- By
**cain** - Follow User

- 105 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'One-way protocols and combinatorial designs' - cain

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### One-way protocols and combinatorial designs

Mike Atkinson

Joint work with

Michael Albert, Hans van Ditmarsch, Robert Aldred, Chris Handley

The plan

- Description of problem
- Modelling the problem
- Solutions

The 2000 Moscow Mathematical Olympiad

- Players Alice, Bob, Crow draw cards from a 7 card deck. A receives 3 cards, B receives 3 cards, C receives 1 card
- How can A, in a single public announcement, tell B what her cards are without C learning a single card of A or B’s holding?

First thoughts

- A could make some very complex announcement (“I hold card 2 or card 4; if I hold card 3 I don’t hold card 5; if I hold any consecutive numbered cards then one is prime,….”)
- B, knowing his own cards, finds A’s announcement useful
- C, knowing only his card, can’t use it

Pitfalls

- Suppose A held 0,1,2; she could say “I hold 0,1,2 or 3,4,5”
- B would successfully learn A’s hand because only one of those possibilities can be consistent with his own hand
- But, for all A knows, C might hold 3 and then C could infer A’s holding (note: A would be safe if C held 6)

Second thoughts

- No matter how complex is A’s announcement it is tantamount to saying “My holding is one of the following …”
- A’s announcement must be effective for B and ineffective for C no matter what B and C hold

First solution

- A says “Modulo 7 my total is x”.
- The 35 possible holdings for A come in 7 groups of 5 corresponding to their sum mod 7
- “Modulo 7 my total is 3” is tantamount to saying “I hold 012, 136, 145, 235, or 046”
- B can now work out C’s card and therefore work out A’s holding
- C can only work out A’s sum modulo 7 and B’s sum modulo 7: he can’t work out any one card of A or B.

Second solution

- A could announce (supposing that she holds 0,1,2) “I hold one of 012,056,034,145,136,235,246”
- Exhaustive check. E.g. suppose B held 345 then he could deduce A holds 012 since all other possibilities intersect his own holding. But C (holding 6) can deduce only that A’s holding is one of 012,034,145,235 and no card of A is revealed.

Other solutions

- All solutions involve an announcement of 5 or 6 or 7 possible holdings
- More than 7 makes it too hard for B
- Less than 5 makes it too easy for C

Reveal as little as possible

- If A wishes to reveal as little as possible she should choose to present 7 possible holdings rather than 5
- How are the “optimal” solutions found?

2

1

4

3

0

6

5

Structure of the solution012,056,034,145,136,235,246- The 7 triples are the lines of the 7 point projective plane

The general problem

- A holds a cards, B b cards, C c cards from a deck of v=a+b+c cards
- A must make one public announcement from which B can infer A’s holding but C cannot infer any card of either A or B
- For which a, b, c is this possible?
- If it is possible, what are the most and least informative announcements?
- Find a suitable announcement!

Communication protocols

- A protocol is a series of messages by various parties to communicate information E.g. A might send a message to B, B might answer with another message, A might send yet another message,…. Eventually the required information is communicated.
- We are studying one-way protocols

The one-way restriction

- Suppose a=2, b=4, c=1 (and v=7)
- No one-way protocol is possible
- There is a 2 message protocol:
- B first announces a number of possible holdings for himself that allows A to deduce B’s holding whereas C learns no card of either A or B
- A now knows C’s card and announces it; this tells C nothing further but allows B to infer A’s holding

The one-way restriction

- Suppose a=2, b=4, c=1 (and v=7)
- No one-way protocol is possible
- There is a 2 message protocol:
- B (holding, say, 1236) could announce he holds one of 3456, 0156, 1245, 1236, 0134, 0235, 0246. A (holding, say, 05) could then infer B’s holding
- A now knows C’s card is 4 and announces it; B can now deduce that A holds 05

Combinatorial conditions

- A collection L of a-subsets of {0,1,..,v-1} is a one-way protocol if and only if
- For all L1,L2 in L , |L1 L2| ≤ a-c-1
- For all c-sets X the set of members of L disjoint from X have empty intersection and their union contains every point not in X

Combinatorial problems

- For given a,b,c find a suitable collection L of a-subsets of {0,1,…,v-1}.
- Find upper and lower bounds on the size of |L|.
- Find general constructions valid for a range of (a,b,c) values.

v!c!

(v-a)!(v-b)!

Bounds on |L|- |L| ≤
- |L| ≥ v(c+1)/a
- Some other bounds also known
- Sometimes the bounds prove that no one-way protocol exists
- Occasionally, they pin down |L| uniquely
- e.g. if b=2, c=1 then |L| = (a+2)(a+3)/6

General construction

- Let D be a set of a integers such that among the (non-zero) differences d1-d2 no value occurs more than e times.
- Let L be the set {i + D |i = 0 … v-1} (arithmetic mod v)
- L realises the parameter set

a,v-2a+e+1,a-e-1

Examples

- Many one-way protocols seem to have no further combinatorial interest
- Those for which |L| is maximal are often more interesting
- v = 13 (all the spades), a = 4, b = 7, c = 2, L is the set of 13 lines of the 13 point projective plane
- v = 11, a = 5, b = 5, c = 1, L is the set of 66 blocks of the Steiner system 4-(5,11,1) whose automorphism group is M11

Examples cont.

- a=4, b=3, c=1. Code the 8 cards as vectors in Z2 Z2 Z2. Let L be the 7 subgroups of order 4 and their complements

Download Presentation

Connecting to Server..