Essential arity gap of Boolean functions

ESSENTIAL ARITY GAP OF BOOLEAN FUNCTIONS SLA VCHO SHTRAKO V Abstract. In this paper we inv estigate the Bool ean functions with maximum essen tial arity gap. Additionally we prop ose a simpler prov e of an impor tant theorem prov ed by M. Couceiro and E. Leh tonen in [3]. They used the Zhe- galkin’s p olynomials as normal f orms for Bo olean functions and describe the functions with essen tial arit y gap equals 2. W e use F ull Conjunctive Normal F or ms instead of these polynomial s whic h allow us to si mplify the pro ofs and to obtain sev eral com binatorial results, concerning the Boolean functions with a given arit y gap. The F ull Conjunct ive Normal F or ms are sum of conjunctions also, in w hi c h all v ariables o ccur. 1. Introduction Essential v a riables of functions are studied by several authors [1, 2, 4]. In this pap er we co nsider the problem for simplification of functions by iden tification of v aria bles. This problem is discusse d in the work o f O. Lupano v, Y u. B reitbart, A. Salomaa, M. Couceiro , E. Lehtonen, e tc. for Bo olean functions and by K. Chimev for ar bitrary discr ete functions . Similar problems for terms a nd universal algebra a re studied by the a uthor and K. Denecke [7]. Essential input v ariables for tre e automata are discussed in [6 ]. The pro blems concerning essential a rity gap of functions are discussed in [3]. Here we study a nd count the Bo olean functions, which hav e max imum arit y gap. Note that if a function f has greater e ssential arity ga p than the essential arity gap of another function g , then f has a simpler automata realization than g . This fact is of a great impo rtance in theore tica l a nd applied computer scienc e and mo deling. 2. E ssential v ariables in Boolean functions Let B = { 0 , 1 } be the se t (ring) of the residuums mo dulo 2. An n -ary Bo ole an function (op er ation) is a mapping f : B n → B for so me natural n umber n , called arity of f . The set of all s uc h functions is denoted by P n 2 . A v a riable x i is called essential in f , o r f essen t ial ly dep ends on x i , if there exist v alues a 1 , . . . , a n , b ∈ B , such that f ( a 1 , . . . , a i − 1 , a i , a i +1 , . . . , a n ) 6 = f ( a 1 , . . . , a i − 1 , b, a i +1 , . . . , a n ) . The set of essential v ariables in a function f is denoted by E ss ( f ) a nd the nu mber of essential v ar iables in f is denoted b y ess ( f ) = | E ss ( f ) | . The v aria bles from X = { x 1 , . . . , x n } which a re not es sen tial in f ∈ P n 2 are called fictive and the set of fictive v ar ia bles in f is deno ted b y F ic ( f ). 2000 Mathematics Subje ct Classific ation. Primar y: 94C10; Secondary: 06E30 ACM-Comp uting Classific ation System (1998) : G.2.0. Key wor ds and phr ases. essent ial v ariable, iden tification minor, essent ial ari t y gap. 1 2 SL. SHTRAKO V Let x i and x j be esse ntial v ariables in f . W e s ay that the function g is o btained from f ∈ P n 2 by identific ation of a variable x i with x j , if g ( x 1 , . . . , x n ) = f ( x 1 , . . . , x i − 1 , x j , x i +1 , . . . , x n ) = f ( x i = x j ) . Briefly , when g is o bta ined fro m f , b y identification of the v ariable x i with x j , w e will wr ite g = f i ← j and g is called the identific ation minor of f . The set of all ident ification mino rs of f will b e denoted by M in ( f ) . F or completeness o f our c o nsideration w e alow to o bta in identification minors when x i or x j are not essential in f , also. Thus if x i do es not o ccur in f , then w e define f i ← j := f . Clearly , ess ( f i ← j ) ≤ ess ( f ), bec ause x i / ∈ E ss ( f i ← j ), even though it may b e essential in f . F or a function f ∈ P n 2 the essen tial arity gap (shortly arity gap ) of f is defined as follows g ap ( f ) := ess ( f ) − max g ∈ M in ( f ) ess ( g ) . It is not difficult to s ee that the functions with ”huge” gap, ar e mor e simple for realization by switching c ircuits and functional schemas in theoretica l and applied computer science. Let us denote by G m p the set of all functions in P n 2 which essentially depend on m v aria bles and hav e ga p equals to p i.e. G m p = { f ∈ P n 2 | ess ( f ) = m & g ap ( f ) = p } , with m ≤ n . An upper b ound of g ap ( f ) for Bo ole a n fu nctions is found by K .Chimev, A. Salomaa and O. Lupanov [2, 4, 5 ]. It is shown t hat g ap ( f ) ≤ 2, when f ∈ P n 2 , n ≥ 2. This result is generalized for ar bitrary finite v alued functions in [3]. It is pr oved that g ap ( f ) ≤ k for all f ∈ P n k , n ≥ k . Let m ∈ N , 0 ≤ m ≤ 2 n − 1 be an integer. It is well kno wn that for every n ∈ N , there is an unique finite sequence ( α 1 , . . . , α n ) ∈ B n such that (1) m = α 1 2 n − 1 + α 2 2 n − 2 + . . . + α n . The equation (1) is kno wn as the presentation o f m in binary p ositional numerical system. One br ie fly writes m = α 1 α 2 . . . α n instead (1). F or a v ariable x and α ∈ B , w e define the follo wing imp ortant function x α =  1 if x = α 0 if x 6 = α. This function is used in many in vestigations, concerning the a pplica tions of discr ete functions in computer science [2]. There ar e many normal forms for repr esentation of functions fro m P n 2 . In this pap er we will use the F ul l Conjunctive Normal F orm (FCNF) for studying the es- sential a rity gap of functions. This normal form is based on the table representation of Bo olean functions. The next t wo theore ms are in the basis o f the Theor y of Bo olea n functions, and they are well known. Theorem 2.1. Each function f ∈ P n 2 c an b e u n iquely r epr esente d in FC NF as fol lows (2) f ( x 1 , . . . , x n ) = a 0 .x 0 1 . . . x 0 n ⊕ . . . a m .x α 1 1 . . . x α n n ⊕ . . . a 2 n − 1 .x 1 1 . . . x 1 n wher e m = α 1 . . . α n , a m ∈ B and ” ⊕ ” , and ” . ” ar e t he op er ations addition and multiplic ation mo dulo 2 in the ring B . ESSENTIAL ARITY GAP OF BOOLEAN FUNCTIONS 3 Theorem 2.2. A variabl e x i is fictive in the function f ∈ P n 2 , if and only if f ( x 1 , . . . , x n ) = = x 0 i .f 1 ( x 1 , . . . , x i − 1 , x i +1 , . . . , x n ) ⊕ x 1 i .f 2 ( x 1 , . . . , x i − 1 , x i +1 , . . . , x n ) , with f 1 = f 2 and x i / ∈ E ss ( f j ) , wher e f j ∈ P n − 1 2 , for j = 1 , 2 . The next lemmas characterize the relation b etw een the iden tification minors of Bo olean functions. Lemma 2.1. L et f , g ∈ P n 2 b e t wo Bo ole an functions re pr esente d by t heir FCNF as follows f = 2 n − 1 − 1 M m =0 a m .x α 1 1 . . . x α n n and g = 2 n − 1 − 1 M m =0 b m .x α 1 1 . . . x α n n , wher e m = α 1 . . . α n . If f i ← j = g i ← j and α i = α j for some i, j with 1 ≤ j < i ≤ n , then a m = b m . Pro of. Without loss of generality we will prov e the le mma , for i = 2 and j = 1. Since f 2 ← 1 = g 2 ← 1 we hav e f ( x 1 , x 1 , x 3 , . . . , x n ) = g ( x 1 , x 1 , x 3 , . . . , x n ) . Hence a m = f ( α 1 , α 1 , α 3 , . . . , α n ) = g ( α 1 , α 1 , α 3 , . . . , α n ) = b m . Lemma 2.2. L et f , g ∈ P n 2 , b e t wo fun ctions, dep ending essential ly on n, n ≥ 3 variables. If f i ← j = g i ← j for al l i, j, 1 ≤ j < i ≤ n , then f = g . Pro of. Let f and g b e functions repr esent ed by their FCNF as in Lemma 2.1. Let m = α 1 . 2 n − 1 + α 2 . 2 n − 2 + . . . + α n be an arbitr a ry integer from { 0 , 1 , . . . , 2 n − 1 } . Since n ≥ 3 there exis t t wo na tur al num b ers i , j with 1 ≤ j < i ≤ n and α i = α j . F rom Le mma 2.1 we obtain a m = f ( α 1 , α 2 , . . . , α n ) = g ( α 1 , α 2 , . . . , α n ) = b m . Consequently , w e have f = g . Example 2.1. L et us c onsider the Bo ole an functions f = x 0 1 x 0 2 ⊕ x 1 1 x 0 2 and g = x 0 1 x 0 2 ⊕ x 0 1 x 1 2 . It is e asy to se e that for al l i, j, 1 ≤ j < i ≤ n we have f i ← j = g i ← j = x 0 1 , but f 6 = g . This example shows that n ≥ 3 is an essen t ial c ondition in L emm a 2.2. 3. Essential Arity Gap o f Boolean Functions F or B o o lean functions ¬ ( x ) denotes the unary operatio n neg a tion, i.e. ¬ x = x 0 =  1 if x = 0 0 if x 6 = 0 . Prop osition 3. 1. F or e ach Bo ole an function f the fol lowing sentenc es ar e held: ( i ) g ap ( f ( x 1 , . . . , x n )) = g ap ( f ( ¬ x 1 , . . . , ¬ x n )) ; ( ii ) g ap ( f ( x 1 , . . . , x n )) = g ap ( ¬ ( f ( x 1 , . . . , x n ))) ; ( iii ) g ap ( f ( x 1 , . . . , x n )) = g ap ( f ( x π (1) , . . . , x π ( n ) )) , wher e π : { 1 , . . . , n } → { 1 , . . . , n } is a p ermutation of the set { 1 , . . . , n } ; ( iv ) ess ( f i ← j ) = ess ( f j ← i ) for all i, j, 1 ≤ j < i ≤ n . 4 SL. SHTRAKO V Note that the last t wo asse r tions ( iii ) and ( iv ) are v alid in mo re general case o f k -v alued fun ctions. F or a n y natur al num b er n, n ≥ 2 we define the follo wing tw o sets: Od n 2 := { α 1 α 2 . . . α n ∈ { 0 , 1 } n | α 1 ⊕ α 2 ⊕ . . . ⊕ α n = 1 } and E v n 2 := { α 1 α 2 . . . α n ∈ { 0 , 1 } n | α 1 ⊕ α 2 ⊕ . . . ⊕ α n = 0 } . Clearly , α 1 α 2 . . . α n ∈ Od n 2 if and only if the num b er of 1’s in α 1 α 2 . . . α n is o dd, and α 1 α 2 . . . α n ∈ E v n 2 when this n um b er is ev en. Prop osition 3. 2. F or any n , n ≥ 4 , if f = M α 1 ...α n ∈ Od n 2 x α 1 1 . . . x α n n or f = M α 1 ...α n ∈ E v n 2 x α 1 1 . . . x α n n , then f ∈ G n 2 . Pro of. Without lo ss of genera lit y let us a ssume that f = L α 1 ...α n ∈ Od n 2 x α 1 1 . . . x α n n . W e hav e to show that ess ( f i ← j ) ≤ n − 2 for all i, j, 1 ≤ j < i ≤ n . Without loss of generality , again we will assume i = 2 and j = 1. Then we hav e f 2 ← 1 = M α 1 ,α 3 ,...α n ∈ Od n − 1 2 x α 1 1 x α 3 3 . . . x α n n = = x 0 1 .  M α 3 ,...α n ∈ Od n − 2 2 x α 3 3 . . . x α n n  ⊕ x 1 1 .  M α 3 ,...α n ∈ Od n − 2 2 x α 3 3 . . . x α n n  = = M α 3 ,...α n ∈ Od n − 2 2 x α 3 3 . . . x α n n . The result is the same, when α 1 . . . α n ∈ E v n 2 . W e ar e going to descr ib e the set G n 2 for n = 2 , 3 , 4 . The results for n = 4 can b e easily extended in the more gener al case n ≥ 4. Theorem 3.1. L et f ∈ P 2 2 . Then f ∈ G 2 2 if and only if f = a 0 . ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ a 1 .x 0 1 x 1 2 ⊕ a 2 .x 1 1 x 0 2 , with a 1 6 = a 0 or a 2 6 = a 0 . Pro of. Let f = a 0 .x 0 1 x 0 2 ⊕ a 1 .x 0 1 x 1 2 ⊕ a 2 .x 1 1 x 0 2 ⊕ a 3 .x 1 1 x 1 2 . The v ariables x 1 and x 2 are essential in f if and only if ( a 0 , a 1 ) 6 = ( a 2 , a 3 ) and ( a 0 , a 2 ) 6 = ( a 1 , a 3 ). Consider the identifi cation minor h := f 2 ← 1 = a 0 .x 0 1 ⊕ a 3 .x 1 1 of f . W e need ess ( h ) = 0 and from Theorem 2.2 it follows a 0 = a 3 . If w e suppo se that a 1 = a 2 = a 0 , then f ( x 1 , x 2 ) = a 0 which contradicts ess ( f ) = 2 . Corollary 3.1. Ther e ar e 6 functions in G 2 2 , i.e. | G 2 2 | = 6 . Pro of. Let a 0 ∈ { 0 , 1 } . F or a 1 and a 2 there are 3 p ossible choices, which s atisfy Theorem 3.1. The bo th cases a 1 = a 2 = a 0 = 0 and a 1 = a 2 = a 0 = 1 are impo ssible b ecause then e ss ( f ) < 2, since Theo rem 2.2. Corollary 3. 2. If f = a 0 .x 0 1 x 0 2 ⊕ a 1 .x 0 1 x 1 2 ⊕ a 2 .x 1 1 x 0 2 ⊕ a 3 .x 1 1 x 1 2 ∈ P 2 2 then es s ( f 2 ← 1 ) = 0 if and only if a 0 = a 3 . The next step is to describ e the functions which essentially dep end o n 3 v aria bles and hav e essential a rity ga p equal to 2. ESSENTIAL ARITY GAP OF BOOLEAN FUNCTIONS 5 Theorem 3. 2. L et f b e a Bo ole an function of thr e e variables. Then f ∈ G 3 2 if and only if it c an b e r epr esente d in one of the fol lowing sp e cial forms: f = x α 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) ⊕ x β 1 x β 2 , (3) or (4) f = x α 3 ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ x ¬ ( α ) 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) , wher e α, β ∈ { 0 , 1 } . Pro of. Note that the presentation of f in (4) is symmetric with resp ect to the v aria bles, but in (3) f is not symmetric with resp ect to the v aria bles x 1 and x 3 . So, the theorem ass erts that f ∈ G 3 2 if and only if f can be r epresented in one o f the forms (3) or (4), after a suitable p e r m utation o f the v aria bles. ” ⇐ ” Clearly , x 1 , x 2 and x 3 are essential v ariables in the functions of the right sides o f (3 ) and (4). T o see that f ∈ G 3 2 it is enough to do immedia te c hecking. Thu s for the function f in (3) w e hav e f 2 ← 1 = x β 1 , f 3 ← 1 =  x β 1 if β = α x β 2 if β 6 = α and f 3 ← 2 =  x β 2 if β = α x β 1 if β 6 = α. The functions f as in (4) are in G 3 2 bec ause x i , x j / ∈ E ss ( f i ← j ) for a ll i, j, 1 ≤ j < i ≤ 3. ” ⇒ ” Assume that f ∈ G 3 2 . Let the F CNF of f is written as follows: f = x 0 3 ( a 0 .x 0 1 x 0 2 ⊕ a 1 .x 0 1 x 1 2 ⊕ a 2 .x 1 1 x 0 2 ⊕ a 3 .x 1 1 x 1 2 ) ⊕ ⊕ x 1 3 ( a 4 .x 0 1 x 0 2 ⊕ a 5 .x 0 1 x 1 2 ⊕ a 6 .x 1 1 x 0 2 ⊕ a 7 .x 1 1 x 1 2 ) = = x 0 3 .g ( x 1 , x 2 ) ⊕ x 1 3 .h ( x 1 , x 2 ) . A. Supp ose that x 1 ∈ E ss ( g 2 ← 1 ) or x 1 ∈ E ss ( h 2 ← 1 ). Then x 1 ∈ E ss ( f 2 ← 1 ) bec ause f 2 ← 1 ( x 3 = 0) = g 2 ← 1 and f 2 ← 1 ( x 3 = 1) = h 2 ← 1 . Hence f ∈ G 3 2 implies x 3 / ∈ E ss ( f 2 ← 1 ) i.e g 2 ← 1 = h 2 ← 1 . Co ns equent ly , a 0 = a 4 and a 3 = a 7 . Then we obtain u = f 3 ← 1 = a 0 .x 0 1 x 0 2 ⊕ a 1 .x 0 1 x 1 2 ⊕ a 6 .x 1 1 x 0 2 ⊕ a 7 .x 1 1 x 1 2 , and v = f 3 ← 2 = a 0 .x 0 1 x 0 2 ⊕ a 2 .x 1 1 x 0 2 ⊕ a 5 .x 0 1 x 1 2 ⊕ a 7 .x 1 1 x 1 2 . There are the follo wing case s : A.a. x 1 / ∈ E ss ( u ). Hence a 0 = a 6 and a 1 = a 7 . A.a.1. If w e supp ose that x 1 / ∈ E ss ( v ), then a 0 = a 2 and a 5 = a 7 implies (according Theor em 2.2 ) that x 1 , x 3 / ∈ E ss ( f ) and f / ∈ G 3 2 . A.a.2. If x 2 / ∈ E ss ( v ), then a 0 = a 5 and a 2 = a 7 . Note that if a 0 = a 7 , then f has to b e a constan t. Hence a 7 = ¬ ( a 0 ). Then we obtain f = a 0 .  x 0 1 x 0 2 x 0 3 ⊕ x 0 1 x 0 2 x 1 3 ⊕ x 0 1 x 1 2 x 1 3 ⊕ x 1 1 x 0 2 x 1 3  ⊕ ⊕ ¬ ( a 0 ) .  x 0 1 x 1 2 x 0 3 ⊕ x 1 1 x 0 2 x 0 3 ⊕ x 1 1 x 1 2 x 0 3 ⊕ x 1 1 x 1 2 x 1 3  = = a 0  x 1 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) ⊕ x 0 1 x 0 2  ⊕ ¬ ( a 0 )  x 0 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) ⊕ x 1 1 x 1 2  ∈ G 3 2 . Clearly , f is presented as in (3). A.b. x 2 / ∈ E ss ( u ). Hence a 0 = a 1 and a 6 = a 7 . A.b.1. If we supp o se that x 2 / ∈ E ss ( v ), then a 0 = a 5 and a 2 = a 7 implies (according Theor em 2.2 ) that x 2 , x 3 / ∈ E ss ( f ) and f / ∈ G 3 2 . 6 SL. SHTRAKO V A.b.2. If x 1 / ∈ E ss ( v ), then a 0 = a 2 and a 5 = a 7 . Again, if a 0 = a 7 , then f has to b e a co nstant . Hence a 7 = ¬ ( a 0 ). Then we obtain f = a 0 .  x 0 1 x 0 2 x 0 3 ⊕ x 0 1 x 0 2 x 1 3 ⊕ x 0 1 x 1 2 x 0 3 ⊕ x 1 1 x 0 2 x 0 3  ⊕ ⊕ ¬ ( a 0 ) .  x 0 1 x 1 2 x 1 3 ⊕ x 1 1 x 0 2 x 1 3 ⊕ x 1 1 x 1 2 x 0 3 ⊕ x 1 1 x 1 2 x 1 3  = = a 0  x 0 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) ⊕ x 0 1 x 0 2  ⊕ ¬ ( a 0 )  x 1 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) ⊕ x 1 1 x 1 2  ∈ G 3 2 . Clearly , f is presented as in (3). B. Let us suppo s e that x 1 / ∈ E ss ( g 2 ← 1 ) and x 1 / ∈ E ss ( h 2 ← 1 ). Then we hav e g ∈ G 2 2 and h ∈ G 2 2 . F ro m Theorem 3.1 it follo ws that g ( x 1 , x 2 ) = a 0 . ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ a 1 .x 0 1 x 1 2 ⊕ a 2 .x 1 1 x 0 2 , and h ( x 1 , x 2 ) = a 4 . ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ a 5 .x 0 1 x 1 2 ⊕ a 6 .x 1 1 x 0 2 . Then we obtain u = f 3 ← 1 = a 0 .x 0 1 x 0 2 ⊕ a 1 .x 0 1 x 1 2 ⊕ a 6 .x 1 1 x 0 2 ⊕ a 4 .x 1 1 x 1 2 , and v = f 3 ← 2 = a 0 .x 0 1 x 0 2 ⊕ a 2 .x 1 1 x 0 2 ⊕ a 5 .x 0 1 x 1 2 ⊕ a 4 .x 1 1 x 1 2 . B.a. x 1 / ∈ E ss ( u ). Hence a 0 = a 6 and a 1 = a 4 . B.a.1. If x 1 / ∈ E ss ( v ), then a 0 = a 2 and a 4 = a 5 . Note that if a 0 = a 4 , then f has to b e a constan t. Hence a 4 = ¬ ( a 0 ). Then we obtain f = a 0 .  x 0 1 x 0 2 x 0 3 ⊕ x 1 1 x 0 2 x 0 3 ⊕ x 1 1 x 0 2 x 1 3 ⊕ x 1 1 x 1 2 x 0 3  ⊕ ⊕ ¬ ( a 0 ) .  x 0 1 x 0 2 x 1 3 ⊕ x 0 1 x 1 2 x 0 3 ⊕ x 0 1 x 1 2 x 1 3 ⊕ x 1 1 x 1 2 x 1 3  = = a 0  x 1 1 ( x 0 2 x 1 3 ⊕ x 1 2 x 0 3 ) ⊕ x 0 2 x 0 3  ⊕ ¬ ( a 0 )  x 0 1 ( x 0 2 x 1 3 ⊕ x 1 2 x 0 3 ) ⊕ x 1 2 x 1 3  ∈ G 3 2 . Clearly , f is presented as in (3). B.a.2. If x 2 / ∈ E ss ( v ), then a 0 = a 5 and a 2 = a 4 . Again, if a 0 = a 4 , then f has to b e a co nstant . Hence a 4 = ¬ ( a 0 ). Then we obtain f = a 0 .  x 0 1 x 0 2 x 0 3 ⊕ x 0 1 x 1 2 x 1 3 ⊕ x 1 1 x 0 2 x 1 3 ⊕ x 1 1 x 1 2 x 0 3  ⊕ ⊕ ¬ ( a 0 ) .  x 0 1 x 0 2 x 1 3 ⊕ x 0 1 x 1 2 x 0 3 ⊕ x 1 1 x 0 2 x 0 3 ⊕ x 1 1 x 1 2 x 1 3  = = a 0  x 0 3 ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ x 1 3 ( x 1 1 x 0 2 ⊕ x 0 1 x 1 2 )  ⊕ ⊕ ¬ ( a 0 )  x 1 3 ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ x 0 3 ( x 1 1 x 0 2 ⊕ x 0 1 x 1 2 )  ∈ G 3 2 . Clearly , f is presented as in (4). B.b. x 2 / ∈ E ss ( u ). Hence a 0 = a 1 and a 6 = a 4 . B.b.1. If w e supp ose that x 1 / ∈ E ss ( v ), then a 0 = a 2 and a 4 = a 5 implies (according Theor em 2.2 ) that x 1 , x 2 / ∈ E ss ( f ) and f / ∈ G 3 2 . B.b.2. If x 2 / ∈ E ss ( v ), then a 0 = a 5 and a 2 = a 4 . Again, if a 0 = a 4 , then f has to b e a co nstant . Hence a 4 = ¬ ( a 0 ). Then we obtain f = a 0 .  x 0 1 x 0 2 x 0 3 ⊕ x 0 1 x 1 2 x 0 3 ⊕ x 0 1 x 1 2 x 1 3 ⊕ x 1 1 x 1 2 x 0 3  ⊕ ⊕ ¬ ( a 0 ) .  x 0 1 x 0 2 x 1 3 ⊕ x 1 1 x 0 2 x 0 3 ⊕ x 1 1 x 0 2 x 1 3 ⊕ x 1 1 x 1 2 x 1 3  = = a 0  x 1 2 ( x 0 1 x 1 3 ⊕ x 1 1 x 0 3 ) ⊕ x 0 1 x 0 3  ⊕ ¬ ( a 0 )  x 0 2 ( x 0 1 x 1 3 ⊕ x 1 1 x 0 3 ) ⊕ x 1 1 x 1 3  ∈ G 3 2 . Clearly , f is presented as in (3). ESSENTIAL ARITY GAP OF BOOLEAN FUNCTIONS 7 Corollary 3.3 . L et f ∈ P 3 2 . Then ess ( f i ← j ) ≤ 1 for al l i, j, 1 ≤ j < i ≤ 3 if and only if f = x α 3 ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ a 1 .x 0 1 x 1 2 x 0 3 ⊕ a 2 .x 1 1 x 0 2 x 0 3 ⊕ ⊕ ¬ ( a 2 ) .x 0 1 x 1 2 x 1 3 ⊕ ¬ ( a 1 ) .x 1 1 x 0 2 x 1 3 , wher e α, a 1 , a 2 ∈ { 0 , 1 } . Pro of. T his Coro llary summarizes all case s considered in Theorem 3.2. F o r instance if α = 1, a 1 = 0 and a 2 = 0 w e obtain f = x 1 3 ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ x 0 1 x 1 2 x 1 3 ⊕ x 1 1 x 0 2 x 1 3 = x 1 3 . This is the case B.b.1 . Corollary 3.4. | G 3 2 | = 10 . Pro of. As we have noted the functions f in the fo rm (4) are symmetric with resp ect to their v ariables . Hence there ar e exactly t wo such functions, obtained for α = 1 and α = 0. These functions are realize d in the case B.a.2. Let us consider the functions f in the form (3) with α = β . Then we hav e f = x 0 1 x 1 2 x α 3 ⊕ x 1 1 x 0 2 x α 3 ⊕ x α 1 x α 2 x 0 3 ⊕ x α 1 x α 2 x 1 3 . It is easy to chec k that in the b oth cases α = 1 and α = 0 the function f is symmetric. Hence there exist exactly t wo functions fro m P 2 2 in the form (3) with α = β . These tw o functions a re realize d in the case A.b.2 . Finally , let us consider the functions in the form (5) f = x α 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) ⊕ x ¬ α 1 x ¬ α 2 . Since f ( α, β , ¬ ( α )) = 0 and f ( ¬ ( α ) , β , α ) = 1 for a ll β ∈ { 0 , 1 } it follows that f is not symmetric with r esp e ct to x 1 and x 3 . F urthermore, it is clear that f is symmetric with res pect to x 1 and x 2 . Hence there a re exactly six functions from P 3 2 in the form (5). When α = 1 we obtain three function by three permutations of the v aria bles and the same num b e r functions for α = 0 . These functions are realized in the cases: A.a.2. , B.a.1. and B.b. 2. Lemma 3.1 . L et f = x 0 4 .g ( x 1 , x 2 , x 3 ) ⊕ x 1 4 .h ( x 1 , x 2 , x 3 ) ∈ P 4 2 . If f ∈ G 4 2 , then ess ( g i ← j ) < 2 and ess ( h i ← j ) < 2 for al l i, j, 1 ≤ i < j ≤ 3 . Pro of. Let us suppo se that the lema is false. Without los s of g e nerality let us assume ess ( g 2 ← 1 ) ≥ 2. If f ∈ G 4 2 , then x 4 / ∈ E ss ( f 2 ← 1 ) b ecause f 2 ← 1 = x 0 4 .g 2 ← 1 ⊕ x 1 4 .h 2 ← 1 and f 2 ← 1 ( x 4 = 0) = g 2 ← 1 . F rom Theo rem 2.2 it follows that g 2 ← 1 = h 2 ← 1 . Let us s et g := 7 M m =0 a (0) m .x α 1 1 x α 2 2 x α 3 3 and h := 7 M m =0 a (1) m .x α 1 1 x α 2 2 x α 3 3 , where m = α 1 α 2 α 3 , and t := a (0) 0 .x 0 1 x 0 2 x 0 3 ⊕ a (0) 1 .x 0 1 x 0 2 x 1 3 ⊕ a (0) 6 .x 1 1 x 1 2 x 0 3 ⊕ a (0) 7 .x 1 1 x 1 2 x 1 3 . Then from g 2 ← 1 = h 2 ← 1 , we obtain g := t ( x 1 , x 2 , x 3 ) ⊕  M α 1 6 = α 2 a (0) m .x α 1 1 x α 2 2 x α 3 3  and 8 SL. SHTRAKO V h := t ( x 1 , x 2 , x 3 ) ⊕  M α 1 6 = α 2 a (1) m .x α 1 1 x α 2 2 x α 3 3  . Note that f 2 ← 1 = t 2 ← 1 = g 2 ← 1 = h 2 ← 1 . If ess ( g 2 ← 1 ) > 2, then from f 2 ← 1 ( x 4 = 0 ) = g 2 ← 1 it follows tha t f / ∈ G 4 2 . Hence ess ( g 2 ← 1 ) = 2. Thus w e have { x 1 , x 3 } = E ss ( g 2 ← 1 ). This implies (6) ( a (0) 0 , a (0) 6 ) 6 = ( a (0) 1 , a (0) 7 ) and ( a (0) 0 , a (0) 1 ) 6 = ( a (0) 6 , a (0) 7 ) . F rom x 4 ∈ E ss ( f ) it follows that there are three num b e rs α 1 , α 2 , α 3 ∈ { 0 , 1 } suc h that a (0) m 6 = a (1) m where m = α 1 α 2 α 3 . Then α 1 6 = α 2 . H ence we have α 1 = α 3 or α 2 = α 3 . Let us assume α 1 = α 3 . Then the iden tification minor u = f 3 ← 1 can b e written as follows u = a (0) 0 .x 0 1 x 0 2 ⊕ a (0) 7 .x 1 1 x 1 2 ⊕ x 0 4 ( a (0) 2 .x 0 1 x 1 2 ⊕ a (0) 5 .x 1 1 x 0 2 ) ⊕ x 1 4 ( a (1) 2 .x 0 1 x 1 2 ⊕ a (1) 5 .x 1 1 x 0 2 ) . Without loss o f gener alit y let us assume that a (0) 2 6 = a (1) 2 , i.e. m = 010 = 2 . (Alternative opp ortunity is m = 5.) Then we hav e a (0) 2 6 = 0 or a (1) 2 6 = 0 . Again, without loss of generality let us as sume a (0) 2 = 1 and a (1) 2 = 0 . Then u ( x 1 = α 1 , x 2 = α 2 ) = a (0) 2 .x 0 4 ⊕ a (1) 2 .x 1 4 . Hence x 4 ∈ E ss ( u ). On the other hand we have u 1 = u ( x 4 = 0) = a (0) 0 .x 0 1 x 0 2 ⊕ a (0) 7 .x 1 1 x 1 2 ⊕ x 0 1 x 1 2 ⊕ a (0) 5 .x 1 1 x 0 2 and u 2 = u ( x 4 = 1 ) = a (0) 0 .x 0 1 x 0 2 ⊕ a (0) 7 .x 1 1 x 1 2 ⊕ a (1) 5 .x 1 1 x 0 2 . Thu s w e hav e: If a (0) 0 = a (0) 7 = 0 or a (0) 0 = a (0) 7 = 1, then E ss ( u 1 ) = { x 1 , x 2 } . Let a (0) 0 6 = a (0) 7 . Then accor ding to (6) w e can assume without lo s s o f genera lit y that a (0) 0 = 1 and a (0) 7 = 0 . No w, we have: If a (0) 5 = 1 or a (1) 5 = 0 , then E ss ( u 1 ) = { x 1 , x 2 } or E ss ( u 2 ) = { x 1 , x 2 } . Finally , if a (0) 0 = 1, a (0) 7 = 0, a (0) 5 = 0 and a (1) 5 = 1 w e have u 1 ( x 1 , x 2 ) = x 0 1 and u 2 ( x 1 , x 2 ) = x 0 2 . So, we hav e sho wn that E ss ( u ) = { x 1 , x 2 , x 4 } . Hence f / ∈ G 4 2 , which is a contradiction. By symmetry , we obta in the same contradiction when α 2 = α 3 and w e ha ve to use the identification minor v = f 3 ← 2 instead o f u = f 3 ← 1 . Lemma 3.2 . L et f = x 0 4 .g ( x 1 , x 2 , x 3 ) ⊕ x 1 4 .h ( x 1 , x 2 , x 3 ) ∈ P 4 2 . If f ∈ G 4 2 , then ess ( g ) = ess ( h ) = 3 . Pro of. Let us suppo se that x 3 / ∈ E ss ( g ) and f ∈ G 4 2 . Let g and h a re repres e nted a s follows g := 7 M m =0 a m .x α 1 1 x α 2 2 x α 3 3 and h := 7 M m =0 b m .x α 1 1 x α 2 2 x α 3 3 , where m = α 1 α 2 α 3 = α 1 . 2 2 + α 2 . 2 + α 3 . Since x 3 / ∈ E ss ( g ), we obtain (7) ( a 0 , a 2 , a 4 , a 6 ) = ( a 1 , a 3 , a 5 , a 7 ) . On the other hand x 3 / ∈ E ss ( g ) implies x 3 ∈ E ss ( h ). Hence, w e hav e (8) ( b 0 , b 2 , b 4 , b 6 ) 6 = ( b 1 , b 3 , b 5 , b 7 ) . ESSENTIAL ARITY GAP OF BOOLEAN FUNCTIONS 9 Without lo ss of genera lit y let us a ssume that b 0 = 1 and b 1 = 0 . Consequently , u = f 2 ← 1 = x 0 4 ( a 0 x 0 1 ⊕ a 6 x 1 1 ) ⊕ x 1 4 ( x 0 1 x 0 3 ⊕ b 6 .x 1 1 x 0 3 ⊕ b 7 x 1 1 x 1 3 ) . F rom u ( x 1 = 0 , x 4 = 1 ) = x 0 3 it follows x 3 ∈ E ss ( u ). If a 0 = 1 , then u ( x 1 = 0 , x 3 = 1) = x 0 4 and if a 0 = 0, then u ( x 1 = 0 , x 3 = 0) = x 1 4 . Hence x 4 ∈ E ss ( u ). The pr o of will be done if w e s how that x 1 ∈ E ss ( u ). Supp ose the o pp osite i.e. x 1 / ∈ E ss ( u ). F rom Theo rem 2.2 it follows that a 0 = a 6 , b 6 = 1 and b 7 = 0. Then we have v = f 3 ← 1 = x 0 4 [ a 0 . ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ a 2 .x 0 1 x 1 2 ⊕ a 4 .x 1 1 x 0 2 ] ⊕ ⊕ x 1 4 [ x 0 1 x 0 2 ⊕ b 2 .x 0 1 x 1 2 ⊕ b 5 .x 1 1 x 0 2 ] . If a 0 = 1, then v ( x 1 = 1 , x 2 = 1) = x 0 4 and if a 0 = 0, then v ( x 1 = 0 , x 2 = 0) = x 1 4 . Hence x 4 ∈ E ss ( v ). On the other side it is clear that v ( x 4 = 1 ) := x 0 1 x 0 2 ⊕ b 2 .x 0 1 x 1 2 ⊕ b 5 .x 1 1 x 0 2 is no t a constant. Assume that x 2 ∈ E ss ( v ). Supp ose that x 1 / ∈ E ss ( v ). Hence a 0 = a 2 = a 4 , b 5 = 1 and b 2 = 0. Thus we obtain w = f 3 ← 2 = a 0 .x 0 4 ⊕ x 1 4 ( x 0 1 x 0 2 ⊕ b 3 .x 0 1 x 1 2 ⊕ b 4 x 1 1 x 0 2 ) . Clearly x 4 ∈ E s s ( w ). O n the o ther hand it is clear that w ( x 4 = 1) := x 0 1 x 0 2 ⊕ b 3 .x 0 1 x 1 2 ⊕ b 4 .x 1 1 x 0 2 is not a constan t. Ass ume that x 2 ∈ E ss ( w ). Supp ose that x 1 / ∈ E ss ( w ). Hence b 3 = 0 and b 4 = 1. Thus finally , w e obtain f = a 0 .x 0 4 ⊕ x 1 4 ( x 0 1 x 0 2 x 0 3 ⊕ x 1 1 x 0 2 x 0 3 ⊕ x 1 1 x 0 2 x 1 3 ⊕ x 1 1 x 1 2 x 0 3 ) . The contradiction is f / ∈ G 4 2 bec ause f 4 ← 2 = a 0 .x 0 2 ⊕ x 1 1 x 1 2 x 0 3 . By analog y we c o nclude that f / ∈ G 4 2 for the all o ther ca ses generated by (7) and (8), which is a co n tradiction. Theorem 3.3. L et f ∈ P 4 2 . Then f ∈ G 4 2 if and only if f = x 0 4 .g ( x 1 , x 2 , x 3 ) ⊕ x 1 4 .h ( x 1 , x 2 , x 3 ) , with (9) g = x α 3 ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ x ¬ ( α ) 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) , and (10) h = x ¬ ( α ) 3 ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ x α 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) , for some α , α ∈ { 0 , 1 } . Pro of. ” ⇐ ” The pro of in this direction is done in Pr opo sition 3.2. ” ⇒ ” Suppo se that so me of the eq uations (9) or (1 0) ar e not satisfied. F rom Lemma 3.1 a nd Lemma 3.2 there are t wo poss ible cases: A. g = x α 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) ⊕ x β 1 x β 2 , and h = x γ 3 ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ x ¬ ( γ ) 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) . Then we hav e the follo wing identification minor of f : u = f 4 ← 1 = x 0 1 x 1 2 x α 3 ⊕ ¬ ( β ) .x 0 1 x 0 2 ⊕ x 1 1 x 1 2 x γ 3 ⊕ x 1 1 x 0 2 x ¬ ( α ) 3 . Since u ( x 1 = 0) = x 1 2 x α 3 ⊕ ¬ ( β ) .x 0 2 it follows that { x 2 , x 3 } ⊆ E ss ( u ). W e will show that x 1 ∈ E ss ( u ), also. Let β = 0. If γ = α , then we ha ve u ( x 2 = 0 , x 3 = γ ) = x 0 1 , and if γ 6 = α , then we hav e u ( x 2 = 1 , x 3 = γ ) = x 1 1 . Let β = 1. If γ = α , then u ( x 2 = 0 , x 3 = ¬ ( γ )) = x 1 1 , and if γ 6 = α , then w e ha ve u ( x 2 = 1 , x 3 = γ ) = x 1 1 . Hence x 1 ∈ E ss ( u ) and f / ∈ G 4 2 in the case A. 10 SL. SHTRAKO V B. g = x α 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) ⊕ x β 1 x β 2 , and h = x γ 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) ⊕ x δ 1 x δ 2 . Since x 4 ∈ E ss ( f ) it follows that g 6 = h . Let us consider the identification minor u o f f , a ls o: u = f 4 ← 1 = x 0 1 x 1 2 x α 3 ⊕ ¬ ( β ) .x 0 1 x 0 2 ⊕ x 1 1 x 0 2 x γ 3 ⊕ δ.x 1 1 x 1 2 . Since u ( x 1 = 0) = x 1 2 x α 3 ⊕ ¬ ( β ) .x 0 2 it follows that { x 2 , x 3 } ⊆ E ss ( u ). W e will prov e that x 1 ∈ E ss ( u ), also. Let β = δ = 0. Then u ( x 2 = 1 , x 3 = α ) = x 0 1 ; Let β = δ = 1. Then u ( x 2 = 0 , x 3 = γ ) = x 1 1 ; Let β = 1 and δ = 0. Then u ( x 2 = 0 , x 3 = γ ) = x 1 1 ; Let β = 0 and δ = 1. Then u ( x 2 = 1 , x 3 = ¬ ( α )) = x 1 1 . Hence x 1 ∈ E ss ( u ) and f / ∈ G 4 2 in the case B. , a lso. This is a con tradiction. Remark 1. Note that g and h have to b e two sp e cial funct ions fr om G 3 2 , r epr esente d by t he e quation (4) of The or em 3.2. Su ch functions c an b e obtaine d in the c ases of the same t he or em B.a.2 and B.b. 2 , only. Corollary 3.5. L et f ∈ P 4 2 . Then f ∈ G 4 2 if and only if f = x 0 4 .g ( x 1 , x 2 , x 3 ) ⊕ x 1 4 .h ( x 1 , x 2 , x 3 ) , with g = x α 3 ( x 0 1 x 0 2 ⊕ x 1 1 x 1 2 ) ⊕ x ¬ ( α ) 3 ( x 0 1 x 1 2 ⊕ x 1 1 x 0 2 ) , and h = ¬ ( g ( x 1 , x 2 , x 3 )) . Corollary 3.6. L et f ∈ P 4 2 . Then f ∈ G 4 2 if and only if f = a 0 .  M α 1 α 2 α 3 α 4 ∈ Od 4 2 x α 1 1 x α 2 2 x α 3 3 x α 4 4  ⊕ ¬ ( a 0 ) .  M α 1 α 2 α 3 α 4 ∈ E v 4 2 x α 1 1 x α 2 2 x α 3 3 x α 4 4  . Corollary 3.7. If f ∈ G 4 2 then x j / ∈ E ss ( f i ← j ) for all i, j ∈ { 1 , 2 , 3 , 4 } i 6 = j . Pro of. The three coro lla ries, ab ov e can b e prov ed by immediate chec king of the bo th functions fr om G 4 2 , obtained in Theorem 3.3. Theorem 3.4. A Bo ole an function f ∈ P n 2 , dep ending on n essential variables with n ≥ 4 , has essential arity gap 2 if and only if f = M α 1 ...α n ∈ Od n 2 x α 1 1 . . . x α n n or f = M α 1 ...α n ∈ E v n 2 x α 1 1 . . . x α n n . Pro of. ” ⇐ ” In this direction the pro o f is done by Prop osition 3.2. ” ⇒ ” W e will pro ceed by induction on n . If n = 4 the theor em is true be c a use of Theorem 3 .3. Supp ose that if 4 ≤ n ≤ l and f ∈ G n 2 , then f = M α 1 ...α n ∈ Od n 2 x α 1 1 . . . x α n n or f = M α 1 ...α n ∈ E v n 2 x α 1 1 . . . x α n n . Let f ∈ G l +1 2 . Hence f can b e presen ted as follows f = x 0 l +1 .g ( x 1 , . . . , x l ) ⊕ x 1 l +1 .h ( x 1 , . . . , x l ) . ESSENTIAL ARITY GAP OF BOOLEAN FUNCTIONS 11 In the same wa y as in Lemma 3 .1 and Lemma 3.2 it ca n b e pr ov ed tha t g , h ∈ G l 2 . By the inductive supp osition g and h are functions o f the forms M γ 1 ...γ l ∈ Od l 2 x γ 1 1 . . . x γ l l or M γ 1 ...γ l ∈ E v l 2 x γ 1 1 . . . x γ l l , with g 6 = h . Note that g and h ar e not co nstants b ecause ess ( f ) = n ≥ 4. Hence E ss ( g i ← j ) = E ss ( h i ← j ) for i , j ∈ { 1 , . . . , l } and i 6 = j . Assume that g = M γ 1 ...γ l ∈ Od l 2 x γ 1 1 . . . x γ l l and h = M δ 1 ...δ l ∈ E v l 2 x δ 1 1 . . . x δ l l . Consequently f = x 0 l +1 . ( M γ 1 ...γ l ∈ Od l 2 x γ 1 1 . . . x γ l l ) ⊕ x 1 l +1 . ( M δ 1 ...δ l ∈ E v l 2 x δ 1 1 . . . x δ l l ) = = M α 1 ...α l +1 ∈ Od l +1 2 x α 1 1 . . . x α l +1 l +1 . The case g = h is impo ssible b ecause e ss ( f ) = l + 1, but the replacemen t of g and h will pro duce the function f = M α 1 ...α l ∈ E v l 2 x α 1 1 . . . x α l l , which do es not depend o n x l +1 . Corollary 3.8 . A Bo ole an function f ∈ P n 2 , which essential ly dep ends on n vari- ables with n > 4 , has essential arity gap 2 if and only if f = x 0 n .g ( x 1 , . . . , x i , . . . , x n − 1 ) ⊕ x 1 n .g ( x 1 , . . . , x i − 1 , ¬ ( x i ) , x i +1 , . . . , x n − 1 ) , wher e g ∈ G n − 1 2 and i ∈ { 1 , . . . , n − 1 } . Pro of. If g = M γ 1 ...γ n − 1 ∈ Od n − 1 2 x γ 1 1 . . . x γ n − 1 n − 1 and h = M γ 1 ...γ n − 1 ∈ Od n − 1 2 x γ 1 1 . . . x γ n − 1 n − 1 , then ¬ ( g ) = h and ¬ ( h ) = g for all l ≥ 4. On the other hand, for each i ∈ { 1 , . . . , n − 1 } , w e have ¬ ( g ) = M γ 1 ...γ n − 1 ∈ Od n − 1 2 x γ 1 1 . . . x γ i − 1 i − 1 ¬ ( x γ i i ) x γ i +1 i +1 . . . x γ n − 1 n − 1 . Corollary 3.9. | G n 2 | = 2 for e ach n, n ≥ 4 . One o f the mo st impo rtant problems concer ning the ess e n tial ar it y ga p is to calculate the num ber of all functions from P n 2 , which depe nd es sent ially o n a t mo st n v ariables and which have the maximum gap i.e. with gap equal to 2. The next theorem gives answer of that problem. It summarize the re sults obtained ab ov e in the pap er. Let us deno te by H n the set o f all functions in P n 2 , which hav e gap equal to 2 i.e. H n := n [ m =2 G m 2 and h n := | H n | . 12 SL. SHTRAKO V Theorem 3.5. The fol lowing c ombinatorial e quations ar e held: ( i ) h 2 = 6 ; ( ii ) h 3 = 28 ; ( iii ) h n = 3 .  n 2  + 5 .  n 3  + 2 n +1 − 2 n − 2 , when n ≥ 4 ; Pro of. ( i ) follows from Cor o llary 3.1 of Theore m 3 .1; ( ii ) Let X 3 = { x 1 , x 2 , x 3 } . There are 6 .  3 2  Bo olean functions with essential arity ga p equal to 2, which dep end es s en tially on 2 v ariables fr om X 3 , a ccording to Corollar y 3.1 of Theo rem 3.1. F rom Coro llary 3.4 o f Theo rem 3.2 it follows that there are 10 Bo olean functions with essential arity ga p equal to 2, which dep end ess en tially on all 3 v ariables from X 3 . Hence h 3 = 6 . 3 + 10 = 28. ( iii ) Let X n = { x 1 , . . . , x n } , n ≥ 4. There ar e 6 .  n 2  Bo olean functions with essential arity gap equal to 2, which dep end essentially on 2 v aria ble s from X n , according to Co rollary 3.1 of Theo rem 3.1. There ar e 10 .  n 3  Bo olean functions with es sent ial ar it y g a p equal to 2, whic h depe nd essentially o n 3 v ariables from X n , according to Corolla ry 3.4 of Theor em 3.2. Finally , for eac h m, 3 < m ≤ n there ar e 2 .  n m  Bo olean functions with essential arity gap equal to 2 , whic h depend essentially on m v ariables from X n , according to Corolla ry 3 .9 of Theorem 3.4. Hence we hav e h n = 6 .  n 2  + 10 .  n 3  + 2 .   n 4  +  n 5  + . . . +  n n   = = 3 .  n 2  + 5 .  n 3  + 2 n +1 − 2 n − 2 . References [1] Y u. Breitbart, On the essential variables of functions in the algebr a of lo gic, Dokl. Acad. Sci. USSR, 172, v. 1, 1967, 9-10 (in Russ i an). [2] K. Chimev, Sep ar able Sets of Ar guments of F unctions , M T A SzT AKI T anulman yok, 180/1986 , 173 pp. [3] M. Couceiro, E. Leht onen, On the eff e ct of variable identific ation on the essential arity of functions on finite sets, Int. Journal of F oundations of Computer Science, vol. 18, Issue 5, (October 2007), 975-986. [4] A. Salomaa, On Essential V ariables of F unctions, Esp e cial ly in the Algebr a of Lo gic , Anna les Academia Scientiarum F ennicae, Ser. A, 333, 1963, 1- 11. [5] O. Lupano v, On a class schemas of functional elements, Pr oblemi Cyb ernetiki, 9, 1963, 333-335 (in Russian). [6] Sl. Shtrak ov, T r e e Automata and Essential Input V ariables , Contributions to General Algebra 13, V erlag Johannes Heyn, Klagenfurt, 2001, pp.309-320. [7] Sl. Shtrak ov, K. Denec ke, Essenti al V ariables and Sep ar able Set s in Universal A lgebr a , T a ylor & F rancis, Multiple-V alued Logic, An In ternational Journal, vol. 8, No.2,2002, 165-182. E-mail addr e ss : shtrakov@aix.swu .bg Dep art ment of Computer Science, South-West Univer sity, 2700 Blagoevgrad, Bul- garia URL : http://h ome.swu.bg/s htrakov