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Let integer [MATH] be maximal such that there are at least [MATH] distinct elements of [MATH] with evaluations that are in [MATH] Then by Lemma 1.4 we have [MATH] But since [MATH] can be spanned by [MATH] vectors, we have |
[EQUATION] By Proposition 2.2 (a), we have [EQUATION] So [MATH] There are [MATH] distinct evaluations and fewer than [MATH] of them fall into [MATH] So a uniform random evaluation is in [MATH] with probability |
[MATH] Since the [MATH] other columns of [MATH] are chosen uniformly and independently, the probability that these [MATH] columns span the whole matrix is at most |
[EQUATION] since [MATH] for some [MATH] to be chosen later. Since [MATH] , we have [MATH] and we can apply Proposition 2.2 to get that |
[EQUATION] for some [MATH] Therefore, by a union bound over all choices of [MATH] columns we have [EQUATION] Note that [MATH] so we have |
[EQUATION] Note that for any constant [MATH] [MATH] is [MATH] Therefore, for fixed constant [MATH] we can choose a sufficiently large [MATH] such that |
[EQUATION] for some constant [MATH] 2.1 Proof of Proposition 2.2 We first give basic inequalities regarding [MATH] that are independent of the choice of [MATH] |
Proposition 2.5 For [MATH] [EQUATION] Proof. It is well known that there are [MATH] non-negative integer solutions to the equation [MATH] Thus by iterating degrees we have |
[EQUATION] On the other hand, if we only consider multi-linear terms, we will get [EQUATION] We now prove part (a): For [MATH] where [MATH] let [MATH] denote the set of monomials of the form [MATH] |
[MATH] Then we have [MATH] Therefore [EQUATION] Now, for fixed [MATH] consider the following process to generate elements in [MATH] we first choose [MATH] elements [MATH] from [MATH] then apply permutation [MATH] over [MATH] to get [MATH] We claim that this process generate each monomial in [MATH] equally many times, i... |
[MATH] can be [MATH] if and only if [MATH] can be [MATH] Moreover, the number of occurrence for each monomial is precisely the number of satisfying permutations, hence only depends on [MATH] Therefore, we have |
[EQUATION] This quantity is a decreasing function of [MATH] Hence we have [EQUATION] Since [MATH] we have [MATH] Therefore for sufficiently large [MATH] |
[EQUATION] We now prove part (b): Define [MATH] We will obtain the inequalities by bounding [MATH] We first bound [MATH] in terms of [MATH] Clearly for each [MATH] there are at most [MATH] choices of [MATH] to yield [MATH] so [MATH] On the other hand, any [MATH] that does not have degree |
[MATH] in [MATH] can be chosen. There are at most [MATH] variables in [MATH] having degree [MATH] so we can choose at least [MATH] |
variables [MATH] since [MATH] This gives us [MATH] We now bound [MATH] in terms of [MATH] Each [MATH] contains at most [MATH] distinct variables, hence [MATH] |
It immediately follows that [MATH] and hence we can choose [MATH] To lower bound [MATH] in terms of [MATH] we show that a large portion of monomials contain many distinct variables and hence each |
[MATH] can be associated with many different [MATH] We first bound the number of monomials that have degree at most [MATH] and are composed of at most [MATH] distinct variables. We can generate such monomials by first choosing [MATH] variables, then using these variables to form a monomial of degree [MATH] and so we ca... |
[MATH] For sufficiently small [MATH] we can argue that this is a small fraction of [MATH] Suppose that [MATH] . Since by hypothesis, [MATH] , we have |
[MATH] and [EQUATION] Summing over all values of [MATH] we obtain that a total fraction at most 3/4 of all monomials in [MATH] have at most [MATH] distinct variables. Therefore, since at least 1/4 of [MATH] contain at least |
[MATH] distinct variables, it must be the case that [MATH] Since [MATH] , we obtain that [MATH] . Hence we derive (b) with [MATH] This completes the proof of Proposition 2.2 (b). |
2.2 Lower bound on the likelihood of bias We now prove Proposition 1.3 on the limits on the extent to which Theorem 1.1 can be improved. The argument is analogous to that of |
for the case of [MATH] Proof of Proposition 1.3 We follow the same division of variables [MATH] into parts [MATH] and [MATH] with [MATH] and |
[MATH] and [MATH] that we used for the upper bound on the bias. Define [MATH] to be the set of all polynomials in [MATH] whose monomials are from the set |
[MATH] (defined earlier) that have degree 1 on [MATH] and degree at most [MATH] on [MATH] By Corollary 2.3 , there is some constant |
[MATH] such that for sufficiently large [MATH] [MATH] and hence [MATH] Therefore, we have [MATH] Now consider the expected bias of polynomials in [MATH] We can write [MATH] chosen uniformly from [MATH] uniquely as |
[EQUATION] where the [MATH] are independently chosen polynomials over monomials [MATH] on [MATH] For [MATH] [EQUATION] Now with probability [MATH] , all the [MATH] for [MATH] are 0 and every [MATH] evaluates to 0, so [MATH] With the remaining probability, [MATH] and hence there is some |
[MATH] and [MATH] such that [MATH] For [MATH] chosen at random from [MATH] , for each fixed value of [MATH] with [MATH] , we have |
[EQUATION] where [MATH] is the constant term of the polynomial [MATH] and is chosen independently of [MATH] . Since [MATH] is uniformly chosen from |
[MATH] for random [MATH] in [MATH] and since [MATH] [MATH] is also uniformly chosen from [MATH] Further, since [MATH] is independent of [MATH] , for every fixed [MATH] the value [MATH] Therefore, |
[MATH] Now since [MATH] we obtain [EQUATION] Therefore [EQUATION] for some [MATH] since [MATH] Since [MATH] , we obtain [MATH] for some constant [MATH] as required. |
Extremal rank properties of truncated Reed-Muller codes In this section we prove Lemma 1.4 Let [MATH] be the natural generating matrix of the [MATH] Reed-Muller code over the field [MATH] for [MATH] an odd prime. That is, |
[MATH] is a [MATH] matrix, where each row is indexed by [MATH] and each column is indexed by a monomial [MATH] and [MATH] Clearly the rank of [MATH] is [MATH] since all the columns are independent. For [MATH] define [MATH] as [MATH] restricted on rows in [MATH] |
For fixed [MATH] [MATH] is precisely the column corresponding to [MATH] in [MATH] restricted to rows indexed by [MATH] . From this prospective, the dimension of the subspace spanned by [MATH] is exactly the (column) rank of the matrix [MATH] So we can restate Lemma 1.4 as the following: |
Lemma 3.1 Let [MATH] be a subset of [MATH] such that [MATH] then [MATH] As noted in the introduction we will derive the more general bound for an arbitrary value of [MATH] , not only for the restricted case that [MATH] that occurs in the above lemma. We start with the special case where [MATH] contains the lexicographi... |
[MATH] if and only if there exists [MATH] such that [MATH] for [MATH] and [MATH] .) We will show that this is in fact an extremal case. |
For integer [MATH] let [MATH] be the subset that contains the smallest [MATH] elements. Define [MATH] as the rank of [MATH] For the completeness of definition, we set [MATH] when [MATH] or [MATH] When [MATH] is a power of [MATH] it is easy to compute the value of [MATH] |
Lemma 3.2 [MATH] Proof. Let [MATH] be a monomial of degree at most [MATH] over [MATH] Notice that [MATH] [MATH] So if [MATH] contains any of the first [MATH] variables, the column indexed by [MATH] will be [MATH] On other other hand, all the columns corresponding to monomials over [MATH] of degree at most [MATH] are li... |
It is more complicated to compute the value for general [MATH] With careful observation, we have the following recursion. Lemma 3.3 |
Let [MATH] be the unique integer so that [MATH] Let [MATH] Then [EQUATION] As a special case, when [MATH] , we have [MATH] Proof. |
For the sake of convenience, let [MATH] be the set of monomials over the last [MATH] variables whose degree equals [MATH] and [MATH] be the set of monomials over the last [MATH] variables whose degree is at most [MATH] |
Consider the block structure of the matrix. Let [MATH] be the submatrix that takes [MATH] as rows and [MATH] as columns. For [MATH] let [MATH] be the submatrix that takes [MATH] as rows and [MATH] as columns. Let [MATH] be the blocks [MATH] then [MATH] is precisely the submatrix that takes [MATH] as rows and [MATH] as ... |
Now let us consider the rows for [MATH] for [MATH] The non-zero parts correspond to monomials that only depend on [MATH] If we group all the monomials by their degree on [MATH] |
then, for [MATH] the row will be of the form [EQUATION] Things are a little different for [MATH] since [MATH] In this case, we define [MATH] as the first [MATH] rows of [MATH] and it is easy to check that the row is of the form |
[EQUATION] Therefore the matrix [MATH] is of the form: [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] |
[MATH] [MATH] [MATH] [MATH] [MATH] We have two important observations: [MATH] is the first [MATH] columns of [MATH] , and [MATH] is the first [MATH] rows of [MATH] |
Therefore, we can apply Gaussian elimination to turn the matrix in to a block-diagonal matrix. We do this in two steps. The following algorithm first eliminates on columns to obtain a triangular matrix, based on the first observation. |
Initialize [MATH] [MATH] For [MATH] set [MATH] /* [MATH] is the coefficient of each block at the beginning. */ for [MATH] do for |
[MATH] do /* Subtract [MATH] times the prefix of length [MATH] of [MATH] -th column block from [MATH] -th column block. */ for [MATH] |
do [MATH] Algorithm 1 Triangular elimination We have the following properties. Claim 3.4 At the beginning of round [MATH] of Algorithm for all [MATH] , we have |
[MATH] and at the end of round [MATH] for all [MATH] , we have [MATH] for [MATH] Proof. Let us consider the function [MATH] which denotes the coefficient of the [MATH] block at the beginning of round [MATH] We prove a strengthened claim: For [MATH] |
[MATH] is a monic degree [MATH] polynomial on variable [MATH] [MATH] for [MATH] , and when [MATH] [MATH] for [MATH] and [MATH] When [MATH] we have [MATH] which is monic and has degree [MATH] Also, |
[MATH] So the base case is true. The update rule given by the algorithm says [EQUATION] When [MATH] , clearly [MATH] is still a monic degree [MATH] polynomial. When [MATH] by the induction hypothesis, |
[MATH] and the degree of [MATH] is strictly smaller than [MATH] Therefore we have that [MATH] is also a monic degree [MATH] polynomial. |
For [MATH] by the induction hypothesis, [MATH] for [MATH] Then both update rules give us [MATH] for [MATH] For the [MATH] case, by Gaussian elimination, |
[MATH] for all [MATH] Combining the two cases we have for [MATH] [MATH] for [MATH] Notice that [EQUATION] Since [MATH] is monic and has degree [MATH] it can have at most [MATH] roots. This implies that [MATH] cannot be its root, hence [MATH] |
Using this claim we complete the proof of Lemma 3.3 After the first step, the matrix is in the form: [MATH] [MATH] [MATH] [MATH] |
[MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] with all other columns having value 0. Now, the second observation says that [MATH] is a submatrix of |
[MATH] Moreover, we have [MATH] for [MATH] So we can eliminate row by row to get a diagonal matrix, whose rank is very easy to compute. The rank of [MATH] is simply [MATH] By the definition of the [MATH] function, the rank of [MATH] is just [MATH] Hence |
[EQUATION] Intuitively, for any set [MATH] of size [MATH] the rank of [MATH] will be larger than that of [MATH] since [MATH] is the most compact way to arrange [MATH] rows. Formally stated, this is the following restatement of Theorem 1.5 |
Theorem 3.5 For arbitrary [MATH] with [MATH] and for any [MATH] [EQUATION] Note that [MATH] does not depend on [MATH] Indeed, all we need of [MATH] is that [MATH] so that the matrix |
[MATH] has at least [MATH] rows. Before we actually prove Theorem 3.5 we first argue that this is all we need to prove Lemma 3.1 Indeed, we can simply set [MATH] then with Lemma 3.2 we just have when [MATH] |
[EQUATION] It turns out that in order to prove Theorem 3.5 it is sufficient to have the following sub-additivity property of the [MATH] function. |
Lemma 3.6 For [MATH] for any [MATH] [EQUATION] Proof of Theorem 3.5 from Lemma 3.6 We use induction on the size of [MATH] . When [MATH] |
[MATH] while [MATH] so the base case is true. Assume that we have proved that for all [MATH] with [MATH] and all degrees [MATH] [MATH] Let [MATH] be the smallest integer so that [MATH] For [MATH] define [MATH] For [MATH] let [MATH] be the submatrix of [MATH] that takes [MATH] as rows and [MATH] as columns. We claim tha... |
[MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] This is because all the other columns can be spanned by this matrix. Indeed, consider a monomial [MATH] where [MATH] is a monomial over [MATH] and [MATH] is over [MATH] Since for all [MATH] [MATH] is fixed fo... |
We also have one important observation about this matrix. For all [MATH] [MATH] is the first [MATH] columns of [MATH] so we can again use Gaussian elimination. Although we may not be able to get a diagonal matrix because we do not know the relationship between the [MATH] we can carefully modify algorithm to obtain a tr... |
Initialize [MATH] [MATH] For [MATH] set [MATH] /* [MATH] is the coefficient of each block at the beginning. */ while [MATH] do [MATH] |
for [MATH] do /* Subtract [MATH] times [MATH] -th column block from [MATH] -th column block */ for [MATH] do [MATH] [MATH] 10 [MATH] |
Algorithm 2 Triangular elimination revised Let [MATH] be the order of indices that Algorithm uses. Then we have the following properties. |
Claim 3.7 At the end of Algorithm 2 we have [MATH] for [MATH] and [MATH] The proof of Claim 3.7 is very similar to that of Claim 3.4 and we omit it here. After the elimination, the matrix is of the form: |
[MATH] Since it is a triangular matrix, we can lower bound its rank as [EQUATION] Recall that columns of [MATH] are [MATH] so [MATH] is actually [MATH] Since [MATH] we can apply the induction hypothesis to get |
[EQUATION] Now we can apply Lemma 3.6 to get [EQUATION] as required. Therefore to complete our proof of the extremal rank properties of these matrices in Theorem 3.5 , and hence Theoremthm:main-extremal and Lemma 1.4 , it only remains to prove the sub-additivity property of Lemma 3.6 for the [MATH] function. |
Properties of the [MATH] function In this section, we generalized the [MATH] function and make some observations about its properties. We then give the proof of Lemma 3.6 in the next section. |
From Theorem 3.3 we can see that it is very easy to compute [MATH] if we write [MATH] in base [MATH] It will be convenient to consider a more general class of functions in other bases. |
Definition 4.1 Fix [MATH] and [MATH] Let [MATH] be a function defined as [MATH] when [MATH] or [MATH] [MATH] when [MATH] [MATH] and for other cases, let [MATH] be the largest integer so that [MATH] then |
[EQUATION] We can verify that [MATH] defined in last section is the same as [MATH] Proposition 4.2 [MATH] Proof. When [MATH] or [MATH] by definition of [MATH] , we have |
[MATH] Also, when [MATH] by Theorem 3.2 , we have [MATH] Next, we claim that [MATH] Since [MATH] we can split all monomials over [MATH] of degree at most [MATH] by their degree in [MATH] That is, |
[EQUATION] which precisely yields the claim. Finally, for input [MATH] not of the above forms, let [MATH] be the largest integer so that [MATH] then Theorem 3.3 implies that |
[EQUATION] So, [MATH] satisfies all the requirements for [MATH] Notice that for fixed [MATH] Definition 4.1 uniquely determines a function over [MATH] so we have [MATH] |
We prove the following important property of the [MATH] function that we will repeatedly apply later. Proposition 4.3 For [MATH] and [MATH] |
[EQUATION] Proof. We first write down [MATH] as a function of [MATH] We represent [MATH] and [MATH] , respectively, in base [MATH] as |
[EQUATION] With this representation, by Definition 4.1 we can explicitly compute [EQUATION] Let [MATH] be the largest integer so that [MATH] Notice that |
[EQUATION] When [MATH] [MATH] so we have [MATH] Hence [EQUATION] We consider two cases based on the value of [MATH] Case [MATH] In this case [MATH] and ( ) is simply |
[EQUATION] Notice that [MATH] so this is an increasing function of [MATH] Hence we have [MATH] as required. Case [MATH] Then, it must be the case that [MATH] for [MATH] Therefore, using the fact that [MATH] for all [MATH] ) is |
[EQUATION] Define [MATH] Then, we can verify that for [MATH] [EQUATION] We claim that [MATH] Indeed, by Definition 4.1 [EQUATION] |
By Proposition 4.2 we have [EQUATION] Comparing the two summations, and using the fact [MATH] we get the claim. So, we can repeatedly apply the claim to get |
[EQUATION] This is increasing in [MATH] so we obtain the required inequality. By telescoping Lemma 4.3 we have Corollary 4.4 For [MATH] [MATH] , for any [MATH] we have |
[EQUATION] Moreover, if we set [MATH] then we can see that for fixed [MATH] [MATH] is monotone in [MATH] Corollary 4.5 For [MATH] , for any [MATH] we have |
[MATH] Proof of Lemma 3.6 In this section we prove the following generalized version of Lemma 3.6 that applies to the generalization [MATH] of [MATH] given in the previous section. |
Lemma 5.1 For any integer [MATH] for integers [MATH] , and for any integer [MATH] [EQUATION] Proof. Define [MATH] as [MATH] For [MATH] define the value of [MATH] of degree [MATH] as |
[MATH] With these notations, Lemma 5.1 can be stated as [EQUATION] We define a total order [MATH] on [MATH] as [MATH] iff either |
[MATH] or [MATH] and [MATH] is lexicographically larger than [MATH] Here we say that [MATH] is lexicographically larger than [MATH] if there exists and [MATH] so that [MATH] for [MATH] and [MATH] |
We prove Lemma 5.1 by induction on this order [MATH] over [MATH] Induction Hypothesis: For all [MATH] with [MATH] [EQUATION] In order to prove the inductive step we divide the proof into four cases depending of the properties of [MATH] |
We divide [MATH] into 4 categories in which we prove the induction step [MATH] with different methods. Define [MATH] CASE I [MATH] |
The set [MATH] is a special class of vectors for which we can directly show the [MATH] without using the induction hypothesis. Indeed, for [MATH] by repeatedly applying Definition 4.1 , we have |
[EQUATION] So we are done if [MATH] For the remaining cases, we prove the inductive step by showing the following claim: Claim 5.2 |
If [MATH] then [MATH] so that [MATH] [MATH] and [MATH] Together with ( ), we have [EQUATION] which is precisely what we need. Our proof is algorithmic, in the sense that we actually provide algorithms to construct [MATH] explicitly. Since we have different operations based on the structure of [MATH] we introduce the st... |
Let [MATH] highest power of [MATH] ) be the largest integer [MATH] so that [MATH] Then for [MATH] we can write [MATH] , where [MATH] and [MATH] We can divide [MATH] into groups so that in each group we have the same [MATH] That is, we divide the interval [MATH] into [MATH] intervals [MATH] where [MATH] |
[MATH] and [MATH] , so that [MATH] [MATH] , and [MATH] for [MATH] Definition 5.3 We define [MATH] as the height of the interval [MATH] An interval [MATH] is called singularized if [MATH] when [MATH] and [MATH] when [MATH] . (That is, at most the first element in the interval has a non-zero remainder.) |
[MATH] is singularized if all its intervals are singularized. (Observe that every [MATH] is singularized.) We show that if [MATH] is not singularized then it can also be improved. |
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