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?
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.
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?
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?
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]
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.
](https://www.ilovephilosophy.com/uploads/default/original/2X/7/7c349bf1053f368a5729d053e1e91215f22ce66c.gif "Brick wall"){kind=small}
[tab]2 *5 =10
