Comments on: “Two Generals’ Problem” doesn’t make sense http://www.brainonfire.net/blog/two-generals-problem-paradox/ Tim McCormack, distilled Mon, 06 Sep 2010 05:11:34 +0000 hourly 1 http://wordpress.org/?v=3.0.1 By: oweseLew http://www.brainonfire.net/blog/two-generals-problem-paradox/comment-page-1/#comment-41502 oweseLew Mon, 06 Sep 2010 05:11:34 +0000 http://www.brainonfire.net/2007/07/31/two-generals-problem-paradox/#comment-41502 Most people that earn money being an affiliate sign up with several Affiliate Programs. In fact, you will have to realize several before you find the ones that will make you as much as possible. About the most important considerations when buying when you choose to promote products as an affiliate is to choose worthwhile products. If you wouldn't buy it or have any use for it chances are your customers won't either. Remember, even though you're selling via the internet and not in person, regardless of whether you truly believe in the products you are promoting will show through in your marketing efforts. Choose products that you truly believe in quotes for quality products to persuade others to buy them. Regards, Most people that earn money being an affiliate sign up with several Affiliate Programs. In fact, you will have to realize several before you find the ones that will make you as much as possible. About the most important considerations when buying when you choose to promote products as an affiliate is to choose worthwhile products. If you wouldn't buy it or have any use for it chances are your customers won't either. Remember, even though you're selling via the internet and not in person, regardless of whether you truly believe in the products you are promoting will show through in your marketing efforts. Choose products that you truly believe in quotes for quality products to persuade others to buy them.

Regards,

]]> By: boarders' paradise http://www.brainonfire.net/blog/two-generals-problem-paradox/comment-page-1/#comment-36670 boarders' paradise Tue, 06 Apr 2010 18:58:54 +0000 http://www.brainonfire.net/2007/07/31/two-generals-problem-paradox/#comment-36670 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? 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 http://www.brainonfire.net/blog/two-generals-problem-paradox/comment-page-1/#comment-36661 scuroNok Tue, 06 Apr 2010 15:51:31 +0000 http://www.brainonfire.net/2007/07/31/two-generals-problem-paradox/#comment-36661 Silence has come :) Does anybody have something to say? Silence has come :) Does anybody have something to say?

]]> By: Tim McCormack http://www.brainonfire.net/blog/two-generals-problem-paradox/comment-page-1/#comment-29360 Tim McCormack Mon, 19 Oct 2009 03:40:26 +0000 http://www.brainonfire.net/2007/07/31/two-generals-problem-paradox/#comment-29360 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? 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 http://www.brainonfire.net/blog/two-generals-problem-paradox/comment-page-1/#comment-27971 Cha0s Mon, 07 Sep 2009 10:34:44 +0000 http://www.brainonfire.net/2007/07/31/two-generals-problem-paradox/#comment-27971 (forgot to check notify) (forgot to check notify)

]]> By: Cha0s http://www.brainonfire.net/blog/two-generals-problem-paradox/comment-page-1/#comment-27970 Cha0s Mon, 07 Sep 2009 09:28:32 +0000 http://www.brainonfire.net/2007/07/31/two-generals-problem-paradox/#comment-27970 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. 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.

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