Math Fun

There are four blues and four browns.
All know that all see that there is at least one of each color: The REAL SET of REAL UNITS.
The day/night/ferry things apply.

All reason:

• If there was only one blue eyed who knew that there was at least one blue eyed (one REAL UNIT of the REAL SET), he would leave the first day. And:
• If there was only one brown eyed who knew that there was at least one brown eyed (one REAL UNIT of the REAL SET), he would leave the first day.

No one leaves the first day. Therefore there are more than 1 of each color, and all know that they might belong to either one, or possibly another color.

Since no one leaves the second day, there are more than 2 of each color.
Since no one leaves the third say, there are more than 3 of each color.
Since all blues see only 3 blues, they leave the 4th day. Since all browns see only 3 browns, they all leave the 4th day.

Wow, FC, you take some serious time out from this place. A good thing. Well, two weeks… I’d hoped it was due to bowing out after realising your error that Sau so sufficiently presented in analogous form, but I guess that was wishful thinking.

You know that’s not an argument, right?
And that it does nothing to defend your strange claim that someone can not be ignorant of something one moment, come to know it, and know it the next moment. Somehow, to you, this is a contradiction - that if they knew it the next moment, they must have known it all along. Yes, that’s what it sounds like you are saying.

Ok. Whilst I hear you say such things, things like this say otherwise:

As I thought, and have said before, your mistake is to say that since 4 blues (or browns) know that everyone else knows that everyone else knows that there is at least 1 blue-eyed islander and at least 1 brown-eyed islander, 1 blue (or brown) must also know that everyone else knows that everyone else knows that there are at least 1 blue-eyed islander and at least 1 brown-eyed islander.

I know why you think this is an ok assumption, because the reality (in this 4blue/brown scenario) is that there ARE 4 blues and 4 browns, and so they can attribute their knowledge to whatever scanerios they are imagining. The problem is that they have to be hypothesising about what 1 blue/brown would know IF there were only 1 blue/brown. In this case, they have to deduce about this 1 blue/brown aside from the knowledge of 4 blues/browns.

The reason you don’t get this appears to be at least one reason why you don’t get the correct solution. You can say you get the solution a million more times, but when you consistently demonstrate that you do not, it just holds no water. You have to do better than just “saying” you get it.

Perhaps it is too difficult for you to understand what I do. It sounds like we both think the same of the other in regard to the validity of one another’s positions. At least one of us has the correct understanding of the problem. How will we decide who this person is? It would be wrong to base my validity on the fact that all other members of this place and the place that hosted the problem side with my explanations, bar the one “I want to be special, even at the expense of sense” person here, and yourself. We must arrive at a much more acceptable measure of validity - any suggestions? Showing you flawless reasoning doesn’t appear to be sufficient so far. I will continue to try.

This question is really the only reason I keep discussing this. It is possibly the most important question I’ve come across in my years of discussing philosophy.

I think the answer starts this way: what would it take to prove you that you’re wrong? For me, I think it would be a logical proof, using explicitly and exhaustively stated premises. I think that I can point to an implied premise that is false, and that thus if all the premises were explicitly and exhaustively stated and did not rely on the implied premise I see, I would be proven wrong.

What about you guys?

Fixed Cross, I have a thought for you: You say that, “If there was only one blue eyed who knew that there was at least one blue eyed (one REAL UNIT of the REAL SET), he would leave the first day,” and we don’t agree. But it is also true that “If there was only one blue, he could not learn that there is at least one blue by looking at the other islanders’ eyes.” And yet, every person on the island knows “there is at least one blue” only because they can see that by looking at the other islanders’ eyes. This distinguishes the knowledge the islanders have before and after the guru speaks: if there were only one blue, he could learn that there is at least one blue by hearing the guru speak.

That is your mistake. You fail at that again and again.
The only proof is one that leaves the absolute lack of alternatives.
The canonical solution is “A solution”. But it is NOT the “Only solution”, and thus is not a proof of anything else being wrong.
You constantly attempt to shoot down a premise by presuming where it is going to lead and thus claim it to be invalid before finding out where it actually leads. You conflate issues in the effort to prevent the beginning of a scenario. That is pure political strategy, not logic.

In order to prove anyone else wrong, you MUST go through their scenario and point out at exactly what point it fails or prove the absolute lack of alternatives to a different argument. But to do that, you must first except the proposed premises that you might LATER shoot down as inapplicable to the final issue. But you refuse to do that. You want to jump ahead and presume your own conclusion; “I think this premise will be a problem later so I am not going to accept it now”. The problem is that you don’t actually know for certain whether it will be or not and refuse to find out. You are “affirming a presumed consequent”.

“I won’t accept the premise of evolution for discussion because that would mean there is no God.”
…Bullshit.
“I refuse to discuss the whereabouts of the defendant because I already know who is guilty.”
…Contempt of court.

1. If I prove a statement A, I know as a corollary that ~~A. If I prove a collection of statements A and A → B, I know as a corollary that B. If a rigorous logical process leads from a collection of premises to a conclusion, I know that the negation of that conclusion can only be true if the premises produce a contradiction. Since you haven’t offered anything to undermine the logic of the canonical solution, the best you’re shooting for is proving the givens to be contradictory.

2. If I say “Here are the givens. Teddy bear. Therefore, solution,” have I really proved anything? How much do you need to “go through the scenario and point out at exactly what point it fails or prove the absolute lack of alternatives to a different argument”? Your solution has everyone pick the same number from thin air (unless they have brown eyes), turn that number by magic into knowledge, and then deduce from that knowledge. I’ll be honest, I don’t know how to explain logically why the color of my eyes doesn’t follow from the number we all start counting at. It’s a non sequitur. The one thing doesn’t follow from the other.

3. Related to point #2, you did not identify what would convince you that you are wrong.

You’re drunk, right?

James, insults, 3-word dismissals, and repeating the same, oft-answered question have not made your position any stronger.

Just to repeat:
the lack of alternatives has been covered by the exhaustive nature of correctly following the logic of the correct solution.

To every blue there are ONLY two possibilities: there are 100 blues or 99. The rest of the solution is examining each of these outcomes just as exhaustively, leaving no stone unturned, and leading to only one solution. It works out very nicely if you really do understand it.

You refusing to discuss anything but your own preferred scenario is not helping your case either (and you haven’t answered the question at all… always trying to jump around it like a frog on a hot plate).

That is absurd.

The canonical solution depends, not only on everyone knowing that everyone knows all of those things we listed, but it also depends on everyone knowing that everyone has no alternative but to be thinking exactly the same scenario.

The problem is that there is a different scenario that also works. So how do they all know that all of the others are using the canonical solution rather than another? Does the guru also tell them that there is one person thinking, “If I was brown and saw 99 blues, then…”. That part wasn’t mentioned.

You are each merely following a proscribed trail. There are alternatives. And because of that, they cannot each know that all of the others are waiting for the same reason.

This is false. The canonical solution is not about what everyone is thinking but about what they can deduce. It is a given that if something can be logically deduced, every islander will deduce it instantly, and the islanders know this about each other. There is no choice of “alternatives”: if we are given A and A → B, and we are perfect logicians who will instantly deduce anything deducible, we will instantly deduce B.

Maybe you should reread your solution… “If number 99 is thinking… that number 98 is thinking…”.

Consider 2+2=4.

As long as everyone accepts the logical definitions of 2, +, = and 4, they are necessarily going to agree with this equation.
If the scenario is that there are “2 here” and “2 there” and nothing besides, and if the sum of those things equals 4, then everyone on an island leaves the island on a ferry at midnight (otherwise nothing changes) then the only things that have any consequence are 2, +, = and 4, and the logic that 2+2=4.

Given that everyone involved is a perfect logician and thus accepts logical definitions, and everyone is aware of the limited number of things that have any consequence, you can know that everyone is ONLY going to be considering these things within these limits. In the above scenario, 2+2=4 is non-negotiable, and the ONLY thing of consequence, so you can KNOW everyone is going to be deducing this in this way, and nothing else. There is no worry of anyone thinking differently.

Now, in the actual puzzle, blue-eyed islanders are ONLY going to be considering the consequences of whether there are 100 blue-eyed islanders or 99. They see 99 and do not know whether they themselves have blue eyes - if they did there would be 100, if not there would be 99. So there are ONLY 2 possibilities to explore:

1. The consequence of there being 100 is that each blue-eyed islander knows they have blue eyes, and leave on the ferry in the knowledge that they have blue eyes.
2. The consequence of there being 99 is that 99 blue-eyed islanders would be ONLY considering the consequences of whether there are 99 blue-eyed islanders or 98. They would see 98 and not know whether they themselves have blue eyes - if they did there would be 99, if not there would be 98. So there would be ONLY 2 possibilities to explore:
   …1) The consequence of there being 99 is that each blue-eyed islander knows they have blue eyes, and leave on the ferry in the knowledge that they
   …have blue eyes.
   …2) The consequence of there being 98 is that 98 blue-eyed islanders would be ONLY considering the consequences of whether there are 98 blue-eyed
   …islanders or 97. They would see 97 and not know whether they themselves have blue eyes - if they did there would be 98, if not there would be 97. So
   …there are ONLY 2 possibilities to explore:
   …1) The consequences of there being 98…, and
   …2) The consequences of there being 97… this is recursive…
   Until:
3. The consequences of there being 1: that they would leave on the ferry in the knowledge that they have blue eyes, and
4. The consequences of there being 0… BUT THERE ARE NOT 0! - the Guru revealed otherwise.

So it’s finally certain that if there were 1, they would leave at the next opportunity. They don’t so there aren’t one. We move back up to:

1. The consequences of there being 2: that they would leave on the ferry at the NEXT opportunity in the knowledge that they have blue eyes, and
2. The consequences of there being 1… BUT THERE ARE NOT 1! - nobody left at the FIRST opportunity.

This is also recursive. It leads back to there definitely being 100 on day 100, and thus all the blues finally know they have blue eyes… AND ALL FROM EXAMINING THE ONLY POSSIBILITIES. There were no other possibilities that were missed out here. And all possibilities were examined.
THIS is why there are no alternatives to the correct solution.

This is the best way in which I’ve explained it yet.

You guys keep trying to reduce the “perfect logician” to being no more than a calculator wherein someone has to push his buttons. In the case of the ultra android, when someone pushes his 2+2 button sequence, it might well reach out slap him as well as add the numbers.

As soon as anyone mentions a rule wherein someone must leave if they deduce their eye color, every perfect logician would instantly deduce what would be required for that to happen, whether they wanted it to or not. That means that they would instantly become aware of every logical scenario that would lead to such a deduction, not merely any one particular scenario that would depend on someone pushing their buttons. And all of them would be thinking about all possible scenarios.

One of the scenarios that would allow the deduction of their color would be dependent upon a guru saying that there was at least one blue and everyone hearing and believing that. But before the guru even had a chance to say that, they would all already know of a scenario that allowed such a deduction without the guru speaking at all, as well as any other possible scenario.

Amongst all of the possible scenarios of which a perfect logician would immediately become aware, is one wherein they all merely start counting in the fashion that I was asking about. They would all realize, before the guru said anything, that if there was anything that could allow them to all begin that counting sequence they would be able to deduce their color. So each would then immediately deduce whether there was such a common situation.

All of that would be taking place before the echo of the rule being mentioned had faded. And since even I can figure out how to deduce such a thing without the guru, it is absurd to think they such perfect logicians could not.

The counting thing doesn’t work; it relies on “counting” actually being a hand-wavey introduction of the exact same information that the guru introduces. The only way any deduction is possible from it is if we assume that where they start “counting” is where they have a base of common knowledge, i.e. they know that if there were only X people on the island, they would know that there are X people on the island with blue eyes. But, as I pointed out to Fixed Cross above, if there were X people on the island, they wouldn’t be able to learn that there are X people on the island with blue eyes merely by counting the blue eyes they see. This is what Silhouette has been talking about for a long time about taking knowledge out of context.

Also, even if the counting thing did work, they could not magically start. There is no logically best number at which to start counting, so the islanders’ flawless logic could not lead them to a common starting point without communicating, which is explicitly excluded in the problem statement.

“Perfect logician” is a potentially ambiguous term, but fortunately the term is defined in the problem statement: “They are all perfect logicians – if a conclusion can be logically deduced, they will do it instantly.” The phrase following the dash should be read as a definition, and no other attributes should be ascribed to a perfect logician.

James, are you arguing that you can prove what the starting number is by deduction? If you think you can, please do. Otherwise, you’re making the argument that while we don’t know, a perfect logician would know. But you must agree that perfect logician is neither omniscient nor a mind reader, correct? Are we returning to a presumption that every islander wants to leave the island? Isn’t it clear that there are simply too many unknowns preventing a perfect islander from deducing a common starting number?

On a tangent, for anyone still following that accepts that solution as the solution: One thing that’s quite fascinating about resistance to this problem is the way it happens regularly around a certain threshold: people are generally willing to accept that on an island with 1, 2, or 3 people, the islanders learn something useful from the guru. But at 4, people refuse to follow the same reasoning, and they struggle to come up with reasons. It happens at the other end, too, such that 100, 99, and 98 people can’t be expected to act without the intervention of the guru, but 97 people can. But there’s a tempting just-so explanation in evolutionary psychology that I’m inclined to accept: that we actually deal with nested knowledge in social situations, but only to a very shallow depth. We might wonder what our parents know, or what our parents know that we know, but rarely will we consider what our parents know that we know that our parents know. It starts to get head-achy around that threshold because our brains aren’t built for processing it, and they aren’t built that way because it’s only in rare and contrived situations when it matters.

Carleas, you don’t know if it works or doesn’t.
You refuse to discuss it.

Nobody is taking perfect logicians as mere calculators. They can think/do anything they like, it’s just that nothing else is relevant to whether they leave or not, except the procedure which I have outlined. There are relevant consequences of there either being 100 blues or 99 blues (to each blue), but nothing else. Everything else, slapping included, is perfectly fine, but irrelevant to the exhaustive procedure of the correct solution.

This is enough to prove why no other solutions are valid in general.
Adequate specific disprovals have also been offered as a bonus, such as there being zero logical reason to choose to start counting from any particular number outside of the procedure of the correct solution. Further to this, there is no logical reason for all islanders to know which of the many combinations of e.g. 97 blue islanders are being referred to if they were to start counting from this or any other number.
No amount of knowing of knowing etc. can replace what the Guru says because none of this knowledge can be transferred, out of context, to situations where it no longer applies - such as the thought experiment of what 1 blue would do if there was only 1 blue (who would obviously only know what 1 blue would know and nothing more). It’s possible for each of 100 blue-eyed perfect logicians to attribute their knowledge to this sole blue, but inconsistent - and that is the problem. There is going to be a problem with any attempt at a solution that starts from irrelevance in order to “better” solve the problem (relevance as defined by the exhaustive analysis of the ONLY consequences that are to have any affect - as completely covered in the correct solution).

This problem has sufficiently been brought to a close.

I hadn’t read the beginning of this thread before, I was just going through it and reading your exchange with Abstract. I got up to the point of reading your hint that T is the next letter when I finally figured it out.
[tab]The letters have no curves.[/tab]

Just figured this one out as well, after looking at your first hint
[tab]sum of digits of two-or-more-digit primes[/tab]

New Problem. Some of you may have heard this in a different form. Avoid spoilers, and avoid looking up the solution.

I’m walking with James to an apartment where 3 girls we know live. Let’s call them Rachel, Monica and Phoebe. As we’re approaching, we see that their livingroom light is on. We know (you can assume this as an undeniable axiom for the case of this problem) that the light would only be on if at least one of them is home. The other thing we both know is that Monica always leaves the house with Phoebe if she leaves at all – either she’s in the house, or she’s out with Phoebe.

James tells me that, based on this information, Rachel by logical necessity MUST be in. His explanation is like this: Suppose that Rachel is out. If Rachel is out, then if Monica is also out Phoebe would have to be in—since someone must be in for the light to be on. However, we know that whenever Monica goes out she takes Phoebe with her, and thus we know as a general rule that if Monica is out, Phoebe is out. So if Rachel is out then the statements “if Monica is out then Phoebe is in” and “if Monica is out then Phoebe is out” would both be true at the same time. As those can’t be true at the same time, Rachel must be in.

Seems like a pretty compelling argument, no? The puzzle is to explain specifically what’s wrong with it.

[tab]Declaration: R, M, P (Rachel, Monica, Phoebe)
State (“in” or “out”) abbreviations: Rin/Rout, Min/Mout, Pin/Pout
Condition 1: if Mout, Pout
Condition 2: at least 1 must be “in”

8 permutations:
(1) Rout, Mout, Pout = wrong (at least 1 must be in)
(2) Rout, Mout, Pin = wrong (if Mout, Pout)
(3) Rout, Min, Pout = possible
(4) Rout, Min, Pin = possible
(5) Rin, Mout, Pout = possible (*only case where Mout possible)
(6) Rin, Mout, Pin = wrong (if Mout, Pout)
(7) Rin, Min, Pout = possible
(8) Rin, Min, Pin = possible

James’s assertation: if Rout:
if Mout, Pin (condition 2)
BUT if Mout, Pout (condition 1)
(Mout, Pin AND if Mout, Pout) =/= true, therefore Rin.
Problem:
This does not factor in the possibility that (given Rachel is out) Monica is “in”: (3) and (4).

[size=50]Also, anything James says is always wrong.[/size]
You got your topology and prime puzzle answers correct btw. I got the first one but not the second, grats.[/tab]