ( 𝑛 , π‘Ÿ ) . This shows that | 𝑆 + 𝐴 𝑔 ( 𝑛 , π‘Ÿ ) | = | 𝑆 Β± 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) Β± 2 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) | = ( 2 𝑛 βˆ’ 1 βˆ’ 2 𝑛 βˆ’ π‘Ÿ ) + ( 2 𝑛 βˆ’ 1 βˆ’ 2 𝑛 βˆ’ π‘Ÿ ) + 2 𝑛 βˆ’ π‘Ÿ βˆ’ 1 + 2 𝑛 βˆ’ π‘Ÿ βˆ’ 1 = 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ .
On the other hand, it is clear that for any 𝐴 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) it results 𝐴 ( 𝑀 ) = 𝑃 for each 𝑀 ∈ 𝑆 + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) and 𝐴 ( 𝑀 ) = 𝑁 for each 𝑀 ∈ 𝑆 βˆ’ 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) . The number ( 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ ) is the biggest number of values 𝑃 that a boolean map 𝐴 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) can assume. Hence, since πœ‚ ( 𝑛 , π‘Ÿ ) ≀ πœ‚ βˆ— ( 𝑛 , π‘Ÿ ) , we have 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ = | 𝑆 + 𝐴 𝑔 ( 𝑛 , π‘Ÿ ) | ≀ πœ‚ ( 𝑛 , π‘Ÿ ) ≀ πœ‚ βˆ— ( 𝑛 , π‘Ÿ ) ≀ 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ ; that is be; πœ‚ ( 𝑛 , π‘Ÿ ) = πœ‚ βˆ— ( 𝑛 , π‘Ÿ ) = 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ . 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 𝑛 = 6 and π‘Ÿ = 2 and let 𝑓 be as given in previous theorem; that is Μƒ 2 Μƒ 𝑓 ∢ 1 0 Β§ 1 2 3 4 ↓ ↓ ↓ ↓ ↓ ↓ ↓ 4 4 0 βˆ’ 1 βˆ’ 1 βˆ’ 1 βˆ’ 5 ( 4 . 6 ) In Figure 2 we have shown the Hasse diagram of the lattice 𝑆 (6,2), where 𝑆 + 1 ( 𝑛 , π‘Ÿ ) is black, 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) is violet, 𝑆 βˆ’ 1 ( 𝑛 , π‘Ÿ ) is red, 𝑆 + 2 ( 𝑛 , π‘Ÿ ) is blue, 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) is brown, 𝑆 βˆ’ 2 ( 𝑛 , π‘Ÿ ) is green. Therefore 𝐴 𝑓 assume the following values.(i)The blue, black, and brown nodes correspond to values 𝑃 of 𝐴 𝑓 ,(ii)The violet, red, and green nodes correspond to values 𝑁 of 𝐴 𝑓 .

847958.fig.002
Figure 2

First to give the proof of Theorem 4.5 we need to introduce some useful results and the concept of basis in 𝑆 ( 𝑛 , π‘Ÿ ) . In the following first lemma we show some properties of the sublattices of 𝑆 ( 𝑛 , π‘Ÿ ) .

Lemma 4.2. Here the following properties hold, where πœƒ = 0 0 β‹― 0 ∣ 0 β‹― 0 and Θ = π‘Ÿ β‹― 2 1 ∣ 1 2 β‹― ( 𝑛 βˆ’ π‘Ÿ ) . (i) ↑ Θ = 𝑆 + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) .(ii) ↓ πœƒ = 𝑆 βˆ’ 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) .(iii) ↑ 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) βŠ† 𝑆 Β± 2 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) .(iv) ↓ 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) βŠ† 𝑆 Β± 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 1 ( 𝑛 , π‘Ÿ ) .(v) ( ( 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) ) 𝑐 = 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) .

Proof. (i) If 𝑀 ∈ ( ↑ Θ ) , then Θ βŠ‘ 𝑀 ; that is, it has the form 𝑀 = π‘Ÿ β‹― 1 ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ , where 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ ∈ { 0 Β§ , 1 , … , 𝑛 βˆ’ π‘Ÿ } ; therefore 𝑀 ∈ 𝑆 + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) . If 𝑀 ∈ 𝑆 + 1 ( 𝑛 , π‘Ÿ ) , it has the form 𝑀 = π‘Ÿ β‹― 1 ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ( 𝑛 βˆ’ π‘Ÿ ) , where 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ∈ { 0 Β§ , 1 , … , 𝑛 βˆ’ π‘Ÿ βˆ’ 1 } ; if 𝑀 ∈ 𝑆 + 2 ( 𝑛 , π‘Ÿ ) , it has the form 𝑀 = π‘Ÿ β‹― 1 ∣ 0 𝑗 2 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ , where 𝑗 2 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ ∈ { 0 Β§ , 1 , … , 𝑛 βˆ’ π‘Ÿ βˆ’ 1 } . In both cases it results that Θ βŠ‘ 𝑀 , that is, 𝑀 ∈ ( ↑ Θ ) .(ii)It is analogue to (i).(iii)The minimum of the sublattice 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) is 𝛼 = 1 0 β‹― 0 ∣ 0 1 β‹― ( 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ) ; since ↑ 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) βŠ† ↑ 𝛼 , it is sufficient to show that ↑ 𝛼 = 𝑆 Β± 2 ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) . The inclusion 𝑆 Β± 2 ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) βŠ† ↑ 𝛼 follows by the definition of 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) and of 𝑆 + 2 ( 𝑛 , π‘Ÿ ) . On the other side, if 𝑀 ∈ ↑ 𝛼 , it follows that 𝛼 βŠ‘ 𝑀 ; that is, 𝑀 = 𝑖 1 β‹― 𝑖 π‘Ÿ ∣ 0 𝑗 2 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ , with 𝑖 1 ≻ 0 Β§ and 𝑗 2 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ ∈ { 0 Β§ , 1 , … , 𝑛 βˆ’ π‘Ÿ βˆ’ 1 } . Therefore 𝑀 ∈ 𝑆 Β± 2 ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) , and this proves the other inclusion.(iv)Let us consider the maximum of the sublattice 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) ; that is 𝑑 1 = π‘Ÿ ( π‘Ÿ βˆ’ 1 ) β‹― 2 0 ∣ 0 β‹― 0 ( 𝑛 βˆ’ π‘Ÿ ) . Since ↓ 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) βŠ† ↓ 𝛽 , it is sufficient to show that ↓ 𝑑 1 = 𝑆 Β± 1 ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 1 ( 𝑛 , π‘Ÿ ) ; this proof is similar to (iii).(v)If 𝑀 ∈ 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) ; it has the form 𝑀 = 𝑖 1 β‹― 𝑖 π‘Ÿ βˆ’ 1 0 ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ( 𝑛 βˆ’ π‘Ÿ ) , with 𝑖 1 ≻ 0 Β§ and 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ∈ { 0 Β§ , 1 , … , 𝑛 βˆ’ π‘Ÿ βˆ’ 1 } ; therefore its complement has the form 𝑀 𝑐 = 𝑖 ξ…ž 1 β‹― 𝑖 ξ…ž π‘Ÿ βˆ’ 1 0 ∣ 0 𝑗 ξ…ž 2 β‹― 𝑗 ξ…ž 𝑛 βˆ’ π‘Ÿ , with 𝑖 1 β€² ≻ 0 Β§ and 𝑗 ξ…ž 2 β‹― 𝑗 ξ…ž 𝑛 βˆ’ π‘Ÿ ∈ { 0 Β§ , 1 , … , 𝑛 βˆ’ π‘Ÿ βˆ’ 1 } ; hence 𝑀 𝑐 ∈ 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) . This shows that ( 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) ) 𝑐 βŠ† 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) .
Now, if 𝑀 ∈ 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) , we write 𝑀 in the form ( 𝑀 𝑐 ) 𝑐 ; then 𝑀 𝑐 ∈ ( 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) ) 𝑐 βŠ† 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) ; therefore 𝑀 ∈ ( 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) ) 𝑐 . This shows that 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) βŠ† ( 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) ) 𝑐 .

At this point let us recall the definition of basis for 𝑆 ( 𝑛 , π‘Ÿ ) , as given in [3] 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 [3] for details). Hence the concept of basis will be fundamental in the sequel of this proof.

Definition 4.3. A basis for 𝑆 ( 𝑛 , π‘Ÿ ) is an ordered couple ⟨ π‘Œ + ∣ π‘Œ βˆ’ ⟩ , where π‘Œ + and π‘Œ βˆ’ are two disjoint antichains of 𝑆 ( 𝑛 , π‘Ÿ ) such that(B1) ( ↓ π‘Œ + ) β‹‚ ( π‘Œ 𝑐 βˆ’ ) = βˆ… , (B2) ( ( ↑ π‘Œ + ) ⋃ ↑ ( π‘Œ 𝑐 βˆ’ β‹‚ ) ) ↓ π‘Œ βˆ’ = βˆ… , (B3) 𝑆 ( 𝑛 , π‘Ÿ ) = ( ( ↑ π‘Œ + ) ⋃ ↑ ( π‘Œ 𝑐 βˆ’ ⋃ ) ) ↓ π‘Œ βˆ’ . 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 [3].

Lemma 4.4. Let ⟨ π‘Š + ∣ π‘Š βˆ’ ⟩ be a basis for 𝑆 ( 𝑛 , π‘Ÿ ) . Then the map ξ‚» ξ€· π‘Š 𝐴 ( π‘₯ ) 𝑃 i f π‘₯ ∈ ↑ + ξ€Έ  ↑ ξ€· π‘Š 𝑐 βˆ’ ξ€Έ , 𝑁 i f π‘₯ ∈ ↓ π‘Š βˆ’ ( 4 . 7 ) is such that 𝐴 ∈ 𝒲 + ( 𝑆 ( 𝑛 , π‘Ÿ ) , 𝟐 ) .

Theorem 4.5. If π‘ž is a fixed integer with 2 𝑛 βˆ’ 1 ≀ π‘ž ≀ 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ , then there exists a boolean map 𝐴 π‘ž ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) such that | 𝑆 + 𝐴 π‘ž ( 𝑛 , π‘Ÿ ) | = π‘ž .

Proof. Let π‘ž be a fixed integer such that 2 𝑛 βˆ’ 1 ≀ π‘ž ≀ 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ . We determine a specific Boolean total map 𝐴 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) such that | 𝑆 + 𝐴 ( 𝑛 , π‘Ÿ ) | = π‘ž . We proceed as follows.
The case π‘Ÿ = 1 is proved in the previous Theorem 4.1. Let us assume then π‘Ÿ > 1 . Since 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) is a finite distributive sublattice of graded lattice 𝑆 ( 𝑛 , π‘Ÿ ) , also 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) is a graded lattice. We denote by 𝑅 the rank of 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) and with 𝜌 1 its rank function. Note that the bottom of 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) is 𝑏 1 = 1 0 β‹― 0 ∣ 1 β‹― ( 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ) ( 𝑛 βˆ’ π‘Ÿ ) and the top is 𝑑 1 = π‘Ÿ ( π‘Ÿ βˆ’ 1 ) β‹― 2 0 ∣ 0 β‹― 0 ( 𝑛 βˆ’ π‘Ÿ ) .
We write π‘ž in the form π‘ž = 2 𝑛 βˆ’ 1 + 𝑝 with 0 ≀ 𝑝 ≀ 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ βˆ’ 2 𝑛 βˆ’ 1 = 2 n βˆ’ 1 βˆ’ 2 𝑛 βˆ’ π‘Ÿ = | 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) | . We will build a map 𝐴 ∈ π‘Š + ( 𝑛 , π‘Ÿ ) such that | 𝑆 + 𝐴 ( 𝑛 , π‘Ÿ ) | = 2 𝑛 βˆ’ 1 + 𝑝 .
If 𝑝 = 0 , we take 𝐴 = 𝐴 𝑓 , with 𝑓 as in Theorem 4.1, and hence we have | 𝑆 + 𝐴 ( 𝑛 , π‘Ÿ ) | = 2 𝑛 βˆ’ 1 .
If 𝑝 = 2 𝑛 βˆ’ 1 βˆ’ 2 𝑛 βˆ’ π‘Ÿ , we take 𝐴 = 𝐴 𝑔 , with 𝑔 as in Theorem 4.1, and we have | 𝑆 + 𝐴 ( 𝑛 , π‘Ÿ ) | = 2 𝑛 βˆ’ 2 𝑛 βˆ’ π‘Ÿ .
Therefore we can assume that 0 < 𝑝 < 2 𝑛 βˆ’ 1 βˆ’ 2 𝑛 βˆ’ π‘Ÿ .
If 0 ≀ 𝑖 ≀ 𝑅 , we denote by β„œ 𝑖 the set of elements 𝑀 ∈ 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) such that 𝜌 1 ( 𝑀 ) = 𝑅 βˆ’ 𝑖 , and we also set 𝛽 𝑖 ∢ = | β„œ 𝑖 | to simplify the notation. We write each β„œ 𝑖 in the following form: β„œ 𝑖 = { 𝑣 𝑖 1 , … , 𝑣 𝑖 𝛽 𝑖 } . If 0 ≀ 𝑙 ≀ 𝑅 , we set π”˜ l Μ‡ ⋃ ∢ = 𝑖 = 0 , … , 𝑙 β„œ 𝑖 .
If 0 ≀ 𝑙 ≀ 𝑅 βˆ’ 2 , we set 𝔅 𝑙 Μ‡ ⋃ ∢ = 𝑖 = 𝑙 + 2 , … , 𝑅 β„œ 𝑖 and 𝔅 𝑅 βˆ’ 1 ∢ = 𝔅 𝑅 ∢ = βˆ… . We can then write 𝑝 in the form βˆ‘ 𝑝 = π‘˜ 𝑖 = 0 | β„œ 𝑖 | + 𝑠 = | π”˜ π‘˜ | + 𝑠 , for some 0 ≀ 𝑠 < | β„œ π‘˜ + 1 | and some 0 ≀ π‘˜ ≀ 𝑅 βˆ’ 1 . Depending on the previous number 𝑠 we partition β„œ π‘˜ + 1 into the following two disjoint subsets: β„œ π‘˜ + 1 = { 𝑣 ( π‘˜ + 1 ) 1 , … , 𝑣 ( π‘˜ + 1 ) 𝑠 } Μ‡ ⋃ { 𝑣 ( π‘˜ + 1 ) ( 𝑠 + 1 ) , … , 𝑣 ( π‘˜ + 1 ) 𝛽 π‘˜ + 1 } , where the first subset is considered empty if 𝑠 = 0 .
In the sequel, to avoid an overload of notations, we write simply 𝑣 𝑖 instead of 𝑣 ( π‘˜ + 1 ) 𝑖 , for 𝑖 = 1 , … , 𝛽 π‘˜ + 1 .
Let us note that 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) = π”˜ π‘˜ Μ‡ ⋃ β„œ π‘˜ + 1 Μ‡ ⋃ 𝔅 π‘˜ .
We define now the map 𝐴 ∢ 𝑆 ( 𝑛 , π‘Ÿ ) β†’ 𝟐 : ⎧ βŽͺ βŽͺ βŽͺ ⎨ βŽͺ βŽͺ βŽͺ ⎩ 𝐴 ( 𝑀 ) = 𝑃 i f 𝑀 ∈ 𝑆 Β± 2 Μ‡  𝑆 ( 𝑛 , π‘Ÿ ) + 1 Μ‡  𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) , 𝑃 i f 𝑀 ∈ π”˜ π‘˜ Μ‡  ξ€½ 𝑣 1 , … , 𝑣 𝑠 ξ€Ύ , 𝑁 i f 𝑀 ∈ 𝔅 π‘˜ Μ‡   𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1  , 𝑁 i f 𝑀 ∈ 𝑆 βˆ’ 1 ( Μ‡  𝑆 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) . ( 4 . 8 ) Let us observe that | 𝑆 + 𝐴 ( 𝑛 , π‘Ÿ ) | = | 𝑆 Β± 2 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) | + | π”˜ π‘˜ Μ‡ ⋃ { 𝑣 1 , … , 𝑣 𝑠 } | = ( 2 𝑛 βˆ’ 1 βˆ’ 2 𝑛 βˆ’ π‘Ÿ ) + 2 𝑛 βˆ’ π‘Ÿ βˆ’ 1 + 2 𝑛 βˆ’ π‘Ÿ βˆ’ 1 + | π”˜ π‘˜ | + 𝑠 = 2 𝑛 βˆ’ 1 + 𝑝 = π‘ž .
Therefore, if we show that 𝐴 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) , the theorem is proved.
We write β„œ π‘˜ in the following way: β„œ π‘˜ = { 𝑀 1 , … , 𝑀 𝑑 } Μ‡ ⋃ { 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ } , where { 𝑀 1 , … , 𝑀 𝑑 } = β„œ π‘˜ β‹‚ ↑ { 𝑣 1 , … , 𝑣 𝑠 } and { 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ } = β„œ π‘˜ ⧡ { 𝑀 1 , … , 𝑀 𝑑 } . Analogously β„œ π‘˜ + 2 = { 𝑧 1 , … , 𝑧 π‘ž } Μ‡ ⋃ { 𝑧 π‘ž + 1 , … , 𝑧 𝛽 π‘˜ + 2 } , where { 𝑧 π‘ž + 1 , … , 𝑧 𝛽 π‘˜ + 2 } = β„œ π‘˜ + 2 β‹‚ ↓ { 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 } and { 𝑧 1 , … , 𝑧 π‘ž } = β„œ π‘˜ + 2 ⧡ { 𝑧 π‘ž + 1 , … , 𝑧 𝛽 π‘˜ + 2 } .
We can see a picture of this partition of the sublattice 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) in Figure 3.
Depending on 𝑠 and π‘˜ , we build now a particular basis for 𝑆 ( 𝑛 , π‘Ÿ ) .
To such aim, let us consider the minimum 𝛼 = 1 0 β‹― 0 ∣ 0 1 β‹― ( 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ) of 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) and the subsets 𝑇 + ∢ = { 𝑣 1 , … , 𝑣 𝑠 , 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ } and 𝑇 βˆ’ ∢ = { 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 , 𝑧 1 , … , 𝑧 π‘ž } . Let us distinguish two cases: ( a 1 ) 𝛼 ∈ ↑ 𝑇 + , ( a 2 ) 𝛼 βˆ‰ ↑ 𝑇 + . We define two different couples of subsets as follows:
In the case ( a 1 ) we set π‘Œ + ∢ = 𝑇 + a n d π‘Œ βˆ’ ∢ = 𝑇 βˆ’ Μ‡ ⋃ { πœƒ } ; in the case ( a 2 ) we set π‘Œ + ∢ = 𝑇 + ⋃ { 𝛼 } and π‘Œ βˆ’ ∢ = 𝑇 βˆ’ Μ‡ ⋃ { πœƒ } .
Step 1. ⟨ π‘Œ + ∣ π‘Œ βˆ’ ⟩ is a couple of two disjoint antichains of 𝑆 ( 𝑛 , π‘Ÿ ) .
In both cases (a1) and (a2) it is obvious that π‘Œ + β‹‚ π‘Œ βˆ’ = βˆ… .
Case (a1)
The elements { 𝑣 1 , … , 𝑣 𝑠 } are not comparable between them because they have all rank 𝑅 βˆ’ ( π‘˜ + 1 ) and analogously for the elements { 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ } that have all rank 𝑅 βˆ’ π‘˜ . Let now 𝑣 ∈ { 𝑣 1 , … , 𝑣 𝑠 } and let 𝑀 ∈ { 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ } ; then 𝑀 βˆ‰ ↓ 𝑣 because 𝜌 1 ( 𝑣 ) < 𝜌 1 ( 𝑀 ) , and 𝑀 βˆ‰ ↑ 𝑣 because { 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ } β‹‚ ↑ { 𝑣 1 , … , 𝑣 𝑠 } = βˆ… by construction. For the elements in π‘Œ βˆ’ different from πœƒ , we can proceed as for π‘Œ + . On the other side, we can observe that πœƒ is not comparable with none of the elements 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 , 𝑧 1 , … , 𝑧 π‘ž since these are all in 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) while πœƒ ∈ 𝑆 βˆ’ 2 ( 𝑛 , π‘Ÿ ) . Thus π‘Œ + is an antichain. Case (a2)
In this case we only must show that 𝛼 is not comparable with none of the elements 𝑣 1 , … , 𝑣 𝑠 , 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ . At first from the fact that 𝛼 βˆ‰ ↑ 𝑇 + it follows 𝑣 𝑖 βˆ‰ ↓ 𝛼 for each 𝑖 = 1 , … , 𝑠 , and 𝑀 𝑗 βˆ‰ ↓ 𝛼 for each 𝑗 = 𝑑 + 1 , … , 𝛽 π‘˜ . Moreover, the elements 𝑣 1 , … , 𝑣 𝑠 and 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ are all in 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) ; hence they have the form 𝑖 1 β‹― 𝑖 π‘Ÿ βˆ’ 1 0 ∣ 𝑗 1 β‹― 𝑗 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ( 𝑛 βˆ’ π‘Ÿ ) , with 𝑖 1 ≻ 0 Β§ , while 𝛼 = 1 0 β‹― 0 ∣ 0 1 β‹― ( 𝑛 βˆ’ π‘Ÿ βˆ’ 1 ) , so 𝛼 βˆ‰ ↓ 𝑣 𝑖 and 𝛼 βˆ‰ ↓ 𝑀 𝑗 for each 𝑖 = 1 , … , 𝑠 and each 𝑗 = 𝑑 + 1 , … , 𝛽 π‘˜ .Step 2. ⟨ π‘Œ + ∣ π‘Œ βˆ’ ⟩ is a basis for 𝑆 ( 𝑛 , π‘Ÿ ) .
We must see that (B1), (B2), and (B3) hold in both cases (a1) and (a2).
Case (a1)
(B1)Let us begin to observe that 𝑇 𝑐 βˆ’ = { 𝑣 𝑐 𝑠 + 1 , … , 𝑣 𝑐 𝛽 π‘˜ + 1 , 𝑧 𝑐 1 , … , 𝑧 𝑐 π‘ž } βŠ† ( 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) ) 𝑐 = 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) . Since πœƒ 𝑐 = Θ , we have then π‘Œ 𝑐 βˆ’ = 𝑇 𝑐 βˆ’ Μ‡ ⋃ { πœƒ 𝑐 } βŠ† 𝑆 Β± 2 Μ‡ ⋃ ( 𝑛 , π‘Ÿ ) { Θ } βŠ† 𝑆 Β± 2 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 1 ( 𝑛 , π‘Ÿ ) . On the other hand, since ↓ π‘Œ + βŠ† ↓ 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) , by Lemma 4.2(iv) we have also that ↓ π‘Œ + βŠ† 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) . ⋃ 𝑆 βˆ’ 1 ( 𝑛 , π‘Ÿ ) . Hence ( ↓ π‘Œ + ) β‹‚ ( π‘Œ 𝑐 βˆ’ ) = βˆ… . This proves (B1).(B2)We show at first that ↑ π‘Œ 𝑐 βˆ’ β‹‚ ↓ π‘Œ βˆ’ = βˆ… . Since π‘Œ 𝑐 βˆ’ βŠ† 𝑆 Β± 2 Μ‡ ⋃ ( 𝑛 , π‘Ÿ ) { Θ } , we have that ↑ π‘Œ 𝑐 βˆ’ βŠ† ↑ ( 𝑆 Β± 2 ⋃ ( 𝑛 , π‘Ÿ ) ) ↑ Θ . By Lemma 4.2(i) and (iii) we have then ↑ π‘Œ 𝑐 βˆ’ βŠ† 𝑆 Β± 2 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) . On the other side, since ↓ 𝑇 βˆ’ βŠ† ↓ 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) , by Lemma 4.2(iv) it follows that ↓ 𝑇 βˆ’ βŠ† 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) . ⋃ 𝑆 βˆ’ 1 ( 𝑛 , π‘Ÿ ) . By Lemma 4.2(ii), we have ↓ πœƒ = 𝑆 βˆ’ 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) . Hence it holds ↓ π‘Œ βˆ’ βŠ† 𝑆 Β± 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) . This proves that ↑ π‘Œ 𝑐 βˆ’ β‹‚ ↓ π‘Œ βˆ’ = βˆ… . We show now that also ↑ π‘Œ + β‹‚ ↓ π‘Œ βˆ’ = βˆ… ; we proceed by contradiction. Let us suppose that there exists an element 𝑧 ∈ ↑ π‘Œ + β‹‚ ↓ π‘Œ βˆ’ , then there are two elements 𝑀 + ∈ π‘Œ + and 𝑀 βˆ’ ∈ π‘Œ βˆ’ such that 𝑀 + βŠ‘ 𝑧 βŠ‘ 𝑀 βˆ’ , hence 𝑀 + βŠ‘ 𝑀 βˆ’ . We will distinguish the following five cases, and in each of them we will find a contradiction.(a) 𝑀 + ∈ { 𝑣 1 , … , 𝑣 𝑠 } and 𝑀 βˆ’ ∈ { 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 } . In this case 𝑀 + and 𝑀 βˆ’ are two distinct elements having both rank 𝑅 βˆ’ ( π‘˜ + 1 ) and such that 𝑀 + βŠ‘ 𝑀 βˆ’ ; it is not possible.(b) 𝑀 + ∈ { 𝑣 1 , … , 𝑣 𝑠 } and 𝑀 βˆ’ ∈ { 𝑧 1 , … , 𝑧 π‘ž } . In this case 𝑀 + has rank 𝑅 βˆ’ ( π‘˜ + 1 ) while 𝑀 βˆ’ has rank 𝑅 βˆ’ ( π‘˜ + 2 ) < 𝑅 βˆ’ ( π‘˜ + 1 ) , and this contradicts the condition 𝑀 + βŠ‘ 𝑀 βˆ’ .(c) 𝑀 + ∈ { 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ } and 𝑀 βˆ’ ∈ { 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 } . This case is similar to the previous because 𝑀 + has rank 𝑅 βˆ’ π‘˜ while 𝑀 βˆ’ has rank 𝑅 βˆ’ ( π‘˜ + 1 ) .(d) 𝑀 + ∈ { 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ } and 𝑀 βˆ’ ∈ { 𝑧 1 , … , 𝑧 π‘ž } . This is similar to the previous because 𝑀 + has rank 𝑅 βˆ’ π‘˜ while 𝑀 βˆ’ has rank 𝑅 βˆ’ ( π‘˜ + 2 ) .(e) 𝑀 + ∈ { 𝑣 1 , … , 𝑣 𝑠 , 𝑀 𝑑 + 1 , … , 𝑀 𝛽 π‘˜ } and 𝑀 βˆ’ = πœƒ . In this case the condition 𝑀 + βŠ‘ 𝑀 βˆ’ implies that 𝑀 + ∈ ↓ πœƒ ; since ↓ πœƒ = 𝑆 βˆ’ 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) this is not possible since 𝑀 + ∈ 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) .(B3)Since 𝛼 is the minimum of 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) we have that 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) βŠ† ↑ 𝛼 , moreover, 𝛼 ∈ ↑ π‘Œ + ; therefore 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) βŠ† ↑ 𝛼 βŠ† ↑ π‘Œ + . Since Θ = πœƒ 𝑐 ∈ π‘Œ 𝑐 βˆ’ , it follows that ↑ Θ βŠ† ↑ π‘Œ 𝑐 βˆ’ . By Lemma 4.2(i) we have then that ( 𝑆 + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) ) = ↑ Θ βŠ† ↑ π‘Œ 𝑐 βˆ’ . By Lemma 4.2(ii) we also have that ( 𝑆 βˆ’ 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) ) = ↓ πœƒ βŠ† ↓ π‘Œ βˆ’ .
To complete the proof of (B3) let us observe that 𝑆 Β± 1 ⋃ ( 𝑛 , π‘Ÿ ) = ( π‘˜ 𝑖 = 0 β„œ 𝑖 ) Μ‡ ⋃ β„œ π‘˜ + 1 Μ‡ ⋃ ( ⋃ 𝑅 𝑗 = π‘˜ + 2 β„œ 𝑗 ) , where ( ⋃ π‘˜ 𝑖 = 0 β„œ 𝑖 ) βŠ† ↑ β„œ π‘˜ βŠ† ↑ π‘Œ + , ( ⋃ 𝑅 𝑗 = π‘˜ + 2 β„œ 𝑗 ) βŠ† ↓ β„œ π‘˜ + 2 βŠ† ↓ π‘Œ βˆ’ .
Moreover, since β„œ π‘˜ + 1 = { 𝑣 1 , … , 𝑣 𝑠 } Μ‡ ⋃ { 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 } , with { 𝑣 1 , … , 𝑣 𝑠 } βŠ† π‘Œ + βŠ† ↑ π‘Œ + and { 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 } βŠ† π‘Œ βˆ’ βŠ† ↓ π‘Œ βˆ’ , then β„œ π‘˜ + 1 βŠ† ( ↑ π‘Œ + ⋃ ↓ π‘Œ βˆ’ ) . This shows that 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) βŠ† ( ↑ π‘Œ + ⋃ ↓ π‘Œ βˆ’ ) . Since the six sublattices 𝑆 + 𝑖 ( 𝑛 , π‘Ÿ ) , 𝑆 Β± 𝑖 ( 𝑛 , π‘Ÿ ) and 𝑆 βˆ’ 𝑖 ( 𝑛 , π‘Ÿ ) for 𝑖 = 1 , 2 are a partition of 𝑆 ( 𝑛 , π‘Ÿ ) , the property (B3) is proved.Case (a2)
(B1)We begin to show that 𝛼 βˆ‰ π‘Œ 𝑐 βˆ’ . In fact, it holds that 𝛼 = ( π‘Ÿ ( π‘Ÿ βˆ’ 1 ) β‹― 2 0 ∣ 0 β‹― 0 ( 𝑛 βˆ’ π‘Ÿ ) ) 𝑐 ; suppose that 𝛼 ∈ π‘Œ 𝑐 βˆ’ , and this shows that we obtain a contradiction. By 𝛼 ∈ π‘Œ 𝑐 βˆ’ it follows 𝛼 𝑐 = 𝑑 1 ∈ π‘Œ βˆ’ . But 𝑑 1 is the top of 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) , so π‘Œ βˆ’ = 𝑑 1 , since π‘Œ βˆ’ is antichain. This means that β„œ π‘˜ + 1 = β„œ 0 and this is not possible, of course. Since π‘Œ + = 𝑇 + Μ‡ ⋃ { 𝛼 } , we have ( ↓ π‘Œ + ) β‹‚ π‘Œ 𝑐 βˆ’ = ( ( ↓ 𝑇 + ) β‹‚ π‘Œ 𝑐 βˆ’ ) ⋃ β‹‚ π‘Œ ( ( ↓ 𝛼 ) 𝑐 βˆ’ ) , and, moreover, as in the proof of (B1) in the case (a1), we also have that ( ↓ 𝑇 + ) β‹‚ π‘Œ 𝑐 βˆ’ = βˆ… ; therefore, to prove (B1) it is sufficient to show that β‹‚ π‘Œ ( ↓ 𝛼 ) 𝑐 βˆ’ = βˆ… ; As in the proof of (B1) in the case (a1), we have π‘Œ 𝑐 βˆ’ βŠ† 𝑆 Β± 2 Μ‡ ⋃ ( 𝑛 , π‘Ÿ ) { Θ } . Since 𝛼 is the minimum of 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) and Θ βˆ‰ ↓ 𝛼 , it follows that the unique element of ↓ 𝛼 that can belong to π‘Œ 𝑐 βˆ’ is 𝛼 , but we have shown before that this is not true.(B2)As in the previous case we have ( ↑ π‘Œ 𝑐 βˆ’ ) β‹‚ ↓ π‘Œ βˆ’ = βˆ… ; moreover ( ↑ π‘Œ + ) β‹‚ ( ↓ π‘Œ βˆ’ ) = ( ( ↑ 𝑇 + ) β‹‚ β‹‚ ( ↑ 𝛼 ) ) ( ↓ π‘Œ βˆ’ ) ( ( ↑ 𝑇 + ) β‹‚ ( ↓ π‘Œ βˆ’ ⋃ β‹‚ ) ) ( ( ↑ 𝛼 ) ( ↓ π‘Œ βˆ’ ) ) . As in the case (a1) we have ( ( ↑ 𝑇 + ) β‹‚ ( ↓ π‘Œ βˆ’ ) ) = βˆ… ; therefore, to prove (B2) also in the case (a2), it is sufficient to show that β‹‚ ( ( ↑ 𝛼 ) ( ↓ π‘Œ βˆ’ ) ) = βˆ… . As in the proof of (B2) in the case (a1) it results that ↓ π‘Œ βˆ’ βŠ† 𝑆 Β± 2 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) , and, by definition of 𝛼 , it is easy to observe that β‹‚ ( ↑ 𝛼 ) ( 𝑆 Β± 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) ) = βˆ… . Hence β‹‚ ( ( ↑ 𝛼 ) ( ↓ π‘Œ βˆ’ ) ) = βˆ… .(B3)It is identical to that of case (a1).Step 3. The map A defined in (4.8) is such that 𝐴 ∈ 𝒲 + ( 𝑛 , π‘Ÿ ) .
Since we have proved that ⟨ π‘Œ + ∣ π‘Œ βˆ’ ⟩ is a basis for 𝑆 ( 𝑛 , π‘Ÿ ) , by Lemma 4.4 it follows that the map 𝐴 ∈ 𝒲 + ( 𝑆 ( 𝑛 , π‘Ÿ ) , 𝟐 ) if the two following identities hold: ξ€· ↑ π‘Œ + ξ€Έ  ξ€· ↑ π‘Œ 𝑐 βˆ’ ξ€Έ = 𝑆 Β± 2 Μ‡  S ( 𝑛 , π‘Ÿ ) + 1 Μ‡  𝑆 ( 𝑛 , π‘Ÿ ) + 2 Μ‡  π”˜ ( 𝑛 , π‘Ÿ ) π‘˜ Μ‡  ξ€½ 𝑣 1 , … , 𝑣 𝑠 ξ€Ύ , ( 4 . 9 ) ↓ π‘Œ βˆ’ = 𝔅 π‘˜ Μ‡   𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1  Μ‡  𝑆 βˆ’ 1 Μ‡  𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) . ( 4 . 1 0 ) We prove at first (4.10). By definition of 𝔅 π‘˜ and 𝑇 βˆ’ it easy to observe that 𝔅 π‘˜ Μ‡ ⋃ { 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 } βŠ† ↓ 𝑇 βˆ’ , and, moreover, by Lemma 4.2(ii) we also have that ↓ πœƒ = 𝑆 βˆ’ 1 ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) ; hence, since ↓ π‘Œ βˆ’ ⋃ = ↓ πœƒ ↓ 𝑇 βˆ’ , it results that 𝔅 π‘˜ Μ‡ ⋃ { 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 } ⋃ 𝑆 βˆ’ 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 2 ( 𝑛 , π‘Ÿ ) βŠ† ↓ π‘Œ βˆ’ . On the other hand, by Lemma 4.2(iv) we have that ↓ 𝑇 βˆ’ βŠ† 𝑆 Β± 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 1 ( 𝑛 , π‘Ÿ ) because 𝑇 βˆ’ is a subset of 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) . At this point let us note that the elements of ↓ 𝑇 βˆ’ that are also in 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) must belong necessarily to the subset 𝔅 π‘˜ Μ‡ ⋃ { 𝑣 𝑠 + 1 , … , 𝑣 𝛽 π‘˜ + 1 } . This proves the other inclusion and hence (4.10).
To prove now (4.9) we must distinguish the cases (a1) and (a2). We set Ξ” ∢ = 𝑆 Β± 2 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ⋃ π”˜ ( 𝑛 , π‘Ÿ ) π‘˜ Μ‡ ⋃ { 𝑣 1 , … , 𝑣 𝑠 } . Let us first examine the case (a1). Since 𝛼 is the minimum of 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) , we have 𝑆 Β± 2 ( 𝑛 , π‘Ÿ ) βŠ† ↑ 𝛼 βŠ† ↑ π‘Œ + . Moreover, since π‘Œ βˆ’ = 𝑇 βˆ’ ⋃ { πœƒ } , it follows that ↑ π‘Œ 𝑐 βˆ’ βŠ‡ ↑ πœƒ 𝑐 = ↑ ( Θ ) = 𝑆 + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) by lemma 3.2(i). Finally, since π”˜ π‘˜ Μ‡ ⋃ { 𝑣 1 , … , 𝑣 𝑠 } βŠ† ↑ 𝑇 + = ↑ π‘Œ + , the inclusion βŠ‡ in (4.9) is proved. To prove the other inclusion βŠ† we begin to observe that ↑ π‘Œ + β‹‚ ( 𝑆 βˆ’ 1 ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) βˆ’ 1 ( 𝑛 , π‘Ÿ ) ) = βˆ… ; therefore the elements of ↑ π‘Œ + that are not in 𝑆 Β± 2 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 1 Μ‡ ⋃ 𝑆 ( 𝑛 , π‘Ÿ ) + 2 ( 𝑛 , π‘Ÿ ) must be necessarily in 𝑆 Β± 1 ( 𝑛 , π‘Ÿ ) , and such elements, by definition of 𝑇 + , must be necessarily in π”˜ π‘˜ Μ‡ ⋃ { 𝑣 1 , … , 𝑣 𝑠 } . This proves that ↑ π‘Œ + βŠ† Ξ” , For ↑ ( π‘Œ 𝑐 βˆ’ ) , we have ↑ ( π‘Œ 𝑐 βˆ’ ) = ↑ ( 𝑇 𝑐 βˆ’ ⋃ { πœƒ } 𝑐 ) = ( ↑ 𝑇 𝑐 βˆ’ ) ⋃ ( ↑ Θ ) , where ↑ ( Θ ) = 𝑆 + 1 Μ‡ ⋃ ( 𝑛 , π‘Ÿ ) 𝑆 + 2 ( 𝑛 , π‘Ÿ ) by Lemma 4.2(i), and, since 𝑇 βˆ’ βŠ† 𝑆 Β±