- Yes
- I’m not quite sure if I agree that no new information was produced by what the guru said. Someone somewhere else took the same problem and challenged their readers to find out what the new information was that the guru provided by saying that, which leaves open the (admittedly bizarre and difficult-to-imaging) possibility that there was some new information in what the guru said, the information just isn’t quite as straight-forward as taking what she said at face value.
- I’m not differentiating from logic that actually applies and logic that hypothetically applies. Either an argument is deductively valid or it is not. I have shown why it is deductively valid for 1 blue person, 2 blue, and I can show for even higher values of blue if need be. I can go all the way to 100. You have NOT shown anything similar for brown-eyed people. You can’t do it for 1, you can’t do it for 2, and I have no reason to think you can do it for 100 until you start doing some logic brother.
So you think that in the case of having 100 blue eyed and 100 brown eyed perfect logicians, one or more of them is going to try to actually apply a logical syllogism that only works for 3 blue eyed?
I’m going to attempt to put this in even more explicit format:
- If one person has blue eyes, he can immediately deduce from what the guru said that he has blue eyes.
- If someone can deduce that he has blue eyes, he leaves that night.
LDC1) If only one person has blue eyes, he leaves the first night.
We’re agreed up to here. That you accepted LDC1 is of the utmost import:
Now, the following logic is done from the point of view of someone with blue eyes who sees one other person with blue eyes – ie the case of 2 blue eyed people:
- I see only one blue-eyed person.
- He is the only one with blue eyes, OR, I have blue eyes (if you have a problem with this premise, say so explicitly)
LDC1) If he is the only one with blue eyes, he will leave the first night.
Now, our logician goes to sleep the first night, not knowing his eye color, and he wakes up the next morning finding that the blue-eyed guy he sees hasn’t left yet
- He did not leave the first night.
- He is not the only one with blue eyes.
- [take 6 and 4 together to produce] I have blue eyes.
He discovers he is the second person with blue eyes during the second day, and leaves that night.
LDC2) If there are two people with blue eyes, they leave the second night.
Now, again, I’m the ONLY one providing rigorous, clear, explicit logic. If you have a problem with any of it, specify which step in the logic doesn’t make sense to you. No more garbage about whether or not the problem is ‘shit’, I’ve done you the favor of laying my argument out completely. I’ve done you the favor of putting my own position in full-view, full-vulnerability. If it is false, you can demonstrate that by explicitly referring to the numbered steps in the argument. No more pussyfooting. No more weasel bullshit. Logic only. That’s what I’m doing, and I’m tired of being the only one.
Notice in my above argument that to prove LDC2, I had to use LDC1. This is a fundamental part of the pattern of logic in this particular problem. LDCn is meant to prove that n people can leave on the nth night. LDC3 requires LDC2, just as LDC2 required LDC1. LDC4 requires LDC3. LDC5 requires LDC4 as a premise. And so on and so forth. Each one higher relies on the one below it, all the way up to LDC100, which is the relevant one in this case, and beyond.
You have not shown any logic for how a brown-eyed person can deduce his eye color. You’ve claimed that ‘the logic is the same’ – surely if that’s true, it should be really easy for you to produce the syllogistic reasoning like I have, and just replace my arguments about blue-eyed-people with brown-eyed-people. If you’re reasoning is solid, you’ll find that you hit a snag because in the case of the brownies, LDC1 isn’t true. And without LDC1, you can’t prove LDC2. And without LDC2, you can’t prove LDC3. And so on up to 100. If your reasoning is solid.
You stopped too soon.
Let’s make it a little easier.
Go through a truth table for the case where there are only 4 on the island, 2 blue and 2 brown.
But the guru merely says, “Go!” and doesn’t mention an eye color.
Who will leave and when?
The people are A0, B0, C1, and D1, where 0 and 1 are the “colors”.
Nobody.
Don’t test me, I’m not testing anybody. I’m providing my own logic, in a clear, precise way. I’m making my argument the most vulnerable it can be by putting every premise explicitly out there. Why don’t you? Put yourself on the line brother. Put it out there. If your idea is as sound as you think it is, why do you hesitate to put it out there like I do?
I’m not testing.
I’m trying to point out the reasoning involved even when an eye color isn’t mentioned.
But in that case of 4, they would have to know that there were at least different colors.
If it had been more than 4, they wouldn’t have to be told that.
You haven’t pointed out the reasoning. You haven’t provided any reasoning. I want you to. I’m asking you to. I’m begging you to. So far, I’m the only one who has produced syllogistic arguments that the logicians can use to prove their own eye color. You haven’t. Do. I’m begging you.
With the 4 people;
1st day;
A0 sees B0 and c1-d1
B0 sees A0 and c1-d1
similar with c1 and d1
So no one leaves that 1st night.
2nd day;
A0 sees B0 and c1-d1 still there.
That means that B0 couldn’t tell that he was a 0 because he was seeing a 0.
So at that point, A0 knows that he is a 0.
But also B0 figures that out too.
They both leave.
Yet no color was mentioned by the guru.
Prove this please.
It is the first night.
The rule is that you must leave if you can deduce.
They know there are two colors involved.
If B0 (or anyone) saw 3, 1’s, he would leave the first night.
No one left the first night, so no one saw 3 of any color.
If there aren’t 3, 1’s, yet you can see 2, 1’s, then you know that you are a 0.
The know there are two colors involved? That wasn’t part of the problem specified. That’s an artificial premise that you added, not part of the problem statement. There’s nothing to say that there aren’t 3 colors of eyes, or 1 color of eyes. I don’t agree that they know that there are two colors involved. We know it, because we know all their eye colors, but how would they know it on the first night?
The problem comes in with the possibility of a third color, a 3 or red.
The second day, you still only know that no one saw 3 of a kind.
But you don’t know whether you are another 0 (blue) or 3 (red).
So even the second night no one leaves.
From there on out, no one can leave.
In the 100x100, they can see that for themselves.
In the 2x2, they have to be told.
Anything over 2x2 and they can see it themselves.
First your argument was that they can leave.
And now it’s that they can’t.
James, I don’t know what you’re talking about anymore. I’m fucking flabbergasted.
It really isn’t that complicated.
And I don’t know what you meant by;
“First your argument was that they can leave.
And now it’s that they can’t.”
…unless you mean due to the red.
You start out by saying that the 0s can leave.
And then you say this:
I don’t know what’s going on in your head. I have no clue.
First I am talking about the fact that they do not have to hear a color mentioned.
The logic is the same.
But then considering that there is a possibility of there being a third color, that logic falls apart.
I’m not following. At all. I have no clue if you think they can leave or they can’t. I don’t know what you’re talking about. Seriously. At all.
Geeezz…
Did you follow the 2x2 example at all?
Did that make sense?