Help a linguist out. I’m in ur webz, solving ur slangs problems.
To pwn.
I said this:
Mathematically it is absolutely irrelevant what Phoneutria said, namely that there is also a line to 6. You just need the information that the angles have the same degree in order to solve the problem mathematically. But which line you prefer is absolutely irrelvant for the mathematical solution.
No. You have not understood it.
To find out that the 12 is the line is already part of the task, namely the part that refers to the common sense. Everything you say about the time on the watch refers to 12, e.g.: “… o’clock”, “10 past …”, “20 past …”, “10 to …”, … and so on, thus it depends on the position of the big hand (minute hand).
Mathematically it is absolutely irrelevant what Phoneutria said, namely that there is also a line to 6. You just need the information that the angles have the same degree in order to solve the problem mathematically. But which line you prefer is absolutely irrelvant for the mathematical solution.
Again:
Duh!
Yes. But the line is irrelevant when it comes to find the mathematical solution!
That was meant ironically, Phoneutria. I had just given him the solution process. 
…
That is not necessary (see above). 
Should tabs - in this thread (!) - not be used because of discretion?
Ah, ok. You’re just really bad at irony 
Tabs are used so when you give the solution, you don’t spoil it for people who want to try to solve by themselves. Carleas knows that he can click the tabs and see the answer, but he wants to try to solve it on his own. Threads like this aren’t really about being the first to solve (since anyone can probably just google for the answers).
Very bad!
And you are just really bad at mathematics.
I called it “descretion”.
Ah, I see.
Carleas! Good luck!
Yes, I know.
In this case, I actually mistakenly thought I was offering the first attempt at a solution. I had the ILP tab open for too long, so I didn’t see the page and a half of responses. And I think Arminius’ solution is much better than mine anyway. Radians?
Probably only for James; tabbed because I don’t think anyone else would care (which is not to assume that James still cares):
[tab]
As with the Pythagorean Theorem, it will be sufficient to show that a solution follows deductively from the premises.
Here is a new, much simpler syllogism for proving the SR portion of the problem.
- The colors each logician can see are part of the set C of known correct answers to the question, “what color is my headband?” (given)
- A color cannot be deduced from a set of colors (given)
- Therefore, a color cannot be added to C based on the other members of C (from 2, general to specific)
- For the problem to be possible, a logician’s headband color must be a member of C (given)
- Therefore, if the color of a logician’s headband is not in C, then the problem would be impossible (from 3, 4)
- The problem is not impossible (given)
- Therefore, the color of every logician’s headband must be a member of C (from 5,6)
To preempt a possible objection, the second given may be contentious. However, I think it’s true, particularly if you consider the distinction between deductive reasoning and educated guessing. Your examples, such as the color wheel and pattern examples, involve at best educated guessing or scientific induction (as opposed to mathematical induction, which is a deductive method). More generally, colors, even ordered colors, do not bear logical relation to one another. One would not look at circle of people with headbands colored green, red, yellow, green, red, yellow, green, blue, and say that is logically inconsistent for them to be sitting that way. This is true even if the floor were giant color wheel or otherwise implied a pattern with the headbands.
All the work here is done by 2 and 6. I think this is basically the ‘certainty’ argument that Phoneutria offered earlier, but jazzed up with sets and provability statements. It comes down to that you can’t derive a member of a set only given the other members, so you can’t deduce any additional correct answer not already known to be a correct answer for at least one logician.
Tangential: this blog post about common knowledge, including the Blue Eyes problem (cast as the Muddy Children problem) and other considerations.[/tab]
I didn’t use degrees or radians.
Carleas,[tab]
You were better off before.
“A color cannot be deduced from a set of colors”?
What the hell is that?? Did you leave out some words or something?
And:
“…the set C of known correct answers…”
What?? Where is it a given that there is a set of known correct answers before they have even began deducing?
I agree but I contend that your example is also merely an educated guess based upon what is seen and the presumption that there is no way to resolve it except by using what is directly seen.
It is not logically deducible that each color is necessarily seen by its wearer.
What you cannot deduce is that you know the closed set. Assuming the set to be already closed is no different than assuming it to be closed by merely one more color. Either way it is an assumption.
The fact that the master said that the problem is solvable does not change which of those assumptions would be more probable and certainly not which would be a certainty.
The bottom line is that everything deduced must be directly or indirectly defined to be correct. All logic resolves to recognizing what has already been defined to be true.
…even worse than the blue-eyed problem. You and those who believe that such puzzles can be resolved in that way are deluding yourselves.
“If this problem was different than it is…”
… is not a valid premise to solve any puzzle.
“it wasn’t common knowledge that even one of them had a muddy forehead.”
…is incorrect.[/tab]
Interesting. What did you set as equivalent? Ratios?
James:
[tab]
Colors bear no logical connection to each other, so no color follows from a set of colors.
For each color c that a logician can see, at least one logician would be right to say “my headband is c”. They are thus known correct answers to the question “what color is my headband” for at least one logician.
Good, because I see part of your objection (what I’ve been calling the Alternatives objection) as dependent on there being an alternative. It seems we agree that no such alternative is on offer.
But I haven’t assumed the set to be closed, I’ve shown that it must be closed in order for the problem to be possible. What line are you saying is an improper assumption? And what other assumption are you saying would make the problem solvable?
I agree the muddy children formulation is not as rigorous as other formulations. My preferred statement of the problem is the tribe who ritually kill themselves if they learn their eye color, because it removes any notion of desire or uncertainty about timing.
Anyway, I’ll note that you’ve avoided the MI syllogism. Is there a specific line you reject there?[/tab]
Yeah, just relative distances as if the hands moved in straight lines.[tab]10h = 120m
h = 12m
xh = xm/12
10xm - xh = xm
10xm - xm/12 = xm
10xm = 13/12 xm
xm = 12/13 * 10
xm = 9.23076923076923[/tab]
Carleas:
[tab]
That seems like a silly thing to be saying.
You propose one set - the visible.
I propose the possibility of others sets - the partially visible.
You have to “deduce from a set of colors” regardless of which way you go with it.
But the whole set isn’t known by anyone but the master. No one can see the whole set of headbands so none can claim to be certain of all colors, at least until they do some deduction.
No. Until you prove that your proposal has no possible contradicting proposal/algorithm (“lack of alternative”), any of the patterns that I have given are as valid as yours. If you can claim that yours is not an assumption, I can claim that mine aren’t either, “because the problem is solvable”.
No. That is what you are failing to accomplish right now.
We already went through that. Your syllogism assumed that any unseen color is a member of an infinite set and therefore not deducible. That was silly. Every color is always a member of an infinite set. That has no bearing on whether it can be deduced. For an example, one can deduce from the assumption (permitted by the “it is solvable” premise) that the only missing color from the primary & secondary colors is the color being worn.
Again, you say that they can assume that each member can see all puzzle colors and I say that they can assume that the only missing color from a known color chart is a part of the puzzle. We can both make that assumption based upon the master’s premise that it is solvable. And there are other examples wherein an assumption can be made so as to make the puzzle solvable - as long as it can be proven that there is no possible contradicting alternative.
The question is whether there is an assumption that can be made that causes there to be only one way to solve the puzzle. Even if you assume that the seen colors are the only colors, you still have to prove that there can not possibly be any recognizable pattern or formula that could be used, even by God. - THAT is what you have not even begun to prove.
I don’t want to bother getting into longer and longer arguments. This one issue is enough to prevent anything else from being relevant.[/tab]
Carleas: and a third objection:[tab]Imagine that you had already discussed this problem many times and believed in the solution that you proposed when you were invited into that exact headband situation. After the first bell, you and everyone could look around the room and see that your proposed solution was applicable. But then because everyone could see the color pattern of that solution, after the second bell, you and they already know your color and leave - even if the colors were randomly situated in the circle.
Because you believed in that answer, it wasn’t the answer.[/tab]
James,
[tab]I don’t understand how having discussed the problem before gives you any information about the color of your headband. Care to present this objection as a syllogism? As I understand it (and I don’t think I do), the syllogism adds the given that one or more logicians had discussed the answer before and believe in the answer, and arrives at the conclusion that the syllogism I’ve provided is no longer sound. Is that right? What line of the syllogism do you see as dependent on the logicians never having seen the answer before? And how does having discussed the logic in advance allow everyone to know your headband color on the second bell?
And why are you avoiding the MI syllogism?[/tab]
Carleas,[tab]
Nahh… I take that one back. I forgot a step in the issue, sorry.
Because you compound your errors and your “MI” is merely a reflection of that. It is an irrelevant concern in the long run. We know that if the color quantities are known by everyone, all they have to do is wait for the lesser group to leave. My issue is that there is a possibility of someone being able to deduce even earlier. That is a possibility until it is proven that there cannot possibly be another algorithm that also works (or is being used by the Master). The fact that your proposal would work IF the master was using it is irrelevant. The members have to prove that the master is using that particular algorithm.
I gave an example of how the same color quantities could be present as in your proposal, yet everyone could leave after the second bell.
Objections:
1) No one can assume anything about the color set based on the assertion that the puzzle is solvable unless they can answer objection (2).
I have given examples of other equally valid assumptions that could be made by ignoring objection (2).
2) The valid solution must prove that the master is using the same solution, which means that there cannot possibly be any other solution to the same color arrangement.[/tab]
Even if we assume that this problem is flawed, the MI syllogism proves the canonical solution to the Blue Eye problem. It is necessary for the solution I’ve proposed for this problem, but it is not irrelevant even if this problem turns out to be unsolvable.
Sure, and if the logicians all knew their colors, they would leave immediately. The trick isn’t to find an assumption that works, but to find a deduction that does. To the extent my syllogism introduces new assumptions, they are tautological (e.g. colors don’t bear any logical relation to each other).
If we assume that there is one correct solution, then the Master is clearly intending to follow and enforce it. So this objection seems to beg the question. If there is a solution, then there’s no need to prove that the Master is following it, that’s a given. If there is no solution, or more than one solution, then the problem breaks down for reasons unrelated to the Master.
I’ve been looking for other premises of the form ‘~X → impossible, ~impossible |- X’ to see if there are other similarly tautological assumptions that lead to a conclusion other than that the color of a logician’s headband is among the colors she can see. “This problem is not impossible” is such a weird premise, I’m open to the idea that the syllogisms it allows are weird too, and that contradictory conclusions are possible (especially since, as someone pointed out, the statement is only true if it’s true). But I haven’t found any, and you haven’t presented anything like a syllogism showing an inconsistent conclusion. The simplest form of my SR argument seems clearly valid:
- If any logician’s headband were not one of the colors she can see, the problem would be impossible.
- The problem is not impossible
Therefore - Each logician’s headband is one of the colors she can see.
The reputable house.
A few families live in a house from which we know the following facts:
• More children than parents live in this house.
• More parents than boys live in this house.
• More boys than girls live in this house.
• More girls than families live in this house.
No family is childless, each has a different number of children. Every girl has at least one brother and at most one sister. One family has more children than all the other families combined.
How many families live in this house and how are they composed?
The dice game.
Each round of a dice game consists of two fair dice; the result of one throw is the product of the thrown numbers. A game consists of 5 rounds.
Bob throws in the second round by 5 more than in the first, in the third round by 6 less than in the second, in the fourth round by 11 more than in the third, and in the fifth round by 8 less than in the fourth.
How many points did he score in each of the 5 rounds?
The reputable house
[tab]2P
1G
1B
2P
1G
2B
2P
2G
2B[/tab]
Edit: I should post the process I guess
[tab]C > P > B > G > F
Because of the limit on the number of sisters a girl can have, from this I deducted that these have to be 2 parent families cuz there’s too many damn parents.
C = B + G
P = F2
G < F2
(B+G) > F*2 > B > G > F
I knew that F was going to be small so after this I tried out some values of F because I got lazy.[/tab]
That is false.
Please read the text one more time.
Dang, left out the “One family has more children than all the other families combined.” part.
Yes.