It is impossible to solve this problwm without constraining the group of colors to only colors that can be known by all participants.
We agree on that.
And likely wise the algorithm must be equally constrained.
But you can’t do that by assuming that the algorithm is the one that all of the members can see.
Fuck you.
love you too, benny
I didn’t pay much attention to what you guys/gals were saying, but you two seemed to be arguing about possible solutions.
I read brute force from phon, and the solution given by Wikipedia does not require it. The argument seems to be easily resolved by considering the solution given on the site.
From what I read of you, your solution was in adherence to that posted.
Can there be alternate solutions to solving the puzzle? No doubt.
Do they all involve brute force? No.
More often than not, the key is in the details. They’ll give you all the tools necessary to solve the puzzle.
In hindsight,
I needn’t respond further.
Ben, full disclosure: you should check the most recent edits on that wiki article. I’ve done a bit of cleaning up there. Which is to say, the wiki article is likely to be exactly as right as my contributions to this conversation (which I am not at all sure are right).
James, I still think what you’re doing only expands the possible solution set. Otherwise, you’re claiming that, by providing one way that they could solve the puzzle without reference to the colors they can see (i.e. your color chart scenario), you’ve proved that there are always at least two solutions to the problem. That claim is every bit as broad as the claim that in a given circumstance, X is the only way to reason out of the problem. It seems clear that if it’s possible to come up with a solution case where there is only one way to reason from it, then that is a legitimate solution the problem, and one the Master could set up. Your claim is a general claim: no such solution exists. You have not proven that (indeed, your argument looks a lot like you’re saying, “‘it’s evident’ that some other algorithm will always exist”).
You have likened your criticism of this problem to the Blue Eye problem, but the big difference here, and why I think your criticism here is well placed, is that here the puzzle is inherently uncertain: we don’t know how many logicians there are or how many colors of headband there are, so the solutions we offer are general, e.g. the problem can be solved if there are three logicians, one with a red, one with a yellow, and one with a blue headband (or is that blue, magenta, and cyan?). We need to assume some facts to make the problem solvable, so we can question which assumptions are legitimate and why, e.g., is it any more legitimate to assume that there are two of each headband than to assume that every logician knows about the the color chart.
Phoneutria, do you reject the idea that there are other solutions to the problem besides the everyone-can-see-their-color (ECSTC) solution? Do you have grounds other than Occam’s Razor for preferring the assumption in the ECSTC case?
Much like the Blue-Eyes puzzle, if a member cannot prove that a proposed algorithm is the only possible algorithm for what he sees, he cannot use that solution.
In logic, assumptions can only be made in order to checkout the hypothesis of them being required facts. Assumptions cannot be used until they are proven to not be assumptions, but necessarily true.
1) You cannot assume that you see your own color on anyone else unless you can prove that there cannot be any other option.
2) You cannot assume any particular algorithm to be the one to use unless you can prove that there cannot be any other possibility (unless the others lead to the exact same bell-timing and result).
Those are the restrictions to the puzzle solution. The Master implied that each member could prove the viability of an algorithm. I haven’t seen the proof of one yet, merely the assumption of one.
Otherwise, you might as well just assume that your color is whatever color you see to your right and just assume that you are right because of it: “Turtles all the way down”.
This isn’t true. If you can prove both pieces of inductive logic you don’t need to prove a third piece that says, “and there is nothing that would confound this logic.”
For example, in the Blue Eye problem, it’s provable that
- if N blue eyes will leave on day N, then N+1 blue eyes will leave on day N+1
- 1 blue eyes will leave on day 1.
If those are both true, nothing else needs to be shown (and I phrase this in the conditional because I’m more interested in the general form of the logical argument than I am in actually rehashing the discussion we had of the Blue Eyes problem).
I will be pedantic, but logic is nothing if not pedantic: you can make as many assumptions as you like in a logical proof, however you need to eliminate them if you want your proof to show that a conclusion follows from the original premises alone. As in the proof by contradiction, introducing an assumption is a standard logical technique.
You make a fair point here, but you gloat too much. You have made the bold claim that the Master’s statement cannot be true, that the Master is a liar. But again, for that to be true, you need to show that for all possible configurations of logicians, colors, and headbands, there is no configuration that leaves only one logical solution which solves it.
I also think you still underestimate the work that the Master’s claim can do. Let’s take a toy logic problem with the same premise, “it’s not impossible”:
A Master and a Student stand in front of a safe that will open if the right number is input on a panel. The Master says to the student, “the number that unlocks this safe is either 8 or it is a number of which it is impossible for a human to conceive. You must deduce the number that unlocks this safe. Don’t worry, it’s not impossible to solve this problem.”
The logician unlocks the safe. How did he do it?
This problem seems trivially solvable, but only because the Master told the student that the problem was not impossible to solve. That’s real logical work, and the logician’s deduction is perfectly valid. If what would otherwise be an assumption is necessary in order for the premise that the problem is not impossible to be true, it isn’t an assumption, it is a corollary.
That certainly IS true. And you are STILL not getting what I am saying.
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 thread should be called “Cheating Fun”.
It’d be cheating if I said, ‘I’m participating in this challenge. Look at the solution I just came up with.’, then pasted the solution from Wiki.
I did not ever claim anything of the sort.
I analysed the solution, considered how the problem could have been solved with the details given, and tipped my hat to a clever puzzle.
My mind is scattered most of the time, and with it, my energy.
I don’t dedicate my limited resources to sweating over a logic puzzle, nor for internet points in response to solving said puzzle.
Feel free though to keep being condescending like phon, and make assumptions about me.
I appreciate it.
Sorry, Ben. You have misunderstood me. I was merely referring to the statement in your post (“These puzzles are on Wikipedia with solutions explained.”) but not to you personally.
If people can get all the solutions and their explanations from other websites, then this thread is more about cheating than mathematics.
What you said was right, Ben. Don’t worry.
The inductive proof’s premises completely determine the answer, without reference to any algorithm:
P1: If N blues leave on day N, N+1 blues leave on day N+1
P2: 1 blue leaves on day 1.
If both of these are true, then we know that 100 blues leave on day 100. It is mathematically certain. Do you intend your criticism to be a rejection of P1?
That isn’t a “proof”.
That is “algorithm A”, a process (or a “syllogism” if you like).
But there is another member who you know to be clearly smarter than you.
He believes that “algorithm B” is more suited even though algorithm A would otherwise work.
Which one are you going to use?
Yes. Specifically because this is the only way to constrain the color group with certainty.
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.
Phon, show us the logic proof that each person must be able to already see his own color.
But this proof doesn’t depend on its own use. You’ll note that the proof does not take itself as a premise, and none of its premises depend on its use. There may be many ways to prove a given conclusion, but that does not change the fact that a proof by induction is still a valid proof of its conclusion.
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.
The bells require that everyone use the same algorithm or one that will produce the same results at the SAME TIME. Each member has to know that the “right” algorithm is the one that they are using.
To know that the right algorithm is the one that you are using, you have to know that there is none other.
) If I use the right algorithm, I will leave at the right time.
) If I use the wrong algorithm, I will leave at the wrong time.
) To know that I am going to leave at the right time, I must know that I do not have the wrong algorithm.
) If another algorithm is the right algorithm, I know that mine is the wrong algorithm.
) To know that I do not have the wrong algorithm, I must know that no other is the right algorithm.
[b]) To know that no other is the right algorithm, I must prove that there is no other algorithm that is right.
Now how are you going to prove that no other algorithm is right?[/b]
The conclusion of your proof must be:
“Therefore, no other right algorithm can exist”.
How many times do I have to ask you this time?