Throughout fix a field Let be finite non-empty subsets of and let denote the grid We use to denote Given functions , we use to denote the Hamming distance between and , i.e. the number of points where they differ.
Let denote the set with the natural partial order For each we denote by the set and by the set For any , we will identify the monomial with and use the monomial notation and the multi-index notation interchangeably.
The following fact is standard.
Each has a unique representation as a polynomial where the degree of in is at most for each . Equivalently, there is a natural one-one correspondence between the space of all functions from to and
Given a downset we associate with it the linear code , called a downset code, defined by
The following lemma allows us to compute the minimum distance for any downset Recall (see e.g. [CLO07]) that a monomial order on monomials in is a total ordering of the monomials that is a well order and moreover satisfies the following for any :
Lemma 2 (Schwartz-Zippel Lemma for ).
Let be arbitrary and let be the leading monomial of w.r.t. a monomial order. Then,
For each there is an such that and can be represented by a linear combination of monomials from In particular, if then such an
Thus, In particular, given it can be found in polynomial time.
Item 1 is an easy consequence of the proof of Macaulay’s theorem [Mac27] (see also [CLO07, Chapter 5.3, Proposition 4]). For completeness, we present a short proof here. Given any polynomial , let denote the set of monomials with non-zero coefficient in . For any set of monomials let denote the set of monomials that divide some monomial in
For any , let Note that is a univariate polynomial of degree that vanishes on . Given any polynomial , the remainder obtained upon dividing by has degree in the variable and evaluates to the same value as at points in . Further, each monomial in divides some monomial in (i.e. ) . Repeating this process, we eventually obtain a polynomial representing the same function as .
Let be the subset of points in where To prove item 1 of the lemma, it suffices to show that every can be represented as a polynomial from Standard linear algebra then implies that which proves item 1.
To prove the above, fix any . By extending in an arbitrary way to , we know that can be represented by some polynomial If does not contain any monomial from then we are done. Otherwise, we choose the largest (w.r.t. ) monomial . Let be the coefficient of in .
Assume that where . Multiplying by , we see that the polynomial vanishes on . Note that is a polynomial such that that also represents the function at points in . Repeating this process, we eventually obtain a polynomial without any monomials from that represents . The polynomial (obtained by dividing by as mentioned above) also represents and furthermore is an element of
For item 2, assume that for each and consider ∎
Let denote with its natural partial order. Let denote the set
Given a downset as above, let For define
The following observation will be useful.
Let be any downset, and let
For each is a downset in Further, we have
For each we have
Clear from the definition, since is a downset.
Let and have weight equal to , and and have weight equal to . Let be any monomial in and be any monomial in . So and . Then by Item 1, and so , that is, . Thus, and hence, has weight at least . So we see that
(Note that the quantity is the distance of the degree- Reed-Solomon code on the set )
As in the result of Kim and Kopparty, we work with the more general problem of decoding Weighted functions (or weighted received word). A weighted function over is a function or equivalently a pair where and Given a weighted function and a function , we define their distance by
A (unweighted) function is identified with the weighted function where is the identically zero function. Note that with this identification, agrees with the standard Hamming distance between and . In particular, the unique decoding problem for immediately reduces to the problem of finding a codeword of distance less than from a given weighted function.
We will also need the following lemma about weighted codewords of a downset code . The proof (in a more general setting) can be found in [KK17, Lemma 2.1].
Assume that are distinct.333In the algorithm, we will only need this for , i.e. for the Reed-Solomon code. Let be any weighted received word. Then, In particular, both and cannot be at distance strictly less than from .
2 The main theorem
is a finite grid in with . We have .
is a downset.
is a weighted received word.
There is a deterministic polynomial time algorithm that given, and a weighted codeword produces a codeword such that if one exists. (If no such codeword exists, the algorithm outputs some arbitrary polynomial.)
For the case when , a result of Forney [Jr.66] yields a deterministic polynomial-time algorithm for this problem. We call this algorithm WeightedRSDecoder and refer the reader to [Jr.66] or [KK17] for a description.
The algorithm is specified as WeightedDownsetDecoder below (Algorithm 1). We prove its correctness by induction on .
For the base case , we simply use the algorithm WeightedRSDecoder and so there is nothing to prove.
Now we assume the correctness of WeightedDownsetDecoder algorithm for indeterminates.
Let be a received weighted word. Suppose there is a with . We can write
where as in Algorithm 1.
We show, by induction on , that the algorithm correctly decodes the polynomial In other words, for each the polynomial computed by Algorithm 1 is the same as the polynomial
Fix . Assume that the algorithm has correctly decoded for each . Let . Note that , for all .
To show that it is enough to show that where is as computed by the algorithm. Then, the induction hypothesis implies that .
Define as in the algorithm by
By induction, we know that for . Hence, we observe that
We now analyze Recall that we have
Fix some . Define Also, define
We claim that
We prove (1) by a case analysis.
This implies that In particular, this implies that since otherwise the Reed-Solomon decoder would have returned instead of This immediately implies (1).
So from now we will assume that By Lemma 4, it follows that
In this case, we immediately have
As in the previous case, we have . But as we have . Thus, by Lemma 4, it follows that . Hence
This completes the proof. ∎
The authors are grateful to Swastik Kopparty and Madhu Sudan for their helpful comments and encouragement.
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