Download presentation

Presentation is loading. Please wait.

Published byColby Hales Modified over 2 years ago

1
1 One-way protocols and combinatorial designs Mike Atkinson Joint work with Michael Albert, Hans van Ditmarsch, Robert Aldred, Chris Handley

2
2 The plan Description of problem Modelling the problem Solutions

3
3 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?

4
4 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

5
5 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)

6
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

7
7 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.

8
8 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.

9
9 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

10
10 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?

11
11 Structure of the solution 012,056,034,145,136,235,246 The 7 triples are the lines of the 7 point projective plane 6 5 4 3 2 0 1

12
12 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!

13
13 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

14
14 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

15
15 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

16
16 Combinatorial conditions A collection L of a-subsets of {0,1,..,v-1} is a one-way protocol if and only if –For all L 1,L 2 in L, |L 1 L 2 | ≤ 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

17
17 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.

18
18 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 v!c! (v-a)!(v-b)!

19
19 General construction Let be a set of a integers such that among the (non-zero) differences d 1 -d 2 no value occurs more than e times. Let L be the set {i + |i = 0 … v-1} (arithmetic mod v) L realises the parameter set a,v-2a+e+1,a-e-1

20
20 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 M 11

21
21 Examples cont. –a=4, b=3, c=1. Code the 8 cards as vectors in Z 2 Z 2 Z 2. Let L be the 7 subgroups of order 4 and their complements

Similar presentations

OK

Bart Jansen 1. Problem definition Instance: Connected graph G, positive integer k Question: Is there a spanning tree for G with at least k leaves?

Bart Jansen 1. Problem definition Instance: Connected graph G, positive integer k Question: Is there a spanning tree for G with at least k leaves?

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google