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Quadratic-Solvability

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and d so that Hd = n. associate each point x Hd with a distinct quadratic-polynomial Px { P1, ..., Pn } ... on any x is a linear combination of A's value on Hd ... – PowerPoint PPT presentation

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Title: Quadratic-Solvability


1
Quadratic-Solvability
  • All you need is gap

2
Def Quadratic-Solvability
  • An instance of Quadratic-Solvability ( QS? )
    is a system of quadratic-polynomials P1, , Pn
    over a field ? and variables y1, , ym (where m
    ? n)
  • Problemto decide whether there exists a common
    root to all P1, , Pn
  • Thm GJ QS? is NP-hard (for any field ?)

3
Def gap-QS D, ?, ?
  • An instance a system of quadratic polynomials
    P1, , Pn over a field ?, and variables y1, ,
    ym,where each Pi depends on at most D variables
  • Problemdistinguish between the following two
    cases
  • yes there exists a root common to all P1, , Pn
  • no any assignment to y1, , yn zeros lt ?
    fraction of P1, , Pn

4
Reducing QS? to gap-QS n, ?, 2?-1
  • ThmHPS gap-QS n, ?, 2?-1 is NP-hard As
    long as ? ? nc for some cgt0
  • Proof Given an instance P1, , Pn of QS?
  • set H so that H ?½
  • and d so that Hd n
  • associate each point x ? Hd with a distinct
    quadratic-polynomial Px ? P1, , Pn
  • For any assignment A, denote by Æ’A Hd ? ? the
    function describing each Pis value on A Namely,
    Æ’A(x), on every x ? Hd, equals PxA

5
Extension
  • Consider now LDEÆ’A namely, the
    low-degree-extension of Æ’A to ? ?d
  • As seen in one of the early home assignments,
    LDEÆ’As value on any x ? ? is a linear
    combination of As value on Hd LDEÆ’A(x) ?y
    ? Hd cxy A(y)
  • Let now Px, for every x ? ?, be Px ?y ? Hd
    cxy Py that is, a quadratic-polynomial being
    the corresponding linear combination of P1, , Pn

6
The Resulting QS Instance
  • Pxx ?? is a set of quadratic-polynomials, over
    the same variables y1, , ymwhich for any
    assignment A satisfies the following
  • If A zeros all Pi, then LDE Æ’A is identically
    zero, and therefore A zeros Px for every x ??
  • In case LDE Æ’A is not identically zero, like a
    good LDF it is almost nowhere zero, hence A zeros
    at most a very small (lt r ?½)fraction of Pxx
    ??

7
Our Goal
  • ThmRS,DFKRS QS O(1), ?, 2/? is NP-hardas
    long as log? ? log?n for any constant ? lt 1
    that is, the difference between Cooks
    characterization of NP and PCP characterizations
    of NP (with constant number of variables each
    local-test depends on) boils down to the number
    of variables each polynomial accesses
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