Computer Science and Information Technology 5(4): 140-147, 2017 http://www.hrpub.org DOI: 10.13189/csit.2017.050404 Algebraic Objects of MBFs and Recursive Computation of the Dedekind Number Tkachenco V.G.1,*, Sinyavsky O.V.2 1Institute of Radio, Television, Electronics, Odessa National Academy of Telecommunications, Ukraine 2Department of Fundamental Sciences, Odessa Military Academy, Ukraine Copyright©2017 by authors, all rights reserved. Authors agree that this article remains permanently open access under the terms of the Creative Commons Attribution License 4.0 International License Abstract In this article the whole set of n-1 rank 2 . Results Monotone Boolean Functions (MBFs) is divided into equivalence classes. It shows how the Dedekind number D(n) Let’s consider the process of recursion of an n-rank MBF can be calculated by using this partition. Five formulas were from an n-1 rank MBF. found to calculate this number as well as the algebraic Let’s map P:( fnij()()− 1, f n −→ 1) fn() of all the n-1 properties of MBF blocks. rank MBF pairs on an n-rank MBF as follows: f n= fn −∨11 x f n −. Let fn−1 be the left or Keywords Monotone Boolean Functions, Free k()()() i nj i () Distributive Lattice, Dedekind Number, Algebraic upper part of fnk () , and the rest of fnj ()−1 – its right or Structures lower part. Let's call a binary notation of a function of its value on all ordered sets of variables. The binary bits will be numbered from right to left from 0 to 21n − , where n is a rank of MBF. Let’s select unity in bit j of the MBF binary notation. Let's 1 .Introduction call subordinated all the MBF’s unities in bits ij> where In 1897 R. Dedekind found ψ(4) number of elements for a the MBF stops being monotone if any of the bits changes in 0. free distributive lattice with four generators, which he Define the dependent units for a given unit in the j-bit. This is published in his article [1]. In algebra parlance the Dedekind the unit itself in the j-bit, as well as the units in the bits ij≤ number D(n) = ψ(n) + 2 is the number of the Monotone that need to be zeroed out when the unit is zeroed in the j-bit Boolean functions (MBFs) dependent on n number of x1, ... , in order for the MBF to remain monotonic. xn variables. D(0) – D(4) were found by Dedekind. D(5) was Example 1. The binary notation of 3-rank MBF is 1111 calculated by Church [2] in 1940, D(6) - by Ward [3] in 1946, 1111, where the bits of this representation are numbered D(7) - by Church [4] in 1965 and D(8) – by Wiedemann [5] from right to left from 0 to 7. Unity from bit 0 is in 1991. The inequivalent MBFs were conducted until the self-dependent and all other unities are subordinated. There seventh rank in [6]. As it turned out, this problem is quite no subordinates for unity from bit 7 but all of them are difficult and not manageable under the traditional method of dependent. Unities in bits 0, 2, 4 and 6 are dependent for generating functions. There is no a simple closed formula for unity from bit 6, and unity in bit 7 is subordinated. Unities in finding the Dedekind number [7]. bits 0 and 4 are dependent for unity from bit 4, and unities However, the recursive method to find the Dedekind from bits 6 and 7 are subordinates. number D(n) by dividing all the elements of the free The binary notation of fnk () consists of two equal parts: distributive lattice of n-1 rank MBFs into equivalence classes 2n−1 which represents the left part (where x = 1) and the is not reported. In some cases, this method reduces the search n of MBFs due to it easier to consider the only n-1 MBFs right part (where xn = 0). Each part represents an n-1 rank (which are significantly less in number) instead of MBF, at that fnij()()−≥11 f n − . This is the considering all n-1 rank MBFs. The aim of this paper is to develop a new method of “greater-than-or-equal-to” for all the elements of the free recursive calculation of the Dedekind number D(n) based on distributive lattice of the n-1 rank MBF. the n-1 MBFs equivalence classes partition and to study the If the upper part of fni ()−1 is replaced with algebraic properties of the MBF blocks. fnij()()−∨11 f n −, the upper part of the binary notation Computer Science and Information Technology 5(4): 140-147, 2017 141 mapped onto another function, i.e. we will get another does not change, because fnij( −≥11) f( n −) . − It is known that conjunction s absorbs conjunction t if all equivalence class. Therefore, the second function fnj ( 1) is its variables are part of conjunction t. Let's say that MBF f always one and the same for all pairs of functions within each absorbs MBF g, if each conjunction g is absorbed by one of equivalence class. The lemma is proved. conjunction f. It is equivalent for the binary notation that Another equivalence class definition can be given based fg≥ , i.e. one function is more than the other. It follows on this lemma and the definition of mapping P: − − from this inequality that f= fg ∨ , and gfg= ∧ . Definition. Function pairs fni ( 1) and fnj ( 1) are Let’s designate monotone a pair of functions, where from the same equivalence class if they form one and the same disjunction fn−∨11 f n − and the second fnij( −≥11) f( n −) . Indeed, the concatenation of binary ij( ) ( ) notations for such pairs is the binary notation of an n-rank function fnj ( −1) does not change. MBF by the definition of monotonicity. Lemma 2. Let’s consider the equivalence class, to which For example, this pair of functions: disjunction functions pair fn( −1) and fn( −1) belongs. Then there is a single fn( −∨11) f( n −) and fn( −1) is always monotone, i.e. i j ij j pair of monotone functions: disjunction the upper part of the MBF is disjunction of fnij( −∨11) f( n −) and fnj ( −1) for any pair of fnij( −∨11) f( n −) . fn−−1, f n 1 within each equivalence class. Assume that all the set of pairs of an n −1 rank MBF are ( ij( ) ( )) divide into equivalence classes. The two pairs fit the same Proof. Let’s consider mapping P, fni( −∨11) x nj f( n −) , equivalence class, if P maps them onto the same n-rank when x = 1 and when x = 0 MBF. n n = Example 2. Let’s take a pair of 4-rank functions – When xn 1 fi (4) =∨∨ xx23 xx 24 xx 34 and fj (4) =∨∨ xx12 xx 13 xx 14, fn( −∨−=−∨−=−∨−1) x f( n 111111) fn( ) f( n) fn( ) f( n ) inji jij. along with another one pair – When xn = 0 xn = 1 fr (4) = xx23 ∨ xx 24 ∨∨∨∨ xx 34 xx 12 xx 13 xx 14 fni( −1) x nj ∨ fn( −= 1) fn i( − 10) ∨ fnj( −= 1) fn j( − 1) and fs (4) =∨∨ xx12 xx 13 xx 14 . These two pairs of functions are mapped by P onto function It follows that fk( n) = fn i( −∨11) x nj f( n −) represents f(5) =∨∨∨ xx12 xx 13 xx 14 xxx 235 ∨ xxx 245 ∨ xxx 345. I.e. a concatenation of two binary functions: fnij( −∨11) f( n −) is the first and fnj ( −1) is the second. Pf( i, f j) = Pf( rs ,5 f) = f( ) . The pair is monotone because −∨ −≥ −. Let’s consider the factor set of function pairs against fnij( 111) fn( ) fn j( ) mapping P and show what form each equivalence class of the Further, taking into account lemma 1, that disjunction factor set takes. fnij( −∨11) f( n −) is constant and the second function It is true that: fnj ( −1) is always one and the same for all pairs of Lemma 1. For any pair of ( fnij( −−1,) f( n 1)) functions, it follows that there can only be the only one disjunction fnij( −∨11) f( n −) and the second function monotone pair within a certain equivalence class. The lemma fnj ( −1) are constant within the whole equivalence class. Proof. is proved. Example 3. Let f(54) = f1( ) xf 52 ∨ ( 4) . If fk( n) = fn i( −11) x nj ∨ f( n −=) f1(4) = xx 23 ∨ xx 24 ∨∨∨∨ xx 34 xx 12 xx 13 xx 14 and =fni( −1) x nj ∨ f( n − 11)( x n ∨=) f2(4) =∨∨ xx 12 xx 13 xx 14, the pair will be monotone, and if =fni( −∨111) x nj fn( −∨) x nj fn( −=) f1(4) =∨∨ xx 23 xx 24 xx 34 and f2(4) =∨∨ xx 12 xx 13 xx 14, =( fni( −∨−∨−11) fn j( )) x nj fn( 1) the pair will be nonmonotone. Here the monotony is broken by the following setups: is correct for any pair of fni ( −1) and fnj ( −1) from the same class. Table 1. Nonmonotonicity f(54) = f1( ) xf 52 ∨ ( 4) It follows that the disjunction of any pair x5 x4 x3 x2 x1 f1 f2 f1x5 f1x5 ∨ f2 fnij( −∨11) f( n −) of one equivalence class is constant for 1 0 0 1 1 0 1 0 1 the entire equivalence class. Otherwise fnk ( ) would be 1 0 1 0 1 0 1 0 1 different and have another equivalence class.