These puzzles are on Wikipedia with solutions explained.
When it comes to logic and philosophy, Wiki is beneath us. They are wrong on many things. Wiki can only tell you what “has been excepted” or what “hasn’t been disputed by anyone reputable enough”.
But the more that happens, the more likely it is that you are not seeing the right algorithm.
And as far as the missing color issue: if you can assume that you can see a color similar to your own color, why can’t you assume that your color is a primary or secondary color (or integer) if all of the others are? It seems to me that those two assumptions are the same kind of assumption.
That’s cheating benny.
If your color is the same as one of the colors you can see, and all the colors you can see are primary, your color is primary.
Assuming all the colors are primary and never repeat, yoy’d have 3 logicians (red, yellow, blue).
The person wearing blue sees a red and a yellow and can make a guess that his color is blue to complete the group of primaries. But since he does not have any information about what the group is or how many times each color can repeat, this has many possible answers. Maybe you are blue, but maybe you are red or yellow, or maybe you’re not a primary color at all and the master is a bastard trying to fool you. There are many possible answers because you don’t have the necessary constraints.
That’s cheating benny.
You have the same constrain as before, “the Master said that it was solvable.”
I agree that it is really only a probability guess. My point is that so are the other algorithms.
All of the algorithms involve the intelligence test question of “What is missing in this picture”, similar to the “what doesn’t belong in this picture”. Both of those are rationality issues, “what is probably missing in this picture”, a part of the “Similarities and Differences” category.
Logical proof, certainty, requires that all potential alternatives be eliminated. That means that you have to prove that it is impossible for any other algorithm to fit the pattern that you see. And because of the bells issue, it also means that you have to eliminate all potential alternate algorithms even if they resolve the same color, because all of the members have to be using the same algorithms at the same time (although the algorithm could change from bell to bell).
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?