A Consensus Protocol in a Prison - Study Guide | N S 0 | Study notes Military Strategy and Training

A Consensus Protocol in a Prison

Jayadev Misra

2/2/2004

The following problem was posed at a recent conference by Leslie Lamport,

who had heard that it was posed in a popular radio program, Car Talk, in the

National Public Radio in U.S. I solved a special case of this problem (assuming

a certain initial state). Anup Rao, a student in my graduate class, solved the

general case in an hour.

Problem A set of prisoners —assume there are at least 2— are asked to play

the following game by the warden. There is a room in the prison which has

two switches; initially, the switches are in arbitrary positions. The warden will

bring one prisoner at a time to the room, and the prisoner must flip one of the

switches. The prisoners do not know the order in which they will be taken to the

room, but they know that every prisoner will visit the room over and over until

the end of the game. The game ends when some prisoner announces, “every

prisoner has been in this room at least once”. If the announcement is correct,

all prisoners go free; if incorrect, they all are executed. The game continues

until the announcement is made.

The prisoners are allowed to confer and decide on a protocol prior to the

start of the game. Once the game starts, they are not allowed to communicate,

nor can they find out who is being taken to the room. The problem is to devise

a protocol for the prisoners.

Solution; special case In the original formulation, there are two switches, A

and B. Simplify the problem by assuming that there is a single switch Swhich

a prisoner may or may not flip. This is a special case of the original formulation:

whenever Sis flipped, flip A, and whenever Sis not flipped flip B. Henceforth,

flip denotes flipping switch S. Let variable sdenote the state of the switch, 0

for off and 1 for on; so, flip changes the value of s.

The prisoners choose a leader who will make the announcement; all other

prosoners are followers. The number of followers, N, is at least 1. Each follower

will vote by flipping the switch appropriately, and the leader will count the

number of votes. The ith follower has a variable viwhich is 1 if he has voted

and 0 otherwise. Clearly, vi= 1 implies that the ith follower has visited the

room. The leader has a variable cfor maintaining the count. Initially, no follower

has voted and cis 0. Assume, for the moment, that the switch is initially off;

so, s= 0. The protocol is described by the steps taken by the leader and the

followers.

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A Consensus Protocol in a Prison Jayadev Misra 2/2/

The following problem was posed at a recent conference by Leslie Lamport, who had heard that it was posed in a popular radio program, Car Talk, in the National Public Radio in U.S. I solved a special case of this problem (assuming a certain initial state). Anup Rao, a student in my graduate class, solved the general case in an hour.

Problem A set of prisoners —assume there are at least 2— are asked to play the following game by the warden. There is a room in the prison which has two switches; initially, the switches are in arbitrary positions. The warden will bring one prisoner at a time to the room, and the prisoner must flip one of the switches. The prisoners do not know the order in which they will be taken to the room, but they know that every prisoner will visit the room over and over until the end of the game. The game ends when some prisoner announces, “every prisoner has been in this room at least once”. If the announcement is correct, all prisoners go free; if incorrect, they all are executed. The game continues until the announcement is made. The prisoners are allowed to confer and decide on a protocol prior to the start of the game. Once the game starts, they are not allowed to communicate, nor can they find out who is being taken to the room. The problem is to devise a protocol for the prisoners.

Solution; special case In the original formulation, there are two switches, A and B. Simplify the problem by assuming that there is a single switch S which a prisoner may or may not flip. This is a special case of the original formulation: whenever S is flipped, flip A, and whenever S is not flipped flip B. Henceforth, flip denotes flipping switch S. Let variable s denote the state of the switch, 0 for off and 1 for on; so, flip changes the value of s. The prisoners choose a leader who will make the announcement; all other prosoners are followers. The number of followers, N , is at least 1. Each follower will vote by flipping the switch appropriately, and the leader will count the number of votes. The ith^ follower has a variable vi which is 1 if he has voted and 0 otherwise. Clearly, vi = 1 implies that the ith^ follower has visited the room. The leader has a variable c for maintaining the count. Initially, no follower has voted and c is 0. Assume, for the moment, that the switch is initially off; so, s = 0. The protocol is described by the steps taken by the leader and the followers.

initially: for all followers, vi := 0; c := 0; s := 0

Follower if vi = 0 {not voted} and s = 0 {switch is off} then flip {s = 1}; vi := vi + 1 fi

Leader if s = 1 {switch is on} then flip {s = 0}; c := c + 1; if c = N then Announce fi fi ————————————

The proof obligations are:

(Safety) Whenever the leader announces, all vis are 1.
(Progress) Eventually, the leader announces.

Proof of Safety: Let v denote the sum of all the vis. Then, 0 ≤ v ≤ N. It is easy to see that c + s = v is an invariant: it holds initially, and after each step by a follower or the leader. Since c = N ∧ s = 0 is a precondition of Announce, we conclude from c+s = v that v = N holds prior to Announce. Hence, all vis are 1.

Proof of Progress: We show that c increases eventually if it is below N. There- fore, eventually c = N. Since c never decreases, c = N will remain true and the leader will Announce. If c < N ∧ s = 1 then on its next visit to the room, the leader will increase c. If c < N ∧ s = 0 some follower has not voted yet; so, s will become 1 either by the voting of this or some other follower. Then, c < N ∧ s = 1 holds, and, from the previous argument, c will be increased.

Solution; general case We have assumed that initially s = 0. Now, we remove this assumption. Then, we have the invariant c + s ≤ v + 1. It is no longer true that c = N ⇒ v = N ; v could be N − 1. However, c can be off by no more than 1 from v. This discrepancy can be handled if every follower votes twice. Now, the precondition of voting is changed to vi < 2, and of Announce to c = 2N.

Proof of Safety: A precondition of Announce is c = 2N ∧ s = 0. From the in- variant c+s ≤ v+1, we get 2N ≤ v+1 as a precondition of Announce. We claim that then every prisoner has visited the room at least once, because if N −1 pris- oners have visited the room and each has voted twice v ≤ 2 N −2, or v +1 < 2 N.

Proof of Progress: Similar to the previous case.

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