Math Fun

It is all the same method.

And that is how you know that it is turtles all the way down.
"Isn’t it obvious?!?! "

Right, but the “it” here is not referring to “any possible configuration of logicians, colors, and headbands,” it’s referring to “the specific configuration of logicians, colors, and headbands in which the Master is speaking.” Again, take it like every other premise: a configuration with zero logicians, or transparent headbands, or whatever other violation of the premises could not be a solution to this problem. In the same, way, a configuration that is unsolvable can exist, it just can’t be the solution to this problem. Part of what this problem asks us to provide is a configuration that is solvable.

y = -|x/2 - 2.5| + 3.5
Sorry, I thought we were actually trying to resolve a disagreement, and not trying to score cheap rhetorical points by getting snide about typos.
Because I think you take my point: there are an infinite number of equations that fit a set of data points but diverge outside the set. That’s a problem for any answers that depends on looking for a pattern in the colors, but not for an answer that depends on constraining the set of colors to the colors a logician can see.

Carl, with that one equation/algorithm alone, you prove my point.

No matter what algorithm you might conceive for whatever color pattern you see, the Master might be using a different one. And you can’t leave until you KNOW. And you cannot know unless you can prove that the one you are using is the ONLY possibility for the colors seen by each member.

You keep trying to merely prove that your algorithm works (if it is the same as the master’s). We already know that it works IF it is used by all parties. How are you going to prove that the master didn’t have a better one? You can’t.

But the solution Phon and I are advocating isn’t an algorithm. You made the distinction yourself: in our solution, each logician is constrained by what she can see, not what she can envision. What each can see is finite set, what each can envision is an infinite set. An infinite set renders the problem unsolvable, but a finite set does not. What we’re saying is that our solution is different in kind from one where an equation or pattern is used, because an equation or pattern is inherently problematic.

And I think to that point, Phon has put it best: the colors each logician can see are certain, in a way that the completion of a pattern is not. Closing the door to all patterns, and constraining ourselves to the visible set of colors, produces a solvable problem. So it must be a solution.

Not true on two accounts.
The assumption that the color must already be in sight is itself not seen, merely speculated/envisioned, a rule to follow, the first in an algorithm to presume.

“If my same color is within my sight and no one left after the first bell (or first opportunity), then my color must be the same as the double color that I can see.”

That is your theory to kick things off, right? Three objections:

1. There is nothing to say that your same color must be within your sight.
2. Perhaps no one left for a different reason.
3. What if everyone left after the first bell but you and perhaps one other?

The members who do not leave after the very first bell are deducing that they CANNOT know their color. What is the proof that they cannot know? The only thing provided is that they cannot envision anything else. But perhaps there is a pattern to be seen.

Again, until you can prove that the decisions being made cannot be anything else, there has been no proof of resolution.

One can envision a standardized, finite set. And there might be many other things that can be envisioned. Only when you can prove that nothing else could possibly qualify, have you found a resolution to the puzzle.

Demand that the group be sufficiently large to randomize any potential pattern. That way, you know that there can be no useable pattern. But how many colors will that take?

Although I agree with that, you haven’t really proven that one either.

You can’t merely declare that your pattern isn’t a pattern. Each and every example of the puzzle will be a pattern of one type or another. Ignoring that doesn’t count.

That part is absolutely wrong.

Imagine that someone has stated that puzzle to include the master saying, “And your same color is within your sight”. And then imagine someone saying it but leaving that one statement out. There is a difference in confidence.

Ah! Common ground!

From here, we can get quite far:

1. The color of each logician’s headband is either seen by that logician or not seen by that logician.
2. “Colors seen by a given logician” is a finite set;
3. “Colors not seen by a given logician” is an infinite set.
4. An infinite set renders the problem unsolvable,
5. The problem is solvable
6. Thus, the color of each logicians headband is not part of the set of colors not seen by that logician.
   Therefore
7. The color of each logician’s headband is a part of the set of colors seen by that logician.

I really wish that I could take that as you being facetious. But no…

True.

True.

False … well, unless you meant the entire set of all colors not seen, then of course, true.

Assuming that was the only option, which … for the 20th time … isn’t.

Seriously?

Sherlock Holmes: “Examining the family, I can’t see who dun’it, therefore the criminal must be outside the family. And those possibilities are infinite. Thus my dear Watson, this mystery cannot be solved!”

I did.

This is a given. It is a premise supplied in the description of the problem. Which is to say yes, seriously.

So it looks like you accept all the premises of my syllogism (except one of the givens, which you are refusing to accept but that’s fine, if you break the problem only by assuming one of the givens is false, I’ll take that as a concession). And your only criticism of the logic seems to be:

As though a syllogism becomes less true if it weren’t true had we taken a different premise, or if there were a different premise that could have been invoked at any given point in the syllogism that would have produced different results. There is no rule of logic that suggests anything like that. Indeed, it is false. A syllogism works regardless of whether it’s believed or whether other syllogisms might have been performed.

If you’d like, you can also run the syllogism with the additional premise, “half the logicians use a different method”:

1. The color of each logician’s headband is either seen by that logician or not seen by that logician. (excluded middle)
   2a) “Colors seen by a given logician” is a finite set; (given)
   2b) “Colors not seen by a given logician” is an infinite set. (given)
2. An infinite set renders the problem unsolvable, (given)
3. The problem is solvable (given)
   4a) Thus, can’t be an infinite set (from 3 and 4)
4. Thus, the color of each logicians headband is not part of the set of colors not seen by that logician. (from 2b, 4a)
5. Half the logicians use a different method. (given)
   Therefore
6. The color of each logician’s headband is a part of the set of colors seen by that logician. (from 5 and 1)

Looks like it doesn’t matter whether or not it’s the only option; as expected, a valid syllogism remains valid in the presence of other options.

we’ve gone in circles several times

Carleas, you are proposing a false dichotomy:
A) One to distinguish from the finite set that I can see.
B) One to distinguish from an infinite set that I cannot see.

The fact that the one that you cannot see is a part of an infinite set doesn’t prevent it from being a member of a finite set that you can deduce (or in your case assume).

Assuming that you can see your same color is no different than assuming anything else that would help resolve the puzzle. Why not just assume that all colors are doubled or that every color is represented with its opposite. That way everyone could go home after the first bell. You can probably make assumptions in order to resolve the puzzle all day long and always use the excuse, “it is solvable”. But they are always just assumptions.

And finding one compatible algorithm doesn’t change anything. You have to prove that there are NONE incompatible, regardless of how many are compatible.

It is just an invalid puzzle.

Go ahead and make your closing statements on the issue.

BUT IT’S NOT CERTAIN AaaAAAGHHhhHh

I have to agree with Phoneutria: there are certain colors we know to be a part of the problem. Any method that would add colors to the problem will allow us to add an infinite number of colors to the problem, which would make the problem unsolvable. Therefore, it must be that the only colors in the problem are the ones I know to be in the problem. (Related: this morning I learned about the Online Encyclopedia of Integer Sequences, which further supports the notion that any reasoning based on pattern is going to render the problem unsolvable; see, for instance, this result for the pattern 2,2,2,2,1,1)

You have this strange skeptical worry about these common knowledge problems, which I think is unwarranted. First, in this problem, the question doesn’t even come up until after we determine which colors are in play. That reasoning does not depend on anyone else’s thoughts, so the objection is a red herring.

Second, the concern is not how logic works. You’re getting caught up on what A knows about what B knows, that maybe they won’t think the same way. But we’re told that they will execute logical operations reliably, and they’ll know something when it’s logically provable. The logical proof for ‘N blues leave on day N’ provable does not depend on any particular method of thought, it is pure mathematical induction.

Now, you might argue that we aren’t proving N blues leave on day N’, but instead ‘N blues leave no later than day N’ (that is not the result of the syllogism, but it seems to be what you keep suggesting). But ‘N blues could not leave earlier than day N’ is also provable:

1. Without the oracle’s statement, no one can learn their eye color. (given)
2. The oracle speaks on day 1. (given)
3. 1 Blue could not leave before day 1 (from 1 and 2)
4. N blues could not leave before day N (assumed for proof by induction)
5. N+1 blues cannot distinguish their situation from the N blues situation without more information (given)
6. The only way N+1 blues can get information from others is by others leaving or staying (given)
7. Staying only conveys information when the blues who stay could leave (given)
8. N blue staying or leaving doesn’t convey information until the end of day N (from 4 and 7)
9. The N+1 blues cannot get information until the end of day N (6 and 8)
10. The N+1 blues cannot leave before day N+1 (5 and 9)
11. IF N blues could not leave before day N, THEN N+1 blues could not leave before day N+1 (4 and 10)
    Therefore
12. N blues could not leave before day N for any natural number N >=1 (from 3 and 11, by induction)

So nothing new in your argument and nothing new in mine.

We disagree on this one (and will on all of its type).

So do you people know that color of the frickin’ headband or not?

Where is Ed3? He’ll figure this shit out. You people are fired.

This really isn’t an Ed3 type of concern. I think he got stuck on another thread that was more in his line (mathematics).

That’s better.

In case you forget again, I’m still here if you come up with a way to express your objection to the syllogisms I’ve provided. Like the last one, which I’ve never offered before (it’s what you might call “something new”) and logically precludes any of the logicians leaving sooner than day N.

???
This is what you wanted me to not forget:

Ummm… okay.

It isn’t anything new simply because you are still ignoring and skipping over the only problems. You keep trying to prove the obvious and ignore the issue.

Don’t worry. I am well aware of your online glaucoma issues. When and if I feel the need to nail your ass again, I will go to the trouble. But this issue isn’t that time.

This part, James. This is what you forgot. This time, as you go away, never admitting defeat, I’d like you to remember who’s going away, and who’s still here.

I’ve given you a syllogism, if it’s not a sound demonstration of the conclusion, then either point to the specific given (if it is valid and not sound) or the specific logical step (if it is invalid) that you think is wrong. I made it easy and numbered them for you. Make clear which are “the only problems” and we can go from there.

To nail anything, you have to get specific. You won’t see the problems with your position until you do more than wave your hands and speak in generalities and repeat rhetorical catch phrases like “turtles all the way down”. If you mean that the reasoning is circular, show that it is. Spell out your reasoning, clearly, directly, without sarcasm or sass. If you’re having trouble articulating it under those conditions, it’s a good bet that the reasoning was never actually there.

I say this because I’ve been wrong before, I’ve known the difficulty of making explicit a flawed position that I’d always accepted without close examination, and at times I’ve stubbornly refused to admit when my argument was vacuous. I even thought for a time that I was wrong on this logicians problem, but Phoneutria’s rebuttals, your lack of exposition, and my own attempts to articulate both your objections and my responses to them leave me pretty well convinced that the logic is sound – in part because of the very arguments you made showing that certain minds of argument would lead to infinite possibilities!

You’re absolutely free to walk away, but don’t pretend I’m engaging in bad faith or refusing to address the arguments you make whenever you’ve deigned to actually make them. And if you make new ones, or even try anew to express arguments you think you’ve already made for your position, I’ll keep offering rebuttals and rearticulating mine. I’m still here.

You went away for 6 months without a word.

I did that already, not going to keep going on about it. Obviously it is one of those blind spots you choose to support.

Do you really want me to start bring sight to the blind?

And you will never prove anything until you actually eliminate all alternatives. The fact that you can only think of one possible solution to a problem doesn’t prove that yours is the only one. And in this kind of problem, if there is another possible solution, then yours isn’t a solution.

That is the part that you can never accept.

But thanks for playing.