Comments on: “Two Generals’ Problem” doesn’t make sense

By: Wireless MAC protocols « Wireless Sensor Networks, Spring 2011

Wireless MAC protocols « Wireless Sensor Networks, Spring 2011 — Tue, 12 Apr 2011 19:44:41 +0000

[...] Here’s a link to the original paper on the limits of distributed computing which introduced a version of what has since become known as the Two General’s Problem. Some of the subtleties of this classic paradox are explored here. [...]

By: boarders' paradise

boarders' paradise — Tue, 06 Apr 2010 18:58:54 +0000

yes, me. But I'm so busy at the moment, I hope I'll get 'round it soon.

In the meanwhile: my text in posting 7 has been completely messed up (I know I posted it right). Tim, can you fix that?

By: scuroNok

scuroNok — Tue, 06 Apr 2010 15:51:31 +0000

Silence has come :) Does anybody have something to say?

By: Tim McCormack

Tim McCormack — Mon, 19 Oct 2009 03:40:26 +0000

Cha0s: Ha, I think I see where my misunderstanding lies! I thought the problem was saying "an algorithm that sometimes allows consensus under these circumstances cannot exist", but what it is really saying (I think) is "there is no algorithm that will always guarantee consensus."

Reactions?

By: Cha0s

Cha0s — Mon, 07 Sep 2009 10:34:44 +0000

(forgot to check notify)

By: Cha0s

Cha0s — Mon, 07 Sep 2009 09:28:32 +0000

FLP = Michael J. Fischer, Nancy A. Lynch, and Michael S. Paterson

Gilbert is correct that this problem is closely related to binary consensus.
I originally encountered it as the coordinated attack problem with the following statement:
Two kings (k1 and k2) on either side of a great (evil) empire wish to bring down the empire. They meet and decide to assess their forces. After doing so (when the problem "begins"), they have a preference as to whether to attack or not. They can send each other messages (to be very specific, lets say that any message sent is sent at 8 AM on whatever day it is sent and that any message that is received arrives at 8 PM on the same day--these details are arbitrary but make no difference in the problem formulation. Even with these restrictions, the problem is insoluble). Finally, any message sent through the empire may be intercepted. There is no guarantee that any message will ever be received.

The following are the formal conditions of the problem:
1. Termination: Each king eventually must decide 0 (do not attack) or 1 (attack).
2. Agreement: The decisions (d1 and d2) of both kings must be the same.
3. Validity: Let i1 and i2 denote the preferences of k1 and k2, respectively. Then:
-if i1 and i2 are 0, d1 and d2 must be 0
-if i1 and i2 are 1, d1 and d2 must be 1

This formulation is equivalent to the first formation: though the information they need to agree on is different, the problem is the same. The following is proof of its insolubility:
Assume there exists a correct algorithm A.
Let E(a,b) be the scenario where i1 = a, i2 = b and all messages are lost. By validity (assuming A is correct), d1=d2=0 in E(0,0) and d1=d2=1 in E(1,1). E(1,1) is indistinguishable from E(1,0) for k1 (remember, no one receives any messages), so k1 decides 1 in E(1,0). Similarly, E(0,0) is indistinguishable from E(1,0) for k2, so k2 decides 0 in E(1,0). Of course this contradicts agreement as d1!=d2 in E(1,0), so A can not exist.

Remember, for A (any algorithm) to be correct, it must work for all scenarios. Here is a scenario where it does not work. Therefore, there can be no algorithm.

Interestingly, a seemingly significantly weaker form of the problem is also insoluble:
Replace condition 3 with:
3a. Weak Validity: If i1=i2=1 and no message is lost, then d1=d2=1.

The proof of this is slightly more complicated, but shows that the problem is quite insoluble.