On the Goal Value of a Boolean Function

Transcription

1 On the Goa Vaue of a Booean Function Eric Bach Dept. of CS University of Wisconsin 1210 W. Dayton St. Madison, WI Lisa Heerstein Dept of CSE NYU Schoo of Engineering 2 Metrotech Center, 10th Foor Brookyn, NY Devorah Ketenik Dept. of CIS Brookyn Coege City University of New York 2900 Bedford Avenue Brookyn, NY Abstract In recent work, Deshpande et a. introduced a new measure of the compexity of a Booean function (Deshpande, Heerstein, and Ketenik 2014). The measure is reated to the maximum utiity vaue of a certain monotone, submoduar utiity function associated with the Booean function. We ca this measure the goa vaue of the function. Proving that a Booean function has sma goa vaue can yied new bounds on the average-depth decision tree compexity of the function, and new approximation agorithms for its sequentia testing probem. We present bounds on the goa vaues of arbitrary and specific Booean functions. We aso compare the goa vaue measure to other, previousy studied, measures of the compexity of Booean functions. Introduction In recent work, Deshpande et a. introduced a new measure of the compexity of a Booean function (Deshpande, Heerstein, and Ketenik 2014). We ca this measure the goa vaue of the function. 1 The measure is reated to the maximum utiity vaue of a certain monotone, submoduar utiity function associated with the Booean function. Proving that a Booean function f has sma goa vaue can ead to a good approximation agorithm for the Stochastic Booean Function Evauation probem for f. Aso, if f has sma goa vaue, it indicates a cose reationship between two other measures of the compexity of f, its average-case decision tree compexity and its average-case certificate compexity. In this work, we expore properties of the goa vaue measure. We review previous resuts on the measure. We prove new bounds on the goa vaues of genera and specific Booean functions. We aso consider the reated 1-goa vaue and 0-goa vaue measures. Finay, for monotone Booean functions, we present a resut reating goa vaue to other, previousy studied, measures of Booean function compexity. We begin by presenting necessary definitions and expaining the known connections between goa vaue, decision tree compexity, and Booean function evauation. We then show that the goa vaue of every Booean function f is at east n, 1 Deshpande et a. caed it the Q-vaue. Since Q has no particuar meaning, we have chosen to use a more intuitive name. if f depends on a n of its input variabes. We note that this ower bound is tight for certain Booean functions, such as AND, OR and XOR functions. We aso present goa vaue bounds for Booean k-of-n functions. Deshpande et a. showed that the goa vaue of every Booean function f is at most ds(f) cs(f), where ds(f) is the minimum number of terms in a DNF formua for the function, and cs(f) is the minimum number of causes in a CNF formua for the function. It is we known and easy to show that ds(f) 2 n 1 and cs(f) 2 n 1. Thus the goa vaue of f is upper-bounded by 2 2n 2. We do not know how cose goa vaue can come to this upper bound, but we show that there is a function whose goa vaue is approximatey 2 n 2. We aso show that the goa vaue of a read-once function is at east the maximum of cs(f) and ds(f). Note that there is ony a quadratic gap between this ower bound and the upper bound of ds(f) cs(f). We show that the poynomia threshod degree of a monotone Booean function is a ower bound on its goa vaue. It can be a very weak ower bound, however, because threshod degree is at most n, whereas goa vaue can be exponentia in n. Terminoogy and Notation Booean function definitions: Let V = {x 1,..., x n }. A partia assignment is a vector b {0, 1, } n. For partia assignments a, b {0, 1, } n, a is an extension of b, written a b, if a i = b i for a b i. We aso say that b is contained in a. Given Booean function f : {0, 1} n {0, 1}, a partia assignment b {0, 1, } n is a 0-certificate of f if f(a) = 0 for a a {0, 1} n such that a b. It is a 1-certificate if f(a) = 1 for a a {0, 1} n such that a b. It is a certificate if it is either a 0-certificate or a 1-certificate. We say that b contains a variabe x i if b i. We wi occasionay treat a certificate b as being the set of variabes x i such that b i, together with the assignments given to them by b. The size of a certificate b is the number of variabes x i that it contains. If b and b are certificates for x with b b, we say that b is contained in b. A certificate b for f is minima if there is no certificate b of f, such that b b and b b. The certificate size of a Booean function f is the maximum, over a x {0, 1} n, of the size of the smaest certificate contained in x. Given a distribution D on {0, 1} n,

2 the expected certificate size of f is the expected vaue, for x drawn from D, of the size of the smaest certificate contained in x. It is we-known that every Booean function f can be represented as a poynomia over GF(2). That is, either f is the constant function f = 0, or f is computed by an expression of the form T 1 + T T k for some k > 0, where each T i is the conjunction (i.e., AND) of the variabes in some subset of V, no two terms contain exacty the same subset S i of variabes, and addition is moduo 2. (If S i =, then T i = 1.) This is sometimes caed the ring-sum expansion of the function. Each Booean function has a unique such representation, up to permutation of the terms T i. A variabe x i appears in the GF(2) poynomia for Booean function f iff function f depends on variabe x i. For function f(x 1,..., x n ) defined on {0, 1} n, and b {0, 1, }, the function f b induced on f by b is defined as foows. Let V b = {x i b i = }. Then V b is the set of input variabes for function f b, and the domain of f b is the set of assignments that assign 0 or 1 to each variabe. For each a in the domain, f b (a) = f(a\b), where a\b denotes the assignment to V produced by setting the variabes in V b according to a, and the variabes in V V b according to b. For k {0, 1}, we use f xi k to denote the function induced on f by the partia assignment setting x i to k, and a other variabes to. A k-of-n function is a Booean function on n variabes whose output is 1 if and ony if at east k of its input variabes are 1. A read-once formua is a monotone Booean formua over the basis of AND and OR gates, whose inputs are a variabes. We can view a read-once formua as a rooted tree with the foowing properties: 1. The eaves are abeed with distinct variabes from the set {x 1,..., x n }, for some n 1 2. Each interna node is abeed AND or OR 3. Each node abeed AND or OR gate has at east 2 chidren. A maxterm of a monotone Booean function f(x 1,..., x n ) is the set of variabes contained in a minima 0-certificate of f. (Equivaenty, it is a set of variabes S such that setting the variabes in S to 0 forces f to 0, but this is not true of any proper subset of S.) A minterm of f is the set of variabes contained in a minima 1-certificate of f. If F is a Booean formua expressing a monotone function, we say that a subset S of its variabes is a maxterm (minterm) of F iff S is a maxterm (minterm) of the function is computes. The canonica DNF formua for a non-constant monotone Booean function f is a formua of the form T 1 T 2... T k where k is the number of minterms of f, and each T i is the conjunction (AND) of the variabes in a distinct minterm of f. It is unique up to permutation of terms. A decision tree computing a Booean function f(x 1,..., x n ) is a binary tree where each binary node is abeed by a variabe in V, and each each eaf is abeed either 0 or 1. The eft chid of an interna node abeed x i corresponds to x i = 0, and the right chid corresponds to x i = 1. The tree computes f if for each x {0, 1} n, the root-eaf path induced by x ends in a eaf whose abe is f(x). Given a decision tree T computing f, and x {0, 1} n, et Depth(T, x) denote the number of interna nodes on the root-eaf path in T induced by x. Then the depth of T is the maximum vaue of Depth(T, x) for a x {0, 1} n. Given a probabiity distribution on {0, 1} n, the expected depth of T is the expected vaue of Depth(T, x), when x is drawn from the distribution. The average depth of T is its expected depth with respect to the uniform distribution on {0, 1} n. The decision tree depth of Booean function f is the minimum depth of any decision tree computing f. The expected decision tree depth of Booean function f, with respect to a distribution on {0, 1} n, is the minimum expected depth of any decision tree computing f, with respect to that distribution. The rank of a (binary) decision tree T is defined as foows (Ehrenfeucht and Hausser 1989): if T is a eaf, then rank(t ) = 0. Otherwise, et rank 0 denote the rank of the subtree rooted at the eft chid of the root of T and rank 1 denote the rank of the subtree rooted at the right chid of the root of T. Then { max{rank0, rank rank(t ) = 1 } if rank 0 rank 1 rank otherwise A k-decision ist is a ist of pairs (t i, v i ) where t i is a term (conjunction) of at most k iteras and v i {0, 1}. The ast term t i is the empty term, which is equa to 1. A decision ist L defines a Booean function such that for any assignment x {0, 1} n, L(x) = v j where j is the minimum index such that t j (x) = 1 (Rivest 1987). We say that a Booean function f is computed by a degree d poynomia threshod function if there is a mutivariate poynomia p(x 1,..., x n ) of degree d, over the rea numbers, such that for a x {0, 1} n, f(x) = sgn(p(x)), where sgn(z) = 1 if z 0 and sgn(z) = 0 if z < 0. The poynomia threshod degree of f is the minimum d such that f is computed by a poynomia threshod function of degree d. Submoduarity Definitions: Let N = {1,..., n}. In what foows, we assume that utiity functions are integervaued. In the context of standard work on submoduarity, a utiity function is a set function g : 2 N Z 0. Given S N and j N, g S (j) denotes the quantity g(s {j}) g(s). We aso use the term utiity function to refer to a function g : {0, 1, } n Z 0 defined on partia assignments. For {0, 1, }, the quantity b xi denotes the partia assignment that is identica to b except that b i =. A utiity function g : {0, 1, } n Z 0 is monotone if for b {0, 1, } n, i N such that b i =, and {0, 1}, g(b xi ) g(b) 0; in other words, additiona information can not decrease utiity. Utiity function g is submoduar if for a b, b {0, 1, } n and {0, 1}, g(b xi ) g(b) g(b x ) i g(b ) whenever b b and b i = b i =. This is a diminishing returns property. We say that Q 0 is the goa vaue of g : {0, 1, } n Z 0 if g(b) = Q for a b {0, 1} n. A goa function for a Booean function f : {0, 1} n {0, 1} is a monotone, submoduar function g : {0, 1, } n

3 Z 0, with goa vaue Q, such that for a b {0, 1, } n, g(x) = Q iff b contains a certificate of f. We define the goa vaue of f to be the minimum goa vaue of any goa function g for f. The goa vaue of f is ceary equa to the goa vaue of the compement of the function f. A 1-goa function for f is a monotone, submoduar function g : {0, 1, } n Z 0, such that for some constant Q 0, g(b) = Q if b is a 1-certificate of f, and g(b) < Q otherwise. We ca Q the 1-goa vaue of g. We define the 1-goa vaue of f to be the minimum goa vaue of any 1-goa function for f. The definition of a 0-goa function for f and the 0-goa vaue of f are anaogous. We denote the goa vaue, 1-goa vaue, and 0-goa vaue of f by Γ(f), Γ 1 (f), and Γ 0 (f) respectivey. We use g(x i = ) to denote the vaue of g(b) for b {0, 1, } n such that b i =, and b j = for j i. Stochastic Booean Function Evauation In the Stochastic Booean Function Evauation (SBFE) probem, we are given a representation of a Booean function f : {0, 1} n {0, 1}, where the representation is drawn from a fixed cass of representations C. For exampe, C might be the cass of read-once formuas. We are aso given a vector of probabiities (p 1,..., p n ) (0, 1) n, and a vector of costs (c 1,..., c n ) Z n 0. We woud ike to determine the vaue of f on an initiay unknown input x {0, 1} n, when cost c i must be paid to get the vaue of x i, and each bit x i of x is equa to 1 with independent probabiity p i. The probem is to find the best (adaptive) order in which to acquire the bits of x, so as to minimize the expected cost incurred before f(x) can be determined. SBFE probems arise in diverse appication areas, incuding medica diagnosis, database query optimization, Operations Research, and machine earning with attribute costs (Ibaraki and Kameda 1984; Krishnamurthy, Bora, and Zanioo 1986; Deshpande and Heerstein 2008; Srivastava et a. 2006; Ünüyurt 2004; Kapan, Kushievitz, and Mansour 2005). Deshpande et a. observed that an instance of the SBFE probem for a Booean function f can be reduced to an instance of a probem caed Stochastic Submoduar Set Cover (SSSC), by constructing a goa function g for f (Deshpande, Heerstein, and Ketenik 2014). Goovin and Krause gave an agorithm for the SSSC probem, caed Adaptive Greedy, that achieves an O(og Q) approximation, when given as input a utiity function g : {0, 1, } n Z 0 with goa vaue Q (Goovin and Krause 2011). Appying it to the goa function g constructed for Booean function f yieds an O(og Q) approximation for the SBFE probem for f. Reca that ds(f) is the minimum number of terms in a DNF formua for f, and cs(f) is the minimum number of causes in a CNF formua for f. Deshpande et a. gave a generic method for constructing a goa function g for an arbitrary Booean function f, such that g has goa vaue ds(f) cs(f). Hence, by running Adaptive Greedy on g, the SBFE probem for f can be approximated to within a factor of O(og(ds(f) cs(f))) when f is given by its CNF and DNF formuas (or by a Booean decision tree). Deshpande et a. aso showed that for certain Booean functions, Γ(f) can be exponentia in n. In particuar, they showed that Γ(f) 2 n/2 both for the function f(x 1,..., x n ) = x 1 x 2 x 3 x 4... x n 1 x n, and for a certain inear-threshod function having coefficients of exponentia magnitude. For such functions, the approach of constructing a goa function g for f, and then running Adaptive Greedy on g, does not yied a good approximation agorithm for the SBFE probem for f. See (Deshpande, Heerstein, and Ketenik 2014) for further detais. Bounds on Decision Tree Depth In (Deshpande, Heerstein, and Ketenik 2013), it was shown that the goa vaue of a Booean function f coud be used to upper bound the expected depth of a decision tree computing f, with respect to the uniform distribution on {0, 1} n. Let DT(f) denote the expected decision tree depth of f, and CERT(f) denote the expected certificate size of f, where both expectations are with respect to the uniform distribution on {0, 1} n. It is easy to show that CERT(f) DT(f). Let g be a goa function for f whose goa vaue is Γ(f). Under the uniform distribution, when run on g, the Adaptive Greedy agorithm of Goovin and Krause outputs a soution (corresponding to a decision tree) that is within a factor of 2(n Γ(f) + 1) of the expected certificate size of f. It foows that DT(f) (2 n Γ(f) + 1)CERT(f). Thus if Γ(f) is sma, expected certificate size and expected depth are cose to each other. However, Γ(f) can be very arge. In (Aen et a. 2014), it was shown that the gap between DT(f) and CERT(f) can be exponentia: there is a function f such that for any constant 0 < ɛ < 1, DT(f) CERT(f) = Ω(2ɛCERT(f) ). This stands in marked contrast to resuts for worst-case depth of decision tree and worst-case certificate size; it is known that the worst-case depth of a decision tree is at most quadratic in the size of the worst-case certificate (Buhrman and Wof 1999). Lower Bound on Goa Vaue In this section, we seek to ower bound the goa vaue for a Booean functions. Let f : {0, 1} n {0, 1} be a Booean function depending on a its variabes. We begin with the foowing emma: Lemma 1. Let f(x 1,..., x n ) be a Booean function. Let, k {0, 1}. Let g be a goa function for f. If there is a minima certificate of f setting x i =, then g(x i = ) g(,..., ) 1. Simiary, if g is a k-goa function for f, and there is a minima k-certificate for f setting x i =, which has size s, then g (x i = ) g (,..., ) 1. Further, if the size of that certificate is s, then the goa vaue of g is at east s. Proof. Let g be a goa function for f, and et Q be its goa vaue. Let g be a k-goa function for f, and et Q be its k-goa vaue. Let d be a minima k-certificate for f. Thus g(d) = Q and g (d) = Q.

4 Let s be the size of certificate d, and without oss of generaity, assume that x 1,..., x s are the variabes contained in d. Consider the sequence d 0, d 1,..., d s where d 0 = (,..., ) and d i = (d 1, d 2,..., d i,,..., ) for 1 i s. Consider a fixed i such that 1 i s. By monotonicity, g(d i 1 ) g(d i ). Suppose g(d i 1 ) = g(d i ). Let ˆd be the partia assignment that is obtained from d s by setting x i to. Since d is a minima certificate for f, ˆd is not a certificate, and so g( ˆd) < Q, and thus g(d) g( ˆd) 1. Since d and ˆd differ ony in the assignment to x i, and ˆd d i 1, by submoduarity, g(d i ) g(d i 1 ) 1 and g(x i = d i ) g(,..., ) 1. The same argument shows that g (d i ) g (d i 1 ) 1 and g (x i = d i ) g(,..., ) 1. From the first inequaity, we get that g (d) s. The emma foows. The proof of the above emma aso shows that Γ(f) d, where d is the size of a minima certificate for f. But d n, and we wi prove that Γ(f) n. First, we prove the foowing emma. Lemma 2. Let f(x 1,..., x n ) be a Booean function depending on a n of its variabes. Then there exists a variabe x i and a vaue {0, 1} such that the induced function f xi depends on a n 1 of its input variabes. Proof. Consider the GF(2) poynomia computing f. The poynomia must contain a n variabes, since it depends on a of them. For each x i, et T (x i ) denote the set of terms of the GF(2) poynomia which contain x i. Define a partia order on those variabes x i such that x i < x j in the partia order if T (x i ) is a proper subset of T (x j ). Now consider an x i such that T (x i ) is a minimum eement in this partia order. We anayze severa different cases: Case 1: The variabe x i appears in a non-constant terms of the poynomia. In this case, setting x i to 1 produces a poynomia that contains the remaining n 1 variabes, and cannot be simpified (since it has no repeated terms). Thus the induced function must depend on a those variabes. Case 2: The variabe x i does not appear in a terms of the poynomia, and there is no x i x j such that T (x i ) = T (x j ). In this case, setting x i to 0 just deetes a terms of the poynomia containing x i, and the resut is the poynomia for the induced function that cannot be simpified. By assumption, for each x j x i, T (x i ) T (x j ), and since T (x i ) is a minimum, there must be a term containing x j but not x i. This term is in the poynomia that remains after setting x i to 0, and so the function setting x i to 0 depends on a remaining x j. Case 3: The variabe x i does not appear in a terms of the poyomia, and there exists an x j x i such that T (x i ) = T (x j ). In this case, set x i to 1 in the poynomia. The resut does not have any repeated terms: if t was a term containing x i in the origina poynomia, then there cannot be a term in the poynomia containing exacty the variabes in t that are not equa to x i, because then it woud contain x j, impying that T (x i ) T (x j ). Thus the resut of setting x i to 1 is to produce a poynomia that is the same as the origina poynomia, except that x i is deeted from any terms containing it (and no simpification is possibe). Since a of the variabes appeared in the origina poynomia, a of the variabes except x i appear in the resuting poynomia. We now have the foowing theorem: Theorem 1. Let f : {0, 1} n {0, 1} be a Booean function. If f depends on exacty n of its input variabes, then Γ(f) n. Further, Γ 1 (f) + Γ 0 (f) n + 1. Proof. We prove the theorem in the case that f depends on a n of its input variabes, so n = n. This is without oss of generaity, because if n < n, we can consider instead the restriction of f to the variabes on which it depends. It is easy to verify that this causes no change in the vaues of Γ, Γ 1, and Γ 0. By repeated appication of Lemma 2, it foows that there exists a sequence of partia assignments, b 0, b 1,..., b n in {0, 1, } n, such that b 0 is the a- assignment, b n has no s, each b i+1 is an extension of b i produced by changing exacty one to a non- vaue, and the function f i that is induced from f by partia assignment b i depends on a n i of its variabes. Without oss of generaity, assume b i assigns vaues in {0, 1} to variabes x n i+1,..., x n, and sets variabes x 1,..., x n i to. Thus f i : {0, 1} n i {0, 1}. Let g be a goa function for f. For i {1,..., n}, et g i : {0, 1, } n i Z 0 be the function induced on g by b i. Thus for a {0, 1, } n i, g i (a) = g(a\b i ), where a\b i denotes the assignment produced by setting the variabes in {x 1,..., x n i } according to a, and the remaining variabes of g according to b i. From the fact that g is a goa function for f, it easiy foows that g i is a goa function for f i, with goa vaue equa to the goa vaue of g. Let x n i+1, where {0, 1}, be the one-variabe assignment that extends b i 1 to produce b i. Since f i 1 depends on x n i+1, there is an assignment b {0, 1} n i+1 such that f i 1 (b xn i+1 0) f i 1 (b xn i+1 1). Hence there is a minima 0-certificate or a minima 1- certificate for f i 1 setting x n i+1 to. Thus by Lemma 1, g i 1 (x n i+1 = ) g i 1 (,..., ) 1, and hence g(b i ) g(b i 1 ) 1. Thus the vaue of g increases by at east 1 for each b i in the sequence b 0, b 1,..., b n, and so g(x n ) n. This proves that Γ(f) n. We now need to prove that Γ 1 (f) + Γ 0 (f) n + 1. Let g be a 1-goa function for f, and et g be a 0-goa vaue function for f. Considering the sequence b 0,..., b n again, the above argument, together with Lemma 1, shows that for each i {1,..., n 1}, g (b i ) g (b i 1 ) 1 or g (b i ) g (b i 1 ) 1. Thus g (b n 1 ) + g (b n 1 ) n 1. Either the assignment x 1 = b n 1 is in a 1-certificate for f n 1, or in a 0-certificate (or both). Without oss of generaity, assume it is in a 1-certificate. Then assignment x 1 = b n 1 is in a 0-certificate for g n 1. Let q be the goa vaue of g and q be the goa vaue for g. Since b n 1 contains neither a 1-certificate nor a 0-certificate for f, g (b n 1 ) q 1 and g (b n 1 ) q 1. It foows that n 1 (q 1)+(q 1), and so q + q n + 1. The foows. Furthermore, these bounds can be tight, as iustrated in the foowing two propositions:

5 Proposition 1. The AND, OR, and XOR functions, and their negations, have goa vaue n. Proof. Their goa vaues must be at east n, by Theorem 1. For AND, consider utiity function g whose vaue equas n on any partia assignment having at east one 0, and whose vaue equas the number of 1 s otherwise. This is a goa function for AND, and there is a dua goa function for OR. For XOR, consider the goa function whose vaue on a partia assignment is equa to the number of variabes set to 0 or 1. Proposition 2. Let f : {0, 1} n {0, 1} be a Booean k- of-n function. Then Γ 1 (f) = k and Γ 0 (f) = n k + 1. Proof. For any k-of-n function f, define the 1-goa function g whose vaue on any subset of bits is just the minimum of k and the number of bits set to 1. This 1-goa function has goa vaue k, so Γ 1 (f) k and by Lemma 1, Γ 1 (f) k. Simiary, define the 0-goa function g whose vaue on any subset of bits is the minimum of n k + 1 and the number of bits set to 0. We can further anayze the goa vaue for k-of-n functions: Theorem 2. Let f : {0, 1} n {0, 1} be a Booean k-of-n function. Then Γ(f) = k(n k + 1). Proof. Let g be a goa function for f with goa vaue Q. Consider a minima 1-certificate b for f. It has exacty k 1 s and no 0 s. Since g(b) = Q it foows from the submoduarity and monotonicity of g that there must be at east one index i 1 such that b i1 = 1 and g({b i1 }) 1 k Q. Let t 1 = g({b i1 }). Thus g(b) g({b i1 }) = Q t 1. It foows again from the monotonicity and submoduarity of g that for some index i 2 i 1 where b i2 = 1, g({b i1, b i2 }) t k 1 (Q t 1 ). Since t 1 1 k Q, and t k 1 (Q t 1) = t 1 (1 1 k 1 ) + 1 k 1 Q, we get that g({b i 1, b i2 }) 1 k (1 1 k 1 )Q + 1 k 1 Q so g({b i1, b i2 }) Q( 2 k ). Settng t 2 = g({b i1, b i2 }), we simiary get i 3 with g({b i1, b i2, b i3 }) 3 k Q and so forth, unti we have g({b i1,..., b ik 1 }) k 1 k Q, where b i 1,..., b ik 1 are k 1 of the k bits of b that are set to 1. Let b ik be the remaining bit of b that is set to 1. Let c be the minima 0- certificate for f such that c ik = 0, and further, c i = 0 for a i where b i =, and c i = for a other i. Let = n k + 1, which is the number of 0 s in c. Let t k 1 = g({b i1,..., b ik 1 }), so t k 1 k 1 k Q. Let d be the assignment such that d ij = 1 for j {1,..., k 1}, and d i = 0 for a other i. Then g(d) = Q and g(d) g({b i1,..., b ik 1 }) = Q t k 1. Let Q = Q t k 1. Let = n k + 1. It foows from an argument simiar to the one above that there are bits c j1,..., c j 1 equa to 0 such that g({b i1,..., b ik 1, c j1,..., c j 1 }) t k Q. Let d be the partia assignment such that b i1,..., b ik 1 = 1, c j1,..., c j 1 = 0 and b i = for the remaining one variabe x i. Thus we have g(d ) t k t k 1 ) and so g(d ) t k 1 (1 1 that t k 1 k 1 k Q = t k ) + 1 Q we get that g(d ) k 1 k (Q Q. Using the fact 1 (1 )Q + 1 Q and so g(d ) k 1 k Q. Further, since d is neither a 0-certificate nor a 1-certificate for f, g(d ) < Q, and thus g(d ) Q 1. It foows that Q k = k(n k + 1). Finay, it foows from previous work (see Lemma 3 in the next section) that there exists a goa function g for f with goa vaue exacty equa to k = k(n k + 1). Upper Bound on Goa Vaue We do not know the maximum possibe goa vaue of a Booean function, if goa vaue is expressed as a function of the number of variabes, n. As previousy mentioned, Deshpande et a. showed that f(x 1,..., x n ) = x 1 x 2 x 3 x 4... x n 1 x n has goa vaue 2.5n. A simiar argument yieds a function with arger goa vaue. Theorem 3. For n a mutipe of 3, there is a Booean function f(x 1,..., x n ) such that Γ(f) 3 n/3 = 2 αn where α = og Proof. Let n be a mutipe of 3, and et f(x 1,..., x n ) = x 1 x 2 x 3 x 4 x 5 x 6... x n 2 x n 1 x n. Our proof is by induction on n. The statement is ceary true for n = 3, because in this case f has a minima 1- certificate of size 3, and so its goa vaue is at east 3. We now show is is true inductivey for arger n. Let g be a goa function for f, with goa vaue Q. Let b be the a 1 s assignment to the variabes in V. Ceary g({b 1, b 2, b 3 }) = Q. By the monotonicity and submoduarity of g, there must be a pair of those three bits for which the vaue of g is at east 2 3 Q. Assume without oss of generaity it is b 1 and b 2 so that g({b 1, b 2 }) 2 3 Q. By monotonicity, adding in x 3 = 0 cannot decrease utiity, so g({b 1, b 2, x 3 = 0}) 2 3 Q aso. Let t = g({b 1, b 2, x 3 = 0}). Consider the induced function f produced from f by setting x 1 = 1, x 2 = 1, and x 3 = 0. Inductivey, this function has goa vaue at east 3 (n 3)/3. It foows that Q t 3 (n 3)/3, and hence Q t + 3 (n 3)/3. Since t 2 3 Q, we get that Q 3n/3. The foowing is impicit in the work of Deshpande et a. Lemma 3. (Deshpande, Heerstein, and Ketenik 2014) For any Booean function f, Γ(f) Γ 1 (f) Γ 0 (f) Proof. The proof of the above proposition is constructive, and is based on a standard OR construction used previousy by (Guiory and Bimes 2011). Let g 1 be a 1-goa function for f and g 2 be a 0-goa function for f. Let Q 1 and Q 0 be the goa vaues for these functions. Then function g(b) = (Q 1 g 1 (b))(q 2 g 0 (b)) is a monotone, submoduar function where g(b) = Q 1 Q 2 iff g 1 (b) = Q 1 (b) or g 0 (b) = Q 0 (b). The emma foows. As shown in Proposition 1, for the AND function, Γ(f) = n, and it is easy to show that Γ 0 (f) = 1 whie Γ 1 (f) = n. Therefore, for the AND (or the OR) function, we have Γ(f) = Γ 1 (f) Γ 0 (f). In contrast, by Lemma 1 and Proposition 1, when f is the XOR function, Γ(f) = Γ 1 (f) = Γ 0 (f) = n. We have the foowing upper bound.

6 Lemma 4. Let f be a Booean function that is not identicay 1 or 0. Then for k {0, 1}, Γ k (f) Γ(f). Proof. Let g : {0, 1, } n Z 0 be a goa function for f, and et Q be its goa vaue. Without oss of generaity, assume k = 1. Let g 1 : {0, 1, } n Z 0 be such that for b {0, 1, } n, g 1 (b) = Q 1 if b contains a 0-certificate of f, and g 1 (b) = g(b) otherwise. Ceary g 1 (b) = Q iff b contains a 1-certificate of f. Using the monotonicity and submoduarity of g, and the fact that any extension of a 0- certificate is aso a 0-certificate, it is straightforward to show that g 1 is monotone and submoduar. Goa Vaue of Read-Once Formuas In this section, we discuss bounds on the goa vaue of functions expressed by read-once formuas. Given a function g : {0, 1, } n Z 0, we say that g gives no vaue to 0 s if for a b {0, 1, }, g(b) = g(b ), where b is the partia assignment produced from b by changing a 0 s in b to s. We say that g gives no vaue to 1 s if the anaogous property hods. The foowing emma wi be usefu. Lemma 5. Let f be a monotone Booean function. Then there is a 1-goa function g for f that gives no vaue to 0 s, whose goa vaue is Γ 1 (g). Simiary, there is a 0-goa function g for f that gives no vaue to 1 s, whose goa vaue is Γ 0 (g). Proof. We prove this for the 1-goa function; the proof is symmetric for 0-goa function. Let g be a 1-goa utiity function for f with 1-goa vaue Γ 1 (f). For b {0, 1, }, et b be the partia assignment produced from b by changing a 0 s in b to s. Define a reated utiity function h such that h(b) = g(b ). It is straightforward to verify that because f is monotone, h is aso a 1-goa utiity function for f, with the same 1-goa vaue as g, namey Γ 1 (f). In what foows, if φ is a DNF formua, we use size(φ) to denote the number of terms of φ. (If φ is the constant formua 0 or 1, size(φ) = 0.) Lemma 6. Let F be a read-once formua on variabe set V = {1,..., n}, and et f be the monotone Booean function that it computes. Let φ be the canonica DNF formua for f. Let h : 2 V Z 0 be a monotone, submoduar set function, such that for some Q > 0, h(s) = Q if S contains a maxterm of f, and h(s) < Q otherwise. Then the foowing 2 properties hod: i) For any variabe x i V, h({x i }) number of terms of φ containing x i ii) h(v ) size(φ). Proof. We do induction on n. In the base case, n = 1. Formua F then consists of a singe node abeed x 1 and φ = x 1. Since g is a 0-goa utiity function, h( ) < h({x 1 }), and therefore h(v ) = h({x 1 }) 1 = number of terms of φ containing x 1 = size(φ). For the induction step, et k 0 and assume that the caim is true for n = k. Suppose F has size n = k + 1. We first prove (i). Let x i be an arbitrary variabe of F. root. OR H H 1... H k There are two cases. Figure 1: x i is in one of the H j Case 1: There is an ancestor of x i in F that is abeed OR Let t be an ancestor of x i in F that is abeed OR. Let H be a subtree of F that is rooted at a chid of t, such that H does not contain x i (that is, the path from t down to x i does not go through H). See Figure 1. Let S denote the variabes in H. Because t is abeed OR, and the terms of φ correspond to the minterms of f, if a term of φ contains any variabes from S, it cannot contain x i. Let V = V S. Let F denote the induced read-once formua produced from F by setting the variabes in H to 0. Setting the variabes in H removes the subtree H from F. In order to produce a read-once formua in vaid form, we foow the foowing procedure: if after removing H, t has ony one chid, t is deeted. If t had a parent, the remaining chid of t wi become the chid of the previous parent of t; if t did not have a parent, the remaining chid becomes the new root. Let φ be the canonica DNF formua for F. Formua φ is the formua produced from φ by deeting a terms containing at east one variabe in H. Since x i does not appear in any of the deeted terms, the number of terms of φ that contain x i is equa to the number of terms of φ that contain x i. We define a set function h : 2 V Z 0 where for a subsets R V, h (R) = h(r S) h(s). Function h is ceary monotone and submoduar. Since F is read-once and the parent of subtree H is abeed OR, set R V contains a maxterm of F iff R H contains a maxterm of F. It foows that h (R) = h (V ) = h(v ) h(s) if R is a maxterm of h, and h (R) < h(v ) h(s) otherwise. By the inductive hypothesis, h ({x i }) (the number of terms of φ containing x i ). Since the number of terms of φ containing x i is equa to the number of terms of φ containing x i, h ({x i }) = h({x i } S) h(s) (number of terms of φ containing x i ). By the submoduarity of h, h({x i }) h({x i } S) h(s), so h({x i }) (number of terms of φ containing x i ). Case 2: A ancestors of x i in F are abeed AND.

7 AND x i H H1... H k Figure 2: merging of the ANDs Because a ancestors of x i are AND, we can merge the ANDs, and without oss of generaity, we can assume that x i is a chid of the root of F, which is abeed AND. See iustration in Figure 2. Let H be a subtree of F rooted at a sibing of x i, and et S denote the variabes in H. Let V = V {x i }. Ceary {x i } is a maxterm of F, and every maxterm of H is a maxterm of F. Consider the read-once formua induced from F by setting x i to 1. (Note: if x i and H were the ony chidren of the root, then the resuting read-once formua is just H.) Ca it F. Since the root is AND, the canonica DNF for f contains x i in every term. Let φ be the canonica DNF for F. It foows that φ is produced from F by deeting x i from each term of φ (and no redundant or subsumed terms are created), and that size(φ) = size(φ ). Consider the restriction of utiity function h to the variabes in V. Ca the restriction h. That is, h : 2 V Z 0 where for a R V, h (R) = h(r). Because h is monotone and submoduar, so is h. Ceary any R V is a maxterm of F iff it is a maxterm of F. Aso, since S is a subset of V, and S contains maxterms of H and hence of F and F, h (V ) = h(v ). For R V, h (R) = h (V ) = h(v ) if R is a maxterm of F, ese h (R) < h(v ). Since F has fewer variabes than F, by the inductive hypothesis, h (V ) size(φ ) = size(φ). Finay, since {x i } is a maxterm of F, and h is a 0-utiity function for f, h(v ) = h({x i }). We thus have h({x i }) = size(φ) = number of terms containing x i, since every term of φ contains x i. This competes the proof of (i) for Case 2. We now prove Property (ii) of the caim, in the genera case. If every variabe x i of F is such that {x i } is a maxterm of F, then F is just the AND of a the variabes, and (ii) foows immediatey from (i). Suppose there is a variabe x i of F where {x i } is not a maxterm of F. By (i), h({x i }) number of terms of φ containing x i. Let F be the read-once formua produced from F by setting x i to 0. Since {x i } is not a maxterm of F, F has at east one variabe. Let φ be the canonica DNF for F. Let V = V {x i }. Let h : 2 V Z 0 where for R V, h (R) = h(r {x i }) h({x i }). DNF formua φ is the formua you get by deeting a terms in φ that contain x i. It is easy to see that h satisfies the conditions of the caim, for formua F. Therefore, by induction, h (V ) the number of terms of φ. Therefore, we have that h (V ) = h(v ) h({x i }), so h(v ) = h (V ) + h({x i }) (number of terms of φ not containing x i ) + (number of terms of φ containing x i ) = size(φ). Theorem 4. If f is a Booean function that is computed by a read-once formua, then Γ 0 (f) ds(f), Γ 1 (f) cs(f), and Γ(f) max{ds(f), cs(f)}. Proof. Let φ be the canonica DNF formua for f. Let size(φ) denote the number of terms in φ. We prove that Γ 0 (f) ds(f). A dua argument shows that Γ 1 (f) cs(f). It then foows from Lemma 4 that Γ(f) max{ds(f), cs(f)}. Let g be a 0-goa function for f, and et Q be its 0-goa vaue. By Lemma 5, we may assume that g gives no vaue to 1 s. We can therefore associate with g a utiity function h defined on the subsets of V, such that for S V, h(s) = g(b), where b is the partia assignment setting the variabes in S to 1, and a other variabes to. Ceary, h(s) = Q iff set S contains a maxterm of h (that is, there is a subset S S such that S is a maxterm of f). By Lemma 6, h(v ) ds(f). Therefore, Q ds(f) and hence Γ 0 (f) ds(f). For a function f represented by a read-once formua, we thus have max{ds(f), cs(f)} Γ(f) ds(f) cs(f). There is at most a quadratic gap between these upper and ower bounds. Goa Vaue and Other Measures of Booean Function Compexity In this section, we reate the goa vaue of a monotone Booean function to decision tree rank and poynomia threshod degree. A decision tree computes a monotone submoduar function h as foows: It has interna nodes abeed with variabes x i V. Each eaf of the tree is abeed with an eement of {0, 1..., d}. In using the tree to compute the vaue of h on input S V, for each interna node abeed x i, you branch right if x i is in S, and eft if not. Lemma 7. If h : 2 V {0, 1,..., d} is a monotone, submoduar set function, then there is a decision tree computing h that has rank d. Proof. It was shown in (Fedman, Kothari, and Vondrák 2013) that for submoduar h (not necessariy monotone), there is a decision tree computing h of rank at most 2d. Here we give a better bound on rank when monotonicity hods. The proof is by induction on n and d. It is ceary true if n = 0 and d = 0, and in particuar, it is true for n + d = 0. Let s > 0, and assume true for n + d < s. We now show true for n + d = s. If h is identicay equa to a singe vaue in {0, 1,..., d}, then the decision tree can consist of a singe node abeed with that vaue. Otherwise, there exists a subset S V such that h(s) h( ). By monotonicity, h(s) > h( ). Let S be such that h(s) is maximized. If h(s) < d, then a decision tree of rank at most d must exist by induction. Suppose h(s) = d.

8 By submoduarity and monotonicity, there exits x i S such that h({x i }) > h( ), so h({x i }) 1. We construct a decision tree for h by first putting x i in the root. The eft subtree of this tree needs to compute the function h : 2 V {xi} {0, 1,..., d} such that h (S) = h(s) for a S V {x i }. This is a function of subsets of n 1 variabes, and function h is monotone and submoduar, and thus by induction, can be computed by a tree of rank at most d. The right subtree needs to compute the function h : 2 V {xi} {0, 1,..., d} such that h (S) = h(s {x i }) for a S V. We now show that there is a decision tree computing h that has rank at most d 1. Consider the function ĥ defined on 2 V {xi} such that for S V, ĥ(s) = h(s {x i}) h({x i }) for a x i in S. Since h is monotone, so is ĥ. Further, since h({x i }) 1 and h(s {x i }) d, it foows that ĥ can be viewed as mapping to {0, 1,..., d 1}. Function ĥ is defined on n 1 variabes, and since it is monotone and submoduar, by induction, it can be computed by a tree of rank at most d 1. If modify the tree by adding the vaue h({x i }) to the vaue in each eaf of the tree, the resuting tree wi compute h. We have thus shown that we can construct a tree computing h such that the two subtrees of the root have rank d and d 1 respectivey. Such a tree has rank d. Theorem 5. Let f be a monotone Booean function, and et d = min{γ 1 (f), Γ 0 (f)}. Then there is a decision tree of rank at most d computing f. There is a poynomia-threshod function of degree at most d computing f. Proof. Consider a 1-goa function g for a monotone Booean function f, whose goa vaue is d. By Lemma 5, we can assume that the vaue of g ony increases on bits that are set to 1, and stays the same on bits that are set to 0. Let V = {x 1,..., x n } Define a set function h : 2 V {0,..., d}, such that for a S V, h(s) is equa to the vaue of g on the partia assignment setting the variabes in S to 1, and a other variabes to 0. By Lemma 7, there is a decision tree computing h with rank at most d. This same tree must compute the function g (if at an interna node abeed x i, you branch eft if x i = 0, and right if x i = 1). For each eaf of the tree, change the vaue of the eaf to 0 if it is abeed with a vaue in 0,..., d 1, and to 1 if it is abeed with the vaue d. Since d is the 1-goa vaue for g, on any assignment in 2 V, the vaue computed by the tree wi be 1 iff the assignment contains a 1-certificate for f. It foows that a tree of rank at most d computes f. By (Bum 1992), a function that can be computed by a tree of rank d has an equivaent d-decision ist. By a simiar argument to that used in (Ehrenfeucht et a. 1989), any function expressibe by a d-decison ist is aso expressibe as a degree-d poynomia threshod function. It foows from the above and from Lemma 4 that the goa vaue of a monotone Booean function is ower bounded by its poynomia threshod degree. Open Questions There remain many open questions concerning the goa vaue of Booean functions. We ist some here: As a function of n, what is the argest possibe goa vaue for a Booean function? What is the expected goa vaue of a random Booean function? Are there any techniques for deriving good ower bounds for Γ(f)? Is our ower bound on Γ(F ) where F is a read-once formua, actuay tight? Is goa vaue poynomiay reated to some known measure of the compexity of a Booean function? References Aen, S.; Heerstein, L.; Ketenik, D.; and Ünüyurt, T Evauation of DNF formuas. CoRR abs/ v3. Bum, A Rank-r decision trees are a subcass of r-decision ists. Information Processing Letters 42(4): Buhrman, H., and Wof, R. D Compexity measures and decision tree compexity: A survey. Theoretica Computer Science 288:2002. Deshpande, A., and Heerstein, L Fow agorithms for parae query optimization. In Proceedings of the 24th Internationa Conference on Data Engineering (ICDE), Deshpande, A.; Heerstein, L.; and Ketenik, D Approximation agorithms for stochastic booean function evauation and stochastic submoduar set cover. CoRR abs/ v2. Deshpande, A.; Heerstein, L.; and Ketenik, D Approximation agorithms for stochastic booean function evauation and stochastic submoduar set cover. In SODA 2014, SIAM. Ehrenfeucht, A., and Hausser, D Learning decision trees from random exampes. Information and Computation 82(3): Ehrenfeucht, A.; Hausser, D.; Kearns, M.; and Vaiant, L A genera ower bound on the number of exampes needed for earning. Information and Computation 82(3): Fedman, V.; Kothari, P.; and Vondrák, J Representation, approximation and earning of submoduar functions using ow-rank decision trees. In Conference on Learning Theory, Goovin, D., and Krause, A Adaptive submoduarity: Theory and appications in active earning and stochastic optimization. Journa of Artificia Inteigence Research 42: Guiory, A., and Bimes, J Active semi-supervised earning using submoduar functions. In UAI,

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