Yes it does. Those colors are the only colors which you can be absolutely certain belong in the color group. Any other colors you choose to include by any method you choose to use, is a guess.
It has everything to do with understanding that the only way to be certain is to use visible colors.
The two statements are to the point that any other solution is uncertain.
I meant if my color is not within my sight of n colors, my solution cannot be certain.
Same probability as any other guess, as careful as your guess might be. That is why you can’t try to guess a color and instead you must use only the colors you can see.
The algorithm I suggested only comes into play once you have determined the constraint for your color group, which is colors that aren’t a guess.
By removing uncertainty.
That this is the only certain method available to all logicians is very clear to me as I have tried extensively to demonstrate. I might be able to put into notation, but it won’t be anything more than what I have already presented in natural language. I can offer no further proof, and as far as I am concerned no further proof is required.
There appears to be 3 basic principles upon which we disagree: 1) Your color must be within sight. 2) There is no need to prove that your unseen algorithm is unchallengeable. 3) The Master cannot be lying because that is a premise to the puzzle.
You claim that the Master’s assertion that the puzzle is solvable allows you to be certain of whatever pattern and algorithm you envision to be the right one, “else it would not be solvable”.
In the following example (1), the Master has declared that the puzzle is solvable. You can envision a pattern. But you cannot visually see the missing color that completes that pattern:
Remember the Master said that it is solvable.
You seem to deny a “premise” and claim that the Master was lying.
A different example (2), again the Master said that it is solvable. So what is the only possible algorithm and when do they each leave?
And a third example (3), again the Master said that it is solvable, so what is the only possible algorithm and when do they each leave?
The combination of the 3 of those proves that: 1) Your color certainly need not be within sight, 2) You certainly must prove that your chosen algorithm is unchallengeable,
And/Or 3) The Master can be lying.
And of course if the Master can be lying, none of the puzzles are ever solvable.
It isn’t a given that all possible configurations are solvable, only that this problem takes place within a solvable configuration. We aren’t told how many logicians there are, or how many colors, or how many of each color. So it isn’t problematic to say that some configurations would leave the logicians unable to deduce their color, while the problem itself is still solvable. Just like a configuration where all the logicians had the same color headband should not be considered because it would violate the premise that there are many colors, so too would an unsolvable configuration not be considered for violating the premise that the problem is not impossible to solve.
To you hypotheticals:
I think that’s an interesting shade of purple you chose; I certainly wouldn’t have used it. My mental idea of the color wheel has a much deeper purple. I might be tempted to fill in the last dot with orange, but are you and I thinking of the same orange? I’d say this problem is unsolvable, because there is still an infinite (though bounded) set of oranges to choose from to complete the pattern.
I’m on the fence about whether this one is solvable. I think there’s a problem in that, though there seems a definite pattern suggested, there are in fact an infinite number of patterns that satisfy this. 2R, 2B, 2Y, 2R, 2B, 2Y, 2R, 2B, 1Y, 1G is a pattern, there doesn’t seem any logical reason to reject it (except that G would have no way to guess that she G and not O or P).
Let’s assume the empty circle is Green, and ask ourselves, what would Yellow see? 2O, 3R, 4B, 5G. Is Yellow thinking, “I must be 1 Purple!” Or maybe “I must be G, so that all of the groups have a number that is divisible by 3 or 4!” It looks to me like this one is unsolvable too. We can therefore rule it out.
Shades of colors were not visible and would violate the constrain of resolvability.
You see the sequence:
[list]1
1.5
2
2.5
3
__
[/list:u]
You merely have to fill in the blank knowing that “it is resolvable”. And you are claiming that the Master is lying. Is that allowed?
If there are an infinite number, surely you can give us one. And you are claiming that the Master is lying. Is that allowed?
If the yellow is the only one who has that problem, all he has to do is wait.
All of the other colors see an obvious pattern if the puzzle is solvable. So all but the yellow leave after the first bell. Then the yellow can see that he is alone, not a green. So what color is he if the puzzle is solvable? Yellow is the only option that fits any solvable pattern.
And you are claiming that the Master is lying. Is that allowed?
The following is the one that demonstrates my main point:
Ambiguous possible patterns/algorithms to choose from;
Which algorithm do you choose? The one that you have been suggesting is only one of several options. Different algorithms lead to different timings of when each person leaves. But the Master said it is solvable. Everyone’s color is within sight.
I say that the Master was lying.
And yet this is the same color grouping that you have recommended as a solvable puzzle.
We don’t know the configuration of the logicians, but the Master does. If you come up with a configurations that isn’t solvable, and of course there are many, then it cannot be a solution.
I’d be interested to see the syllogism (not the algorithm) you are using. Be as explicit as you can be, sassy “turtles all the way down” hand waiving just looks like you can’t actually trace the logic you’re alleging to be there, which is my suspicion. You seem to be making two claims which produce some tension:
There is no logical basis for a logician to conclude that her headband is one of the colors she can see.
There is a logical basis for a logician to conclude that her headband completes some pattern that she can see in the other headbands of the other logicians around the room.
If there are cases where alternate syllogisms introduce ambiguity, they do not defeat the problem. All there needs to be is one situation where the type of ambiguity you see in your examples does not exists. If your examples are actually ambiguous, they aren’t solvable, so they aren’t part of the solution set. If they are solvable by some other means than the Logicians being able to see their own color, they are still part of the solution set, and you will have only shown that there are another class of solutions in the solution set.
2.5
Or were you not using the function y = -|x/2 - 2.5| + 3.5 … ?
The Master said that it is solvable, so anything that solves it, solves it.
Obviously it is solvable, just as the Master said. You just can’t figure out how it is. Do I need to explain the solution to you again? It is similar to yours: 1) “Assume that the Master is right” 2) “Imagine a possible pattern for the example.” 3) “Be 100% certain that you are right because you can’t think of any other way to solve the puzzle.” 4) “Argue endlessly by any means possible that you cannot possibly be wrong.” 5) “Don’t answer any question that might incriminate you. After all, it is merely about Ego defense.”
You have yet to prove that you have found such an example, because you refuse to answer any questions.
You know, certain things in science haven’t been fully proven yet, so you shouldn’t go on an airplane, ever. You never know, this whole thing keeping airplanes in the air might be turtles all the way down.
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.
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