Title: Parent Identifying Codes The case of multiple parents
1Parent Identifying CodesThe case of multiple
parents
2Definitions
- An alphabet Q, Qq
- A code C?Qn , of length n
- A parameter t
- A set of codewords P p1,p2, ,pt ? C
- (called parents)
3Descendants
- The descendants of P
- D(P)
Q0,1,2,3,4,5 t3 n5
4Parent Identifying Codes (def.)
- A code C ? Qn has the identifiable parent
property (IPP) of order t if for any n-word s ?Qn
either - It is not a descendant of any set of t codewords
in C - There exists p?C such that
- p can be identified from s
5IPP - Example
- IPP definition
- ? s?Qn either
- Not a descendant
- ? p?C such that
t 2 n 4 Q 1,2,3,4
? p c4
P1 c1,c4 P2 c2,c3 P3 c1,c2
P1 c1,c4 P2 c3,c4
? Not a descendant
? C does not have 2-IPP
6IPP - Example 2
- Any repetition code has IPP of any order t
c1
c2
C
cq
7Binary IPP codes
- For t gt1
- There are no binary t-IPP codes of size gt 2
x
y
z
s Maj(x,y,z)
- Generally, we require q gt t
8IPP codes - Motivation
Distributor
9Fingerprinting (cont.)
- At most t traitors
- The traitors compare their copies
- Any non-constant symbol can be changed
- Narrow case
- The new symbol is taken from one of the parents
- Zero-Error traitor tracing using IPP codes
- Wide case
- No restriction on the new symbol
- Zero-Error traitor tracing is impossible
10Maximum cardinality IPP codes
- Define
- ft(n,q) max C?Qn C has t-IPP
- Large Qq
11Previous Work
- Chor, Fiat, Naor Traitor tracing (1994)
- Boneh, Shaw Fingerprinting (1998)
- Hollman, Van-Lint, Linnartz, Tolhuizen (1998)
- 2-IPP codes,
- f2(3,q) (3-o(1))q
- ?(q1.5) lt f2(4,q) lt O(q2)
- c(q/4)n/3 f2(n,q) 3q?n/3?
12Previous Work (cont.)
- Alon, Fischer, Szegedy (2001)
- q2-? lt f2(4,q) lt ? q2
- Barg , Cohen, Encheva, Kabatiansky, Zemor (2001)
Alon, Cohen, Krivelevich, Litsyn (2002) - t gt 2, fixed q large n
- Partially Hashing
- Staddon, Stinson, Wei (2001)
- Traceability codes
13General bound for ft(n,q)
- Define
- s(t)
- That is
- Theorem 1
- There exist two functions c1(t) and c2(t), such
that for every n,q - (c1(t)?q) ? ft(n,q) ? c2(t)?q? ?
14The lower bound (BCEKZ, 2001)
- Minimal configuration
- Let ?(X1,X2,,Xm) be a collection of subsets of
codewords such that Xi? t and -
-
15Minimal Configurations
- Suppose C does not have t-IPP
- A parent of s cannot be identified
- There are Pi s such that s ? D(Pi ), ?Pi ??
- Remove Pi until it is a minimal configuration
- ? There is a minimal configuration Pi
consisting of parents sets, with a common
descendant
16The lower bound (cont.)
- Recall
- Lemma 1
- Suppose ? is a minimal configuration, then
-
17Lemma 1
- Proof
- There exist B(X) b1,b2,,bm
- such that
-
- All bi s are different
- B(X)\ bi ? Xi
- m 1 ? t
18Lemma 1 (cont.)
QED Lemma 1
19Hashing
i
U
- Definition
- A code C?Qn is u - hashing if for any subset
U?C, Uu, there is some coordinate - i? 1,,n such that
- ? x?U, y?U, x?y xi?yi
-
20Hashing
i
x
?Pj
- Lemma 2
- (s(t)1)- hashing ? t-IPP
- Proof
The son s
Suppose the code does not have t-IPP
?Contradicting example ( s, Pj s?D(Pj), ?Pj ?
)
?Minimal configuration (? ?Pj ? s(t)1 )
?By (s(t)1) - hashing, there is x??Pj (for
which xi si)
QED Lemma 2
21The lower bound (cont.)
- Probabilistic construction
- QED
- Pick a code at random
- Delete one codeword from any set that violates
the (s(t)1)-hashing property, to obtain a t-IPP
code - Calculate the expectation of the number of such
sets
22Theorem 1 - The upper bound
- We have to show ft(n,q) ? c2(t)?q ? ?
- Proof idea
- Reduce the problem to a problem on codes of
length s(t) - Bound the size of such codes
- ft(n,q) ? ft(s(t)?? ?,q) ?
- ? ft(s(t), q? ? ) ? c2(t)?q ? ?
Lemma 3
Lemma 4
23Reducing the problem to ft(s(t),q)
- Lemma 3
- For any a,t,n,q ft(na,q) ? ft(n,qa)
- Proof
- Every a symbols from Q ? One symbol from Qa
- The short (n) code with large (qa) alphabet
also has t-IPP.
n
a
a
a
a
24Bounding ft(s(t),q)
- Lemma 4
- ft(s(t),q) ? s(t) q ( ? ft(s(t),q) O(q)
) - Proof (t is even)
- Suppose we have a code C of size C gt s(t) q
that has the t-IPP - Stage 1
- Remove codewords until in each coordinate every
symbol appears at least 1 times
25Bounding ft(s(t),q) (cont.)
- Each coordinate is responsible for deleting at
most q codewords - We remove at most s(t) q codewords
26Bounding ft(s(t),q) (cont.)
- Contradict C has t-IPP
- Pick a set P of 1 distinct parents
- Construct a descendant s of P s?D(P)
- Every parent p? P can be replaced by
codewords - ? None of the parents p? P is essential
27Example (t8 ? s(t)( 1)2-1 24, 1 5)
s
28Bounding ft(s(t),q)
s
- Every parent p contributes at most
coordinates that no other parent does - p can be replaced by other codewords
- (? There is a parents set of size t without p)
- No essential parent
- C does not have t-IPP
QED Lemma 4, Theorem 1
29Bounding ft(s(t),q) - Remarks
- Similar calculations for odd t
- It is enough if every symbol appears twice
- Lemma 4 can be improved
- Real Lemma 4
- ft(s(t),q) ? s(t) q
30The case t 2
- s(2)3
- By Theorem 1
- c(q/4)n/3 f2(n,q) 3q?n/3?
- ns(2)3 The longest linear size code
- Hollman, VanLint, Linnartz, Tolhuizen 1998
f2(3,q) (3-o(1))q
31 f2(3,q)
- By Theorem 1 (Real Lemma 4) f2(3,q) ? 3q
- Constructing a 2-IPP code C ? Q3 of size (3-o(1))q
- Split the alphabet Q
- Q2, Q3 of size
- Q1 - the rest of the alphabet (
)
- C1 concatenating a symbol from Q1 and a couple
of symbols from Q2?Q3 (? C1 Q1 )
- C2, C3 Cyclic shifts of the codewords in C1
- C C1? C2 ? C3 (? C 3Q1(3-o(1))q)
32 f2(3,q) (cont.)
- C has 2-IPP
- We show how to identify s ?Q3 s parent
- We can tell from which Ci every symbol in s comes
- If the the symbols came from
- 1 set ? There is a symbol from Q1
- This symbol comes from a unique parent
- This parent can be identified
- 2 sets ? 2 coordinates from one set
- Only one parent came from that set
- This parent can be identified
- 3 sets ? not a descendant of at most 2 codewords
in C
33The case t 3
- s(3)5
- By Theorem 1
- When n s(3) 5
- f3(5,q) ? (5-o(1))q
-
34 f3(5,q)
- Theorem 2
- f3(5,q) (5-o(1))q
- Proof idea
- Similar construction to the one for f2(3,q)
- Split the alphabet Q into Q1,,Q5
- Q2,,Q5 are small (q3/4), Q1 is large ( (1-o(1)q
) - Construct a code B ? Q2 ? Q3?Q4 ? Q5 of size
Q1 - (with some properties)
- C1 - Concatenating a symbol from Q1 to a codeword
from B - Take all 5 cyclic shifts
35 The case ns(t)1 (? f2(4,q), f3(6,q))
- The first super-linear code for t2 is n4
- Theorem 1 ? (q1.33) f2(4,q) 3q2
- HLLT(1998) ? (q1.5) lt f2(4,q) lt O(q2)
- AFS (2001) ????
- The first super-linear code for t3 is n6
- Theorem 1
- Theorem 3
- (similar technique used by AFS for t2, including
Szemeredi Regularity Lemma)
36The case n ? s(t)
- When n ? t ft(n,q)q
- When nt1
- e.g. f3(4,q) ( o(1))q
-
37The case t 3
38Open Problems
- What is the exact power for n ? 0 (mod s(t))?
- Is ft(s(t),q) (s(t)-o(1))q ?
- Exact constants for n lt s(t)
- Explicit code constructions