Math Fun

Yes, arminius, after you put your imaginary like there, your premise was complete. Also very cute with my lil spider there.

Huh???
I posted the answer long ago. Are you reading anything??

You do not have to do anything of that, because the solution and the solution process are already given.

[tab]H = Hours. M = Minutes.
H/12 x 360 + M/60 x 360/12 = 30 H + 0.5 M.
Position of the big hand: M/60 x 360 = 6 M.
Position of the little hand: H/12 x 360 + M/60 x 360/12 = 30 H + 0.5 M.
The sum of both angles is 360°.
So: 30 H + 0.5 M + 6 M = 360.
For H = 10:
300 + 6.5 M = 360 => M = 60/6.5 = 9.231 minutes.
Thus: 9 minutes, 13.8 seconds.
Time on watch: 9 minutes and 13.8 seconds past 10.[/tab]

Thanks guys. I understand now. But I’m still stuck on the infinite divisibility of the units of time thing. Also, what if the watch, which isn’t digital, stopped before the gear system which turns the hands stopped before the gear teeth were completely seated? You know each each ‘tick-tock’ is the turn of the gear wheel… so what if the tension created by the winding, which powers the gears, was at zero percent before the watch completed its final tick?

What time would it then be? You see the infinite divisibility of time units I’m talking about now in a different way. The watch’s gear teeth need to be seated in order for a unit of its time to be recognized. It could have stopped somewhere between 10:10 and 10:10.1 for all we know. We have Zeno’s wrist watch.

They used spiral gears - smooth, no slack.

And even if you wanted to quantize the whole thing, you would still solve it in the same way but then truncate the answer to the nearest quantum step.

Not a solution, but: EDIT: oops, missed a page of discussion on this one Anyway, this is my untainted first stab.
[tab]I’m assuming the equal angle is between the hands of the clock and the vertical. I’m also assuming that the hands are meant to be moving fluidly, such that at exactly 10 O’clock, the hour hand points straight at 10 and the minute hand points straight at 12. At 10:15, the minute hand points straight at three, and the hour hand points at the spot 1/4 of the way between 10 and 11.
So, we can narrow the answer down to between 10 and 10:15.

To find the answer, we need to convert time to radians, take the speed of each hand in radians/second, such that the speed of the minute hand 12x faster than the speed of the hour hand, and then find where the values cross (using the absolute value and counting up to pi and back down).

Does that at least get the question right?[/tab]

James, I hope to have time to give you an updated syllogism later today.

If the said watch was digital, then there would be no geometrical aspect in the said task. It would become senseless, because there would be no geometrical circle but merely numbers. The task is about realising the facts given in the text, the recognition of the geometric facts, and the finding of the algebraic solution.

I have already given the solution process.

[tab]H = Hours. M = Minutes.
H/12 x 360 + M/60 x 360/12 = 30 H + 0.5 M.
Position of the big hand: M/60 x 360 = 6 M.
Position of the little hand: H/12 x 360 + M/60 x 360/12 = 30 H + 0.5 M.
The sum of both angles is 360°.
So: 30 H + 0.5 M + 6 M = 360.
For H = 10:
300 + 6.5 M = 360 => M = 60/6.5 = 9.231 minutes.
Thus: 9 minutes, 13.8 seconds.
Time on watch: 9 minutes and 13.8 seconds past 10.[/tab]

So you put solutions in tabs, but you don’t know why?

He means this fact

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.

1. 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)
2. A color cannot be deduced from a set of colors (given)
3. Therefore, a color cannot be added to C based on the other members of C (from 2, general to specific)
4. For the problem to be possible, a logician’s headband color must be a member of C (given)
5. Therefore, if the color of a logician’s headband is not in C, then the problem would be impossible (from 3, 4)
6. The problem is not impossible (given)
7. 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]