AbstractFor a set C of words of length 4 over an alphabet of size n, and for a, b∈C, let D(a, b) be the set of all descendants of a and b, that is, all words x of length 4 where xi∈{ai, bi} for all 1⩽i⩽4. The code C satisfies the Identifiable Parent Property if for any descendant of two code-words one can identify at least one parent. The study of such codes is motivated by questions about schemes that protect against piracy of software. Here we show that for any ε>0, if the alphabet size is n>n0(ε) then the maximum possible cardinality of such a code is less than εn2 and yet it is bigger than n2−ε. This answers a question of Hollmann, van Lint, Linnartz, and Tolhuizen. The proofs combine graph theoretic tools with techniques in additive number theory

Alon, Noga

Fischer, Eldar

Szegedy, Mario

Elsevier - Publisher Connector 

Journal of Combinatorial Theory, Series A 95, 349359 (2001)
Parent-Identifying Codes
Noga Alon1
Department of Mathematics, Raymond and Beverly Sackler Faculty of Exact Sciences,
Tel Aviv University, Tel Aviv, Israel
E-mail: nogamath.tau.ac.il
Eldar Fischer
NEC Research Institute, 4 Independence Way, Princeton, New Jersey 08540; and DIMACS
E-mail: fischerresearch.nj.nec.com
and
Mario Szegedy
School of Mathematics, Institute for Advanced Study, Princeton, New Jersey 08540
E-mail: szegedyias.edu
Communicated by the Managing Editors
Received May 12, 2000; published online June 18, 2001
For a set C of words of length 4 over an alphabet of size n, and for a, b # C, let
D(a, b) be the set of all descendants of a and b, that is, all words x of length 4
where xi # [ai , bi] for all 1i4. The code C satisfies the Identifiable Parent
Property if for any descendant of two code-words one can identify at least one
parent. The study of such codes is motivated by questions about schemes that
protect against piracy of software. Here we show that for any =>0, if the alphabet
size is n>n0(=) then the maximum possible cardinality of such a code is less than
=n2 and yet it is bigger than n2&=. This answers a question of Hollmann, van Lint,
Linnartz, and Tolhuizen. The proofs combine graph theoretic tools with techniques
in additive number theory.  2001 Academic Press
doi:10.1006jcta.2001.3177, available online at http:www.idealibrary.com on
349
0097-316501 35.00
Copyright  2001 by Academic Press
All rights of reproduction in any form reserved.
1 Research supported in part by a USAIsraeli BSF grant, by the Israel Science Founda-
tion, and by the Hermann Minkowski Minerva Center for Geometry at Tel Aviv University.
Part of this work was done while visiting the School of Mathematics, Institute for Advanced
study, Princeton, NJ 08540.
1. INTRODUCTION
Let |N |=n and CN4. For a, b # C define the set D(a, b) of all
descendants of a, b as
D(a, b)=[x # N4 | xi # [ai , b i] for 1i4 ].
We say that the code C has the Identifiable Parent Property (IPP) if for
every descendant one can always identify at least one of the parents, that
is, for every x # a, b # C D(a, b) there is a p # C such that if a, b # C and
x # D(a, b) then p # [a, b]. Equivalently, as mentioned in [3], C has the
IPP if and only if:
(IPP1) For every distinct a, b, c # C there is an 1i4 such that
ai , bi , ci are all distinct, and
(IPP2) for every a, b, c, d # C with [a, b] & [c, d ]=< there is an
1i4 such that [ai , bi] & [ci , di]=<.
Define
f (n)=max[ |C |: CN4 has IPP].
The study of f (n) is motivated by questions about schemes that protect
against piracy of software. The authors of [3] proved that
(1+o(1)) n32 f (n)n2 (1)
and raised the problem of closing the gap between the upper and lower
bounds. Here we show that for every =>0 there is an n0=n0(=) such that
for every n>n0 ,
f (n)=n2 (2)
and yet
f (n)n2&=. (3)
2. THE UPPER BOUND
It is convenient to distinguish the alphabets that are used in each coor-
dinate. Let Ni be the alphabet used in coordinate i (1i4). |Ni |=n, and
Ni are pairwise disjoint. Thus CN1_N2_N3_N4 . By omitting all
members of C that have a coordinate that does not belong to any other
code word we omit at most 4n words, and may assume now that:
350 ALON, FISCHER, AND SZEGEDY
(V) Each letter l # N1 _ N2 _ N3 _ N4 appears in at least two
members of C (or does not appear at all).
Fact 2.1. No two members of C have three common coordinates.
Proof. If a, b # C with a1=b1 , a2=b2 , a3=b3 , then by assumption (V)
there is a c # C, c{a such that c4=a4 . But then [a, b, c] violate
(IPP1). K
Fact 2.2. If there are distinct i1 , i2 # [1, 2, 3, 4] and two distinct words
a, c # C with ai1=ci1 , ai2=ci2 , then there are no distinct words b, d # C such
that bj1=dj1 , bj2=dj2 , where [ j1 , j2]=[1, 2, 3, 4]"[i1 , i2].
Proof. Assume the opposite. Then a, b, c, d violate (IPP2) if all words
are distinct. If, say, a=b, then [a, c, d ] violate (IPP1). K
Fact 2.3. For every distinct i1 , i2 # [1, 2, 3, 4],
|[x # C | (_y # C)(( y{x) 7 ( yi1=xi1) 7 ( yi2=x i2))]|2n&1.
Proof. Assume the fact does not hold for say, i1=1, i2=2. Construct
a bipartite graph G with color classes N3 and N4 as follows: for each
x # [x # C | (_y # C)(( y{x) 7 (x1= y1) 7 (x2= y2))] the pair x3x4 is an
edge of G. By assumption and Fact 2.2, G has more than 2n&1 edges,
hence it has a cycle. Therefore, since it is bipartite, it contains a path of
length 3. Let x4(x3= y3)( y4=z4) z3 be that path, where these coordinates
arise from appropriate x, y, z # C. Let x$ # C be such that x$1=x1 , x$2=x2 ,
x${x. If x$=z then [x, y, z] violate (IPP1), otherwise [x$, y, x, z] violate
(IPP2) with the grouping [x, z], [x$, y]. K
To prove the upper bound, we also need the following result, proved in
Alon et al. [1] by applying the regularity lemma of Szemere di [5].
Lemma 2.4 [1, Proposition 4.4]. For every #>0 and every integer k
there exists a $=$(k, #)>0 such that every graph G on n vertices containing
less than $nk copies of the complete graph Kk on k vertices, contains a set
of less than #n2 edges whose deletion destroys all copies of Kk in G.
We can now prove the required upper bound for f (n).
Theorem 2.5. For every =>0 there exists n0=n0(=) such that f (n)<=n2
for every n>n0 .
351PARENT-IDENTIFYING CODES
Proof. Let CN1_N2_N3_N4 have the IPP, |C|= f (n), with Ni
being pairwise disjoint and satisfying |N1 |=|N2 |= |N3 |=|N4 |=n. By
Facts 2.1 and 2.3 we can omit from C at most 6 } 2n+4n=16n members to
get a code C$, |C$| f (n)&16n that has IPP in which no two code words
share more than one coordinate. Let H be the 4-partite graph on the
classes of vertices N1 , N2 , N3 , N4 obtained by taking the edge-disjoint
union of all K4 copies [x1 , x2 , x3 , x4] for every x # C$.
This graph has at least ( f (n)&16n) 6 edges, and as it is the edge-disjoint
union of f (n)&16n copies of K4 , one has to delete at least f (n)&16n of its
edges to destroy all copies of K4 contained in the graph. If we assume that
f (n)>=n2, this implies, for sufficiently large n, that we have to delete at
least =2n
2 edges of H to destroy all copies of K4 .
By Lemma 2.4 (with k=4, #==2), this implies that H contains at least
$n4 distinct copies of K4 for a constant $=$(=)>0. Among these K4 copies,
only f (n)n2 correspond each to one x # C. Similarly, the number of K4
copies that contain at least two edges arising from the same x # C is at
most O(n3), since there are at most n2 ways to choose x, at most 15 ways
to choose two of its edges, and this determines already at least three ver-
tices of the K4 . It follows that H contains a copy of K4 in which every edge
comes from a different x # C. In particular, if a1 , a2 , a3 , a4 are the vertices
of this K4 , then there exist distinct x, y, z, w # C such that
x1=a1 , y3=a3 , z2=a2 , w1=a1 ,
x2=a2 , y4=a4 , z3=a3 , w4=a4 .
But then x, y, z, w violate (IPP2), contradicting the fact that C has IPP.
Thus f (n)=n2 for n>n0(=), completing the proof. K
Remark 2.6. The proof and the known bounds in the proof of the
regularity lemma actually show that
f (n)=O \ n
2
(log* n)15+ ,
where
log* n=min[k | log2 log2 } } } log2
k times
n1].
3. THE LOWER BOUND
Our main tool here is an arithmetic lemma proven using the method of
Behrend [2], and its extension by Ruzsa [4], with some modifications.
352 ALON, FISCHER, AND SZEGEDY
A linear equation with integer coefficients
: ai xi=0 (4)
in the unknowns x i is homogeneous if  ai=0. If XN=[1, 2, ..., n], we
say that X has no non-trivial solution to (4), if whenever xi # X and
 ai xi=0, it follows that all xi are equal.
Note that if X has no non-trivial solution to (4), then the same holds for
any shift (X+u) & N (where u is positive, negative or zero).
We need the following simple fact, which follows from the convexity of
the function g(t)=t2.
Fact 3.1. Let p1 , p2 , ..., pk be k strictly positive reals whose sum is 1,
and suppose ki=1 pi ri=r, where r1 , r2 , ..., rk are reals. Then
:
k
i=1
pir2i r
2,
and the inequality is strict unless r1=r2= } } } =rk=r.
Proof. Put ri=r+=i , then
:
k
i=1
pi (r+= i)=r+ :
k
i=1
pi=i=r
and hence ki=1 pi =i=0. It thus follows that
:
k
i=1
p ir2i = :
k
i=1
pi (r+=i)2= :
k
i=1
pir2+2r :
k
i=1
pi =i
+ :
k
i=1
pi =2i =r
2+ :
k
i=1
pi=2i r
2,
and the last inequality is strict unless all numbers =i are 0. K
Lemma 3.2 (Main Lemma). For q=W2- log n X there exist:
(1) a set X1 N, |X1 |n2O(log
34 n) with no non-trivial solution to
2x+3y+qz&(q+5) w=0; (5)
(2) a set X2 N, |X2 |n2O(log
34 n) with no non-trivial solution to
5x+(q+3) y&3z&(q+5) w=0; (6)
353PARENT-IDENTIFYING CODES
(3) a set X3 N, |X3 |n2O(log
34 n) with no non-trivial solution to
5x+qy&2z&(q+3) w=0. (7)
Proof. To prove part (1) we apply the method of Behrend [2]. Let d
be an integer (to be chosen later) and define
X1={ :
k
i=0
x i d i | x i<
d
q+5
(0ik) 7 :
k
i=0
x2i =B= ,
where k=wlog nlog dx&1 and B is chosen to maximize the cardinality of
X1 . If x, y, z, w # X1 satisfy (5) and
x= :
k
i=0
xi d i, y= :
k
i=0
y i d i, z= :
k
i=0
zi d i, w= :
k
i=0
wi d i,
then
2xi+3yi+qzi=(q+5) wi
for every 0ik. But then, by Fact 3.1 (with k=3, p1= 2q+5 , p2=
3
q+5 ,
and p3=
q
q+5),
2x2i +3y
2
i +qz
2
i (q+5) w
2
i
for every 0ik, and each such inequality is strict unless x i= yi=zi=wi .
As  x2i = y
2
i = z
2
i = w
2
i , this implies that xi= yi=zi=wi for
0ik, showing that X1 has no non-trivial solution to (5). The size of X1
satisfies
|X1 |
n
d 2(q+5)k+1 (k+1)(d 2(q+5)2)

n
(q+5) log n log d d 4 log n
.
Take d=w2- log n log qx (>>q) to conclude that
|X1 |
n
2O(- log n log q)
. (8)
In order to prove Part (2) we apply the method of Ruzsa [4]. By
Behrend’s method (that is, by an obvious modification of the constants in
354 ALON, FISCHER, AND SZEGEDY
the argument given in the proof of Part (1) above) there exists Q
[1, 2, ..., q5] satisfying |Q|q2O(- log q) with no non-trivial solution to
5x= y+3z+w. Define
X2={ :
k
i=0
x i (q+4) i | xi # Q= ,
where k=wlog n log(q+4)x&1. Note that
|X2 |=|Q| k+1
n
2O(log n- log q)
. (9)
Suppose now that there is a non-trivial solution x, y, z, w # X2 of (6),
where
x= :
k
i=0
xi (q+4) i, y= :
k
i=0
yi (q+4) i,
z= :
k
i=0
zi (q+4) i, w= :
k
i=0
wi (q+4) i.
Then
:
k
i=0
5x i (q+4) i+(q+3) :
k
i=0
yi (q+4) i
= :
k
i=0
3zi (q+4) i+(q+5) :
k
i=0
wi (q+4) i.
Let j be the minimum index such that not all [xi , yi , zi , wi] are equal.
Then
:
k
i= j
5xi (q+4) i+(q+3) :
k
i= j
yi (q+4) i
= :
k
i= j
3zi (q+4) i+(q+5) :
k
i= j
wi (q+4) i.
Reducing modulo (q+4) j+1 we conclude that
5xj (q+4) j#yj (q+4) j+3zj (q+4) j+wj (q+4) j (mod (q+4) j+1).
But both sides are less than (q+4) j+1, as x j , y j , zj , wj 15q, hence this is
an equality (and not only a modular equality):
5xj (q+4) j= yj (q+4) j+3zj (q+4) j+wj (q+4) j.
355PARENT-IDENTIFYING CODES
Dividing by (q+4) j we get 5x j= yj+3zj+wj , contradicting the assump-
tion that Q has no non-trivial solution to this equation. Thus X2 has no
non-trivial solution to (6), as needed.
The proof of Part (3) is analogous to that of Part (2). Here we start with
Q/[1, 2, ..., 15q] having no non-trivial solution to 5x= y+2z+2w and
satisfying |Q|q2O(- log q). Then we take X3=[ki=0 xi (q+1)
i | x i # Q]
where k=wlog n log(q+1)x&1.
As before,
|X3 |
n
2O(log n- log q)
. (10)
If we assume that x, y, z, w # X3 form a non-trivial solution to (7), and
define xi , yi , zi , wi and j as before, we conclude, by reducing modulo
(q+1) j+1, that
5xj (q+1) j#yj (q+1) j+2zj (q+1) j+2wj (q+1) j (mod (q+1) j+1).
As before, this is actually an equality, implying that 5xj= yj+2zj+2wj and
supplying the desired contradiction.
This completes the proof of the lemma. Since
q=W2- log nX
we obtain, from (8), (9), and (10), that
|X1 |, |X2 | , |X3 |
n
2O((log n)
34)
. K
Corollary 3.3. There exists a set X/[1, ..., n] satisfying
|X |
n
2O((log n)
34)
such that X has no non-trivial solution to (5), no non-trivial solution to (6),
and no non-trivial solution to (7).
Proof. Take two integers &nu2n and &nu3n randomly,
uniformly and independently. X=X1 & (X2+u2) & (X3+u3) has no non-
trivial solution to any of the above equations, and each x # X1 has prob-
ability 0(2&O((log n)34)) to lie in the intersection. The result thus follows
from the linearity of the expectation. K
356 ALON, FISCHER, AND SZEGEDY
Theorem 3.4. The function f (n) satisfies
f (n)
n2
2O((log n)
34)
. (11)
Proof. It is more convenient to show that
f (n2- log n+6n)
n2
2O((log n)
34)
,
which clearly gives (11).
Put q=W2- log nX and let X be as in the corollary. Define
C=[( p, p+2x, p+5x, p+(q+5) x) | 1pn, x # X].
Then C/N 4 for N=[1, 2, ..., (q+6) n]. Clearly
|C|
n2
2O((log n)
34)
.
We claim that C has the IPP. Indeed, no two words in C share more
than one coordinate. Thus, if a, b, c # C are distinct they cannot violate
(IPP1) since otherwise for every 1i4 there exists a pair among a, b, c
sharing the same coordinate in place i, implying by the pigeonhole prin-
ciple that some pair of words shares at least 2 coordinates, which is
impossible.
It remains to check (IPP2). Suppose that
a=( p1 , p1+2x, p1+5x, p1+(q+5) x),
b=( p2 , p2+2y, p2+5y, p2+(q+5) y),
c=( p3 , p3+2z, p3+5z, p3+(q+5) z),
d=( p4 , p4+2w, p4+5w, p4+(q+5) w)
satisfy [a, b] & [c, d ]=< and yet [ai , bi] & [ci , di]{< for all 1i4.
Choose gi # [ai , bi] & [ci , di] for each i. No word can share 3 coordinates
with g=(g1 , g2 , g3 , g4). Indeed, if for example, a1= g1 , a2= g2 and
a3= g3 then, as gi # [ci , di] for every i, either c or d have to agree with a
on at least 2 coordinates, which is impossible.
Since gi # [ai , bi] and gi # [ci , d i] for every i, each of the 4 words
a, b, c, d agrees with g=(g1 , g2 , g3 , g4) on exactly 2 coordinates. More-
over, the indices of those common coordinates of a and g, and those of b
and g, are disjoint (as together they have to cover all 4 coordinates); and
357PARENT-IDENTIFYING CODES
the same occurs with those of c and g with respect to those of d and g. It
follows that up to symmetry there are 3 possible cases.
Case 1.
a1= g1 , b3= g3 , c2= g2 , d1= g1 ,
a2= g2 , b4= g4 , c3= g3 , d4= g4 .
Case 2.
a1= g1 , b2= g2 , c2= g2 , d1= g1 ,
a3= g3 , b4= g4 , c3= g3 , d4= g4 .
Case 3.
a1= g1 , b2= g2 , c1= g1 , d3= g3 ,
a3= g3 , b4= g4 , c2= g2 , d4= g4 .
In Case 1, by noting that
(g2& g1)+(g3& g2)+(g4& g3)&(g4& g1)=0
and that
g2& g1=a2&a1=2x, g3& g2=c3&c2=3z,
g4& g3=b4&b3=qy, g4& g1=d4&d1=(q+5) w,
we conclude that
2x+3z+qy&(q+5) w=0.
Thus x= y=z=w by the construction of X that has no non-trivial
solution to (5). But then it follows that a=d, in contradiction to [a, b] &
[c, d ]=<.
Similarly, Case 2 leads by the fact that X has no non-trivial solution to
(6), to the fact that x= y=z=w and hence again to the contradiction
a=d. Case 3 leads to x= y=z=w as X has no non-trivial solution to (7),
giving the contradiction a=c. This completes the proof. K
ACKNOWLEDGMENTS
We thank Benny Sudakov, Sergei Konyagin, and Endre Szemere di for helpful discussions.
Note added in proof. As observed by S. Konyagin, the lower bound given in Theorem 3.4
can be slightly improved to
f (n)
n2
2O((log n)
23)
by proving the first part of Lemma 3.2 using the method applied in the proofs of its second
and third part.
358 ALON, FISCHER, AND SZEGEDY
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359PARENT-IDENTIFYING CODES


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