On the Degree of Univariate Polynomials Over the Integers

On the Degree of Univariate Polynomials

Over the IntegersGil Cohen

Weizmann Institute

Amir Shpilka and Avishay Tal

Joint work with

The Question

What is the minimal degree ofa polynomial of the form

?

The Question

0

1

2

0123 4567Mmm…

0

What is the minimal degree ofa non-constant polynomial of the

form ?

The Question

0

1

2

0123 4567Mmm…

What is itto Us?

George Boole Alan Turing

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).

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)

[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 )

Minimal degree .

Observations

When , the minimal degree is (e.g., ).

By the pigeonhole principle, the minimal degree .

[GR97] For , minimal degree .

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?

Theorem I. the minimal degree is at least .

Main Result I

Main Result I

𝑚

minimal degree

𝑛

1

𝑛−𝑛 .525

𝑛−𝑛

log log𝑛

𝑛1

𝑛/3

2

/22 𝑛−1

Threshold

Main Result IIminimaldegree

𝑛

1

𝑛−𝑛

log log𝑛

𝑛1

𝑛−1𝑚?

Hey, I’m over here!

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

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.

Main Result III (Upper Bounds)

Theorem III. For there exists an s.t.

Holds .

Hence, by Theorem II

Theorem IIngredients

• 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

Newton Polynomials

Do polynomials over the integers have integral coefficients?

𝑓 (𝑥 )=− 12𝑥2+

32𝑥+1

This is the right intuition, but the wrong basis!

Newton Polynomials

For every , define

The set of polynomials

is called the Newton Basis for degree polynomials.

Newton Polynomials

For a degree polynomial

In our case the ’s are all integers!

𝛾𝑘=∑𝑗=0

𝑘

(−1 )𝑘− 𝑗 ⋅(𝑘𝑗) 𝑓 ( 𝑗 )

Newton Polynomials

Going back to the example

𝑓 (𝑥 )=3(𝑥2 )−(𝑥1 )+(𝑥0 )

• 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

Lucas Theorem

Theorem [Luc78].

Lucas Theorem

Theorem [Luc1878].

Lucas Theorem

Theorem [Luc1878]. Let , and be a prime.Denote

Then,

• 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

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]:

• 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

Linear Recurrence Relations

If has degree , then determine .

Lemma [GR97]: If has degree , then for all

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 .

Theorem IProof of a weaker

version

Theorem. Let be non-constant. Then .

Proof of Theorem I (weak version)

Proof. By contradiction.

There exists a prime .

By [GR97], for

Proof of Theorem I (weak version)

By [GR97], for

Proof of Theorem I (weak version)

By [GR97], for

By Lucas Theorem, for

For , (𝑝0 )=(𝑝𝑝 )=1Define the polynomial

Proof of Theorem I (weak version)

By Lucas Theorem, for

Define the polynomial

Proof of Theorem I (weak version)

By Lucas Theorem, for

Define the polynomial

Since

𝑔 : {0 ,…,𝑛−𝑝 }→ {0 ,1,2 }

Proof of Theorem I (weak version)

𝑔 : {0 ,…,𝑛−𝑝 }→ {0 ,1,2 }

Proof of Theorem I (weak version)

𝑔 : {0 ,…,𝑛−𝑝 }→ {0 ,1,2 }If is not a constant, then

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

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.

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 }}

Theorem IIIproof idea

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 )

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 .

Open Questions

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

Thank You!