aa r X i v : . [ c s . CC ] J un An Optimal Tester for k -Linear Nader H. Bshouty
Dept. of Computer ScienceTechnion, Haifa, 32000June 9, 2020
Abstract
A Boolean function f : t , u n Ñ t , u is k -linear if it returns the sum (over the binaryfield F ) of k coordinates of the input. In this paper, we study property testing of the classes k -Linear, the class of all k -linear functions, and k -Linear ˚ , the class Y kj “ j -Linear. We give anon-adaptive distribution-free two-sided ǫ -tester for k -Linear that makes O ˆ k log k ` ǫ ˙ queries. This matches the lower bound known from the literature.We then give a non-adaptive distribution-free one-sided ǫ -tester for k -Linear ˚ that makes thesame number of queries and show that any non-adaptive uniform-distribution one-sided ǫ -testerfor k -Linear must make at least ˜Ω p k q log n ` Ω p { ǫ q queries. The latter bound, almost matchesthe upper bound O p k log n ` { ǫ q known from the literature. We then show that any adaptiveuniform-distribution one-sided ǫ -tester for k -Linear must make at least ˜Ω p? k q log n ` Ω p { ǫ q queries. Property testing of Boolean function was first considered in the seminal works of Blum, Luby andRubinfeld [9] and Rubinfeld and Sudan [38] and has recently become a very active research area.See for example, [1, 2, 3, 4, 7, 8, 10, 12, 13, 14, 15, 16, 17, 18, 19, 21, 25, 27, 30, 31, 33, 32, 34, 39]and other works referenced in the surveys and books [23, 24, 35, 36].A Boolean function f : t , u n Ñ t , u is said to be linear if it returns the sum (over thebinary field F ) of some coordinates of the input, k -linear if it returns the sum of k coordinates,and, k -linear ˚ if it returns the sum of at most k coordinates. The class Linear (resp. k -Linearand k -Linear ˚ ) is the classes of all linear functions (resp. all k -linear functions and Y ki “ k -Linear).Those classes has been of particular interest to the property testing community [7, 8, 9, 10, 11, 21,22, 24, 28, 35, 36, 37, 39]. Let f and g be two Boolean functions t , u n Ñ t , u and let D be a distribution on t , u n . Wesay that f is ǫ -far from g with respect to (w.r.t.) D if Pr D r f p x q “ g p x qs ě ǫ and ǫ -close to g w.r.t. D if Pr D r f p x q “ g p x qs ď ǫ . 1n the uniform-distribution and distribution-free property testing model, we consider the prob-lem of testing a class of Boolean function C . In the distribution-free testing model (resp. uniform-distribution testing model), the tester is a randomized algorithm that has access to a Booleanfunction f : t , u n Ñ t , u via a black-box oracle that returns f p x q when a string x is queried.The tester also has access to unknown distribution D (resp. uniform distribution) via an oraclethat returns x P t , u n chosen randomly according to the distribution D (resp. according to theuniform distribution). A distribution-free tester , [26], (resp. uniform-distribution tester ) A for C is an tester that, given as input a distance parameter ǫ and the above two oracles to a Booleanfunction f ,1. if f P C then A accepts with probability at least 2 { f is ǫ -far from every g P C w.r.t. D (resp. uniform distribution) then A rejects withprobability at least 2 { A an ǫ -tester for the class C or an algorithm for ǫ -testing C . We say that A is one-sided if it always accepts when f P C ; otherwise, it is called two-sided tester. The querycomplexity of A is the maximum number of queries A makes on any Boolean function f . If thequery complexity is q then we call the tester a q -query tester or a tester with query complexity q .In the adaptive testing (uniform-distribution or distribution-free) the queries can depend onthe answers of the previous queries where in the non-adaptive testing all the queries are fixed inadvance by the tester.In this paper we study testers for the classes k -Linear and k -Linear ˚ . Throughout this paper we assume that k ă ? n . Blum et al. [9] gave an O p { ǫ q -query non-adaptiveuniform-distribution one-sided ǫ -tester (called BLR tester) for Linear. Halevy and Kushilevitz, [28],used a self-corrector (an algorithm that computes g p x q from a black box query to f that is ǫ -closeto g ) to reduce distribution-free testability to uniform-distribution testability. This reduction givesan O p { ǫ q -query non-adaptive distribution-free one-sided ǫ -tester for Linear. The reduction canbe applied to any subclass of Linear. In particular, any q -query uniform-distribution ǫ -tester for k -Linear ( k -Linear ˚ ) gives a O p q q -query distribution-free ǫ -tester.It is well known that if there is a q -query uniform-distribution ǫ -tester for Linear and a q -query uniform-distribution ǫ -tester for the class k -Junta then there is an O p q ` q q -query uniform-distribution O p ǫ q -tester for k -Linear ˚ . Since k -Linear “ k -Linear ˚ zp k ´ q -Linear ˚ , if there is a q -query uniform-distribution ǫ -tester for k -Linear ˚ then there is an O p q q -query uniform-distributiontwo-sided ǫ -tester for k -Linear. Therefore, all the results for testing k -Junta are also true for k -Linear ˚ and k -Linear in the uniform-distribution model.For lower bounds on the number queries for two-sided uniform-distribution testing k -Linear (seethe table in Figure 1): For non-adaptive testers Fisher, et al. [21] gave the lower bound Ω p? k q .Goldreich [22], gave the lower bound Ω p k q . In [8], Blais and Kane gave the lower bound 2 k ´ o p k q .Then in [7], Blais et al. gave the lower bound Ω p k log k q . For adaptive testers, Goldreich [22], gavethe lower bound Ω p? k q . Then Blais et al. [7] gave the lower bound Ω p k q and in [8], Blais and Kane The class of boolean functions that depends on at most k coordinates k ´ o p k q . Then in [39], Saglam gave the lower bound Ω p k log k q . This boundwith the trivial Ω p { ǫ q lower bound gives the lower boundΩ ˆ k log k ` ǫ ˙ (1)for the query complexity of any adaptive uniform-distribution (and distribution-free) two-sidedtesters.For upper bounds for uniform-distribution two-sided ǫ -testing k -Linear, Fisher, et al. [21] gavethe first adaptive tester that makes O p k { ǫ q queries. In [11], Buhrman et al. gave a non-adaptivetester that makes O p k log k q queries for any constant ǫ . As is mentioned above, testing k -Linear canbe done by first testing if the function is k -Junta and then testing if it is Linear. Therefore, usingBlais [5, 6] adaptive and non-adaptive testers for k -Junta we get adaptive and non-adaptive uniform-distribution testers for k -Linear that makes O p k log k ` k { ǫ q and ˜ O p k . { ǫ q queries, respectively.For upper bounds for two-sided distribution-free testing k -Linear, as is mentioned above, fromHalevy et al. reduction in [28], an adaptive and non-adaptive distribution-free ǫ -tester can beconstructed from adaptive and non-adaptive uniform-distribution ǫ -testers. This gives an adaptiveand non-adaptive distribution-free two-sided testers for k -Linear that makes O p k log k ` k { ǫ q and˜ O p k . { ǫ q queries, respectively. See the table in Figure 1. In this paper we prove
Theorem 1.
For any ǫ ą , there is a polynomial time non-adaptive distribution-free one-sided ǫ -tester for k -Linear ˚ that makes O ˆ k log k ` ǫ ˙ queries. By the reduction from k -Linear to k -Linear ˚ , we get Theorem 2.
For any ǫ ą , there is a polynomial time non-adaptive distribution-free two-sided ǫ -tester for k -Linear that makes O ˆ k log k ` ǫ ˙ queries. For one-sided testers for k -Linear we prove Theorem 3.
Any non-adaptive uniform-distribution one-sided ǫ -tester for k -Linear must make atleast ˜Ω p k q log n ` Ω p { ǫ q queries. This almost matches the upper bound O p k log n ` { ǫ q that follows from the reduction of Gol-dreich et. al [26] and the non-adaptive deterministic exact learning algorithm of Hofmeister [29]that learns k -Linear with O p k log n q queries.For adaptive testers we prove Theorem 4.
Any adaptive uniform-distribution one-sided ǫ -tester for k -Linear must make at least ˜Ω p? k q log n ` Ω p { ǫ q queries. The table in 1 summarizes all the results in the literature and our results for the class k -Linear.3pper/ One-Sided/ Adaptive/ Uniform/Lower Two-Sided Non-Adap. Dist. Free Result O { Ω ReferenceUpper Two-Sided Adaptive Uniform k { ǫ [21]Upper Two-Sided Adaptive Uniform k log k ` k { ǫ [6]Upper Two-Sided Adaptive Dist. Free k log k ` k { ǫ [28]Upper Two-Sided Non-Adap. Uniform k log k ( ǫ Const.) [11]Upper Two-Sided Non-Adap. Uniform k . { ǫ [5]Upper Two-Sided Non-Adap. Dist. Free k . { ǫ [28]Upper Two-Sided Non-Adap. Dist. Free k log k ` { ǫ OursLower Two-Sided Non-Adap. Uniform 1 { ǫ TrivialLower Two-Sided Non-Adap. Uniform ? k ` { ǫ [21]Lower Two-Sided Non-Adap. Uniform k ` { ǫ [22]Lower Two-Sided Non-Adap. Uniform k log k ` { ǫ [7]Lower Two-Sided Adaptive Uniform ? k ` { ǫ [22]Lower Two-Sided Adaptive Uniform k ` { ǫ [7, 8]Lower Two-Sided Adaptive Uniform k log k ` { ǫ [39]Upper One-Sided Non-Adaptive Dist. Free k log n ` { ǫ [26]Lower One-Sided Non-Adaptive Uniform ˜Ω p k q log n ` { ǫ OursLower One-Sided Adaptive Uniform ˜Ω p? k q log n ` { ǫ OursFigure 1: A table of results for the testability of the class k -Linear. In this section we give overview of the techniques used for proving the results in this paper. k -Linear ˚ The tester for k -Linear ˚ first runs the tester BLR of Blum et al. [9] to test if the function f is ǫ -close to Linear w.r.t. the uniform distribution, where ǫ “ Θ p {p k log k qq . BLR is one-sidedtester and therefore, if f is k -linear then BRG accepts with probability 1. If f is ǫ -far from Linearw.r.t. the uniform distribution then, with probability at least 2 {
3, BLR rejects. Therefore, if thetester BLR accepts, we may assume that f is ǫ -close to Linear w.r.t. the uniform distribution. Let g P Linear be the function that is ǫ -close to f . If f is k -linear ˚ then f “ g . This is because ǫ ă { {
2. BLRmakes O p { ǫ q “ O p k log k q queries.In the second stage, the tester tests if g (not f ) is k -linear ˚ . Let us assume for now that we canquery g in every string. Since g P Linear, we need to distinguish between functions in k -Linear ˚ andfunctions in Linear z k -Linear ˚ . We do that with two tests. We first test if g P k -Linear ˚ and thentest if it is in k -Linear ˚ assuming that it is in 8 k -Linear ˚ . In the first test, the tester “throws”,uniformly at random, the variables of g into 16 k bins and tests if there is more than k non-emptybins. If g is k -linear ˚ then the number of non-empty bins is always less than k . If it is k -linear forsome k ą k then with high probability (w.h.p.) the number of non-empty bins is greater than k .Notice that if f is k -linear ˚ then the test always accepts and therefore it is one-sided. This tests4akes O p k q queries to g .The second test is testing if g is in k -Linear ˚ assuming that it is in 8 k -Linear ˚ . This is done byprojecting the variables of g into r “ O p k q coordinates uniformly at random and learning (findingexactly) the projected function using the non-adaptive deterministic Hofmeister’s algorithm, [29],that makes O p k log r q “ O p k log k q queries. Since g P k -Linear ˚ , w.h.p., the relevant coordinatesof the function are projected to different coordinates, and therefore, w.h.p., the learning gives alinear function that has exactly the same number of relevant coordinates as g . The tester acceptsif the number of relevant coordinates in the projected function is at most k . If g P k -Linear ˚ , thenthe projected function is in k -Linear ˚ with probability 1 and therefore this test is one-sided. Thistest makes O p k log k q queries.We assumed that we can query g . We now show how to query g in O p k log k q strings so wecan apply the above two tests. For this, the tester uses self-corrector, [9]. To compute g p z q , theself-corrector chooses a uniform random string a P t , u n and computes f p z ` a q ` f p a q . Since f is O p {p k log k qq -close to g w.r.t. the uniform distribution, we have that for any string z Pt , u n and an a P t , u n chosen uniformly at random, with probability at least 1 ´ O p {p k log k qq , f p z ` a q ` f p a q “ g p z ` a q ` g p a q “ g p z q . Therefore, w.h.p., the self-corrector computes correctlythe values of g in O p k log k q strings. If f P k -Linear then g “ f and f p z ` a q ` f p z q “ f p z q “ g p z q ,i.e., the self-corrector gives the value of g with probability 1. This shows that the above two testsare one-sided.Now, if f is k -linear ˚ then f “ g . If f is ǫ -far from every function in k -Linear ˚ w.r.t. D then itis ǫ -far from g w.r.t. D .In the final stage the tester tests whether f is equal to g or ǫ -far from g w.r.t. D . Hereagain the tester uses self-corrector. It asks for a sample tp z p i q , f p z i qq| i P r t su according to thedistribution D of size t “ O p { ǫ q and tests if f p z p i q q “ f p z p i q ` a p i q q ` f p a p i q q for every i P r t s ,where a p i q are i.i.d. uniform random strings. If f p z p i q q “ f p z p i q ` a p i q q ` f p a p i q q for all i then itaccepts, otherwise, it rejects. If f is k -linear then f p z p i q q “ f p z p i q ` a p i q q ` f p a p i q q for all i andthe tester accepts with probability 1. Now suppose f is ǫ -far from g w.r.t. D . Since f is ǫ -closeto g w.r.t. the uniform distribution and ǫ ď { { f p z p i q ` a p i q q ` f p a p i q q “ g p z p i q ` a p i q q ` g p a p i q q “ g p z p i q q . Therefore, assuming the latter happens,then, with probability at least 1 ´ ǫ we have f p z p i q q “ g p z p i q q “ f p z p i q ` a p i q q ` f p a p i q q . Thus, w.h.p,there is i such that f p z p i q q “ f p z p i q ` a p i q q ` f p a p i q q and the tester rejects. This stage is one-sidedand makes O p { ǫ q queries. k -Linear As we mentioned in the introduction, the one-sided q -query uniform-distribution ǫ -tester for k -Linear ˚ gives a two-sided uniform-distribution O p q q -query ǫ -tester for k -Linear. This is because, in the uni-form distribution, the linear functions are 1 { ǫ ă { f is ǫ -close to a k -linear function g then it is p { ´ ǫ q -far from p k ´ q -Linear ˚ . This is not truefor any distribution D , and therefore, cannot be applied here.The algorithm in the previous subsection can be changed to a two-sided tester for k -Linear asfollows. The only part that should be changed is the test that g is in k -Linear ˚ assuming that itis in 8 k -Linear ˚ . We replace it with a test that g is in k -Linear assuming that it is in 8 k -Linear ˚ .The tester rejects if the number of relevant coordinates in the function that is learned is not equal to k . This time the test is two-sided. The reason is that the projection to O p k q variables doesnot guarantee (with probability 1) that all the variables of f are projected to different variables.5herefore, it may happen that f is k -linear and the projection gives a p k ´ q -linear ˚ function. We first show the result for non-adaptive testers. Suppose there is a one-sided non-adaptive uniformdistribution 1 { A p s, f q for k -Linear that makes q queries, where s is the random seed of thetester and f is the function that is tested. The algorithm has access to f through a black boxqueries.Consider the set of linear functions C “ t g p q u Y t g p ℓ q “ x n ` ¨ ¨ ¨ ` x n ´ ℓ ` | ℓ “ , . . . , k ´ u Ďp k ´ q -Linear ˚ where g p q “
0. Any k -linear function is 1 { C w.r.t. theuniform distribution. Therefore, using the tester A , with probability at least 2 {
3, we can distinguishbetween any k -linear and any function in C . By running the tester A O p log k q times, and accept ifand only if all accept, we get a tester A that asks O p q log k q queries and satisfies1. If f P k -Linear then with probability 1, A p s, f q accepts.2. If f P C then, with probability at least 1 ´ {p k q , A p s, f q rejects.By an averaging argument (i.e., fixing coins for A ) and since | C | “ k , there exists a deterministicnon-adaptive algorithm B that makes q “ O p q log k q queries such that1. If f P k -Linear then B p f q accepts.2. If f “ C then B p f q rejects.Let a p i q , i “ , . . . , q be the queries that B makes. Let M be a q ˆ n binary matrix where the i -th row of M is a p i q and x f P t , u n where x fi “ i is a relevant coordinate in f . Then the vectorof answers to the queries of B p f q is M x f . If M x f “ M x g for some g P C , that is, the answers of thequeries to f are the same as the answer of the queries to g , then B p f q rejects. Therefore, for every f P k -Linear and every g P C we have M x f “ M x g . Now since t x f | f P k ´ Linear u is the set of allstrings of weight k , the sum (over the field F ) of every k columns of M is not equal to 0 and not equalto the sum of the last ℓ columns of M , for all ℓ “ , . . . , k ´
1. In particular, if M i is the i th columnof M , for every i , . . . , i k ´ ℓ ď n ´ k ` M i ` ¨ ¨ ¨ ` M i k ´ ℓ ` M n ´ ℓ ` ` ¨ ¨ ¨ ` M n “ M n ´ ℓ ` ` ¨ ¨ ¨ ` M n and therefore M i ` ¨ ¨ ¨ ` M i k ´ ℓ “
0. That is, the sum of every less or equal k ´ n ´ k ` M is not equal to zero. We then show that such matrix has at least q “ Ω p k log n q rows. This implies that q “ Ω pp k { log k q log n q .For the lower bound for adaptive testers we take C “ t g p ℓ q u for some ℓ P t , , . . . , k ´ u andget a q ˆ n matrix M that the sum of every k ´ ℓ columns of M is not zero. We then show thatthere exists ℓ ď k ´ q “ ˜Ω p? k log n q rows. k -Linear ˚ and k -Linear In this section we give the non-adaptive distribution-free one-sided tester for k -Linear ˚ and thenon-adaptive distribution-free two-sided tester for k -Linear.6 .1 Notations In this subsection, we give some notations that we use throughout the paper.Denote r n s “ t , , . . . , n u . For S Ď r n s and x “ p x , . . . , x n q . For X Ă r n s we denote by t , u X the set of all binary strings of length | X | with coordinates indexed by i P X . For x P t , u n and X Ď r n s we write x X P t , u X to denote the projection of x over coordinates in X . We denoteby 1 X and 0 X the all-one and all-zero strings in t , u X , respectively. For a variable x i and a set X , we denote by p x i q X the string x over coordinates in X where for every j P X , x j “ x i . For X , X Ď r n s where X X X “ H and x P t , u X , y P t , u X we write x ˝ y to denote theirconcatenation, i.e., the string in t , u X Y X that agrees with x over coordinates in X and agreeswith y over coordinates in X . For X Ď r n s we denote X “ r n sz X “ t x P r n s| x R X u .For example, if n “ X “ t , , u , X “ t , u , y is a variable and z “ p z , z , z , z , z , z , z qP t , u then p y q X ˝ z X ˝ X Y X “ p y , z , y , , y , , z q . Consider the tester
Test-Linear ˚ k for k -Linear ˚ in Figure 2. The tester uses three procedures.The first is Self-corrector that for an input x P t , u n chooses a uniform random z P t , u n and returns f p x ` z q ` f p z q . The procedure BLR that is a non-adaptive uniform-distribution one-sided ǫ -tester for Linear. BLR makes c { ǫ queries for some constant c , [9]. The third procedureis Hoffmeister’s Algorithm p N, K q , a deterministic non-adaptive algorithm that exactly learns K -Linear ˚ over N coordinates from black box queries. Hoffmeister’s Algorithm makes c K log N queries for some constant c , [29].To test k -Linear we use the same tester but change step 11 to:(11) If the output is not in k -Linear then rejectWe call this tester Test-Linear k . In this section we prove
Theorem 5. Test-Linear k is a non-adaptive distribution-free two-sided ǫ -tester for k -Linear thatmakes O ˆ k log k ` ǫ ˙ queries. Theorem 6. Test-Linear ˚ k is a non-adaptive distribution-free one-sided ǫ -tester for k -Linear ˚ thatmakes O ˆ k log k ` ǫ ˙ queries.Proof. Since there is no stage in the tester that uses the answers of the queries asked in previousones, the tester is non-adaptive.In Stage 1 the tester makes O p { ǫ q “ O p k log k q queries. In stage 2.1, O p k q queries. In stage 2.2, O p k log r q “ O p k log k q queries and in stage 3, O p { ǫ q queries. Therefore, the query complexity ofthe tester is O p k log k ` { ǫ q . 7 est-Linear ˚ k Input : Oracle that accesses a Boolean function f Output : Either “Accept” or “Reject”
ProceduresSelf-corrector g p x q : “ f p x ` z q ` f p z q for uniform random z P t , u n . BLR
A procedure that ǫ -tests Linear using c { ǫ queries. Hofmeister’s Algorithm p N, K q for learning K -Linear ˚ over N coordinates using c K log N queries. Stage 1. BLR . Run BLR on f with ǫ “ {p p k ` c k log p k qqqq . If BLR rejects then reject.
Stage 2.1. Testing if g is in Linear z k -Linear ˚ . Choose a uniform random partition X , . . . , X k . Count Ð . Choose a uniform random z P t , u n .6 . For i “ k . if g p z X i ˝ X i q “ Count Ð Count ` . If Count ą k then reject. Stage 2.2. Testing if g is in k -Linear assuming it is in k -Linear ˚ . Choose a uniform random partition X , . . . , X r for r “ k . Run Hofmeister’s algorithm p N “ r, K “ k q in orderto learn F “ g pp y q X ˝ p y q X ˝ ¨ ¨ ¨ ˝ p y r q X r q . If the output is not in k -Linear ˚ then reject {˚ In Test-Linear k (for testing k -Linear) we replace (11) with: {˚
11. If the output is not in k -Linear then reject Stage 3. Consistency test . Choose a sample x p q , . . . , x p t q according to D of size t “ { ǫ .13 . For i “ t .14 . If f p x p i q q “ g p x p i q q then reject.15 . Accept.Figure 2: An optimal two-sided tester for k -Linear.We will assume that k ě
12. For k ă
12, (see the introduction and Table 1) the non-adaptivetester of k -Junta with the BLR tester and the self-corrector gives a non-adaptive testers that makes O p { ǫ q “ O p k log k ` { ǫ q queries. Completeness : We first show the completeness for
Test-Linear k that tests k -Linear. Suppose f P k -Linear. Then for every x we have g p x q “ f p x ` z q ` f p z q “ f p x q ` f p z q ` f p z q “ f p x q .Therefore, g “ f . In stage 1, BLR is one-sided and therefore it does not reject. In stage 2.1, since X , . . . , X k are pairwise disjoint, the number of functions g p x X i ˝ X i q , i “ , , . . . , k , thatare not identically zero is at most k and therefore stage 2.1 does not reject. In stage 2.2, with8robability at least 1 ´ ` k ˘ {p k q ě {
3, the relevant coordinates of f fall into different X i andthen F “ g pp y q X ˝ p y q X ˝ ¨ ¨ ¨ ˝ p y r q X r q “ f pp y q X ˝ p y q X ˝ ¨ ¨ ¨ ˝ p y r q X r q is k -linear. Then,Hofmeister’s algorithm returns a k -linear function. Therefore, with probability at least 2 { f “ g .Now for the tester Test-Linear ˚ k , in stage 2.2, with probability 1 the function F is in k -Linear ˚ .In fact, if t relevant coordinates falls into the set X i then the coordinate i (that correspond to thevariable y i ) will be relevant in F if and only if t is odd. Therefore, the tester does not reject.Notice that Test-Linear ˚ k is one-sided and Test-Linear k is two-sided. Soundness : We prove the soundness for
Test-Linear k . The same proof also works for Test-Linear ˚ k . Suppose f is ǫ -far from k -Linear w.r.t. the distribution D . We have four cases Case 1 : f is ǫ -far from Linear w.r.t. the uniform distribution. Case 2 : f is ǫ -close to g P Linear and g is in Linear z k -Linear ˚ . Case 3 : f is ǫ -close to g P Linear and g is in 8 k -Linear ˚ z k -Linear. Case 4 : f is ǫ -close to g P Linear, g is in k -Linear and f is ǫ -far from k -Linear w.r.t. D .For Case 1, if f is ǫ -far from Linear then, in stage 1, BLR rejects with probability 2 { f is ǫ -close to g , for any fixed x P t , u n with probability at least1 ´ ǫ (over a uniform random z ), f p x ` z q ` f p z q “ g p x ` z q ` g p z q “ g p x q . Since stages 2.1 and 2.2makes p k ` c k log r q queries (to g ), with probability at least 1 ´ p k ` c k log r q ǫ ě { g p x q is computed correctly for all the queries in stages 2.1 and 2.2.For Case 2, consider stage 2.1 of the tester. If g is in Linear z k -Linear ˚ then g has more than8 k relevant coordinates. The probability that less than or equal to 4 k of X , . . . , X k containsrelevant coordinates of g is at most ˆ k k ˙ k ď ˆ e k k ˙ k k ď . If X i contains the relevant coordinates i , . . . , i ℓ then g p x X i ˝ X i q “ x i ` ¨ ¨ ¨ ` x i ℓ and therefore,for a uniform random z P t , u n , with probability at least 1 { g p z X i ˝ X i q “
1. Therefore, if atleast 4 k of X , . . . , X k contains relevant coordinates then, by Chernoff bound, with probabilityat least 1 ´ e ´ k { ě {
12, the counter “
Count ” is greater than k . Therefore, for Case 2, if g is inLinear z k -Linear ˚ then, with probability at least 1 ´ p { ` { ` { q “ {
3, the tester rejects.For Case 3, consider stage 2.2. If g is in 8 k -Linear ˚ z k -Linear then g has at most 8 k relevantcoordinates. Then with probability at least 1 ´ ` k ˘ {p k q ě {
6, the relevant coordinates of g fall into different X i and then Hofmeister’s algorithm returns a linear function with the samenumber of relevant coordinates as g . Therefore stage 2.2 rejects with probability at least 2 { g is in k -Linear and f is ǫ -far from k -Linear w.r.t. D , then f is ǫ -far from g w.r.t. D . Then for uniform random z and x „ D , Pr D ,z r f p x q “ g p x qs ě Pr D ,z r f p x q “ g p x q| g p x q “ f p x ` z q ` f p z qs Pr D ,z r g p x q “ f p x ` z q ` f p z qs“ Pr D r f p x q “ g p x qs Pr z r g p x q “ f p x ` z q ` f p z qsě ǫ p ´ ǫ q ě ǫ { . Therefore, with probability at most p ´ ǫ { q t “ p ´ ǫ { q { ǫ ď {
3, stage 3 does not reject.9
Lower Bound
In this section we prove
Theorem 7.
Any non-adaptive uniform-distribution one-sided { -tester for k -Linear must makeat least ˜Ω p k log n q queries. Theorem 8.
Any adaptive uniform-distribution one-sided { -tester for k -Linear must make atleast ˜Ω p? k log n q queries. We first show the result for non-adaptive testers.Suppose there is a non-adaptive uniform-distribution one-sided 1 { A p s, f q for k -Linearthat makes q queries, where s is the random seed of the tester and f is the function that is tested.The algorithm has access to f through a black box queries.Consider the set of linear functions C “ t g p q u Y t g p ℓ q “ x n ` ¨ ¨ ¨ ` x n ´ ℓ ` | ℓ “ , . . . , k ´ u Ďp k ´ q -Linear ˚ where g p q “
0. Any k -linear function is 1 { C w.r.t. theuniform distribution. Therefore, using the tester A , with probability at least 2 { A can distinguishbetween any k -linear function and functions in C . We boost the success probability to 1 ´ {p k q by running A , log p k q{ log 3 times, and accept if and only if all accept. We get a tester A that asks O p q log k q queries and satisfies1. If f P k -Linear then with probability 1, A p s, f q accepts.2. If f P C then, with probability at least 1 ´ {p k q , A p s, f q rejects.Therefore, the probability that for a uniform random s , A p s, f q accepts for some f P C is atmost 1 {
2. Thus, there is a seed s such that A p s , f q rejects for all f P C (and accept for all f P k -Linear). This implies that there exists a deterministic non-adaptive algorithm B p“ A p s , ˚qq that makes q “ O p q log k q queries such that1. If f P k -Linear then B p f q accepts.2. If f P C then B p f q rejects.Let a p i q , i “ , . . . , q be the queries that B makes. Let M be a q ˆ n binary matrix that it’s i -throw is a p i q . Let x f P t , u n where x fi “ i is relevant coordinate in f . Then the vector of answersto the queries of B p f q is M x f . If M x f “ M x g for some g P C , that is, the answers of the queries to f are the same as the answers of the queries to g , then B p f q rejects. Therefore, for every f P k -Linearand every g P C we have M x f “ M x g . Now since t x f | f P k ´ Linear u is the set of all strings of weight k , the sum (over the field F ) of every k columns of M is not equal to 0 (zero string) and not equalto the sum of the last ℓ columns of M , for all ℓ “ , . . . , k ´
1. In particular, if M i is the i th columnof M , for every i , . . . , i k ´ ℓ ď n ´ k ` M i ` ¨ ¨ ¨ ` M i k ´ ℓ ` M n ´ ℓ ` ` ¨ ¨ ¨ ` M n “ M n ´ ℓ ` ` ¨ ¨ ¨ ` M n and therefore M i ` ¨ ¨ ¨ ` M i k ´ ℓ “
0. That is, the sum of every less or equal k columns of the first n ´ k ` M is not equal to zero. We then show in Lemma 10 that such matrix has atleast q “ Ω p k log n q rows. This implies that q “ Ω pp k { log k q log n q .Let π p n, k q be the minimum integer q such that there exists a q ˆ n matrix over F that thesum of any of its less than or equal k columns is not 0. We have proved10 emma 9. Any non-adaptive uniform-distribution one-sided { -tester for k -Linear must make atleast Ω p π p n ´ k ` , k q{ log k q queries. Now to show that Ω p π p n ´ k ` , k q{ log k q “ Ω p k log n q we prove the following result. Thislemma follows from Hamming’s bound in coding theory. We give the proof for completeness Lemma 10. (Hamming’s Bound) We have π p n, k q ě log t k u ÿ i “ ˆ ni ˙ “ Ω p k log p n { k qq . Proof.
Let M be a π p n, k qˆ n matrix over F that the sum of any of its less than or equal k columnsis not 0. Let m “ t k { u and S “ t M i ` ¨ ¨ ¨ ` M i t | t ď m and 1 ď i ă ¨ ¨ ¨ ă i t ď n u Ď t , u π p n,k q be a multiset. The strings in S are distinct because, if for the contrary, we have two strings in S that satisfies M i ` ¨ ¨ ¨ ` M i t “ M j ` ¨ ¨ ¨ ` M j t then M i ` ¨ ¨ ¨ ` M i t ` M j ` ¨ ¨ ¨ ` M j t “ t ` t ď k , which is a contradiction. Therefore, 2 π p n,k q ě | S | “ ř mi “ ` ni ˘ and π p n, k q ě log | S | . For the lower bound for adaptive testers we take C “ t g p ℓ q u for some ℓ P t , , . . . , k ´ u and getan adaptive algorithm A that makes q queries and satisfies1. If f P k -Linear then with probability 1, A p s, f q accepts.2. If f “ g p ℓ q then, with probability at least 2 { A p s, f q rejects.This implies that there exists a deterministic adaptive algorithm B “ A p s , ˚q that makes q queriessuch that1. If f P k -Linear then B p f q accepts.2. If f “ g p ℓ q then B p f q rejects.Then, by the same argument as in the case of non-adaptive tester, we get a q ˆ n matrix M that thesum of every k ´ ℓ columns of the first n ´ ℓ columns of M is not zero. Let Π p n, k q be the minimuminteger q such that there exists a q ˆ n matrix over F that the sum of any of its k columns is not 0.Then, we have proved that Lemma 11.
Any adaptive uniform-distribution one-sided { -tester for k -Linear must make atleast Ω p max ď ℓ ď k Π p n ´ k, ℓ qq queries. In the next subsection, we show that there exists 1 ď ℓ ď k such that Π p n, ℓ q “ ˜Ω p? k log n q . Π In this section we prove
Lemma 12.
We have max ď ℓ ď k Π p n, ℓ q “ ˜Ω p? k log n q . L an L - good matrix M is a matrixthat for every ℓ P L the sum of every ℓ columns of M is not zero. A k -good matrix is a t k u -goodmatrix. We say that the matrix M is almost L -good if there is a “small” number ( poly p k q ) ofcolumns of M that can be removed to get an L -good matrix. The concatenation M ˝ M (thematrix that contains the rows of both matrices) of almost L -good matrix M with an almost L -good matrix M is an almost L Y L -good matrix.Let K “ t ? k {p k q u and r K s “ t , , . . . , K u . The idea of the proof is to construct an almost r K s -good matrix M by concatenating t “ O p log k q matrices M ˝ M ˝ ¨ ¨ ¨ ˝ M t where M i is k i -good p Π p n, k i q ˆ n q -matrices for some k i ď k . Then after removing small number ( poly p k q ) columns of M we get a r K s -good matrix M with ř ti “ Π p n, k i q rows and n ´ poly p k q columns. By Hamming’sbound, Lemma 10, M contains at least Ω p K log n q rows. Therefore, ř ti “ Π p n, k i q “ Ω p K log n q .So there is i such that Π p n, k i q “ Ω p K log n { log k q “ Ω p? k log n { log k q “ ˜Ω p? k log n q .We now give more intuition to how to construct an almost r K s -good matrix from k i -goodmatrices. Denote by N d “ t i : d ffl i u X r K s . Let k “ k . We first show that if M is k -goodmatrix then there exists a set of integers L Ď r K s such that M is almost L -good matrix and d : “ gcd pr K sz L q ffl k . The intuition is that if, for the contrary, there are many pairwise disjointsets of columns that sum to 0 that the great common divisor of their sizes divides k , then the unionof some of them gives k -set of columns that sum to 0 and then we get a contradiction. Therefore d “ L Ě N d and M is almost N d -good. We then take k : “ d t k { d u and a k -goodΠ p n, k q ˆ n matrix M . Then, as before, M is almost N d -good matrix with d ffl k . Therefore, d ffl d . Now the concatenation of both matrices M ˝ M is almost N d Y N d “ N lcm p d ,d q .Since d ffl d we must have d : “ lcm p d , d q ě d . We then take k “ d t k { d u and a k -goodΠ p n, k q ˆ n matrix M and concatenate it with M ˝ M to get an almost N lcm p d ,d ,d q -goodmatrix with lcm p d , d , d q ě d “ p d , d q ě d . After, t “ O p log k q iterations, we get a p ř ti “ Π p n, k i qq ˆ n matrix M ˝ M ˝ ¨ ¨ ¨ ˝ M t that is almost N d -good for some d ě t d ą K andtherefore, almost r K s -good.We note here that we can get the bound Ω p? k p log log k q log n { log k q by choosing k “ lcm p , , , ¨ ¨ ¨ , m i q ď k , and then k i “ d i ´ lcm p , , , ¨ ¨ ¨ , m i q ă k where m i “ O p log p k qq . See [20].We now give the full proof. We start with some preliminary results, Lemmas 13-18. Lemma 13.
Let W Ď r m s and w “ gcd p W q . There is a subset W Ď W of size O ˆ log mw log log mw ˙ ă log mw such that gcd p W q “ gcd p W q .Proof. Define the set D “ W { w “ t b { w | b P W u . Then D Ď r t m { w u s and gcd p D q “
1. Let D Ď D be a minimum size set with gcd p D q “ W “ wD Ď W . Let D “ t d , . . . , d t u and g i “ gcd p D zt d i uq for i “ , . . . , t . Since D is minimum g i ą
1. We also have for i “ j ,1 “ gcd p D q “ gcd p gcd p D zt d i uq , gcd p D zt d j uqq “ gcd p g i , g j q and therefore g , . . . , g t are pairwise relatively prime. Since for all i ą g i “ gcd p D zt d i uq| d wehave ś ti “ g i | d . Therefore, t m { w u ě d ě ś ti “ g i ě ś ti “ i “ t ! “ | D | ! “ | W | ! and the resultfollows. Lemma 14.
Let d, d , k, y ě be integers that satisfy d | y, d | k, d | d and gcd p y, d q “ d . There is ď λ ă d { d such that d |p k ´ λy q . roof. Let ˆ y “ y { d, ˆ k “ k { d and ˆ d “ d { d . Then gcd p ˆ y, ˆ d q “
1. Consider the set B “ t ˆ k ´ i ˆ y | i “ , . . . , ˆ d ´ u . If for 0 ď i ă i ď ˆ d ´ k ´ i ˆ y “ p ˆ k ´ i ˆ y mod ˆ d q then p i ´ i q ˆ y “ p d q . Since gcd p ˆ y, ˆ d q “ i “ p i mod ˆ d q and therefore i “ i . This shows that theelements in B are distinct modulo ˆ d and therefore there is 0 ď λ ă ˆ d “ d { d such that ˆ k ´ λ ˆ y “ p d q . Then k ´ λy “ p d q . Lemma 15.
Let k be an integer. Let J “ t j , . . . , j ℓ u be a set of integers such that ď j , . . . , j ℓ ď? k { ℓ and d : “ gcd p j , . . . , j ℓ q| k . There exist non-negative integers ď λ , . . . , λ ℓ ´ ď ? k and ď λ ℓ ď k such that λ j ` λ j ` ¨ ¨ ¨ ` λ ℓ j ℓ “ k. Proof.
We prove the result by induction on ℓ . For ℓ “
1, given J “ t j u , 1 ď j ď ? k and d “ j | k we let λ “ k { d . Then λ ď k and λ j “ k .Assume that the result is true for ℓ ´
1. We prove the result for ℓ .Given d : “ gcd p j , . . . , j ℓ q| k . Let d “ gcd p j , . . . , j ℓ q . We have two cases: d “ d and d ą d .If d “ d then d | k and for i ą j i ď ? k { ℓ ď ? k {p ℓ ´ q . By the induction hypothesis there are0 ď λ , . . . , λ ℓ ´ ď ? k and 0 ď λ ℓ ď k such that λ j ` λ j ` ¨ ¨ ¨ ` λ ℓ j ℓ “ k. We choose λ “ d ą d . We have d | j , d | k , d | d and gcd p j , d q “ d . By Lemma 14, there is λ such that 0 ď λ ă d { d and d | k : “ k ´ λ j . Since λ ă d { d ď j ď ? k , we also have k “ k ´ λ j ě k ´ ? k ? k { ℓ “ ℓ ´ ℓ k and therefore j , . . . , j ℓ ď ? k { ℓ ď ? k {p ℓ ´ q . Since d | k , by the induction hypothesis there exist0 ď λ , . . . , λ ℓ ´ ď ? k ď ? k and 0 ď λ ℓ ď k ă k such that λ j ` λ j ` ¨ ¨ ¨ ` λ ℓ j ℓ “ k . Then λ j ` λ j ` ¨ ¨ ¨ ` λ ℓ j ℓ “ k. Let M be a q ˆ n binary matrix. Recall that M i is the i th column of M . For every j ě
1, let ℓ j p M q denotes the maximum number of disjoint j -subsets A , A , . . . of r n s such that ř j P A i M j “ i . We say that M is p j, ℓ q - good if ℓ j p M q ď ℓ and p j, ℓ q - bad if it is not p j, ℓ q -good, i.e., ℓ j p M q ą ℓ . For L, J
Ď r n s , we say that M is p L, ℓ q -good if it is p j, ℓ q -good for all j P L and p J, ℓ q -bad if it is p g, ℓ q -bad for all j P J . When ℓ “ j -good, L -good, j -bad and J -bad.For two q ˆ n and q ˆ n matrices M and M , respectively, the concatenation of M and M is M ˝ M “ r M ˚ | M s ˚ where ˚ is the transpose of a matrix. That is, M ˝ M is the p q ` q q ˆ n matrix that results from the rows of M follows by the rows of M .The following result is obvious Lemma 16. If M is p L, ℓ q -good and M is p L , ℓ q -good then M ˝ M is p L Y L , ℓ q -good. Lemma 17.
Let M be a q ˆ n matrix. If M is pr d s , ℓ q -good then q “ Ω p d log pp n ´ p ℓd { qq{ d qq .Proof. For every j P r d s we have ℓ j p M q ď ℓ . That is, for every j , there are at most ℓ disjoint j -setsof columns that sum to zero. We remove those columns (for all j P r d s ) and get a pr d s , q -goodmatrix. The number of columns that are removed is at most ř dj “ ℓj ď ℓd {
2. Using Hamming’sbound, Lemma 10, the result follows.We now prove 13 emma 18.
Let m , q , w and t “ mqw be integers. Let J “ t j , . . . , j w u Ď r m s . Let M be a p J, t q -bad matrix. Then for any λ , . . . , λ w P r q s we have that M is p λ j ` ¨ ¨ ¨ ` λ w j w q -bad.Proof. Let r “ λ j ` ¨ ¨ ¨ ` λ w j w . We need to show that there are r columns of M that sum to 0.Since M is p j , t q -bad and λ ď t , there are λ pairwise disjoint j -sets A , , A , , ¨ ¨ ¨ , A ,λ suchthat ř j P A ,i M j “ i P r λ s . Since M is p j , t q -bad and λ ď t ´ λ j , there are λ pairwisedisjoint j -sets A , , A , , ¨ ¨ ¨ , A ,λ sets that are also pairwise disjoint with A , , A , , ¨ ¨ ¨ , A ,λ such that ř j P A ,i M j “ i P r λ s . We continue with this procedure until we find a collection A of disjoint sets that contains, for every i ď w ´ λ i j i -sets that corresponds to columns of M that sum to 0. Now since λ w ď t ´ p λ j ` ¨ ¨ ¨ ` λ w ´ j w ´ q , there are λ w pairwise disjoint j w -sets A w, , A w, , ¨ ¨ ¨ , A w,λ w sets that are also pairwise disjoint with all the sets in A such that ř j P A w,i M j “ i P r λ w s . Let A “ A Y t A w,i | i P r λ w su . Obviously, | Y A | “ λ j ` ¨ ¨ ¨ ` λ w j w and ř j PY A M j “ k -good matrix M is p J, poly p k qq -bad then gcd p J q ffl k . Lemma 19.
Let K “ t ? k {p k q u , κ “ k . , J Ď r K s and k { ď k ď k . Let M be a matrix thatis k -good and p J, κ q -bad. Then gcd p J q ffl k .Proof. Let d “ gcd p J q and suppose, for the contrary, that d | k . By Lemma 13, there is J Ď J ofsize w : “ | J | ď log p K { d q ă log k such that d “ gcd p J q . Let J “ t j , . . . , j w u . By Lemma 15,there exist 0 ď λ , . . . , λ w ď k such that λ j ` ¨ ¨ ¨ ` λ w j w “ k . By Lemma 18, M is k -bad. Acontradiction.Let K “ t ? k {p k q u and κ “ k . . Let N d be the set of integers in r K s that are not divisibleby d . Lemma 20.
Let J be the maximum subset of r K s such that M is p J, κ q -bad. Then M is p N gcd p J q , κ q -good.Proof. Since J is the maximum set, M is pr K sz J, κ q -good. Since J Ď r K sz N gcd p J q we have r K sz J Ě N gcd p J q and therefore M is p N gcd p J q , κ q -good.We now show how to construct from a p N d , κ q -good matrix a p N d , κ q -good matrix with d ě d . Lemma 21.
Let M be a q ˆ n matrix that is p N d , κ q -good. There exist k ď k , q “ q ` Π p k , n q , d ě d and a q ˆ n matrix M that is p N d , κ q -good.Proof. Consider k “ d t k { d u and let ˆ M be a Π p n, k q ˆ n matrix that is k -good. Let J be themaximum subset of r K s such that ˆ M is p J , κ q -bad. By Lemma 19, gcd p J q ffl k “ d t k { d u andtherefore gcd p J q ffl d . By Lemma 20, ˆ M is p N gcd p J q , κ q -good. Define M “ M ˝ ˆ M .First, the number of rows of M is q “ q ` Π p k , n q . Now, by Lemma 16, M is p N gcd p J q Y N d , κ q -good. Since N gcd p J q Y N d “ N d for d “ lcm p gcd p J q , d q we have that M is p N d , κ q -good. Sincegcd p J q ffl d we have d “ lcm p gcd p J q , d q ě d . This implies the result.We are ready now to prove the final result Lemma 22.
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