Math Fun

[phoneutria]

I'm sorry if this sounds repetitive in itself and in addition to all we've already posted, but I think this sounds straightforward.


There are infinite hues of colors in the spectrum of visual light.
Therefore my color options are infinite.

I cannot know with certainty the color of an object that I can't see by picking one color from an infinite number of colors.
The solution to the puzzle requires that I know the color of my headband with certainty.
Therefore picking a color from an infinite number of colors will not solve the puzzle.

In my field of vision I see n colors, all which are certainly a part of the puzzle.
I cannot know with certainty the color of an object that I can't see by picking one color from an infinite number of colors.
Therefore the color of my headband is included in n.

Likewise all other participants must be certain of the color of their headbands. Therefore n is the same for all.


let me know what you think, James.

[Carleas]

Come on Carl, all you actually said was:
If the rule is that "N blues leave on day N", then 100 blues leave on day 100.

I could also say:
If the rule is that "N blues leave on N/2", then 50 blues leave on day 100.

Those don't prove anything.

That is not what I said. Again:
An inductive proof relies on two premises: 1) the base case, and 2) the inductive step. The base case is that a statement is true for N=1. The inductive step is that if a statement is true for N, then it is true for N+1. Given both premises, we know the statement to be true for any natural number N >= 1. This is the structure of a inductive proof, it is a common and well accepted tool of mathematical reasoning.

Here, the premises are, respectively:
1) If there is 1 Blue, they leave on day 1, i.e. C(1)=L(1)
2) If it's true that N Blues will leave on day N, then it is true that N+1 Blues will leave on day N+1, i.e. C(N) = L(N) -> C(N+1) = L(N+1)

Given those two statement, we know that C(N) = L(N) for all N >= 1

And again, I think your problem with the syllogism is actually with one or both of the premises. At least, I hope it is, because rejecting the entire form of inductive proof is ... not likely to be a productive way to discuss a logical problem. I'd have to ask what your standard for a logical proof is. Are you going to require me to demonstrate modus ponens? If not, the argument for the validity of inductive proof is just the repeated application of modus ponens:
X is true for 1
If it's true for a number, it's true for that number plus 1
1+1 is 2
X is true for 2
2+1 is 3
X is true for 3
...
X is true for N-1
N-1+1 is N
X is true for N

Are you really rejecting proof by induction?

[James S Saint]

Are you really rejecting proof by induction?

No. You are not proving anything by induction, avoiding the issue, not solving the puzzle, and not answering direct questions (again - as always).

You have algorithm A that you can see would work if everyone used it.
Another member, smarter than you, has algorithm B that he can see would work for everyone.

Which one are you each going to use?

There are infinite hues of colors in the spectrum of visual light.
Therefore my color options are infinite.

Before any deducing is involved, true.

I cannot know with certainty the color of an object that I can't see by picking one color from an infinite number of colors.

Assuming that seeing is the only option available, true.

The solution to the puzzle requires that I know the color of my headband with certainty.

Critically true.

Therefore picking a color from an infinite number of colors will not solve the puzzle.

"Therefore, attempting that algorithm would not solve the puzzle", true.

In my field of vision I see n colors, all which are certainly a part of the puzzle.

True.

I cannot know with certainty the color of an object that I can't see by picking one color from an infinite number of colors.

A repeat.

Therefore the color of my headband is included in n.

Nonsequitor.

"In my field of vision, I see n colors involved, and possibly my own color." - insisting on using colors seen.
"In my mind I envision n colors involved, and possibly my own color." - allowing for a color unseen.

What is the difference?

"In my head, I envision the use of one particular algorithm.
I envision no other algorithm to use.
Therefore the one that I envision is the right one.
Therefore, it is turtles all the way down."

"I can envision God creating the Earth in 6 days.
I envision no other algorithm to understand how the Earth got here.
Therefore, the Earth got here by God creating it in 6 days.
Therefore, scientists are lying."

Do you understand the problem with just assuming that the only algorithm that you can currently imagine is the right one?

[Carleas]

You have algorithm A that you can see would work if everyone used it.
Another member, smarter than you, has algorithm B that he can see would work for everyone.

Which one are you each going to use?

This misses the point. If we've proven that C(N) = L(N), it doesn't matter how it was proven, we know it to be true. It's as true as any given. Certainly if we were given a piece of information you wouldn't be concerned whether or not it might also be derived from other givens, right?

Are you criticizing inductive proof generally of being unable to prove things because there might be other ways to prove to them? If not, you can only mean to reject a premise of the proof, either the base case or the inductive step.

[James S Saint]

Carl, you are missing the point.

Trying to prove that one method works is not the point.

What you have to prove is that someone ELSE is using that method.

[Carleas]

To the extent that your response has anything to do with what I'm saying, it is a rejection of the premises, not of the method of inductive proof. All I'm trying to do right now is establish that proof by induction is a valid form of logical reasoning, and that a conclusion reached from the two necessary givens is known to be true without any additional given that disclaims that another form of argumentation will show otherwise.

The following does not make any reference to anyone doing anything. It is pure mathematical logic, and it is valid without having to include a separate premise about what the number 1 was thinking.
(1) C(1)=L(1)
(2) C(N) = L(N) -> C(N+1) = L(N+1)
Therefore,
(3) C(N) = L(N) for all N >= 1

This is a logical conclusion, and no other given is needed to say that if (1) and (2) are true, then (3) is true. Agreed?

[James S Saint]

All I'm trying to do right now is establish that proof by induction is a valid form of logical reasoning,

Why are you bothering to do that (other than a strawman attempt)? I have never implied that inductive reasoning is not a valid form of reasoning. But trying to use induction as a means of proof (the lack of any possible alternative) is tricky.

and that a conclusion reached from the two necessary givens is known to be true without any additional given that disclaims that another form of argumentation will show otherwise.

The truth of your syllogism IS NOT THE ISSUE!!!

Of course what you gave is true, as was the contrary statement that I gave. Those prove nothing concerning the puzzle.

What you have to prove is that someone ELSE is using that method.

You have algorithm A that you can see would work if everyone used it.
Another member, smarter than you, has algorithm B that he can see would work for everyone.

Which one are you each going to use?

[phoneutria]

Are you really rejecting proof by induction?

No. You are not proving anything by induction, avoiding the issue, not solving the puzzle, and not answering direct questions (again - as always).

You have algorithm A that you can see would work if everyone used it.
Another member, smarter than you, has algorithm B that he can see would work for everyone.

Which one are you each going to use?

There are infinite hues of colors in the spectrum of visual light.
Therefore my color options are infinite.

Before any deducing is involved, true.

I cannot know with certainty the color of an object that I can't see by picking one color from an infinite number of colors.

Assuming that seeing is the only option available, true.

The solution to the puzzle requires that I know the color of my headband with certainty.

Critically true.

Therefore picking a color from an infinite number of colors will not solve the puzzle.

"Therefore, attempting that algorithm would not solve the puzzle", true.

In my field of vision I see n colors, all which are certainly a part of the puzzle.

True.

I cannot know with certainty the color of an object that I can't see by picking one color from an infinite number of colors.

A repeat.

Therefore the color of my headband is included in n.

Nonsequitor.

"In my field of vision, I see n colors involved, and possibly my own color." - insisting on using colors seen.
"In my mind I envision n colors involved, and possibly my own color." - allowing for a color unseen.

What is the difference?

"In my head, I envision the use of one particular algorithm.
I envision no other algorithm to use.
Therefore the one that I envision is the right one.
Therefore, it is turtles all the way down."

"I can envision God creating the Earth in 6 days.
I envision no other algorithm to understand how the Earth got here.
Therefore, the Earth got here by God creating it in 6 days.
Therefore, scientists are lying."

Do you understand the problem with just assuming that the only algorithm that you can currently imagine is the right one?

it is not a non sequitur. it requires the wntire sequence that I posted to arrive at that conclusion, and it also shows how necessarily all other participants will use the same algorythm.

[James S Saint]

"In my field of vision, I see n colors involved, and possibly my own color." - insisting on using colors seen.
"In my mind I envision n colors involved, and possibly my own color." - allowing for a color unseen.

What is the difference?

"In my head, I envision the use of one particular algorithm.
I envision no other algorithm to use.
Therefore the one that I envision is the right one.
Therefore, it is turtles all the way down."

"I can envision God creating the Earth in 6 days.
I envision no other algorithm to understand how the Earth got here.
Therefore, the Earth got here by God creating it in 6 days.
Therefore, scientists are lying."

Do you understand the problem with just assuming that the only algorithm that you can currently imagine is the right one?

I take it that your answer is "no".

In my field of vision I see n colors, all which are certainly a part of the puzzle.

The fact that you only see those n colors and that they are obviously a part of the puzzle has nothing to do with what your own color is.

I cannot know with certainty the color of an object that I can't see by picking one color from an infinite number of colors.

The fact that you cannot pick from an infinite list of possible colors has nothing to do with what your own color is.

Therefore the color of my headband is included in n.

Huh??
You make two statements that have nothing to do with your color and then conclude something about your color from them? No.

What you probably wanted to say was;
) If my color is not within my sight of n colors, I must blindly pick it from an infinite list.

That is not a true statement, but necessary for your syllogism.

As I first demonstrated, if you see one of each of the primary and secondary colors except one, you can pick that one with the same probability as picking the algorithm that you suggest using. You cannot "see" that algorithm any more than you can see the missing color. You have to imagine either in order to use them.

The problem is that if you are merely going by what you imagine, whether that be a color or an algorithm, you cannot be certain that you are imagining the "right" one (the Master already has a "right" algorithm in mind that he is going to force the members to obey even if they do not understand it).

So how do you prove that you know which algorithm the Master is using?

Just because yours would work, doesn't mean that his wouldn't work better.
You have to prove that his and yours are necessarily identical or compatible.


I am not asserting that such is impossible. You have to prove the absence of a better algorithm than the only one you know. I am merely pointing out that no one has done that and until they do, NO proposed algorithm is legitimately proven to resolve the puzzle.

[Ben JS]

Feel free though to keep being condescending like phon, and make assumptions about me.

What could be more condescending than to go into a thread and not even bother to read the current conversation, but just state that you can google the answers, as though we were somehow unaware of this magical technological tool you've found, even though we are all here on the same fucking internet?
If you just need to say something, anything, so much that you'll do such a thing, you may as well just come in and tell some jokes. At least we'll get to laugh.

I ask myself,

'Why do they have to argue unnecessarily?'

I know I don't like getting into arguments. I'm also interested in truth and learning.

Therefore, I think to myself, 'Well, if I show that the solutions are available, they can end their argument & learn from their initial reactions to the puzzle.'

Simple as that.

Have you considered the solution, phon? And then considered how you initially responded?

-

My interest isn't in being liked.

If it were, I'd refrain from comments like - 'Fuck you'

Doesn't paint a great picture of me.

I like that I've pissed you off - makes me think my actions are relevant.

An offended phon is a rare occurrence.

Keep fighting the good fight, lady

[Carleas]

Therefore, I think to myself, 'Well, if I show that the solutions are available, they can end their argument & learn from their initial reactions to the puzzle.'

Ben, there are claimed solutions available, but James raises a good objection that the solutions don't address. And the solutions written on the page could be wrong. After all, I edited the solutions on the Wiki page, and I could be wrong.

The Wikipedia solution is not authoritative in itself, and that page is un-cited. We can't just take its word for it, we need to hash it out. And even if we could take Wikipedia as authoritative, hashing it out would help us understand the answer.

You might not like arguing, but it is likely to produce more accurate beliefs than the blind trust of quasi-authorities.

What you have to prove is that someone ELSE is using that method.

No, I don't, and this is something that distinguishes the Blue Eye problem from the Master Logician problem.

In the Blue Eye problem, we are given as a premise that those with blue eyes will leave when they know their eye color. We are also given that when they could know their eye color, they will know their eye color. We know that one Blue would know their eye color on day one, and we know that if N Blues would know their eye color on day N, then N+1 would know their eye color on day N+1. If we accept those, we will know that the 100 Blues will leave on day 100. There is no reference to any choice of algorithm, such a choice is not a part of the deduction: it is a pure, mathematical deduction.

The Master problem is different. There is the deduction, but also there is a logical leap, and I think there your criticism is more of a problem: can we logically rule out the possibility that any other logician will pick a different way to resolve the problem that without more information, the problem is impossible. Is there a way to establish that the colors seen are the only colors in the problem given that the Master says the problem is not impossible? Here, I think the question is well placed: how do you know someone else won't resolve the impossibility by reference to a color wheel, or by finding a pattern in the wavelengths of light reflected by the various headbands, or otherwise plug in a different assumption to complete the syllogism?

Another issue with the Master problem, also unique to the Master problem, is that we don't know the full setup. If there are N possible colors and N logicians, and each logician is wearing one of the colors such that they can each see N-1 colors, it might be that they could conclude that their headband is the missing color. That doesn't affect whatever deductions could be made in other setups, but it is relevant since the setup is part of what we're trying to find.

"In my field of vision, I see n colors involved, and possibly my own color." - insisting on using colors seen.
"In my mind I envision n colors involved, and possibly my own color." - allowing for a color unseen.

One possible response to this is that the infinity of possible algorithms based on a given set of colors would itself exclude the possibility that it is any color besides those a logician can see. There are an infinite number of extensions to any pattern, of equations that fit a set of points. But if the color set is closed, constrained to the colors each logician can see, there is no such infinity.
The set of colors a logician can see is unique,
The sets of colors she can envision is non-unique
It is impossible to deduce which set is actually the correct set from a non-unique set of sets
The Master said it isn't impossible
Thus, it cannot be limited to the set of colors the logicians can envision
Thus, it must be limited to the set of colors a logician can see.

[phoneutria]

I can look at your whole post in a bit, but let me just pick this for a second:

What you probably wanted to say was;
) If my color is not within my sight of n colors, I must blindly pick it from an infinite list.

Not blindly, with uncertainty.
The key word for that sentence is certainty. Any method you use to find the color by trying to understand what the group of colors is, is a calculated guess. Using the colors you can see is the only way you can be absolutely certain, therefore it is the only option for all the logicians.

[phoneutria]

In my field of vision I see n colors, all which are certainly a part of the puzzle.

The fact that you only see those n colors and that they are obviously a part of the puzzle has nothing to do with what your own color is.

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.

I cannot know with certainty the color of an object that I can't see by picking one color from an infinite number of colors.

The fact that you cannot pick from an infinite list of possible colors has nothing to do with what your own color is.

It has everything to do with understanding that the only way to be certain is to use visible colors.

Therefore the color of my headband is included in n.

Huh??
You make two statements that have nothing to do with your color and then conclude something about your color from them? No.

The two statements are to the point that any other solution is uncertain.

What you probably wanted to say was;
) If my color is not within my sight of n colors, I must blindly pick it from an infinite list.

That is not a true statement, but necessary for your syllogism.

I meant if my color is not within my sight of n colors, my solution cannot be certain.

As I first demonstrated, if you see one of each of the primary and secondary colors except one, you can pick that one with the same probability as picking the algorithm that you suggest using. You cannot "see" that algorithm any more than you can see the missing color. You have to imagine either in order to use them.

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.

The problem is that if you are merely going by what you imagine, whether that be a color or an algorithm, you cannot be certain that you are imagining the "right" one (the Master already has a "right" algorithm in mind that he is going to force the members to obey even if they do not understand it).

So how do you prove that you know which algorithm the Master is using?

By removing uncertainty.

Just because yours would work, doesn't mean that his wouldn't work better.
You have to prove that his and yours are necessarily identical or compatible.

I am not asserting that such is impossible. You have to prove the absence of a better algorithm than the only one you know. I am merely pointing out that no one has done that and until they do, NO proposed algorithm is legitimately proven to resolve the puzzle.

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.

[James S Saint]

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.

[Carleas]

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:
1) 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.

2) 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).

3) 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.

[James S Saint]

1) 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.

Shades of colors were not visible and would violate the constrain of resolvability.
You see the sequence:

1
1.5
2
2.5
3
__

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?

2) 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 -- not true).

If there are an infinite number, surely you can give us one.
And you are claiming that the Master is lying. Is that allowed?

3) 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.

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?

[James S Saint]

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.

[phoneutria]

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.

These configurations could not exist according to the premise of the puzzle, because the puzzle is solvable, and these hypoteticals are not.

What you are doing is the same as going what if I had two dads huh huh?

[phoneutria]

.

[Carleas]

And you are claiming that the Master is lying. Is that allowed?

No, I'm not. Perhaps I wasn't clear:

Just like a configuration where all the logicians had the same color headband [cannot be a solution] because it would violate the premise that there are many colors, so too an unsolvable configuration [cannot be the solution because it] violates the premise that the problem is not impossible to solve.

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.

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.

1) 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.

2) 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.

You see the sequence:
1
1.5
2
2.5
3
__

2.5
Or were you not using the function y = -|x/2 - 2.5| + 3.5 ... ?

[EDIT: equation corrected]

[phoneutria]

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.

This one is solvable with certainty only one way.

[James S Saint]

2) 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 -- not true).

If there are an infinite number, surely you can give us one .. ?

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.

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."

2) 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.

You have yet to prove that you have found such an example, because you refuse to answer any questions.

You see the sequence:
1
1.5
2
2.5
3
__

2.5
Or were you not using the function y = -|x/2 - 2.5| + 3.5 ... ?

So you just can handle math at all?

[James S Saint]

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.

This one is solvable with certainty only one way.

Prove that there "is only one way".

[phoneutria]

already did

[James S Saint]

As I said several times, "intuitively obvious" is not a proof.

You and Carleas amount to Flat-Earthers.