On the effect of variable identification on the essential arity of functions on finite sets

Science 18(5) (2007) 975–986. DOI: 10.1142/S012905410700508X. ccopyright World Scientific Publishing Company,http://www.worldscientific.com/worldscinet/ijfcs.

ON THE EFFECT OF VARIABLE IDENTIFICATION ON THE ESSENTIAL ARITY OF FUNCTIONS ON FINITE SETS

MIGUEL COUCEIRO AND ERKKO LEHTONEN

Abstract. We show that every function of several variables on a finite set of kelements withn > kessential variables has a variable identification minor with at leastn−kessential variables. This is a generalization of a theorem of Salomaa on the essential variables of Boolean functions. We also strengthen Salomaa’s theorem by characterizing all the Boolean functions f having a variable identification minor that has just one essential variable less thanf.

1. Introduction

Theory of essential variables of functions has been developed by several authors [2, 5, 6, 7, 14, 16]. In this paper, we discuss the problem how the number of essential variables is affected by identification of variables (diagonalization). Salomaa [14]

proved the following two theorems: one deals with operations on arbitrary finite sets, while the other deals specifically with Boolean functions. We denote the number of essential variables off by essf.

Theorem 1.1. Let Abe a finite set withkelements. For everyn≤k, there exists ann-ary operationf onAsuch thatessf =nand every identification of variables produces a constant function.

Thus, in general, essential variables can be preserved when variables are identified only in the case thatn > k.

Theorem 1.2. For every Boolean functionf withessf ≥2, there is a functiong obtained from f by identification of variables such that essg≥essf−2.

Identification of variables together with permutation of variables and cylindrifi- cation induces a quasi-order on operations whose relevance has been made apparent by several authors [3, 8, 9, 10, 12, 15, 18]. In the case of Boolean functions, this quasi-order was studied in [4] where Theorem 1.2 was fundamental in deriving cer- tain bounds on the essential arity of functions.

In this paper, we will generalize Theorem 1.2 to operations on arbitrary finite sets in Theorem 3.1. We will also strengthen Theorem 1.2 on Boolean functions in Theorem 4.1 by determining the Boolean functions f for which there exists a functiongobtained fromf by identification of variables such that essg= essf−1.

Key words and phrases. Functions on finite sets; Boolean functions; essential variables; variable identification; arity gap; minors of functions.

1

2. Variable identification minors

LetAandBbe arbitrary nonempty sets. AB-valued function of several variables on A is a mappingf :Aⁿ →B for some positive integern, called the arity of f. A-valued functions onAare calledoperations onA. Operations on{0,1}are called Boolean functions.

We say that thei-th variable is essential in f, orf depends onx_i, if there are elementsa₁, . . . , a_n, b∈Asuch that

(1) f(a1, . . . , ai, . . . , an)6=f(a1, . . . , ai−1, b, ai+1, . . . , an).

The number of essential variables in f is called theessential arity of f, and it is denoted by essf. Thus the only functions with essential arity zero are the constant functions.

For ann-ary functionf, we say that anm-ary functiongis obtained fromf by simple variable substitution if there is a mappingσ:{1, . . . , n} → {1, . . . , m}such that

(2) g(x₁, . . . , x_m) =f(x_σ(1), . . . , x_σ(n)).

In the particular case that n = m and σ is a permutation of {1, . . . , n}, we say thatgis obtained fromf bypermutation of variables. For indicesi, j∈ {1, . . . , n}, i 6=j, ifxi and xj are essential in f, then the function f_i←j obtained fromf by the simple variable substitution

(3) f_i←j(x₁, . . . , x_n) =f(x₁, . . . , x_i−1, x_j, x_i+1, . . . , x_n)

is called a variable identification minor of f, obtained by identifyingxi with xj. Note that essfi←j < essf, because xi is not essential in fi←j even though it is essential inf.

We define a quasiorder on the set of allB-valued functions of several variables on A as follows: f ≤ g if and only if f is obtained from g by simple variable substitution. If f ≤g andg≤f, we denotef ≡g. Iff ≤g butg6≤f, we denote f < g. It can be easily observed that iff ≤g then essf ≤essg, with equality if and only iff ≡g.

For a B-valued function f of several variables on A, we denote the maximum essential arity of a variable identification minor off by

(4) ess^<f = max

g<f essg,

and we define thearity gap off by gapf = essf −ess^<f.

3. Generalization of Theorem 1.2

Theorem 3.1. Let A be a finite set of k≥2 elements, and let B be a set with at least two elements. Every B-valued function of several variables onA with n > k essential variables has a variable identification minor with at least n−k essential variables.

In the proof of Theorem 3.1, we will make use of the following theorem due to Salomaa (Theorem 1 in [14]), which is a strengthening of Yablonski’s [16] “funda- mental lemma”.

Theorem 3.2. Let the function f :M1× · · · ×Mn→N depend essentially on all of its n variables, n ≥2. Then there is an index j and an element c ∈Mj such that the function

(5) f(x₁, . . . , x_j−1, c, x_j+1, . . . , x_n) depends essentially on all of its n−1 variables.

We also need the following auxiliary lemma.

Lemma 3.3. Letf be ann-ary function withessf =n > k. Then there are indices 1≤i < j ≤k+ 1such that at least one of the variablesx1, . . . , xk+1 is essential inf_i←j.

Proof. Sincex1 is essential inf, there are elementsa1, . . . , an, b∈Asuch that (6) f(a₁, a₂, . . . , a_n)6=f(b, a₂, . . . , a_n).

Thus there are indices 1≤i < j≤k+ 1 such thatai=aj. Ifi6= 1, then it is clear thatx1 is essential inf_i←j. If there are no such iandj withi6= 1, then i= 1< j and we have thatb=alfor some 1< l≤k+1,l6=j. Form= 1, . . . , n, letcm=am

if m /∈ {1, j, l} and letcm =a1 if m ∈ {1, j, l}. Then f(c1, c2, . . . , cn) is distinct from at least one of f(a1, a2, . . . , an) and f(b, a2, . . . , an). If f(c1, c2, . . . , cn) 6=

f(a1, a2, . . . , an), thenxl is essential inf_1←j. Iff(c1, c2, . . . , cn)6=f(b, a2, . . . , an),

thenxlis essential inf1←l.

Proof of Theorem 3.1. By Theorem 3.2, there existk+1 constantsc₁, . . . , c_k+1∈A such that, after a suitable permutation of variables, the function

(7) f(c1, . . . , ck+1, xk+2, . . . , xn)

depends on all of itsn−k−1 variables. There are indices 1≤i < j≤k+ 1 such that ci =cj, and by Lemma 3.3 there are indices 1 ≤ l < m ≤k+ 1 such that at least one of the variables x1, . . . , xk+1 is essential in fl←m. With a suitable permutation of variables, we may assume thati= 1, j= 2, 1≤l≤3,m=l+ 1.

If one of the variablesx₁, . . . , x_k+1 is essential in f_1←2, then we are done. Oth- erwise we have that for alla_k+2, . . . , a_n∈A,

(8) f(c1, c1, c3, c4, . . . , ck+1, ak+2, . . . , an) =

f(c3, c3, c3, c4, . . . , ck+1, ak+2, . . . , an).

Thus the variables xk+2, . . . , xn are essential in f_2←3. If one of the variables x1, . . . , xk+1 is essential in f2←3, then we are done. Otherwise we have that for allak+2, . . . , an ∈A,

(9) f(c3, c3, c3, c4, . . . , ck+1, ak+2, . . . , an) =

f(c₃, c₄, c₄, c₄, . . . , c_k+1, a_k+2, . . . , a_n), and so the variables xk+2, . . . , xn are essential in f_3←4 and also at least one of x1, . . . , xk+1is essential inf_3←4. This completes the proof of Theorem 3.1.

We would like to remark that our proof is considerably simpler than Salomaa’s original proof of Theorem 1.2.

4. Strengthening of Theorem 1.2

It is well-known that every Boolean function is represented by a unique multi- linear polynomial over the two-element field. Such a representation is called the Zhegalkin polynomial (or theReed–Muller polynomial) off [11, 13, 17]. It is clear that a variable is essential inf if and only if it occurs in the Zhegalkin polynomial of f. We denote by degpthe degree of polynomialp. Ifpis the Zhegalkin polynomial off, then we denote the Zhegalkin polynomial off_i←j byp_i←j. Note that the only polynomials of degree zero are the constant polynomials.

Theorem 4.1. Let f be a Boolean function with at least two essential variables.

Then the arity gap off is two if and only if the Zhegalkin polynomial off is of one of the following special forms:

• x_i1+x_i2+· · ·+x_in+c,

• x_ix_j+x_i+c,

• x_ix_j+x_ix_k+x_jx_k+c,

• x_ix_j+x_ix_k+x_jx_k+x_i+x_j+c,

wherec∈ {0,1}. Otherwise the arity gap of f is one.

We prove first an auxiliary lemma that takes care of the functions of essential arity at least four whose Zhegalkin polynomial has degree two.

Lemma 4.2. If f is a Boolean function with at least four essential variables and the Zhegalkin polynomial off has degree two, then the arity gap off is one.

Proof. Denote the Zhegalkin polynomial of f by p. We need to consider several cases and subcases.

Case 1. Assume first thatp is of the form

(10) p=x_ix_j+x_ix_k+x_jx_k+x_ia_i+x_ja_j+x_ka_k+a,

whereai,aj,ak are polynomials of degree at most 1 andais a polynomial of degree at most 2 such that there are no occurrences of variablesxi,xj,xk inai,aj,ak,a.

Subcase 1.1. Assume that dega_i = dega_j = dega_k = 0. Then a contains a variablex_l distinct fromx_i,x_j, x_k, and we can writea =x_la⁰+a⁰⁰, wherea⁰ and a⁰⁰ do not containx_l. Thenf_l←i is represented by the polynomial

(11) pl←i=xixj+xixk+xjxk+xia⁰+a⁰⁰,

where all essential variables off except forxloccur, because no terms cancel, and hence gapf = 1.

Subcase 1.2. Assume that at least one ofai, aj,ak has degree 1, say degai = 1.

Thenai contains a variablexl distinct fromxi, xj,xk, and soai=xl+a⁰_i, where a⁰_i has degree at most 1 and does not containx_l. Consider

(12) pj←k=xk(1 +aj+ak) +xiai+a.

If all essential variables of f except forxj occur in p_j←k, then gapf = 1 and we are done. Otherwise we need to analyze three different subcases.

Subcase 1.2.1. Assume that variablexk occurs inp_j←k but there is a variablexl

that occurs inajandak but not inai nor inasuch thatxldoes not occur in pj←k

(due to some cancelling terms inaj andak). Writeaj=xl+a⁰_j,ak=xl+a⁰_k, and consider

p_j←l = xixl+xixk+xlxk+xiai+xl+xla⁰_j+xkxl+xka⁰_k+a

= x_ix_l+x_ix_k+x_ia_i+x_l+x_la⁰_j+x_ka⁰_k+a.

(13)

Every essential variable off except forxj occurs inpj←l, and hence gapf = 1.

Subcase 1.2.2. Assume thatxk does not occur inpj←k. In this caseaj =ak+ 1.

Consider

(14) p_j←i=xi(1 +ai+aj) +xkak+a.

If any term ofa_j is cancelled by a term ofa_i, it still remains as a term of a_k, and hence all variables occurring in a_i, a_j, a_k occur in p_j←i. If both x_i and x_k also occur inp_j←i, then all essential variables of f except forx_j occur in p_j←i, and so gapf = 1.

Ifxk does not occur inp_j←i, thenak = 0 and soaj= 1. Then (15) pl←i=xixj+xixk+xjxk+xi+xia⁰_i+xj+a,

and every essential variable off except forxl occurs inp_l←i. Thus gapf = 1.

Ifxi does not occur inp_j←i, thenaj =ai+ 1, and henceai =ak. Consider then (16) p_i←k =x_k(1 +a_i+a_k) +x_ja_j+a=x_k+x_ja_j+a.

Again all essential variables off except forxi occur inpi←k, and so gapf = 1.

Subcase 1.2.3. Assume that bothxiandxk occur inpj←k but there is a variable xloccurring inaiand inaj but neither inaknor inasuch thatxldoes not occur in pj←k (due to some cancelling terms inai andaj). Writeai=x_l+a⁰_i,aj =x_l+a⁰_j, and consider

p_j←l = xixl+xixk+xlxk+xixl+xia⁰_i+xl+xla⁰_j+xkak+a

= xixk+xlxk+xia⁰_i+xl+xla⁰_j+xkak+a.

(17)

Every essential variable off except forxj occurs inp_j←l, and so gapf = 1.

Case 2. Assume then thatp is of the form

(18) p=x_ix_j+x_ix_ka_ik+x_ia_i+x_ja_j+x_ka_k+a,

whereaikis a polynomial of degree 0;ai,aj,ak are polynomials of degree at most 1; and ais a polynomial of degree at most 2 such that variables xi, xj,xk do not occur inaik,ai,aj,ak,a. Note thataik andak cannot both be 0, for otherwisexk

would not occur inp. Consider

(19) p_j←i=xi(1 +ai+aj) +xixkaik+xkak+a.

By the above observation that a_ik and a_k are not both 0, x_k occurs in p_j←i. If all essential variables of f except forx_j occur inp_j←i, then gapf = 1 and we are done. Otherwise we distinguish between two cases.

Subcase 2.1. Assume that x_i does not occur in p_j←i. In this case a_j =a_i+ 1, aik= 0, and ak 6= 0. Consider

pi←k = xjxk+xkaik+xkai+xjaj+xkak+a

= xjxk+xk(ai+ak) +xj+xjai+a.

(20)

Bothx_j and x_k occur inp_i←k, because the termx_jx_k cannot be cancelled. If any term ofa_i is cancelled by a term ofa_k, it still remains inx_ja_i. Thus, all essential variables off except forx_i occur inp_i←k, and hence gapf = 1.

Subcase 2.2. Assume thatxi occurs inp_j←i but there is a variable xl occurring in ai andaj but not in aik, ak, nor ina such thatxl does not occur in p_j←i (due to some cancelling terms inai andaj). Consider

(21) p_k←l=x_ix_j+x_ix_la_ik+x_ia_i+x_ja_j+x_la_k+a.

Ifaik= 1, then the terms xixl inxiai and inxixlaik cancel each other. These are the only terms that may be cancelled out. Nevertheless, xl occurs also inaj, and so all essential variables off except forxk occur inpk←l. Therefore gapf = 1 also

in this case.

Proof of Theorem 4.1. Denote the Zhegalkin polynomial off byp. It is straight- forward to verify that if p has one of the special forms listed in the statement of the theorem, thenf does not have a variable identification minor of essential arity essf −1 but it has one of essential arity essf−2. For the converse implication, we will prove by induction on essf that ifpis not of any of the special forms, then there is a variable identification minorgoff such that essg= essf−1, i.e.,f has arity gap 1.

If essf = 2 and p is not of any of the special forms, then p = x_ix_j +c or p=x_ix_j+x_i+x_j+c where c ∈ {0,1}, and in both casesp_j←i =x_i+c. In this case gapf = 1.

If essf = 3, thenp has one of the following forms:

• x_ix_jx_k+x_ix_j+x_ix_k+x_jx_k+a_ix_i+a_jx_j+a_kx_k+c,

• x_ix_jx_k+x_ix_k+x_jx_k+a_ix_i+a_jx_j+a_kx_k+c,

• x_ix_jx_k+x_ix_j+a_ix_i+a_jx_j+a_kx_k+c,

• x_ix_jx_k+a_ix_i+a_jx_j+a_kx_k+c,

• xixj+xixk+xjxk+xk+c,

• xixj+xixk+xjxk+xi+xj+xk+c,

• xixj+xixk+aixi+ajxj+akxk+c,

• xixk+aixi+ajxj+akxk+c,

where ai, aj, ak, c∈ {0,1}. It is easy to verify that in each case p_j←i contains the termxixk, and hence bothxi andxk are essential inf_j←i, and so gapf = 1.

For the sake of induction, assume then that the claim holds for 2≤essf < n, n≥4. Consider the case that essf =n. Since the case where degp= 1 is ruled out by the assumption thatpdoes not have any of the special forms and the case where degp= 2 is settled by Lemma 4.2, we can assume that degp≥3. Choose a variablexmfrom a term of the highest possible degree inp, and write

(22) p=x_mq+r,

where the polynomials q and rdo not contain xm. We clearly have that degq= degp−1, and q andrrepresent functions with less than nessential variables. Of course, every essential variable off except forxmoccurs inq orr. We have three different cases to consider, depending on the comparability under inclusion of the sets of variables occurring inqandr.

Case 1. Assume that there is a variablex_i that occurs inqbut does not occur inr, and there is a variablex_j that occurs inrbut does not occur inq. Write (23) q=xiq⁰+q⁰⁰, r=xjr⁰+r⁰⁰,

whereq⁰,q⁰⁰,r⁰,r⁰⁰ do not containx_i,x_j. Then

(24) p=xmxiq⁰+xmq⁰⁰+xjr⁰+r⁰⁰, and we have that

(25) p_j←i=xmxiq⁰+xmq⁰⁰+xir⁰+r⁰⁰,

where no terms can cancel. Hence all essential variables of f except for x_j are essential inf_j←i and so gapf = 1.

Case 2. Assume that every variable occurring in r occurs inq. In this case q represents a functionqof essential arity essf−1, containing all essential variables off except forxm. We also have that degq= degp−1≥2.

Subcase 2.1. If essf ≥ 5, then essq ≥ 4, and we can apply the inductive hypothesis, which tells us that there are variablesx_i and x_j such that essq_i←j = essq−1. Hencef_i←j is represented by the polynomialp_i←j =x_mq_i←j+r_i←j, and all essential variables of f except forx_i occur inp_i←j, since no terms can cancel betweenx_mq_i←j andr_i←j. Thus gapf = 1.

Subcase 2.2. If essf = 4, then essq = 3, and we can apply the inductive hypothesis as above unlessq=xixj+xixk+xjxk+c orq=xixj+xixk+xjxk+ xi+xj+c. If this is the case, consider first the case where q contains a variable xl∈ {xi, xj, xk} that does not occur inr. Consider then

(26) p_m←l=x_lq+r.

Thenxlqcontains the term xixjxk, which cannot be cancelled. Namely, all other terms of xlq have degree at most 2, and since there are at most two variables occurring in r, the terms of ralso have degree at most 2. Thus, all variables of f except forxm occur inpm←l, and so the arity gap off is 1.

Consider then the case that qand rcontain the same variables, i.e.,x_i, x_j, x_k. If degr ≤ 2, then it is easily seen that p_m←i contains the term x_ix_jx_k, and all essential variables of f except for x_m are essential in f_m←i. Otherwise, we can apply the inductive hypothesis on the function r represented by r and we obtain variablesx_αandx_β such that essr_α←β= essr−1. It can be easily verified that no identification of variables bringsq into the zero polynomial, so xm and two other variables will occur in p_α←β =xmq_α←β+r_α←β. We have that gapf = 1 also in this case.

Case 3. Assume that every variable occurring in q occurs in r but there is a variablexl that occurs inrbut does not occur in q. If degr= 1, thenr=xl+r⁰ wherer⁰ does not containxl. Thenp_m←l=xlq+xl+r⁰, where the only term that may cancel out isxl, and this happens ifqhas a constant term 1. Nevertheless,xl

occurs inrm←lbecause degq≥2. Of course, all other essential variables off except forxm also occur inpm←l, so gapf = 1. We may thus assume that degr≥2.

Subcase 3.1. Assume first that essf = 4 (in which casercontains three variables and q contains at most two variables) and r = x_ix_j +x_ix_k +x_jx_k +c or r = x_ix_j +x_ix_k +x_jx_k +x_i +x_j +c. Since we assume that degp ≥ 3, we have that degq ≥ 2 and hence q contains at least two variables. Thus exactly two variables occur inqand so also degq= 2. Thenq=x_αx_β+b₁x_α+b₂x_β+dwhere α, β∈ {i, j, k}andb₁, b₂, d∈ {0,1}. Letγ∈ {i, j, k} \ {α, β}. Thenp_m←γ contains the termxixjxk, and hence all essential variables off except forxmoccur inp_m←γ, and so gapf = 1.

Subcase 3.2. Assume then that essf >4 or essf = 4 but rdoes not have any of the special forms. In this case we can apply the inductive hypothesis on the function r represented by r. Let xi and xj be such that essr_j←i = essr−1. If q_j←i6= 0, thenxmand all other essential variables off except forxj occur inp_j←i, and we are done—the arity gap of f is 1. We may thus assume thatqj←i = 0.

Writeqandrin the form

q = xixja1+xia2+xja3+a4, (27)

r = x_ix_jb₁+x_ib₂+x_jb₃+b₄, (28)

where the polynomials a1, a2, a3, a4, b1, b2, b3, b4 do not contain xi, xj. Define the polynomialsq1, . . . ,q7as follows (cf. the proof of Theorem 4 in Salomaa [14]):

q1 consists of the terms common toa1,a2, anda3.

q_i,i= 2,3, consists of those terms common toa₁ anda_i which are not inq₁. q₄ consists of those terms common toa₂ anda₃ which are not inq₁.

q_4+i,i= 1,2,3, consists of the remaining terms ina_i.

Define the polynomials andr₁, . . . ,r₇ similarly in terms of theb_i’s. Note that for anyi6=j, q_i andq_j do not have any terms in common, and similarlyr_i andr_j do not have any terms in common. Hence,

q = xixj(q1+q2+q3+q5) + xi(q1+q2+q4+q6) + xj(q1+q3+q4+q7) +a4, (29)

r = x_ix_j(r₁+r₂+r₃+r₅) + xi(r1+r2+r4+r6) + x_j(r₁+r₃+r₄+r₇) +b₄. (30)

Identification ofx_i withx_j yields

q_j←i = x_i(q₁+q₅+q₆+q₇) +a₄, (31)

rj←i = xi(r1+r5+r6+r7) +b4. (32)

Since we are assuming thatqj←i = 0, we have that q1 =q5 =q6 =q7=a4 = 0.

On the other hand,q6= 0, soq2, q3,q4 are not all zero. Thus (33) q=xixj(q2+q3) +xi(q2+q4) +xj(q3+q4).

All essential variables off except forxj are contained inrj←i.

Subcase 3.2.1. Assume that there is a variablextoccurring inb4 that does not occur inr1,r5,r6,r7. Consider

(34) pm←t=xtq+r=xlq+xixjb1+xib2+xjb3+b4.

Cancelling may only happen between a term ofxtqand a term ofr. No term ofb4

can be cancelled, because every term of x_tq containsx_i or x_j but the terms of b₄ do not contain either. The variables that do not occur in b₄ occur in some terms ofb₁,b₂,b₃that do not containx_t. Thus, all essential variables off except forx_m occur inp_m←t, and so in this casef has arity gap 1.

Subcase 3.2.2. Assume that all variables of rexcept forx_i, x_j occur already in r1+r5+r6+r7. Consider

p_m←i = xixj(q2+q4+r1+r2+r3+r5) + xi(q2+q4+r1+r2+r4+r6) + xj(r1+r3+r4+r7) +b4. (35)

Subcase 3.2.2.1. Assume first thatxi does not occur inp_m←i in (35). Then q2+q4+r1+r2+r3+r5 = 0,

(36)

q₂+q₄+r₁+r₂+r₄+r₆ = 0, (37)

and since ther_i’s do not have terms in common, we have that (38) r₁+r₂=q₂+q₄, r₃=r₄=r₅=r₆= 0.

Then all variables ofrexcept forxi,xj occur already inr1+r7. Consider p_m←j = xixj(q3+q4+r1+r2+r3+r5) +

x_i(r₁+r2+r4+r6) +

xj(q3+q4+r1+r3+r4+r7) +b4

= x_ix_j(q₂+q₃) + xi(r1+r2) +

xj(q2+q3+r2+r7) +b4. (39)

All variables ofr1 are there on the fifth line of (39). If a term of r7 is cancelled by a term of q2+q3 on the sixth line, it still remains on the fourth line, so all variables ofr₇ are also there. We still need to verify that the variablesx_i andx_j are not cancelled out from (39). Ifq₂+q₃6= 0 then we are done. Assume then that q₂+q₃= 0, in which case q₄6= 0. Since

(40) r1+r2+r4+r6=r1+r2=q2+q4=q46= 0, we havexi in (39). Since

(41) q3+q4+r1+r3+r4+r7=q4+r1+r7

andr1+r7contains all variables ofrexcept forxi,xj, butq4does not,q4+r1+r76= 0, so we also havexj in (39). Thus, the arity gap off equals 1 in this case.

Subcase 3.2.2.2. Assume then thatxi occurs in p_m←i in (35). Nothing cancels out on the third line of (35), and therefore the variables ofr1andr7occur inpm←i. Terms of r5 may be cancelled out by terms of q2 +q4 on the first line of (35) but such terms will remain on the second line. Thus the variables of r5 occur in pm←i. A similar argument shows that the variables ofr6 also occur in pm←i. In order forf to have arity gap 1, we still need to verify thatx_j occurs in p_m←i. If q₂+q₄+r₁+r₂+r₃+r₅6= 0, then we are done. We may thus assume that (42) q2+q4+r1+r2+r3+r5= 0.

By the assumption thatx_i occurs inp_m←i, the second line of (35) does not vanish, i.e.,

(43) 06=q2+q4+r1+r2+r4+r6=r3+r4+r5+r6.

If the third line of (35) does not vanish either, i.e.,r1+r3+r4+r76= 0, then we have both xi and xj and we are done. We may thus assume thatr1+r3+r4+r7= 0, i.e., r1=r3 =r4 =r7 = 0. Then all variables ofrexcept for xi,xj occur already inr5+r6. Equation (42) implies thatr2+r5=q2+q4. Consider

p_m←j = xixj(q3+q4+r1+r2+r3+r5) + xi(r1+r2+r4+r6) +

xj(q3+q4+r1+r3+r4+r7) +b4

= x_ix_j(q₂+q₃) +

xi(q2+q4+r5+r6) + x_j(q₃+q₄) +b₄. (44)

Assume first thatq2+q3= 0, in which caseq46= 0. If a term ofr5+r6is cancelled by a term ofq4on the fifth line of (44), it will still remain on the sixth line. Therefore we have inpm←j all variables ofrexcept forx_i and x_j. Sincer5+r6 contains all variables ofrexcept forx_i,x_j butq₂+q₄=q₄does not, the fifth line of (44) does

not vanish, and so we havexi. We also havexj because q3+q4=q4 6= 0 on the sixth line. In this casef has arity gap 1.

Assume then thatq2+q36= 0. Then the fourth line of (44) does not vanish and both x_i and x_j occur inp_m←j. If any term ofr₅+r₆ is cancelled by a term ofq₂ on the fifth line of (44), it still remains on the fourth line, and if it is cancelled by a term of q₄, it remains on the sixth line. Thus all variables of roccur in p_m←j, andf has arity gap 1 again. This completes the proof of Theorem 4.1.

5. Concluding remarks

We do not know whether the upper bound on arity gap given by Theorem 3.1 is sharp. For base sets Awith k≥3 elements, we do not know whether there exists an operationf onAwith essf ≥k+ 1 and gapf ≥3. We know that for allk≥2, there are operations on ak-element set Awith arity gap 2. Consider for instance the quasi-linear functions of Burle [1]. A function f is quasi-linear if it has the form

(45) f =g(h1(x1)⊕h2(x2)⊕ · · · ⊕hn(xn)),

whereh1, . . . , hn:A→ {0,1},g:{0,1} →Aare arbitrary mappings and⊕denotes addition modulo 2. It is easy to verify that if those hi’s that are nonconstant coincide (andg is not a constant map), thenf has arity gap 2.

In general, if there is an operationf on ak-element setAwith with gapf =m, then there are operations of arity gapmon all setsBof at leastkelements. Namely, it is easy to see that any operationgonB of the form

(46) g=φ(f(γ(x1), γ(x2), . . . , γ(xn))),

whereγ:B→Ais surjective andφ:A→B is injective, satisfies essg= essf and gapg= gapf.

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(M. Couceiro)Department of Mathematics, Statistics and Philosophy, University of Tampere, FI-33014 Tampereen yliopisto, Finland

E-mail address:[email protected]

(E. Lehtonen)Institute of Mathematics, Tampere University of Technology, P.O. Box 553, FI-33101 Tampere, Finland

E-mail address:[email protected]