Talk:Common knowledge (logic)/Archives/2013

First order logic

Beware of the use first order logic which has a precise meaning in logic. As second order logic and higher order logic, see section Comparison with other logics in article first order logic for a classification. Pierre de Lyon (talk) 21:54, 19 February 2008 (UTC)

New knowledge

For k>1, how would the outsider's statement add anything not already known? 75.118.170.35 (talk) 16:17, 12 August 2009 (UTC)

Common versus Mutual Knowledge

I think there is an important mistake in the Steven Pinker reference from Stuff of Thought. It currently states "...the notion of common knowledge (dubbing it mutual knowledge, as it is often done in the linguistics literature)". In fact, Pinker and linguists specifically differentiate mutual knowledge - that which we all know - from common knowledge - that which we all know that we all know (and we all know that we all know that we all know, ad infinitum). Unless I'm misunderstanding it, I'd like to change it. Dashing Leech (talk) 19:39, 18 December 2010 (UTC) — Preceding unsigned comment added by Dashing Leech (talk • contribs) 19:34, 18 December 2010 (UTC)

Modal logic (syntactic characterization) --- finite or infinte?

Am I missing something? First, you say that there are n modal operators Ki (with i = 1, ..., n). Then you define a conjunction of the Ki with i in G (a possibly infinite set) and complain that this conjunction is not finitary. But you just finished saying that there are only finitely many Ki. Which is it?

In my opinion, G should always be countable (although that is not crucial as far as I can see) but the conjunction should always be of only finitely many of the Ki. (I think G should be countable in any real world situation, but I see no harm in allowing uncountable G. However, for common knowledge to be meaningful, I can't see how an infinite conjunction can be allowed.)

I'll leave it to someone else to edit this, using the correct symbols. Dagme (talk) 20:02, 21 May 2011 (UTC)

Can someone please explain why this logic doesn't work?

Firstly, let us assert that people don't *want* to leave the island, so they will work to avoid knowing it if they possibly can, and that this is common knowledge too.

Secondly, let us assert the the number of blue eyed people on the island is 'large' - let us assume 10.

What prevents the reasoning: "What if the blue-eyed people I see don't leave the island, despite the taboo? Then I would leave, in the mistaken belief that I had blue eyes! I should not leave the island because I won't know my own eye colour. Since there are a large number of blue-eyed people on the island, they can come to the same conclusion, so they may choose not leave the island, despite of the taboo - because the taboo is not broken. No-one must leave." — Preceding unsigned comment added by 2.26.62.147 (talk) 10:42, 29 June 2011 (UTC)

All of the above is indeed correct. However, for purposes of the problem, the islanders simply cannot help but count the number of blue-eyed people and work out all the deductions. And they all know this about each other, and they all know that they all know this about each other, etc. So no "deliberate ignorance" is possible. 71.162.46.164 (talk) 17:37, 11 February 2012 (UTC)

The blue eyes puzzle is wrong

I feel like I'm being trolled on a mass scale.

Everyone uses examples of 1 and 2 people, where it works, and then jumps to it working for k > 2, when it just doesn't. If k > 1, *everyone* already sees blue eyes, and the announcement doesn't introduce any new knowledge in any way.

We are to assume the islanders can count the number of blue eyes people to induce the number of days to wait as k, but they have some inexplicable internal block that says despite everyone seeing at least one blue eye person (in k>2) they don't know it until the oracle states it explicitly?

I've spent the last few days trying to "understand" the solution, wracking my brain, and now I'm calling bullshit. I really just think people are being dishonest here and introducing a secretive vague absurdly ridiculous rule (everyone can see blue eyes, yet doesn't *technically* know that everyone else knows there are blue eyes) to justify a CLEARLY flawed example.

I also appreciate the correct answer, which is that if(k > 1) everyone blue-eyes person leaves on the kth day /from when they arrived/. But it doesn't seem to be anything more than a deceptive puzzle, and has little to do with the actual topic of common knowledge, and any attempt to amend the puzzle to account for this vague extra rule would seem to make it lose its appeal and just be convoluted.

"Oh yeah, everyone sees blue eyes but no one's really sure that everyone else sees them, oh but, wait - no, they do know that later when they are counting days, because someone announced it... dur de dur"

The premise of not being able to assume that the other agents are paying attention to their surroundings, really breaks down when they all pay attention to their surroundings to induce the number of days they have to wait to leave.

--173.228.31.160 (talk) 11:42, 14 July 2012 (UTC)

• Sorry, but of course it introduces new knowledge. It works with 3 guys, just as well as with 2 or a million. But let p1,p2,p3 be your guys. Then, and try following this. Before the statement, this is the case: p1 can imagine that p2 sees p1 having red eyes (p1 does not know his own color). Furthermore, the same holds if we swap p2 and p1. Thus: p1 can imagine that p2 believes p2's colors to be red and furthermore that he sees p1 having red eyes. This is the case for two agents. Then we proceed. If we take our first statement, and changes p1,p2 with p2,p3, we get that p2 can imagine that p3 sees p2 having red eyes (p2 does not know his own color). Furthermore, the same holds if we swap p2 and p3. Now, we already know that p1 accepts that p2 can believe that both p1 and p2 has red eyes. According to the sentence above, p1 can also accept that p2 believes that p3 believes that p3 has red eyes. Hence, p1 accepts that p2 can believe that p3 believes that everyone has red eyes. Now, if some deus says that some guy has blue eyes, p1 cannot anymore accept that statement. Hence something new has been learned. -- Pål GD (80.203.133.93 (talk) 09:51, 9 August 2012 (UTC))

Hey 173.228.31.160,

in case you are still trying to wrap your brain around this problem I will try point out something important here (in case the very nice post of 80.203.133.93 did not already clarify it):

You wrote: "If k > 1, *everyone* already sees blue eyes, and the announcement doesn't introduce any new knowledge in any way."

Let's say for K=5 (p1-5): Before the announcement p1 can think that p2 can think that p3 can think that p4 can think that p5 can think that there are only green eyed people. Understand? Even though everyone already *sees* blue eyes, everyone can *think* that another person can *think* ... (and so on for the correct amount of times) ... that another person *thinks* there are only green-eyed people. But that is not possible anymore after the announcement! This is a very relevant example for common knowledge (everyone knows that everyone else knows).

Importantly, p1 CANNOT e.g. think that p5 can think that there are only green-eyed people (because p1 thinks that p5 sees at least 3 blue-eyed people) Or e.g. p1 CANNOT think that p2 can think that p5 can think that there are only green-eyed people (because p1 thinks that p2 thinks that p5 sees at least 2 blue-eyed people).

It has to always go through the whole "recursion chain"! (the order does not matter btw)

--Felix Tritschler (talk) 22:37, 24 September 2012 (UTC)

There is a problem with the way the induction is commonly presented, in that k is used both for the day and the number of blue-eyed islanders. Separate variables should be used, because you are given an island with n blue-eyed residents and never change that number. You can do induction only on the what each knows about the others' knowledge on day k. This article could clarify this in its discussion of the puzzle. --98.69.157.202 (talk) 11:03, 3 November 2012 (UTC)