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:
- 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.
- 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: - The consequences of there being 1: that they would leave on the ferry in the knowledge that they have blue eyes, and
- 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:
- 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
- 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]
You got it Sil. Here’s a bit more to chew on maybe:
[tab]As you said, James Hadn’t considered the possibility of Rout Min. What James actually proved is that, if Rout, then Min necessarily. Because Rout AND Mout produces Pout AND Pin, as he showed.
The misunderstanding James had was to think that A → B and A → ¬B is a contradictory scenario. B and ¬B is contradictory, but A → B and A → ¬B is not contradictory at all (not unless you’re also saying A). You can reduce logical implication thusly:
A → B is the same as saying B v ¬A
So, (A → B and A → ¬B) is reducible to (B v ¬A) ^ (¬B v ¬A). The way to make that conjunction true, is of course to make ¬A true. No contradiction at all, it just requires A not to be true.
So, A → B and A → ¬B isn’t a contradiction, it’s just a disproof of A.[/tab]
[tab]Some more that I was just thinking:
James’ thought process was obviously that A → B and A → ¬B is a contradiction.
To be a true contradiction, it would have to be (A → B) and ¬(A → B), and so he may have thought that ¬(A → B) = (A → ¬B)
Let’s look at what ¬(A → B) really equals:
As we said, A → B = B v ¬A
¬(A → B) = ¬(B v ¬A) = ¬B ^ A
A → ¬B = ¬B v A
¬B v A =/= ¬B ^ A (although they do look similar)
So ¬(A → B) =/= (A → ¬B)[/tab]
Sleeping Beauty volunteers to undergo the following experiment. On Sunday she is given a drug that sends her to sleep. A fair coin is then tossed just once in the course of the experiment to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday, given a second dose of the sleeping drug, and awakened and interviewed again on Tuesday. The experiment then ends on Tuesday, without flipping the coin again. The sleeping drug induces a mild amnesia, so that she cannot remember any previous awakenings during the course of the experiment (if any). During the experiment, she has no access to anything that would give a clue as to the day of the week. However, she knows all the details of the experiment.
Each interview consists of one question, “What is your credence now for the proposition that our coin landed heads?”
[hint: A viable answer is to remove the abstract term ‘credence’ and turn the situation into something more…tangible. If you so wish.]
[tab]When she’s interviewed, there’s a .333[bar] chance that it’s Monday and the coin has landed heads, a .333[bar] chance that it’s Monday and the coin has landed tails, and a .333[bar] chance that it’s Tuesday and the coin has landed tails. This means that her credence for the proposition that the coin landed heads is logically one third full each interview.[/tab]
For the following:
0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
Insert any mathematical operators (except ones that involve writing a number) into the left hand side of each of the above equations, such that they become correct.
No manipulation of any number shapes or outside-of-the-box thinking is required to do this. Use of brackets/parentheses is encouraged for clarity.
Still missing:
8 8 8=6
9 9 9=6
The rest:
tab!=6
(1+1+1)!=6
2+2+2=6
(3x3)-3=6
(4-(4/4))!=6
5+(5/5)=6
6+6-6=6
7-(7/7)=6[/tab]
I think I need to incorporate another operation…
Good effort! 
8s are the hardest, btw.
[tab]Good job thinking of using factorials - that was the least obvious one to me.
You’re right about needing to incorporate another operation to finish things off.[/tab]
Don’t open the next tab if you want to see if you can do it all without any help…
[tab]Here’s a clue, squirt. Another operation would neaten up the 4s a little while you’re at it.[/tab]
I figured out a cheat-y way that solves all of them:
[tab]( ( d/dx(8) )! + ( d/dx(8) )! + ( d/dx(8) )! )!
That works no matter what number you use, because the derivative of any constant is 0, and the factorial of 0 is 1. It’s a bit of a cheat because d/dx is arguably adding another number (dx).
It also makes the problem boring.
Still no legit solution to 8 and 9.[/tab]
Haha!
Wow, it’s amazing the creativity that difficult problems can drive us to.
The function you’re looking for really is fairly elementary.
Another hint, should you need it:
[tab]Start with the 9s. Think about what you can do to a number like that, especially in light of a number I’ve already mentioned in my previous hint.[/tab]
Hmmm…
[tab]I think you’re suggesting taking the square root, which I wrote off because I saw it as incorporating another number: 2.
If that’s what you’re suggesting, 9 is easy: ( 9/sqrt(9) ) +sqrt(9) = 6
and 8 - sqrt( sqrt( 8+8 ) = 6
But why not cube root + cube root + cube root?
And if we’re doing that, why not x root of c + c + c, where x is whatever irrational number root results in 6?
Or log base y where y is a number that when raised to the 6th power equals c + c + c?
I don’t know why I treated the square root as adding a number, but not the factorial. As I think about it, I find the line between adding another number and just adding an operator to be fuzzy.
After all, you can always define a new operator. So we could define the unary operator B such that Bc=(c x 0) + 6.
B(c + c + c) = 6 for all c.[/tab]