Math Fun

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:

  1. If any logician’s headband were not one of the colors she can see, the problem would be impossible.
  2. The problem is not impossible
    Therefore
  3. 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 < F
2

(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.

I feel like I must be missing something:[tab]3 families, each with 2 parents:
p = 2 2 2 = 6
g = 2 0 2 = 4
b = 1 1 3 = 5
c = 3 1 5 = 9[/tab]

[tab]1 3 = 4
2 7 = 9
3 4 = 7
4 1 = 5
5 1 = 6[/tab]

No. It doesn’t. You keep thinking that just because you have an operable algorithm, you have the only possible solution. For all of that type of problem, you MUST prove that your proposed solution is the only possible solution, else it isn’t a solution.

All of your deductions include an assumption that must be validated by being the only assumption possible in order to cause the puzzle to be solvable.

Proving that you have the one and only possible solution proves that the master is using it.

Definitely not true.

Well done, James. =D>

What about the solution process?

That is false. Please read the text one more time.

The reputable house

Duh myself :slight_smile:
[tab]2P
1B

2P
2G
1B

2P
2G
3B[/tab]

Well done, Phoneutria. =D>

What about the solution process?

The dice game

[tab]a
b = a+5
c = b -6
d =c + 11
e = d - 8

c= (a+5) - 6
d = [(a+5) - 6] + 11
e = {[(a+5) - 6] + 11} - 8
max total=12

Round d can only be 11+1.

a=2
b= 7
c=1
d=12
e=4[/tab]

I just shuffled things around a little bit. I knew I couldn’t have been too far off because of the limit on the number of sisters.

ESP. :smiley:

Damn … looking at my scratch work, it looks like I rewrote the whole puzzle … ](*,) [tab]2 *5 =10
3 *5 =15
3 *3 =9
4 *5 =20
2 *6 =12[/tab]

That is false.

Please read the text one more time. :slight_smile:

Well done, James. =D>

What about the solution process? :slight_smile: