on the degree of univariate polynomials over the integers
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On the Degree of Univariate Polynomials Over the Integers. Gil Cohen Weizmann Institute. Joint work with. Amir Shpilka and Avishay Tal. The Question. The Question. What is the minimal degree of a polynomial of the form ?. Mmm … 0. The Question. What is the minimal degree of - PowerPoint PPT PresentationTRANSCRIPT
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On the Degree of Univariate Polynomials
Over the IntegersGil Cohen
Weizmann Institute
Amir Shpilka and Avishay Tal
Joint work with
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The Question
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What is the minimal degree ofa polynomial of the form
?
The Question
0
1
2
0123 4567Mmm…
0
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What is the minimal degree ofa non-constant polynomial of the
form ?
The Question
0
1
2
0123 4567Mmm…
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What is itto Us?
George Boole Alan Turing
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The degree of a (Boolean multivariate) polynomial is a natural complexity measure for the function it represents [MP68, NS91, Pat92, GR97]
Original Motivation
Related to other complexity measures.
[NS91] Gave a tight lower bound on the degree of a polynomial of the form (assuming dependency in all variables).
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Original Motivation
[GR97] proved an lower bound for symmetric Boolean multivariate polynomials. Can be thought of as univariate polynomials of the form .
[GR97] asked what could be said when the range is ?
Γ (𝑛)=𝑂 (𝑛0 . 525)
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[ST11] Improved lower bounds on the minimal degree of polynomials implies better algorithms for learning symmetric (Boolean!) juntas.
Later Motivation
In general, formal derivatives increase the range, and at this point good lower bounds might be useful.
𝑓 𝑦 (𝑥 )= 𝑓 (𝑥+𝑦 )− 𝑓 (𝑥 )
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Minimal degree .
Observations
When , the minimal degree is (e.g., ).
By the pigeonhole principle, the minimal degree .
[GR97] For , minimal degree .
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Observations
𝑚
minimal degree
1
𝑛−𝑛 .525
𝑛1
[GR97]
𝑓 (𝑥 )=𝑥⌈𝑛+1𝑚+1
⌉𝑛/3
2
/22 𝑛−1
𝑛
What is the Real Behavior?
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Theorem I. the minimal degree is at least .
Main Result I
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Main Result I
𝑚
minimal degree
𝑛
1
𝑛−𝑛 .525
𝑛−𝑛
log log𝑛
𝑛1
𝑛/3
2
/22 𝑛−1
Threshold
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Main Result IIminimaldegree
𝑛
1
𝑛−𝑛
log log𝑛
𝑛1
𝑛−1𝑚?
Hey, I’m over here!
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Main Result II ()
A dichotomous behavior - no intermediate degree.
Theorem II. or
Holds .
I’m listening..Oh, fine by me!
Probably an artifact of the
proof
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Upper Bounds
Best they constructed: degree .
[GR97] Asked for upper bounds on the minimal degree of a non-constant polynomial ?
Program search gave as well.
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Main Result III (Upper Bounds)
Theorem III. For there exists an s.t.
Holds .
Hence, by Theorem II
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Theorem IIngredients
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• Newton polynomials• Lucas theorem• The gap between consecutive primes• Linear recurrence relations
Ingredients for Theorem I
• Newton polynomials• Lucas theorem• The gap between consecutive primes• Linear recurrence relations
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Newton Polynomials
Do polynomials over the integers have integral coefficients?
𝑓 (𝑥 )=− 12𝑥2+
32𝑥+1
This is the right intuition, but the wrong basis!
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Newton Polynomials
For every , define
The set of polynomials
is called the Newton Basis for degree polynomials.
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Newton Polynomials
For a degree polynomial
In our case the ’s are all integers!
𝛾𝑘=∑𝑗=0
𝑘
(−1 )𝑘− 𝑗 ⋅(𝑘𝑗) 𝑓 ( 𝑗 )
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Newton Polynomials
Going back to the example
𝑓 (𝑥 )=3(𝑥2 )−(𝑥1 )+(𝑥0 )
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• Newton polynomials• Lucas theorem• The gap between consecutive primes• Linear recurrence relations
Ingredients for Theorem I
• Newton polynomials• Lucas theorem• The gap between consecutive primes• Linear recurrence relations
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Lucas Theorem
Theorem [Luc78].
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Lucas Theorem
Theorem [Luc1878].
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Lucas Theorem
Theorem [Luc1878]. Let , and be a prime.Denote
Then,
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• Newton polynomials• Lucas theorem• The gap between consecutive primes• Linear recurrence relations
Ingredients for Theorem I
• Newton polynomials• Lucas theorem• The gap between consecutive primes• Linear recurrence relations
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The Gap Between Consecutive Primes
Let be the -th prime number.
What is the asymptotic behavior of ?
Theorem [Cra36]: Assuming Riemann Hypothesis
Conjecture [Cra36]:
Unconditionally [BHP01]:
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• Newton polynomials• Lucas theorem• The gap between consecutive primes• Linear recurrence relations
Ingredients for Theorem I
• Newton polynomials• Lucas theorem• The gap between consecutive primes• Linear recurrence relations
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Linear Recurrence Relations
If has degree , then determine .
Lemma [GR97]: If has degree , then for all
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Linear Recurrence Relations
g (𝑥 )=7 ⋅ f (𝑥+5 )−2 ⋅ 𝑓 (𝑥−1 )+3
A linear recurrence of is a linear combination of shifts of .
Of course, .
Lemma: The degree of a linear recurrence of with summands .
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Theorem IProof of a weaker
version
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Theorem. Let be non-constant. Then .
Proof of Theorem I (weak version)
Proof. By contradiction.
There exists a prime .
By [GR97], for
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Proof of Theorem I (weak version)
By [GR97], for
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Proof of Theorem I (weak version)
By [GR97], for
By Lucas Theorem, for
For , (𝑝0 )=(𝑝𝑝 )=1Define the polynomial
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Proof of Theorem I (weak version)
By Lucas Theorem, for
Define the polynomial
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Proof of Theorem I (weak version)
By Lucas Theorem, for
Define the polynomial
Since
𝑔 : {0 ,…,𝑛−𝑝 }→ {0 ,1,2 }
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Proof of Theorem I (weak version)
𝑔 : {0 ,…,𝑛−𝑝 }→ {0 ,1,2 }
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Proof of Theorem I (weak version)
𝑔 : {0 ,…,𝑛−𝑝 }→ {0 ,1,2 }If is not a constant, then
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Proof of Theorem I (weak version)
𝑔 : {0 ,…,𝑛−𝑝 }→ {0 ,1,2 }Otherwise, since
Hence is linear. As takes integer values and its range is smaller than its domain, must be constant – a contradiction!
QED
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How to get a stronger bound?
Modulo a prime is nicer to analyze though we loose information.
Natural idea: use many primes!
How does one combine all pieces of information from different primes?
The set of primes should have some structure.
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A cube of primes
7=711=7+4
17=7+4+613=7+6
Lemma: cube with primes in .
Cube
𝑃𝑝 ;𝛿1 ,…,𝛿𝑘={𝑝+∑
𝑖=0
𝑘
𝑎𝑖 𝛿𝑖:∀ 𝑖𝑎𝑖∈ {0 ,1 }}
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Theorem IIIproof idea
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Theorem III – proof idea
(20) (21)(22)
(30) (31)(32)
(00) (01)(02)
(10) (11)(12)
(40) (41)(42)
𝑛=4𝑑=2
∀ 𝑗∈ [𝑛 ]|∑𝑖=0
𝑑
𝑎𝑖( 𝑗𝑖 )|≤𝑚𝑎0 ,…,𝑎𝑑∈𝑁
𝑓 (𝑥 )=∑𝑖=0
𝑑
𝑎𝑖(𝑥𝑖 )
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Proving the existence of a not too-high-degree polynomial with a given range boils down to proving the existence of a short vector in an appropriate lattice.
Theorem III – proof idea
To avoid trivialities, we prove the existence of such vectors that are linearly independent. By the structure of the lattice this implies that one of them has degree .
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Open Questions
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1. Break that barrier!2. For the latter will improve the algorithm
for learning symmetric juntas. 3. What is it with that third in Thm II?4. Better upper bounds – if exist..5. Find more applications
Open Questions
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Thank You!