the round complexity of two-party random selection saurabh sanghvi and salil vadhan harvard...

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The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

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Page 1: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Round Complexity of Two-Party Random Selection

Saurabh Sanghvi and Salil VadhanHarvard University

Page 2: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Random Selection Problem Several mutually distrusting parties wish to

select jointly at random an element of a fixed universe.

Goal: Protocol such that even if a party cheats, the outcome will not be too “biased”.

Applications: Design a protocol where a trusted third-party makes the selection, then replace third-party with random selection protocol.

Page 3: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

Types of Random Selection

Blu82, Lin01, KO04 Dam94, DGW94, GGL98, GSV98, CCM98, DHRS04

CGMA85, GMW87, KOS03

BL89, Sak89, AN90, ORV94, GGL98, RZ98, Fei99

Computational Information-Theoretic

2 parties

N parties

Our focus

Page 4: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

2-party Information-Theoretic Random Selection Protocols

Examples of Uses Convert honest-verifier ZKPs to general

ZKPs [Dam94, DGW94, GSV98] Perform oblivious transfer in bounded-

storage model [CCM98, DHRS04] Perform general fault-tolerant

computation [GGL98] Each evaluated by different criteria…

Page 5: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

Defining Random Selection

Alice

Coins rA

Bob

Coins rB.

.

.

Output:

Our complexity measure: # of rounds

(k)

Page 6: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

Evaluating a Protocol Statistical Criterion (SC) – 9 constants s.t. as

long as one party is honest:

8 T µ {0,1}n of density · Pr[ Output 2 T ] · 1-

Equivalent to the statistical difference of the protocol’s output with uniform being 1-(1).

Extension of “resilience” in leader election/collective coin flipping

Achievable? Yes! [GGL98] (with 2n rounds)

What is the necessary and sufficient round complexity?

“cheating sets”

Page 7: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

Our results Upper bound:

9 protocol satisfying the Statistical Criterion with 2log* n + O(1) messages

Lower bound: log*n-log*log*n – O(1) messages are

necessary.

Tantalizingly similar to results in leader election, collective coin-flipping [RZ98, RSZ99, Fei99]

Page 8: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

Our Protocol – Iterated Random Shift

Given n, Alice and Bob want to select from U={0,1}n.

Let m = n3. Recursively apply:

Inspired by leader election protocols [RZ98] and proof that BPP 2 2P [Lau83]

b1, …, bm à U

a1, …, am à U

Recurse on U’ = {ai+bj}…

Page 9: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Main Lower Bound Theorem: Any random selection protocol

satisfying the Statistical Criterion must have at least log*n – log*log*n – O(1) rounds.

Recall Statistical Criterion: 9 constants s.t. 8 T µ {0,1}n of density · Pr[ Output 2 T ] · 1-

First nonconstant lower bound on round complexity for any random selection protocol not imposing additional constraints (e.g., on communication size or “simulatability”).

Page 10: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

Proof Strategy

Suppose protocol has ¿ log* n rounds.

Show that one of the players can force the output into a “cheating” set of density o(1) with probability 1-o(1).

Strategy: induction on game tree…

Page 11: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Two-Round CaseBob’s message

Alice’s message

Can think of any two-round protocol as: Bob sends Sµ{0,1}n to Alice (according to some dist.

on P({0,1}n)) Alice selects output according to some dist. on S.

m1

S={f(m1, ²)}

m2Alice selects m2, output is

x=f(m1,m2)

(“Alice selects x2S”)

Bob selects m1, restricting output to

S={f(m1,²)}

(“Bob selects set S”)

Page 12: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Two-Round Case: Cheating Bob

Bob’s message

Alice’s message

Case 1: 9 “small” set (of size o(n)). Bob violates SC by selecting that set as his cheating set..

1) Bob’s cheating set

3) Alice’s chosen output 2 Bob’s cheating set with prob.

1

2) Bob deterministically

chooses this branch

Page 13: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

2) Bob plays honestly

The Two-Round Case: Cheating Alice

Bob’s message

Alice’s message

Case 2: Bob must give Alice a “big” (i.e., ω(1) elements) set.

Random cheating set of density o(1) intersects w.h.p. ) Alice cheats successfully.

1) Alice’s cheating set = random set of red elements

3) Alice selects output from intersection

Page 14: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Three-Round Case

Now, Alice chooses a set of sets, from which Bob chooses a set, from which Alice chooses the output.

Alice

Bob

Alice

m1

m2

S = f(m1, m2, ²) output = f(m1, m2, m3)

m3

Page 15: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Three-Round Case

Case 1: If Alice can choose a branch whereby all sets are “big”, then she can violate the statistical criterion.

Alice

Bob

Alice

1) Alice’s random cheating set = set of red elements

4) Alice can choose output in her cheating

set

2) Alice deterministically chooses branch

3) Bob plays honestly

Page 16: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Three-Round Case

Thus, every branch has at least one “small” set.

Not immediately helpful to Bob…

Alice

Bob

Alice

Page 17: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Three-Round Case

Key question: Down a given branch chosen by Alice, how many disjoint, small sets are there?

Bob benefits if there are many.

Alice

Bob

Alice

Page 18: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Three-Round Case

Case 2: All initial Alice messages let Bob choose from many disjoint small sets.

Randomly chosen set of o(1) density contains a small set w.h.p. ) Bob cheats successfully.

Alice

Bob

Alice

1) Bob’s random cheating set = set of red elements

4) Alice must choose output in his cheating set

3) Bob selects set contained in cheating set

2) Alice randomly picks a branch

Page 19: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Three-Round Case

What if there is a branch with few disjoint small sets?

Need to argue Alice can take advantage.

Alice

Bob

Alice

Page 20: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

The Three-Round Case

Case 3: A branch with no large disjoint subcollection Set intersecting all small sets + random set

) Alice cheats successfully

Alice

Bob

Alice

1) Alice’s cheating set = intersect-set + … … a random set

2) Alice deterministically selects branch

3) Bob plays honestly

4) Whether Bob chose big or small set, Alice selects from

cheating set

Implies a small set intersects every set in collection (e.g., union of maximal disjoint

subcollection)

Page 21: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

3 -> log*n-log*log*n-O(1) To generalize, induct on the game tree…

label every node A-WIN, B-WIN, or TIE: WIN – player can violate SC by choosing

cheating set randomly. TIE – both players can violate SC with a

cheating set of the form R U S, where R is random and S is a small set of non-random elements.

The result stops at ~log* n rounds because |S| grows as a tower in the # of rounds.

Page 22: The Round Complexity of Two-Party Random Selection Saurabh Sanghvi and Salil Vadhan Harvard University

Conclusions We provide matching upper and lower bounds

(up to a constant factor) for the round complexity of protocols satisfying a natural criterion.

Open Problems/Future Work Leverage results for open problems in well-studied

multiparty protocols (leader election, collective coin-flipping, and collective sampling).

Study the impact of additional constraints required in literature (e.g., simulatability or message length).