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2 Solution of Kakeya’s problem in finite fields

2 Solution of Kakeya’s problem in finite fields

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16.2 Solution of Kakeya’s problem in finite fields



227



Fig. 16.1 As the middle (flexible) point moves around the smaller circle, the needle

rotates through 360◦ .



instance, one can rotate a unit needle inside a unit disk, which has area π/4.

By using a deltoid one requires only π/8 area (see Fig. 16.1).

The Kakeya conjecture in more dimensions states that any subset of Rn

that contains a unit line segment in every direction has Hausdorff dimension

equal to n. This conjecture remains open in dimensions three and higher, and

gets more difficult as the dimension increases.

To approach this question, Wolff (1999) proposed a simpler finite field

analogue of the Kakeya conjecture. If Fn is a vector space over a finite field

F, define a Kakeya set to be a subset K ⊆ Fn which contains a line in every

direction, namely for any v ∈ Fn there exists a vector w ∈ Fn such that the

line {w + tv : t ∈ F} is contained in K; here, vector w is the origin and

vector v the direction of the line. The finite field Kakeya conjecture stated

that there exists a constant c > 0 depending only on the dimension n such

that every Kakeya set K ⊆ Fn has cardinality |K| ≥ c|F|n .

This finite field version of the conjecture has had a significant influence

on the subject, in particular inspiring work on the sum-product phenomenon

in finite fields, which has since proved to have many applications in number

theory and computer science. Modulo minor technicalities, the progress on

the finite field Kakeya conjecture was, however, essentially the same as that

of the original “Euclidean” Kakeya conjecture.

Recently Dvir (2009) used a surprisingly simple application of the polynomial method to prove the finite field Kakeya conjecture.

Lemma 16.5. Let f ∈ F[x1 , . . . , xn ] be a polynomial of degree at most q − 1

over a finite field with q = |F| elements. If f vanishes on a Kakeya set K,

then f is the zero polynomial.

Proof. The argument is similar to that in the proof of Lemma 16.2. Suppose

d

for a contradiction that f is nonzero. We can write f =

i=0 fi , where

0 ≤ d ≤ q − 1 is the degree of f and fi is the i-th homogeneous component;

thus fd is nonzero. Since f vanishes on K, d cannot be zero. Hence, fd is a

nonzero polynomial.



228



16 The Polynomial Method



Let v ∈ Fn \ {0} be an arbitrary direction. As K is a Kakeya set, K

contains a line {w + tv : t ∈ F} for some w ∈ Fn , thus f (w + tv) = 0 for

all t ∈ F. The left-hand side is a polynomial gw,v (t) in t of degree at most

q − 1, and must be the zero polynomial by the factor theorem, that is, all its

coefficients are zero. In particular, the coefficient of td , which is fd (v), must

be zero. Since v was arbitrary, it follows that the polynomial fd (x) vanishes

on all points in Fn . But since dq n−1 ≤ (q − 1)q n−1 < q n , Lemma 16.2 implies

that fd must be a zero polynomial.

Theorem 16.6 (Dvir 2009). Let K ⊂ Fn be a Kakeya set. Then

|F | + n − 1

n



|K| ≥







|F|n

.

n!



Proof. Let q = |F| and suppose that |K| < n+q−1

. Then, by Lemma 16.1,

n

there exists a nonzero polynomial f ∈ F[x1 , . . . , xn ] of degree at most q − 1

that vanishes on K, which contradicts Lemma 16.5.



16.3 Combinatorial Nullstellensatz

The following special case of Hilbert’s Nullstellensatz has found numerous

applications in combinatorics.

Theorem 16.7 (Nullstellensatz). Let f ∈ F[x1 , . . . , xn ], and let S1 , . . . , Sn

be nonempty subsets of F. If f (x) = 0 for all x ∈ S1 × · · · × Sn , then there

are polynomials h1 , . . . , hn ∈ F[x1 , . . . , xn ] such that deg(hi ) ≤ deg(f ) − |Si |

and

n

f (x1 , . . . , xn ) =



(xi − s) .



hi (x1 , . . . , xn )

i=1



s∈Si



Proof (due to Alon 1999). Define di = |Si | − 1 for all i, and consider polynomials

(xi − s) = xidi +1 −



gi (xi ) =

s∈Si



di



j=0



aij xji .



Observe that if xi ∈ Si then gi (xi ) = 0, that is,

xidi +1 =



di

j=0



aij xji .



(16.1)



Let f be the polynomial obtained by writing f as a linear combination of

monomials and replacing, repeatedly, each occurrence of xtii (1 ≤ i ≤ n),

where ti > di , by a linear combination of smaller powers of xi , using the

relations (16.1). The resulting polynomial f is clearly of degree at most di in



229



16.3 Combinatorial Nullstellensatz



xi for each 1 ≤ i ≤ n, and is obtained from f by subtracting from it products

of the form hi gi , where deg(hi ) ≤ deg(f ) − deg(gi ) = deg(f ) − |Si |. So,

n



f (x) = f (x) −



hi (x)gi (xi ) .



(16.2)



i=1



Moreover, f(x) = f (x) for all x ∈ S1 × · · ·×Sn , since the relations (16.1) hold

for these values of x. Since, by our assumption, f (x) = 0 for all these values,

we obtain that f (x) = 0 for all x ∈ S1 × · · · × Sn as well, and Exercise 16.5

implies that f (x) = 0 for all x ∈ Fn . Together with (16.2), this implies that

f = ni=1 hi gi , as desired.

Using the Nullstellensatz we can derive the following generalization of the

DeMillo–Lipton–Schwartz–Zippel lemma.

Theorem 16.8 (Combinatorial Nullstellensatz). Let f (x1 , . . . , xn ) be a polynomial of degree d over a field F. Suppose that the coefficient of the monomial

xt11 · · · xtnn in f is nonzero and t1 +· · ·+tn = d. If S1 , . . . , Sn are finite subsets

of F with |Si | ≥ ti + 1, then there are exists a point x in S1 × · · · × Sn for

which f (x) = 0.

Proof. We may assume that |Si | = ti + 1 for all i. Suppose the result is

false, and define gi (xi ) = s∈Si (xi − s). Let h1 , . . . , hn be the polynomials

guaranteed by the Nullstellensatz. Hence,

deg(hi ) ≤ deg(f ) − deg(gi ) = deg(f ) − (ti + 1)

and f (x) =



n

i=1



(16.3)



hi (x)gi (xi ), that is,

n



f (x) =

i=1



xiti +1 hi (x) + (terms of degree < deg(f )) .

n



By assumption, the coefficient of i=1 xtii on the left-hand side is nonzero,

while it is impossible to have such a monomial on the right-hand side, a

contradiction.

In a similar vein is the following result.

Theorem 16.9 (Chevalley–Warning). Let p be a prime, and f1 , . . . , fm polym

nomials in Fp [x1 , . . . , xn ]. If i=1 deg(fi ) < n then the number of common

zeros of f1 , . . . , fm is divisible by p. In particular, if there is one common

zero, then there is another one.

Although this theorem can be derived from the Combinatorial Nullstellensatz, we give a slightly more direct proof due to Alon (1995).



230



16 The Polynomial Method



Proof. By Fermat’s Little Theorem (see Exercise 1.15), ap−1 ≡ 1 mod p for

all a ∈ Fp , a = 0. Hence, the number N of common zeros of f1 , . . . , fm

satisfies

m



1 − fj (x1 , . . . , xn )p−1 .



N=



(in Fp )



(16.4)



x1 ,...,xn ∈Fp j=1



By expanding the right-hand side we get a linear combination of monomials

of the form

n

i=1



xtii with



n



m



ti ≤ (p − 1)

i=1



deg(fj ) < (p − 1)n.

j=1



Hence, in each such monomial there is an i with ti < p − 1. But then (see

Exercise 16.1)

xtii = 0 (in Fp ),

xi ∈Fp



implying that the contribution of each monomial to the sum (16.4) is 0 modulo p, completing the proof of the theorem.

We illustrate the potential applications of the Combinatorial Nullstellensatz on several examples.

16.3.1 The permanent lemma

Let A = (ai,j ) be an n × n matrix over a field F. The permanent Per(A) of

A is the sum

Per(A) =

a1,i1 a2,i2 · · · an,in

(i1 ,i2 ,...,in )



of n! products, where (i1 , i2 , . . . , in ) is a permutation of (1, 2, . . . , n).

Theorem 16.10. If Per(A) = 0, then for any vector b ∈ Fn , there is a subset

of columns of A whose sum differs from b in all coordinates.

This is just a special case of the following lemma for all Si = {0, 1}.

Lemma 16.11 (Permanent Lemma). Let b ∈ Fn and S1 , . . . , Sn be subsets

of F, each of cardinality at least 2. If Per(A) = 0, then there exists a vector

x ∈ S1 × · · · × Sn such that Ax differs from b in all coordinates.

Proof. The polynomial

n



n



i=1



j=1



aij xj − bi



f=



231



16.3 Combinatorial Nullstellensatz



is of degree n and the coefficient of x1 x2 · · · xn in it is Per(A) = 0. The

result now follows directly from the Combinatorial Nullstellensatz with all

ti = 1.

16.3.2 Covering cube by affine hyperplanes

An affine hyperplane is a set of vectors H = {x ∈ Rn : a, x = b} with

a ∈ Rn and b ∈ R. How many such hyperplanes do we need to cover {0, 1}n?

If we have no further restrictions on the covering, then just two hyperplanes

H0 = {x ∈ Rn : e1 , x = 0} and H1 = {x ∈ Rn : e1 , x = 1} are enough,

where e1 = (1, 0, . . . , 0) is the first unit vector. But what if we, say, require

that the all-0 vector 0 remains uncovered? In this case n hyperplanes Hi =

{x ∈ Rn : ei , x = 1}, i = 1, . . . , n are still enough. It turns out that this is

already optimal.

Theorem 16.12. Suppose that the hyperplanes H1 , H2 , . . . , Hm in Rn avoid 0,

but otherwise cover all 2n − 1 vertices of the cube {0, 1}n. Then m ≥ n.

Proof. As 0 is not contained in Hi , we have that each hyperplane is of the

form Hi = {x ∈ Rn : ai , x = 1} for some ai ∈ Rn . Assume that m < n, and

consider the polynomial

m



f (x) =



n



(1 − ai , x ) −



i=1



(1 − xi ) .

i=1



The degree of this polynomial is clearly n (since we assumed that m < n) and

the coefficient at x1 · · · xn is (−1)n+1 = 0. When applied with Si = {0, 1}

and ti = 1, the Combinatorial Nullstellensatz implies that there must be a

point x ∈ {0, 1}n for which f (x) = 0. We have x = 0, as f (0) = 1 − 1 = 0.

But then ai , x = 1 for some i (as x is covered by some Hi ), implying that

f vanishes at this point, a contradiction.

16.3.3 Regular subgraphs

A graph is p-regular if all its vertices have degree p. The following sufficient

condition for a graph to contain a regular subgraph was derived by Alon,

Friedland and Kalai (1984) using the Combinatorial Nullstellensatz.

Theorem 16.13. Let G = (V, E) be a graph. Assume that G has no self-loops

but multiple edges are allowed. Let p be a prime number. If G has average

degree bigger than 2p−2 and maximum degree at most 2p−1, then G contains

a spanning p-regular subgraph.

Proof. Associate each edge e of G with a variable xe and consider the polynomial

p−1



1−



f=

v∈V



av,e xe

e∈E







(1 − xe )

e∈E



232



16 The Polynomial Method



over GF (p), where av,e = 1 if v ∈ e and av,e = 0 otherwise. Note that

deg(f ) = |E|, since the degree of the first product is at most (p − 1)|V | <

|E|, by the assumption on the average degree 2|E|/|V | of G. Moreover, the

coefficient of e∈E xe in f is (−1)|E|+1 = 0.

We can therefore apply the Combinatorial Nullstellensatz with Se = {0, 1}

and te = 1 for all e ∈ E and obtain a (0, 1)-vector x = (xe : e ∈ E) such that

f (x) = 0. Consider the spanning subgraph H consisting of all edges e ∈ E for

which xe = 1. As f (0) = 0, we conclude that x = 0 and H is non-empty. The

second summand in f (x) is therefore zero and it follows from Fermat’s Little

Theorem (Exercise 1.15) that

e∈E av,e xe ≡ 0 mod p for every vertex v.

Therefore, in the subgraph H all degrees are divisible by p, and since the

maximum degree is smaller than 2p, all positive degrees are precisely p, as

needed.

16.3.4 Sum-sets

The Cauchy-Davenport Theorem, which has numerous applications in Additive Number Theory, is the following. Given two sets A and B of elements of

some field F, their sum-set is the set A + B = {a + b : a ∈ A, b ∈ B}.

Theorem 16.14 (Cauchy–Davenport). If p is a prime, and A, B are two

non-empty subsets of Zp , then

|A + B| ≥ min{p, |A| + |B| − 1} .

We will see in Sect. 25.3.1 that this theorem is just a special case of Kneser’s

theorem. Here we show how to derive this theorem from the Combinatorial

Nullstellensatz.

Proof. If |A| + |B| > p the result is trivial, since in this case for every x ∈ Zp

the two sets A and x − B intersect, implying that A + B = Zp . Assume,

therefore, that |A| + |B| ≤ p and suppose that |A + B| ≤ |A| + |B| − 2. Let

C be a subset of Zp satisfying A + B ⊆ C and |C| = |A| + |B| − 2. Define

a polynomial f (x, y) = c∈C (x + y − c) and observe that by the definition

of C,

f (a, b) = 0 for all (a, b) ∈ A × B.

(16.5)

Put t1 = |A| − 1, t2 = |B| − 1 and note that the coefficient of xt1 y t2 in f is

the binomial coefficient

t 1 + t2

t1



=



|A| + |B| − 2

|A| − 1



which is nonzero in Zp , since |A| + |B| − 2 < p (see Exercise 1.14). We can

therefore apply the Combinatorial Nullstellensatz (with n = 2, S1 = A and

S2 = B) and obtain a poinr (a, b) ∈ A×B for which f (a, b) = 0, contradicting

(16.5) and completing the proof.



16.3 Combinatorial Nullstellensatz



233



The Cauchy–Davenport theorem was extended in many ways. Let us mention one important result in that direction. For a subset A ⊆ Zp and a natural

number 1 ≤ k ≤ |A|, let sk (A) be the number of elements b ∈ Zp that can be

represented as a sum b = a1 + · · · + ak of k distinct elements of A.

Theorem 16.15 (Dias da Silva and Hamidoune 1994). If p is a prime then,

for every subset A ⊆ Zp and every 1 ≤ k ≤ |A|,

sk (A) ≥ min{p, k|A| − k 2 + 1} .

16.3.5 Zero-sum sets

Using the pigeonhole principle, one can show that any sequence a1 , . . . , an of

n integers contains a non-empty consecutive subsequence ai , . . . , ai+m whose

sum is divisible by n.

To show this, make n pigeonholes labeled from 0 up to n − 1 and place the

n sequences

(a1 ), (a1 , a2 ), . . . , (a1 , a2 , . . . , an )

into the pigeonholes corresponding to the remainder when the sum is divided

by n. If any of these sequences is in the pigeonhole 0 then the sum of its

numbers is divisible by n. If not, then the n sequences are in the other n − 1

pigeonholes. By the pigeonhole principle some two of them, (a1 , a2 , . . . , ar )

and (a1 , a2 , . . . , as ) with r < s, must lie in the same pigeonhole, meaning that

the sum ar+1 + ar+2 + · · · + as is divisible by n.

A question of a similar flavor is the following one. Given a natural number

n, what is the smallest N such that any sequence of N integers contains a

subsequence of n (not necessarily consecutive) numbers whose sum is divisible

by n? That is, this time we want to find a subsequence of a given length n.

The sequence 0n−1 1n−1 of n − 1 copies of 0 and n − 1 copies of 1 shows that

N ≥ 2n − 1. It turns out that this lower bound is also an upper bound for

the sequence length N .

Theorem 16.16 (Erdős–Ginzburg–Ziv 1961). Any sequence of 2n − 1 integers contains a subsequence of cardinality n, the sum of whose elements is

divisible by n.

There are several different proofs of this theorem – the interested reader

can find them, as well as some interesting extensions of this result to higher

dimensions, in the paper of Alon and Dubiner (1993). The original proof was

based on the Cauchy–Davenport theorem.

First proof of Theorem 16.16. We will first prove the theorem only for the

case when n = p is a prime number, and then show how the general case

reduces to it.

Let a1 ≤ a2 ≤ . . . ≤ a2p−1 be integers. If ai = ai+p−1 for some i ≤ p − 1,

then ai + ai+1 + · · · + ai+p−1 = pai = 0 (in Zp ) and the desired result



234



16 The Polynomial Method



follows. Otherwise, define Ai := {ai , ai+p−1 } for i = 1, . . . , p − 1. By repeated

application of the Cauchy-Davenport theorem, we conclude that

|A1 + A2 + · · · + Ap−1 | ≥ min{p, |A2 + · · · + Ap−1 | + 1}

≥ min{p, |A3 + · · · + Ap−1 | + 2}

···

≥ min{p, |Ap−1 | + p − 2} = p ,

and hence, every number from Zp is a sum of precisely p − 1 of the first 2p − 2

elements of our sequence. In particular, the number −a2p−1 is such a sum,

supplying the required p-element subset whose sum is 0 in Zp .

The general case may be proved by induction on the number of primes

in the prime factorization of n. Put n = pm where p is a prime, and let

a1 , . . . , a2n−1 be the given sequence. By the result for the prime case, each

subset of 2p − 1 members of the sequence contains a p-element subset whose

sum is 0 modulo p. Therefore, we can find pairwise disjoint p-element subsets

I1 , . . . , I of {1, . . . , 2n − 1}, where

aj ≡ 0 mod p

j∈Ii



for each i = 1, . . . , . Moreover, ≥ 2m − 1 since otherwise the number of left

elements would still be 2pm − 1 − (2m − 2)p = 2p − 1, and we could choose

the next subset I +1 . Now define a sequence b1 , . . . , b2m−1 where

bi =

j∈Ii



aj

p



(recall that each of these sums is divisible by p). By the induction hypothesis

this new sequence has a subset {bi : i ∈ J} of |J| = m elements whose sum

is divisible by m, and the union of the corresponding sets {aj : j ∈ Ii } with

i ∈ J supplies the desired n-element subset of our original sequence, whose

sum is divisible by n = pm.

Theorem 16.16 can also be derived using the Chevalley–Warning theorem

about the zeroes of multivariate polynomials (see Theorem 16.9).

Second proof of Theorem 16.16 (Alon 1995). We will prove the theorem only

for a prime n = p; the general case reduces to it (see the first proof of

Theorem 16.16).

Let a1 , a2 , . . . , a2p−1 be integers, and consider the following system of two

polynomials in 2p − 1 variables of degree p − 1 over Fp :

2p−1

i=1



ai xp−1

= 0,

i



2p−1

i=1



xp−1

= 0.

i



235



Exercises



Since 2(p − 1) < 2p − 1 and x1 = x2 = · · · = x2p−1 = 0 is a common

solution, Theorem 16.9 implies the existence of a nontrivial common solution

(y1 , . . . , y2p−1 ). Since p is a prime, Fermat’s Little Theorem (see Exercise 1.15)

tells us that xp−1 = 1 in Fp for every x ∈ Fp , x = 0. So, if we take I = {i : yi =

0} then the first equation ensures that i∈I ai = 0, while the second ensures

that |I| ≡ 0 mod p, and hence, that |I| = p because |I| ≤ 2p − 1. This

completes the proof of the theorem for prime n.



Exercises

16.1. Let p ≥ 2 be a prime and consider the field Fp . From Algebra we know

that there is an a ∈ Fp such that Fp \ {0} = ai : i = 0, 1, . . . , p − 2 . Use this

fact together with Fermat’s Little Theorem to prove that, for every t ≤ p − 2,

n

t

z i = (z n+1 − 1)/(z − 1).

x∈Fp x = 0 in Fp . Hint:

i=0

16.2 (Low local degree polynomials). Let f be a nonzero polynomial in n

variables over a field F. Suppose that the maximum exponent of each of its

variables in f does not exceed d. (Hence, the degree of f may be up to dn

which makes the bound given by Lemma 16.3 trivial, if |S| ≤ n.) Show that

we still have the following upper bound: For any subset S ⊆ F of size |S| ≥ d,

|{x ∈ S n : f (x) = 0}| ≥ (|S| − d)n .

Hint: Argue by induction on n. In the induction step take a point (a1 , . . . , an ) ∈ Fn on

which f (a1 , . . . , an ) = 0, and consider two polynomials:

f0 (x1 , . . . , xn−1 ) := f (x1 , . . . , xn−1 , an )

f1 (xn ) := f (a1 , . . . , an−1 , xn ).



16.3. Show that the bound in Exercise 16.2 is the best possible. Hint: Consider

the polynomial f (x1 , . . . , xn ) =



d

(x

i=1 1



− i) · · ·



d

(x

i=1 n



− i) .



16.4. Use Exercise 16.2 to show the following: If S a subset of F that has

d + 1 elements, then any nonzero polynomial of local degree d has a nonzero

point in S n .

16.5. Prove the following “granulated” version of the result established in

Exercise 16.2. Let f ∈ F[x1 , . . . , xn ] be a polynomial, and let ti be the maximum degree of xi in f . Let Si ⊆ F with |Si | ≥ ti . If f (x) = 0 for all n-tuples

x ∈ S1 × · · · × Sn , then f (x) = 0 for all x ∈ Fn .

16.6. Let f (x1 , . . . , xn ) be a multivariate polynomial over a field F with the

degree sequence (d1 , . . . , dn ), which is defined as follows: let d1 be the maximum exponent of x1 in f , and let f1 (x2 , . . . , xn ) be the coefficient of xd11 in f ;

then, let d2 be the maximum exponent of x2 in f1 , and f2 (x3 , . . . , xn ) be the

coefficient of xd22 in f1 ; and so on. Suppose that f is not the zero polynomial,



236



16 The Polynomial Method



and let S1 , . . . , Sn ⊆ F be arbitrary subsets. For r i ∈ Si chosen independently

and uniformly at random, show that

Pr [f (r1 , . . . , rn ) = 0] ≤



d2

dn

d1

+

+ ···+

.

|S1 | |S2 |

|Sn |



16.7. (R. Freivalds 1977). Suppose that somebody gives us three n × n matrices A, B, C with real entries and claims that C = A · B. We are too busy

to verify this claim exactly and do the following. We take a random vector

r of length n whose entries are integers chosen uniformly from the interval

{0, 1, . . . , N − 1}, and check whether A · (B · r) = C · r. If this is true we

accept the claim, otherwise we reject it. How large must N be set to make the

probability of false acceptance smaller than 1/100? Hint: Consider the matrix

X = A · B − C. If C = A · B then X has a row x = 0. Take a scalar product of this

row with the random vector r, observe that Pr [X · r = 0] ≤ Pr [x · r = 0], and apply

Exercise 16.2.



17. Combinatorics of Codes



In this chapter we will discuss some extremal properties of error-correcting

codes. We will also use expander graphs to construct very easily decodable

codes.



17.1 Error-correcting codes

Error-correcting codes enable one to store or transmit a message in such a

way that one can later recover the message even if it is partially corrupted,

that is, up to some number t of bits are flipped by noise. Messages are strings

of some fixed length k over some alphabet A. In order to be able to recover

corrupted messages, we encode our messages by strings of length n > k over

A (or some other alphabet). That is, we take a particular subset C ⊆ An

(called a code) and assign to each message w its own codeword x = xw ∈ C.

During the transmission some bits may be flipped (by noise, by an adversary or whatever). So, the receiver is presented with a corrupted version

x ∈ An of the original codeword x ∈ C. The receiver knows the set of all

codewords C, as well as the encoding algorithm. He also knows that the received vector x can differ from the original x codeword in at most t bits.

What conditions must the code C fulfill in order that we may recover the

original codeword x?

Here the notion of Hamming distance comes into the play. The Hamming

distance dist(x, y) between two strings x and y of the same length is just

the number of positions in which these two strings differ. The minimum

distance, dist(C), of a subset C ⊆ An is the minimal Hamming distance

dist(C) between any pair of distinct strings in C.

The key observation is that, if dist(C) ≥ 2t + 1, then the receiver can

(at least in principle) reconstruct the original codeword x from the received,

possibly corrupted vector x . The point is that x is the only codeword in the

Hamming ball Bt (x ) = {y ∈ An : dist(x , y) ≤ t} of radius t around the

S. Jukna, Extremal Combinatorics, Texts in Theoretical Computer Science.

An EATCS Series, DOI 10.1007/978-3-642-17364-6_17,

© Springer-Verlag Berlin Heidelberg 2011



237



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