1. Algorithmic Construction of Sets for k-Restrictions
Dana Moshkovitz
Joint work with Noga Alon and Muli Safra
Tel-Aviv University

2. Talk Plan Problem definition: k-restrictions Applications: …
group testing
generealized hashing
Set-Cover Hardness
Background
Techniques and Results

3. Techniques Greedine$$ k-wise approximating distributions Concatenation
multi-way splitters via the topological Necklace Splitting Theorem

4. Problem Definition

5. On Forgetful Hot-Tempered Pirates and Helpless Goldsmiths
One day the hot-tempered pirate asks the goldsmith to prepare him a nice string in m.

6. But the capricious pirate has various contradicting local demands he may pose when he comes to collect it…
this pattern!
should differ!

7. What will the goldsmith do?

8. make many strings, so every demand is met!

9. Formal Definition [~NSS95]
Input: alphabet , length m. demands f1,…,fs:k{0,1},
Solution: Am s.t
for every 1i1<…<ikm, 1js,
there is aA s.t. fj(a(i1),…,a(ik))=1.
Measure: how small |A| is  k

10. Applications

11. Goldsmith-Pirate Games Capture Many Known Problems
universal sets
hashing and its generalizations
group testing
set-cover gadget
separating codes
superimposed codes
color coding …

12. Application I Universal Set
every k configuration is tried.
circuit 1 . 1 . 1 . . .
. . . m

13. Application II Hashing
Goal: small set of functions [m][q]
For every kq in [m], some function maps them to k different elements
small set of functions u1 u2 u3 u4 . um k r1 r2 . rq

14. Generalized Hashing Theorem
Definition (t,u)-hash families [ACKL]: for all TU, |T|=t, |U|=u, some function f satisfies f(i)≠f(j) for every iT, jU-{i}.
Theorem: For any fixed 2≤t<u, for any >0, one can construct efficiently a (t,u)-hash family over alphabet of size t+1, whose
rate (i.e logqm/n) ≥ (1-)t!(u-t)u-t/uu+1ln(t+1)

15. Application III Group Testing [DH,ND…].
m people
at most k-1 are ill
can test a group: contains illness?
Goal: identify the ill people by few tests.
. . . ? ? ? ? ? ?

16. Group-Tests Theorem
Theorem: For every >0, there exists d(), s.t for any number of ill people d>d(), there exists an algorithm that outputs a set of at most (1+)ed2lnm group-tests in time polynomial in the population’s size (m).

17. Application IV Orientations [AYZ94]
Input: directed graph G
Question: simple k-path?
if G were DAG…

18. Application IV Orientations [AYZ94]
Need several orientations, s.t wherever the path is, one reflects it.
Pick an orientation
Delete ‘bad’ edges
Now G is a DAG… 3 5 1 4 2 1 2 3 4 5

19. Application V Set-Cover Gadget
sets
Gadget: a succinct set-cover instance so that:
a small, illegal sub-collection is not a cover. 
elements
legal cover: set and its complement
small: its total weight ≤ …
sets and complements differ in weight    

20. Approximability of Set-Cover
approximation ratio (upto low-order terms)
known app. algorithms [Lov75,Sla95,Sri99]
ln n
if NPDTIME(nloglogn) [Feige96]
if NPP [RS97]

21. Random and Pseudo-Random Solutions
Background
Random and Pseudo-Random Solutions

22.  = minI,j PraD[ fj(a(I))=1 ]
Density m
D:m[0,1] - probability distribution.
density w.r.t D is:
 = minI,j PraD[ fj(a(I))=1 ]    m    . k

23. Probabilistic Strategy
Claim: t=-1(klnm+lns+1) random strings from D form a solution,
with probability≥½.
Let Cj be some constraint.
Prr[rCj]  1-
Choose a random set of strings A.
 Prr[A∩Cj =]  (1-)|A|
 Prr[A∩Cj =]  e-|A|
The probability some constraint has A∩Cj=, is sC(m,k)e-|A|.

24. Deterministic Construction!

25. First Observation m
support(D) is a solution if density positive w.r.t D. 
every demand is satisfied w.p ≥
|support(uniform)|=qm k

26. Second Observation A k-wise, O()-close to D is a solution.m k 
every demand is satisfied w.p  (1-..)
A k-wise, O()-close to D is a solution.
Theorem [EGLNV98]: Product dist. are efficiently (poly(qk,m,-1)) approximatable

27. So What’s the Problem? It’s much more costly than a random solution!
Random solution: ~ klogm/ for all distributions!
k-wise -close to uniform: O(2kk2 log2m /2) [AGHP90]
for other distributions, the state of affairs is usually much worse…
e.g for the uniform distribution 2k-2k2log2m

28. Background Sum-Up
Random strings are good solutions for k-restriction problems
if one picks the ‘right’ distribution…
k-wise approximating distributions are deterministic solutions
of larger size…
Our goal: simulate deterministically the probabilistic bound

29. Our Results

30. Outline k=O(1) Greedy on approximation k=O(logm/loglogm) Concatenation
assumes invariance under permutations +
k=O(logm/loglogm)
Concatenation
works for some problems +
multi-way splitters
larger k’s

31. same as random solution!
Greedine$$
same as random solution! m
Claim: Can find a solution of size --1(klnm+lns) in time poly(C(m,k), s, |support|)
Proof:
Formulate as Set-Cover:
elements: <position,constraint>
sets: <support vector>
Apply greedy strategy. k

32. Concatenation m m’ m’ N N
hash family
inefficient solution

33. Concatenation Works For Permutations Invariant Demands

34. Theorem Theorem: Fix some eff. approx. dist. D.
Given a k-rest. prob. with density  w.r.t D,
obtain a solution of size arbitrarily close to
(2klnk+lns)/ × k4logm
in time poly(m,s,kk,qk,-1).

35. Dividing Into BLOCKS m

36. Splitters, [NSS95] What are they? How to construct?
several block divisions
any k are splat by one
k-restriction problem!
How to construct?
needs only (b-1) cuts
use concatenation

37. Multi-Way Splitters m
For any I1⊎…⊎It[m], |⊎Ij|k, some partition to b blocks is a split.
k-restriction problem! b k

38. Necklace Splitting [A87]
b thieves
t types
How many splits?

39. Necklace Splitting [A87]

40. Necklace Splitting Theorem
Theorem (Alon, 1987): Every necklace with bai beads of color i, 1it, has a b-splitting of size at most (b-1)t.
tight!
Corollary: A multi-way splitter of size
b(b-1)t+1 C(m, (b-1)t)
is efficiently constructible.
Noga’s result:
Continuous splitting: Given a t-coloring of [0,1], divide it into b pairwise disjoint Lebesgue measurable subfamilies, s.t each captures 1/b measure.
Reduction from discrete splitting to continuous splitting: Given a necklace color [0,1] accordingly. Problem: single bids can be splat too. Solution: show we can get rid of ‘bad cuts’. Draw a multi-graph: vertex for each thief, edge for each color i s.t bid of this color is splat between the two thieves. Note: All degrees are even! [Each color is splat evenly between the thieves] Hence, there is an Euler cycle in that multi-graph. Now – slide the cuts along this cycle.
Prove the continuous version: (1) Show it holds for every prime (the case b=2 was already proven). (2) Note this implies the general case: say we prove for (t,k) and (t,l). For (t,kl): First split to k parts. Gather all of them, re-scale and split to l parts.
Topological Proof for (1): via a generalization of the Borsuk-Ulam theorem [Any continuous function from an n-sphere into Euclidean n-space maps some pair of antipodal points to the same point.]
C(k2,
·|Hashm,k2,k|
concatenation

41. The b=t=2 Case

42. Sum-Up Beat k-wise approximations for k-restriction problems.
Multi-way splitters via Necklace Splitting.
 Substantial improvements for:
Group Testing
Generalized Hashing
Set-Cover

43. Further Research
Applications: complexity, algorithms, combinatorics, cryptography…
Better constructions? different techniques?