Math Fun

No, I’m not. First, I’m dealing with a completely abstract problem whose entire set of parameters is known and laid out explicitly. This isn’t a scientific inquiry, it’s a logical inquiry.

And the thing that I’m assuming is not a theory of the problem, it’s a fundamental aspect of the given logical structure. That the islanders don’t have enough information to deduce their eye color prior to the guru speaking is the whole point of the problem. It’s solution is counter-intuitive because they don’t have enough information before the guru speaks, and it looks like the guru isn’t providing any new information.

To be clear, and to state again the assumption you’re complaining about in the Blue Eyes problem: an islander cannot learn his eye color solely from observing the eye color of another islander. I maintain that it is a given of the problem, in the same way that the normal strictures of logic are a part of the set of givens of this problem.

That is why those two puzzles are not legitimate puzzles. And realize that every “logic/math structure” must be independently proven before accepted by logicians/mathematicians.

Having a given in a math puzzle that 2+2=3 disallows the puzzle from being properly resolved regardless of it being a “given”.

And there is no actual distinction between good science and good logic other than one ends by saying “see!” in the physical sense rather than merely the mental sense.

Petitio Principii, If P → Q

That’s false. The most basic aspects of logic are simply defined, they are not and cannot be proven using logic. A system cannot prove itself.

It just isn’t provable that Y doesn’t follow from X alone.

Right, but “2+2=3” and “my eye color does not follow as a logical consequence of your eye color” are not comparably flawed starting premises.

[EDIT: typo]

A declared definition IS irrefutable LOGIC. There is no true or false determination. It is a given, thus “proven independently of all else”; “Logically true by definition.”

But take it up in the other thread.

You seem to be rejecting the statement “my eye color does not follow as a logical consequence of your eye color” as a given of the form you describe. Hopefully you agree that the statement “X does not follow from Y alone” is such a statement, something inherent in the structure of logic that doesn’t need to be independently proven?

The other thread does not address the issue we’re discussing here. The premise of that thread (that my argument is of the form P → Q , Q |- P) is a misunderstanding of my argument. The idea that the methods used in this purely logical problem are the same as the methods used in experimental physics is a misunderstanding of the difference between formal logic and observation, between mathematical induction and scientific induction.

Here you can see a large rectangle, consisting of 10 squares:

rechteck_mit_zehn_quadraten.gif
How big are the sides of each square at least, if they are all different in size and integer (thus: in whole numbers)?

I believe that YOU are misunderstanding your argument (by not seeing that you are skipping a relevant beginning of it). But discuss it on the proper thread.

To rephrase:
What is the least size of each square if they are all different integer sizes?

You are allowed to choose the least size of the sides of each square, but the size should be integer, thus a whole number.

[tab]Choosing the least size is actually part of the task. Knowing you, I am sure that you are going to find the right number for the least size.[/tab]

James, your thread is not about our conversation around the Blue Eye problem. You started a thread at best tangentially related to the discussion, and I’m not interested in that tangent (not least because it misrepresents my positions and our disagreement).

But, to repeat, and ask pointedly: do you agree that the following two propositions are true:

  1. X does not follow from Y alone
  2. my eye color does not follow as a logical consequence of your eye color alone

I’m pretty sure this isn’t what you mean, but it’s true:
[tab]1, because that’s the smallest (positive) integer.

Perhaps the question is better posed like this: ]
Suppose the smallest square is 1x1. How big is the largest square?

Is that the same problem?[/tab]

Be careful, because the following tab contains parts of the solution process:
[tab]Choosing the most qualified integer is alraedy part of the task. At first you have to derive equations from the diagram, then you must pay attention to some subtleties, then you have to solve the equations, and after that you should choose the most qualified integer … and so on.

For reasons of simplification you should choose “1” as the most qualified integer for one variable.[/tab]

For ease of reference:
rechteck_mit_zehn_quadraten.gif
No solution yet, but…
[tab]I did a bit of work on this before I realized that they are all different sizes, which means the whole shape is not a square (otherwise B = C).

So I have to start over, which I will do later. My basic plan of attack is to list out the equations (e.g. A + B = C + E + F, etc.), then let J = 1 and solve for each other variable, then multiply them all by whatever gets them all to be integers (I was wrong earlier, J can’t be 1, because A - (B + H) < J, so if J = 1, either A, B, or H could not be a integer).[/tab]

Just do it.

That is you - always just denying the other route and refusing to even discuss it while trying to press your own point again and again.

If you want to start a thread about the logic of the Blue Eye problem, I’m happy to engage. Instead, you started a thread about a “common mistake” in science and philosophy. And, rather than actually show that these mistakes are present in the canonical solution to the Blue Eye problem, rather than actually laying out your case, rather than quoting what I or anyone who’s endorsed the canonical solution has actually said (instead of making up a quote that is trivially fallacious), you just state that it “dawned” on you and then explore an intro-level logical fallacy.

Whatever prompted you to have your dawning, that thread is not about the logic of the Blue Eye problem. So far, this is the only thread in which that problem has been directly addressed.

This problem is pretty tedious, I think I must be doing it wrong.
[tab]Here are some equations:

A = C+G
B = H+D
B = I+F
C = G+E
D = H+J
E = G+I
F = J+D
H = I+J
A+B = C+E+F
A+C = B+H+I+F
A+G = B+H+I

Should be enough. Let’s try and solve for a variable in terms of J.

A+C = B+H+I+F
(C+G)+C = (I+F)+H+I+F
2(G+E)+G=2(I+F)+(I+J)
2(G+(G+I))+G = 3I+2F+J
5G+2I = 3I+2F+J

F = J+D
F = 2J+H
F = 3J+I

5G+2I = 3I+2(3J+I)+J
5G+2I = 5I+7J
5G = 3I+7J

A+G = B+H+I
G = (H+D)+H+I-(C+G)
2G = 2H+(H+J)+I-(G+E)
3G = 3H+J+I-(G+I)
4G = 3H+J
4G = 3(I+J)+J
4G = 3I+4J
G = (3I+4J)/4

5((3I+4J)/4) = 3I+7J
(15/4)I+5J = 3I+7J
(15/4)I-3I = 2J
(3/4)I = 2J

If J = 1 → I = 1.5
→ H = 2.5
D = 3.5
B = 6
F = 4.5

A+C = 14.5
A+G = 10

C-G=4.5
E = 4.5
G = 3
C= 7.5
A = 7…

Shit.[/tab]
Yep, definitely did it wrong :stuck_out_tongue:

Why don’t you properly show your full and clear understanding of the problem that I am pointing out (on the right thread). And then you can clarify the distinction between it and your reasoning. Why do you always avoid doing that every time? We have already annoyed this thread to death with our differences about the Blued problem.

You and I so often seem to take the exact same approach. The difference seems to be that when I find that it isn’t exactly true, I look at different means of tackling the problem. That relates to why your answers to the Blued problem and the Master’s logician meeting puzzle are not valid … you ignore the possibility of alternative methods while the puzzles forbid any.

No more appropriate thread exists. There are many, many threads discussing many, many fallacies that I’m not invoking in my solution to the Blue Eyes problem. I’m not going to go into any of them and explain how that particular logical fallacy doesn’t apply to my solution to the Blue Eyes problem.

And as with your approach to the Blue Eyes problem, you’ve yet to show your work.

Your continued evasion merely proves my point.

OK, found the a problem:
[tab]A = C+G
B = H+D
B = I+F
C = G+E
D = H+J
E = G+I
F = J+D
H = I+J
A+B = C+E+F
A+C = B+H+I+F
A+C = B+H+J+F and/or
A+C = B+H+I+E
A+G = B+H+I

I’ll have to power through solving for something in terms of J later.[/tab]

Carleas???
[tab]10 Squares Puzzle Solution.png.png[/tab]