polynomial versions of integer partitions and their zerosrboyer/talks/talk_rutgers_2012.pdf ·...

Polynomial Versions of Integer Partitions andTheir Zeros

Robert Boyer

Rutgers University – Experimental Math Seminar

February 9, 2012

Robert Boyer Polynomial Versions of Integer Partitions and Their Zeros

Joint Work

William M. Y. Goh

Daniel T. Parry

1 Partition Polynomials: Asymptotics and Zeros,Contemporary Mathematics, Volume 457 (2008), 99-112(with Bill Goh)

2 On the Zeros of Plane Partition Polynomials, (with DanielParry), Electronic Journal of Combinatorics, Volume 18 (2)(2012), # P30 (with Daniel Parry) 26 pages

3 Phase Calculations for Planar Partition Polynomials,accepted by Rocky Mountain Journal of Mathematics (withDaniel Parry)


Main Examples

Partition Polynomials:n∑

k=1

pk(n)xk

Plane Partition Polynomials:n∑

k=1

ppk (n)xk

Odd Partition Polynomials:n∑

k=1

poddk (n)xk


Polynomial Versions of Different Partition Numbers

Example 1 Partition Polynomialp(n): partitions of npk (n): partition of n with exactly k parts

Fn(x) =∑n

k=1 pk (n)xk

F4(x) = x4 + x3 + 2x2 + xsince4 = 4, = 3 + 1, = 2 + 2, = 2 + 1 + 1, = 1 + 1 + 1 + 1


Plane Partitions

Example 2 Plane Partition Polynomial

• Plane partition of n is an array πi ,j of positive integers withsum n whose rows and columns are decreasing.• Its trace is the sum of its diagonal entries.


Plane Partitions

Example 2 Plane Partition Polynomial

• Plane partition of n is an array πi ,j of positive integers withsum n whose rows and columns are decreasing.• Its trace is the sum of its diagonal entries.

PL(n): plane partitions of n

PLk (n): plane partitions of n with trace k

Qn(x) =∑n

k=1 PLk (n)xk

Q4(x) = x4 + 2x3 + 6x2 + 4x


Plane Partitions of 4 and Their Traces

PL1(4) = 4, PL2(4) = 6, PL3(4) = 2, PL4(4) = 1• Trace = 4,4

• Trace = 3

3 1 ,31

,

• Trace = 2

2 2 ,22

, 2 1 1 ,2 11

,211

,1 11 1

,

• Trace = 1

1 1 1 1 ,1 1 11

,1 111

,

1111

Conclude4 3 2


Last Polynomial Example

Example 3 Odd Partition Polynomial

podd(n): partitions of n all of whose parts are oddpodd ,k (n): partitions of n with exactly k parts all of which areoddFodd ,n(x) =

∑nk=1 podd ,k(n)xk


Last Polynomial Example

Example 3 Odd Partition Polynomial

podd(n): partitions of n all of whose parts are oddpodd ,k (n): partitions of n with exactly k parts all of which areoddFodd ,n(x) =

∑nk=1 podd ,k(n)xk

Other natural examples: use partitions whose parts lie in anarithmetic progression, partitions whose parts are congruent toseveral residue classes modulo a, etc.


Zeros of the basic examples of degree 1000


Partition Polynomial of degree 1000

10.50

-0.5 0-1

-0.5

0.5

-1

1


Plane Partition Polynomial of degree 1000

K1.0 K0.5 0 0.5 1.0

K1.0

K0.5

0.5

1.0


Odd Partition Polynomial of degree 1000

K1.0 K0.5 0 0.5 1.0

K1.0

K0.5

0.5

1.0


Zeros Accumulate on Unit Circle

Theorem

Zeros accumulate to the unit circle as n → ∞


Zeros Accumulate on Unit Circle

Theorem

Zeros accumulate to the unit circle as n → ∞

• Proof uses general techniques.


Theorem of Erdös and Turán

Theorem

Let q(x) be the polynomial∑n

k=0 akxk of degree n withnon-zero constant term a0 6= 0. For 0 ≤ θ1 < θ2 ≤ 2π,

|# {z : arg z ∈ [θ1, θ2], q(z) = 0} −θ2 − θ1

2πn

∣

∣

∣

∣

< 16

√

n ln( |a0| + |a1| + · · · + |an|√

a0an

)

.


Asymptotics for the Partition Numbers

p(n) ∼ 1

4n√

3exp

(

π

√

2n3

)

PL(n) ∼ ζ(3)7/36eζ′(−1)

211/36√

πn25/36exp

(

3 3√

ζ(3)(n

2

)2/3)

.

podd(n) ∼ 2−5/4 3−1/4 n−3/4 exp

(

π

√

n3

)


Arguments of Zeros Are Uniformly Distributed

• qn(x) =

n∑

k=1

pk (n)xk−1

• deg(qn) = n − 1, aj = pj+1(n), a0 = an−1 = 1

•n−1∑

k=0

ak =

n∑

k=1

pk (n) = p(n) ∼ 1

4n√

3exp

(

π

√

2n3

)


Arguments of Zeros Are Uniformly Distributed,continued

• Average number of zeros of qn(x) in the sectorθ1 < arg x < θ2

=1

n − 1# {z : arg z ∈ [θ1, θ2], qn(z) = 0}



• Average number of zeros of qn(x) in the sectorθ1 < arg x < θ2

=1

n − 1# {z : arg z ∈ [θ1, θ2], qn(z) = 0}

By Theorem of Erdös and Turán:

1n − 1

∣

∣

∣

∣

# {z : arg z ∈ [θ1, θ2], qn(z) = 0} − θ2 − θ1

2π(n − 1)

∣

∣

∣

∣

is bounded by

16

√

ln( |a0| + |a1| + · · · + |an−1|√

a0an−1

)

= 16

√

ln[p(n)]√

n − 1



16

√

ln[p(n)]√

n − 1∼ 16

1√n − 1

√

ln[

1

4n√

3eπ

√2n/3

]

= 161√

n − 1

√

− ln(4n√

3) + π√

2n/3

Conclude: the arguments of the zeros of the partitionpolynomials are uniformly distributed.


Average Number of Zeros

10.50

-0.5 0-1

-0.5

0.5

-1

1

Elementary arguments show:average number of zeros both inside and outside the unit circlego to 0 as n → ∞.



10.50

-0.5 0-1

-0.5

0.5

-1

1


Conclusion: zeros are uniformly distributed around the unitcircle.



10.50

-0.5 0-1

-0.5

0.5

-1

1


Conclusion: zeros are uniformly distributed around the unitcircle.

Problem: Ignore the contribution of zeros inside the unit disk.


Zero Attractor

Definition

Informally, the Zero Attractor for {Fn(x)} is the set of limits ofthe zeros of Fn(x).

Formally, the Zero Attractor for {Fn(x)} is the limit of the setsZ (Fn) in the space of compact subsets of the complex plane inthe Hausdorff metric.


Structure of Asymptotics

Assumption: asymptotics of Fn(x) inside unit disk D have thefollowing form:D is the union of disjoint open regions R1, R2, . . . (up to theirboundaries) and

Fn(x) ∼ a1,n(x)en1/2L1(x), x ∈ R1,

Fn(x) ∼ a2,n(x)en1/2L2(x), x ∈ R2, · · ·

where aj ,n(x) 6= 0 on Rj , both aj ,n(x), Lj(x) are analytic on Rj ,

and aj ,n(x) = o(en1/2Lj (x)).

RegionsR1 = {x : ℜL1(x) > ℜL2(x),ℜL3(x), · · · }R2 = {x : ℜL2(x) > ℜL1(x),ℜL3(x), · · · }, · · ·


Informal Principle to Find Zero Attractor

• For adjacent regions, say R1 and R2, their common boundaryis a subset of the level set ℜL1(x) = ℜL2(x).


Informal Principle to Find Zero Attractor

• For adjacent regions, say R1 and R2, their common boundaryis a subset of the level set ℜL1(x) = ℜL2(x).

• The zeros of Fn(x) cannot converge to a point interior to anyof the regions R(j); that is, the zeros accumulate along theircommon boundary.

•Conclusion: The zero attractor =⋃

j≥1

boundary(Rj)


Summary

• Need polynomial analogues of the partition numberasymptotics to find their zero attractor.

• Zeros converge to the boundaries of the regions where thesense of the asymptotics change.


Plots of Zeros

Asymptotics and Their Regions


Example 1Partition Polynomials

Example 2Plane Partition Polynomials

Example 3Odd Partition Polynomials


Partition Polynomial of degree 1000

10.50

-0.5 0-1

-0.5

0.5

-1

1


Partition Polynomial Zero Attractor


Partition Polynomial Zero Attractor in Upper HalfPlane


Partition Polynomial Zero Attractor Closeup


Partition Polynomial Zero Attractor with Degree70,000 Zeros


Partition Polynomial Zeros of Degree 25,000, 30,000,40,000, 50,000, 60,000, 70,000


Dilogarithm

Dilogarithm: Li2(x) =

∞∑

m=1

xm

m2

Functions for Partition Polynomial Asymptotics

Lk(x) =1k

√

Li2(xk )


Asymptotics of Partition Polynomials

Theorem

Fn(x) ∼√

1 − x√

L1(x)1

n3/4exp

(

2√

nL1(x))

, x ∈ R1

Fn(x) ∼ a2(x)√

L2(x)(−1)n

n3/4exp

(

2√

nL2(x))

x ∈ R2

Fn(x) ∼ a3,n(x)√

L3(x)1

n3/4exp

(

2√

nL3(x))

, x ∈ R3


Zero Attractor of Partition Polynomials

Three Regions:

R1 = {x : ℜ[L1(x)] > max(ℜ[L2(x)],ℜ[L3(x)])},R2 = {x : ℜ[L2x)] > max(ℜ[L1(x)],ℜ[L3(x)])},R3 = {x : ℜ[L3(x)] > max(ℜ[L1(x)],ℜ[L2(x)])}.


Zero Attractor of Partition Polynomials

Three Regions:

R1 = {x : ℜ[L1(x)] > max(ℜ[L2(x)],ℜ[L3(x)])},R2 = {x : ℜ[L2x)] > max(ℜ[L1(x)],ℜ[L3(x)])},R3 = {x : ℜ[L3(x)] > max(ℜ[L1(x)],ℜ[L2(x)])}.

Zero attractor consists of portions of the three level sets

ℜ[L1(x)] = ℜ[L2(x)], ℜ√

Li2(x) =12ℜ√

Li2(x2)

ℜ[L1(x)] = ℜ[L3(x)], ℜ√

Li2(x) =13ℜ√

Li2(x3)

ℜ[L2(x) = ℜ[L3(x)],12ℜ√

Li2(x2) =13ℜ√

Li2(x3)


Special Points on Zero Attractor

Special Points:

• Three curves intersect at (−0.692206, 0.6913717)

• Solution to ℜL1(eit) = 13ℜL3(eit) is 2.06672966

(where “green curve” intersects unit circle)

• Solution to 12ℜL2(eit) = 1

3ℜL3(eit) is 2.36170417(where “blue curve” intersects unit circle)



K1.0 K0.5 0 0.5 1.0

K1.0

K0.5

0.5

1.0


Zero Attractor for Plane Partition Polynomials

K1.0 K0.5 0 0.5 1.0

K1.0

K0.5

0.5

1.0


K1.0 K0.5 0 0.5 1.0

K1.0

K0.5

0.5

1.0


Closeup of Zero Attractor for Plane PartitionPolynomials with degree 800 zeros


• Special points on zero attractor

on unit circle ei2.989863546

on negative axis −0.8250030529


Trilogarithm

Trilogarithm: Li3(x) =∞∑

m=1

xm

m3

Functions for Plane Partition Polynomial Asymptotics

Lk(x) =1k

3√

Li3(xk ), k = 1, 2, . . .

Level Set for Zero Attractor

ℜL1(x) = ℜL2(x)

Interval in Zero Attractor [−0.8250030529, 0]


Asymptotics of Plane Partition Polynomials

Theorem

(a) Let x ∈ R1 \ [x∗, 0],

Qn(x) ∼ 12√

1 − x

√

L1(x)

6πn4/3exp

(

32 n2/3L1(x)

)

.

(b) Let x ∈ R2,

Qn(x) ∼ (−1)n 24√

1 − x2 8

√

1 − x1 + x

√

L2(x)

6πn4/3exp

(

32n2/3 L2(x)

)

.



K1.0 K0.5 0 0.5 1.0

K1.0

K0.5

0.5

1.0


Odd Partition Polynomial Zero Attractor

K1.0 K0.5 0 0.5 1.0

0.2

0.4

0.6

0.8

1.0


Odd Partition Zero Attractor Closeup

K0.10 K0.05 0.00 0.05 0.10

0.92

0.94

0.96

0.98

1.00


Odd Partition Zero Attractor with Composite Zeros ofDegree 2n, n = 12, · · · , 15

K0.10 K0.05 0.00 0.05 0.10

0.8

0.9

1.0


Asymptotics of Odd Partition Polynomials

Functions for the Asymptotics

L2j+1(x) =1

2j + 1

√

Li2(x2j+1), j = 0, 1, 2, . . .

L2j(x) =1j

√

Li2(−x j), j = 1, 2, . . .

Level Sets for Zero Attractor

ℜL1(x) = ℜL4(x), ℜL2(x) = ℜL4(x)


Generating Functions G(x , q)

Partition Polynomials

∞∏

ℓ=1

11 − xqℓ

=

∞∑

n=0

Fn(x) qn




∞∏

ℓ=1

11 − xqℓ

=

∞∑

n=0

Fn(x) qn

Plane Partition Polynomials

∞∏

ℓ=1

1(1 − xqℓ)ℓ

=

∞∑

n=0

Qn(x) qn




∞∏

ℓ=1

11 − xqℓ

=

∞∑

n=0

Fn(x) qn

Plane Partition Polynomials

∞∏

ℓ=1

1(1 − xqℓ)ℓ

=

∞∑

n=0

Qn(x) qn

Odd Partition Polynomials

∞∏

ℓ=1

11 − xq2ℓ−1 =

∞∑

n=0

Fodd ,n(x) qn


Circle Method

Cauchy Integral Formula: for any radius 0 < r < 1

Fn(x) =1

2πi

∫

|u|=r

G(x , u)

un+1 du


Circle Method


Fn(x) =1

2πi

∫

|u|=r

G(x , u)

un+1 du

Subdivide the circle |u| = r into subarcs relative to rationalpoints distributed around the circle using Farey fractions.


Circle Method


Fn(x) =1

2πi

∫

|u|=r

G(x , u)

un+1 du


Farey fractions of order N

FN = {h/k : (h, k) = 1, 1 ≤ k ≤ N}


Circle Method


Fn(x) =1

2πi

∫

|u|=r

G(x , u)

un+1 du


Farey fractions of order N

FN = {h/k : (h, k) = 1, 1 ≤ k ≤ N}

Special intervalsLet h1/k1 < h/k < h2/k2 be three consecutive Farey fractionsof order N:

(

h + h1

k + k1− h

k,h + h2

k + k2− h

k

)

with corresponding circular arc ξ(N)h,k : re2πiθ


Circle Method - Continued

Circle Radius, Farey Fraction OrderBoth the radius r and the order N are functions of n and x

r = r(x , n) N = N(x , n).

Order of Farey fractions: N = N(x , n) = δn1/2

Radius of integration circle: r(x , n) = exp(

−ℜLm(x)

2πn1/2

)

,

(x ∈ R(m), m = 1, 2, 3)


Circle Method - Continued

Circle Radius, Farey Fraction OrderBoth the radius r and the order N are functions of n and x

r = r(x , n) N = N(x , n).

Order of Farey fractions: N = N(x , n) = δn1/2

Radius of integration circle: r(x , n) = exp(

−ℜLm(x)

2πn1/2

)

,

(x ∈ R(m), m = 1, 2, 3)

Question: Which Farey arcs contribute as n → ∞? (“majorarcs")


G(x , u) in neighborhood of e2πih/k

By using series expansion of the logarithm and the identity

e−τ =1

2πi

∫ σ+i∞

σ−i∞Γ(s)τ−s ds, σ > 1,ℜτ > 0

ln[G(x , e−w+2πih/k)] =1

2πi

∫ σ+i∞

σ−i∞Qh,k(x , s) τ−s ds, σ > 1,ℜw > 0


G(x , u) in neighborhood of e2πih/k

By using series expansion of the logarithm and the identity

e−τ =1

2πi

∫ σ+i∞

σ−i∞Γ(s)τ−s ds, σ > 1,ℜτ > 0

ln[G(x , e−w+2πih/k)] =1

2πi

∫ σ+i∞

σ−i∞Qh,k(x , s) τ−s ds, σ > 1,ℜw > 0

where

Qh,k(x , s) =

∞∑

m,ℓ=1

xℓ

ℓe2πimℓh/k (mℓ)−s

=

∞∑

ℓ=1

xℓ

ℓs+1

∞∑

m=1

e2πimℓh/k

ms , σ > 1


• Given1

2πi

∫ σ+i∞

σ−i∞Qh,k(x , s) τ−s ds, σ > 1,,

we want to shift the vertical line to the left half plane.

• The residue of Qh,k (x , s) at s = 1 will give the dominantcontribution to the integral.

• Need useful form of Qh,k(x , s) to get the residues.


Special Functions

Polylogarithm: Lis(x) =∞∑

n=1

xn

ns

Zeta Function: ζ(s) =∞∑

n=1

1ns

Hurwitz Zeta Function: ζ(s, a) =∞∑

n=0

1(n + a)s

Lerch Phi Function: Φ(x , s, a) =

∞∑

n=0

xn

(n + a)s


Expansion of Qh,k(x , s) in terms of special functions

Qh,k (x , s) =ζ(s)

ks+1 Lis+1(xk ) +

k−1∑

r=1

x r

ks+1 Φ(xk , s + 1, r/k) Lis(e2πirh/k )



Qh,k (x , s) =ζ(s)

ks+1 Lis+1(xk ) +

k−1∑

r=1

x r


ζ(s) has a simple pole at s = 1 with residue 1. This gives theonly contribution:



Qh,k (x , s) =ζ(s)

ks+1 Lis+1(xk ) +

k−1∑

r=1

x r


ζ(s) has a simple pole at s = 1 with residue 1. This gives theonly contribution:

Theorem

The residue of Qh,k(x , s) at s = 1 is

1k2 Li2(x

k )


Regions R1, R2, R3

Theorem

ℜLk(x) < max[ℜL1(x),ℜL2(x),ℜL3(x)], k ≥ 4, x 6= 0.


Regions R1, R2, R3

Theorem

ℜLk(x) < max[ℜL1(x),ℜL2(x),ℜL3(x)], k ≥ 4, x 6= 0.

Hence the regions Rm, with m = 1, 2, 3 have the stronger form

Rm ={

x : ℜLm(x) > ℜLj(x), j 6= m}

.


Estimation of Integrals from Circle Method

• Assume x ∈ R(m), m = 1, 2, or 3

The main contribution to the integral∫

ξ(n)h,k

G(x , u)

un+1 du

is gotten from the integral Ih,k ,n(x) where

Ih,k ,n(x) =1

2πn1/2

∫ 2πn1/2/[k(k+k ′′)]

2πn1/2/[k(k+k ′)]exp

[

n1/2 Φ(x , z)]

dz, x ∈ Rm


Estimation of Integrals from Circle Method

• Assume x ∈ R(m), m = 1, 2, or 3

The main contribution to the integral∫

ξ(n)h,k

G(x , u)

un+1 du

is gotten from the integral Ih,k ,n(x) where

Ih,k ,n(x) =1

2πn1/2

∫ 2πn1/2/[k(k+k ′′)]

2πn1/2/[k(k+k ′)]exp

[

n1/2 Φ(x , z)]

dz, x ∈ Rm

Φ(x , z) =Lm(x)2

ℜLk(x) − iz+ (ℜLm(x) − iz), x ∈ Rm


Saddle Point MethodAssume that ℜB(x) has a unique maximum on (a, b) andℜB(x0) < 0.

∫ b

aet B(x) dx ∼ etB(x0)

√

2π

−tB′′(x0)

∂

∂zΦ(x , z) =

i Lm(x)2

(ℜLm(x) − iz)2 − i = 0 =⇒ z0 = −ℑLm(x)

∂2

∂2zΦ(x , z) =

i Lm(x)2

(ℜLm(x) − iz)3 (−2)(−i) = −2Lm(x)2

(ℜLm(x) − iz)3

ℜ ∂2

∂2zΦ(x , z0) = −2ℜ 1

Lm(x)< 0


Conclusion of Estimation Using Saddle Point Method

Ih,k ,n(x) ∼ 12πn1/2

√π√

Lm(x)1

n1/4en1/2Lm(x)

=1

2√

π

√

Lm(x)1

n3/4en1/2Lm(x)


Conclusion of Estimation Using Saddle Point Method

Ih,k ,n(x) ∼ 12πn1/2

√π√

Lm(x)1

n1/4en1/2Lm(x)

=1

2√

π

√

Lm(x)1

n3/4en1/2Lm(x)

• A more detailed analysis is needed to get the full asymptoticexpansion of the polynomials.


I applied the above method to the followingexamples but I do not have proofs.


Polynomials for partitions whose parts arecongruent to either 1 or 2 modulo 3


Degree 400


Zeros from several polynomials


Attractor with zeros in second quadrant


Zero Attractor

L1(x) =√

2Li2(x)

L2(x) =12

√

2Li2(x2)

L3(x) =13

[

2Li2(x3) + xΦ(x3, 2, 1/3) · e2πi/3

+ x2Φ(x3, 2, 2/3) · (e2πi/3)2

+ xΦ(x3, 2, 1/3) · (e2πi/3)2 + x2Φ(x3, 2, 2/3)(e4πi/3)2]1/2


Zero Attractor

Zero attractor• level set ℜL1 = ℜL3 which is near the imaginary axis andconnects 0 with e1.59829i (solution to ℜL1(eit) = ℜL3(eit).

• level set ℜL2 = ℜL3 which is near −1 and connects thepoints −0.897454 on the negative axis with e2.92246i (which aresolutions to ℜL2(t) = ℜL3(t) for t ∈ (−1, 0) andℜL2(eit) = ℜL3(eit)).


Parts congruent to either 0 or 2 modulo 3


Parts congruent to either 0 or 2 modulo 3 – Closeup


Zero Attractor

L1(x) =√

2Li2(x), L2(x) =12

√

2Li2(x2)

L3(x) =13

[

Li2(x3) + xΦ(x3, 2, 1/3) + x2Φ(x3, 2, 2/3)

+ xΦ(x3, 2, 1/3) · (e2πi/3)2 + x2Φ(x3, 2, 2/3)(e4πi/3)2]1/2


Zero Attractor

Zero attractor• [−0.55008, 0]

• level set ℜL1 = ℜL3 connects 0 to e1.60212i (solution toℜL1(eit) = ℜL3(eit));

• level set ℜL2 = ℜL3 connects −0.55008 to e2.84288i (solutionto ℜL2(eit) = ℜL3(eit))


Parts congruent to either 1 or 3 modulo 4 - Degree1500


Parts congruent to either 1 or 3 modulo 4 - Degrees3000, 4500, 6000 – Closeup to imaginary axis


Zero Attractor Closeup with zeros of degree 1500,3000, 45000, 6000


For 1 ≤ b < a, let

Kb,a;h,k(x) =1

ak2 Li2(xk ) +

k−1∑

r=1, k |ra

x r e2πirbh/k

ak2 Φ(xk , 2, r/k)

Lh,k(x) =[

K1,4;h,k(x) + K3,4;h,k(x)]1/2

L1(x) = L0,1(x), L2(x) = L1,2(x), L4(x) = L1,4(x).


Zero Attractor

• imaginary axis interval [0, 0.97474] i (solution toℜL1(it) = ℜL4(it) for 0 < t < 1) [ℜL1(x) > max{ℜLk (x) : k ≥ 2}on disk of radius ≃ 0.97474]

• level set ℜL2 = ℜL4 (in second quadrant) that connects0.97474 i to the point on the unit circle e1.60867i (solution toℜL2(eit) = ℜL4(eit);

• level set ℜL1 = ℜL4 (in second quadrant) that connects0.97474 i to the point on the unit circle e1.53291i (solution toℜL1(eit) = ℜL4(eit).


Parts congruent to either 1 or 4 modulo 5 – Degree6000


Zero Attractor and Zeros for Parts congruent to either1 or 4 modulo 5


Closeup of Zero Attractor with zeros of degree 3000,4000, 5000, 6000


Functions for Asymptotics and Zero Attractor

L1(x) =[

K1,5;0,1(x) + K4,5;0,1(x)]1/2

L5(x) =[

K1,5;3,5(x) + K4,5;3,5(x)]1/2

• e1.581091115i is the solution to ℜ[L1(eit) = ℜ[L5(eit)]

• Level set ℜL1(x) = ℜL5(x)


polynomial versions of integer partitions and their zerosrboyer/talks/talk_rutgers_2012.pdf ·...

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