Math Fun

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Re: Math Fun

Postby Fixed Cross » Fri May 03, 2013 9:51 pm

" So, if there are 2, 1 gets 100 and 2 gets 0.
If there are 3, the oldest has to only earn 1 vote. He gives 1 coin to #2 and 0 to #1, and 99 for himself.
If there are 4, the oldest has to earn 2 votes.
He gives 1 to #1, 2 to #2, and 0 to #3, leaving 97 for himself. "

Why does he have to give #2 more than 1 coin?
I don't see a problem with your solution of the chessboard, it seems indefeatable. Not so with the aryan guru and her so called logical necessity. Obviousness doesn't figure into logic, it's too far short of certainty. But the riddle is 'ambiguous' - it's not meant to be taken literally.
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Re: Math Fun

Postby Silhouette » Sat May 04, 2013 12:44 am

Fixed Cross wrote:" So, if there are 2, 1 gets 100 and 2 gets 0.
If there are 3, the oldest has to only earn 1 vote. He gives 1 coin to #2 and 0 to #1, and 99 for himself.
If there are 4, the oldest has to earn 2 votes.
He gives 1 to #1, 2 to #2, and 0 to #3, leaving 97 for himself. "

Why does he have to give #2 more than 1 coin?
I don't see a problem with your solution of the chessboard, it seems indefeatable. Not so with the aryan guru and her so called logical necessity. Obviousness doesn't figure into logic, it's too far short of certainty. But the riddle is 'ambiguous' - it's not meant to be taken literally.

Um, if there's 3, the oldest has to earn 2 votes to get at least 50%.

The eldest of 4 has to give the youngest 2 coins due to the clause: "Each pirate, having a strong sense of self-interest, bases their vote on the question, "Would I get less if this proposal failed?"
If the answer to that question is "yes," they vote for it, else they vote against it."
The youngest would ask himself that question, and answer "no", he wouldn't get less, he would get the same. So he would vote against it. The eldest needs his vote to both survive and get the most coins out of the proposal, so he has to change the proposal so that the youngest answers "yes" to that question, and thus votes FOR his proposal.

As for that old guru one, u clearly didn't keep up - we all proved the need for her several times over in several ways. That one's well and truly solved, explored and resolved.
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Re: Math Fun

Postby James S Saint » Sat May 04, 2013 3:32 am

Fixed Cross wrote:I don't see a problem with your solution of the chessboard, it seems indefeatable.

So you figure that if you painted the board all the same color leaving only the grid, you would be able to solve it???
The solution involves odd and even numbers and symmetry.
Anything that you can't do to one side of the board, you can't do to the other.
Clarify, Verify, Instill, and Reinforce the Perception of Hopes and Threats unto Anentropic Harmony :)
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From THIS age of sleep, Homo-sapien shall never awake.

The Wise gather together to help one another in EVERY aspect of living.

You are always more insecure than you think, just not by what you think.
The only absolute certainty is formed by the absolute lack of alternatives.
It is not merely "do what works", but "to accomplish what purpose in what time frame at what cost".
As long as the authority is secretive, the population will be subjugated.

Amid the lack of certainty, put faith in the wiser to believe.
Devil's Motto: Make it look good, safe, innocent, and wise.. until it is too late to choose otherwise.

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Re: Math Fun

Postby Carleas » Sat May 04, 2013 4:26 am

On this one, I think James has a point. While the square coloring demonstrates the fact that it can't be done, the explanation for why can't rely on the color of the board. The coloring is a stand-in for odds/evens.
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Re: Math Fun

Postby Flannel Jesus » Sat May 04, 2013 7:37 am

The first time I actually heard the explanation that I offered, I heard it in the context of a colorless board. The problem was just a grid of squares, all white, with 2 squares removed (the grid was not 8x8, it was a bit smaller I think, maybe 6x4 or something).

The solver first showed a couple diagrams of attempts to cover all the squares, all of them failing.

And then he said, "Instead of trying every possible arrangement of dominos, there's this faster way of solving it: imagine the squares as colored black and white, like a chessboard." And then he offered the same solution.

So, the solution, though it refers to the colors of the chessboard, doesn't explicitly necessarily depend on them being those colors. Even if they weren't those colors, even if they were all white, we could still, in our solution, artificially imagine them as colored and the solution would still hold. That our solution involves a checkered grid doesn't mean that the board has to be a checkered grid. We merely have to imagine the board as checkered to see that it still holds, even on a fully white board.

"The solution involves odd and even numbers and symmetry."

This could mean any number of things.
The board in question, with opposite corners removed, is symmetrical. It's symmetrical across both diagonals.
There are symmetrical boards that have ways to tile the dominos to cover all squares, and asymmetrical boards that have ways to do the same.
There are symmetrical boards that are impossible to domino, and assymetrical boards that are also impossible to tile.

When you said "odd and even" -- again, I'm not sure what you mean by that, but it's potentially very closely related to the chessboard solution.
If you were to number the tiles in a spiraling fashion, 1-64, you would find that the black tiles are even and the white tiles are odd (or vice versa, depending on where you start, etc.). You can't tile the board if you remove 2 even squares only, or 2 odd squares only, but this becomes synonymous then with 2 black squares only, or 2 white squares only. "Odd/even" is interchangable with "black/white" if you number them in a spiral order.

Carleas, do you not consider it an explanation for why? If I've used the coloring to prove to you that no possible arrangement could cover all the squares...isn't that showing why? It's a fully general solution, and if someone takes such a board up to me and asks "Why can't I cover them all with dominos?" I don't feel anything is wrong with the answer, "Well, notice that when you cover 2 squares with a domino, it must cover 1 black and 1 white square. So no matter how many dominos you put on, there will always be an equal number of black and white squares covered. The board you've brought to me has less whites than blacks (or vice versa), and so that's why it can't be done." I don't see how it fails at 'why'.

The one problem, classically, with this solution that most people have is not that it doesn't work as a solution, not that it doesn't explain why, but that it's creative. In the world of mathematics, logic and proofs, this proof was too...informal. It worked for them, it was a great explanation, made it clear why it's impossible, but they wanted a less creative solution. They raced to come up with the most non-creative solution.

I wouldn't think you guys would have a problem with creative solutions though...
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Re: Math Fun

Postby Fixed Cross » Sat May 04, 2013 10:35 am

James and Carleas - obviously the colors are only illustrative of the odd even thing. But a very solid illustration.

Silhouette - ah yes, thanks.
No, the riddle has not been solved at all in the way you claim. No one showed me how my demonstration of 4 blues and 4 browns fails. I understood that Sauwelios thinks my and James' position were interchangeable, and that he therefore did not even try to understand my explanations. Maybe you're doing the same. To be honest I did not read most of James's stuff there because he was dealing with it in a very different manner and making much more complicated claims. I could not keep up with that at the same time as verifying/falsifying my own position.
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Re: Math Fun

Postby Fixed Cross » Sat May 04, 2013 10:55 am

James S Saint wrote:
Carleas wrote:
James S Saint wrote:If there were only 100 and you were one of them and starting a count from 100;
1) they count the number of blues they see above the number they started counting from = -1
2) they count the days of no one leaving. = 0
3) if they count less blues above the start number (-1) to the count of non-leaving days (0), they deduce they are the last blue. = Leave


Only if "starting a count from X" means "know it to be common knowledge that there are X blues".

NO!!
It does NOT!!

It means that they start counting from the same number REGARDLESS OF WHY!!!

You keep denying the hypothesis.
"IF they ALL start with the SAME number,.."

That is all they have to do.
WHY they do it or HOW is NOT the question being asked.

Just answer the question that is asked.

Well but, they will not ever do that if they don't understand how* they could start from the same number. And apparently, that isn't obvious. Interestingly, even in the canonical solution it is required that all start from the same number. Maybe the whole problem is that none of them understands the logical requirements of the canonical solution.** This has crossed my mind before.


* I am familiar with this phenomenon - I used to suffer from it when I was a child. Often I would not be able to work with a tool or toy if I did not understand how it worked.

** They would in the canonical case simply not know that they do not know. Which speaks for it being called canonical.
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Re: Math Fun

Postby Carleas » Sat May 04, 2013 4:44 pm

Flannel Jesus wrote:Carleas, do you not consider it an explanation for why? If I've used the coloring to prove to you that no possible arrangement could cover all the squares...isn't that showing why? It's a fully general solution.

Yes and no. I think it's a flawless proof that it is impossible. But my understanding is that for any chessboard from which 1 white and 1 black tile are removed, there is a solution. Because this seems like a generalization of the problem, I have the intuition that a truly general solution would solve both, i.e. a general solution should work for any chessboard from which coordinates (x1,y1) and (x2,y2) are removed. Color gives a good approximation for solutions to the narrow subset where (x1,y1) and (x2,y2) are both the same color, but tells us next to nothing about the situation where they are different colors.

So, maybe "why" is the wrong term, but I do think there is a more general solution, and that the hand-waivey odds/evens and symmetry explanation, while incomplete, hints at that more general solution
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Re: Math Fun

Postby Flannel Jesus » Sat May 04, 2013 4:55 pm

Carleas wrote:But my understanding is that for any chessboard from which 1 white and 1 black tile are removed, there is a solution.

I think that's also the case, and I also wouldn't know how to prove that.
My explanation, you're right, merely rules out the solution to any chessboard in which the removed pieces are the same color, it doesn't prove that if they're of different colour that there will necessarily be a solution.

I consider that a different problem from the problem being asked, though, as the problem that's being asked is indeed only about a chessboard in which the pieces removed are of the same color, and why there's no possible solution in that case.
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Re: Math Fun

Postby Sauwelios » Sat May 04, 2013 4:59 pm

Fixed Cross wrote:No one showed me how my demonstration of 4 blues and 4 browns fails. I understood that Sauwelios thinks my and James' position were interchangeable, and that he therefore did not even try to understand my explanations.

Your "understanding" may be a misunderstanding. Where did you make a demonstration? Note that general claims about hypotheticals are not a demonstration. Post or link to your step-by-step solution, and I will explain to you which steps, if any, are false—just as I've done with James'.
"Someone may object that the successful revolt against the universal and homogeneous state could have no other effect than that the identical historical process which has led from the primitive horde to the final state will be repeated. But would such a repetition of the process--a new lease of life for man's humanity--not be preferable to the indefinite continuation of the inhuman end? Do we not enjoy every spring although we know the cycle of the seasons, although we know that winter will come again?" (Leo Strauss, "Restatement on Xenophon's Hiero".)
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Re: Math Fun

Postby Carleas » Sat May 04, 2013 5:06 pm

Flannel Jesus wrote:I consider that a different problem from the problem being asked, though, as the problem that's being asked is indeed only about a chessboard in which the pieces removed are of the same color.

Yes, I'm really saying "this solution isn't the best solution to this problem, because there's this other problem it doesn't solve." Which, I admit, is a bullshit criticism about a solution to this problem. But I still think a solution that addresses the whole class of problems would be the better solution.
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Re: Math Fun

Postby Flannel Jesus » Sat May 04, 2013 5:11 pm

I personally doubt that there's any single proof which covers both. Even the mathematicians and logicians working on the problem are creating proofs for the independent problem of "Why is there no solution if the squares removed are of the same color?", while not proving that there are always solutions for squares removed of different color.

You might find a proof for one, and in the same paper a proof for the other...but just putting it on the same paper doesn't make it the same proof.
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Re: Math Fun

Postby James S Saint » Sat May 04, 2013 6:03 pm

Due to the symmetry, it doesn't matter if you begin with rows or columns, so let's choose rows.
In the top row, there are an odd number of spaces to fill thus
There is no alternative but to share a domino between the first row and the second.
The domino could be placed on either the right or left side of the board but in either case
There is no alternative but to cause an odd number of spaces on the same side of our domino as the missing space was on
We now have the same situation of even on one side an odd on the other but on the second row thus
We have no alternative but to share a domino with the third row and restricted to the odd side which
Leads to the same situation with the fourth row with the odd number shifted opposite as the last
As we go from row to row, we cause the next row to be odd on opposite side of our domino depending on how we started
That pattern has no alternative but to continue down the rows. With each row, the odd number of spaces shifts from right to left.

In order to solve the problem, because the last row has a missing space opposite from the first row,
We must have the odd number of spaces shifted to the opposite side from where we began with the second row
But there are an odd number of rows from the second to the last yielding no alternative but to have
the odd side be the same on the last row as on the second row.

Thus there is no solution.
Clarify, Verify, Instill, and Reinforce the Perception of Hopes and Threats unto Anentropic Harmony :)
Else
From THIS age of sleep, Homo-sapien shall never awake.

The Wise gather together to help one another in EVERY aspect of living.

You are always more insecure than you think, just not by what you think.
The only absolute certainty is formed by the absolute lack of alternatives.
It is not merely "do what works", but "to accomplish what purpose in what time frame at what cost".
As long as the authority is secretive, the population will be subjugated.

Amid the lack of certainty, put faith in the wiser to believe.
Devil's Motto: Make it look good, safe, innocent, and wise.. until it is too late to choose otherwise.

The Real God ≡ The reason/cause for the Universe being what it is = "The situation cannot be what it is and also remain as it is".
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Re: Math Fun

Postby Fixed Cross » Sat May 04, 2013 7:24 pm

Sauwelios wrote:
Fixed Cross wrote:No one showed me how my demonstration of 4 blues and 4 browns fails. I understood that Sauwelios thinks my and James' position were interchangeable, and that he therefore did not even try to understand my explanations.

Your "understanding" may be a misunderstanding. Where did you make a demonstration? Note that general claims about hypotheticals are not a demonstration. Post or link to your step-by-step solution, and I will explain to you which steps, if any, are false—just as I've done with James'.

You asked me the exact same thing before. I then offered three posts, none of which got a response from you. You later told me you didn't understand them. I haven't simplified the logic, so I can't help you there.

Where did you demonstrate false steps in James reasoning? And did James concur that he stood corrected? I must have missed all that, but given James compliant attitude now perhaps it did happen.
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Re: Math Fun

Postby Sauwelios » Sat May 04, 2013 7:43 pm

Fixed Cross wrote:
Sauwelios wrote:
Fixed Cross wrote:No one showed me how my demonstration of 4 blues and 4 browns fails. I understood that Sauwelios thinks my and James' position were interchangeable, and that he therefore did not even try to understand my explanations.

Your "understanding" may be a misunderstanding. Where did you make a demonstration? Note that general claims about hypotheticals are not a demonstration. Post or link to your step-by-step solution, and I will explain to you which steps, if any, are false—just as I've done with James'.

You asked me the exact same thing before. I then offered three posts, none of which got a response from you. You later told me you didn't understand them. I haven't simplified the logic, so I can't help you there.

Those posts were just rambling. And as I said, general claims about hypotheticals are not a demonstration.


Where did you demonstrate false steps in James reasoning? And did James concur that he stood corrected? I must have missed all that, but given James compliant attitude now perhaps it did happen.

I demonstrated the false steps in his reasoning in that post of mine in which I made some phrases blue and some red. James need not concur: he has not shown how his reasoning is impervious to my critique.
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Re: Math Fun

Postby Fixed Cross » Sat May 04, 2013 7:51 pm

Sauwelios wrote:
Fixed Cross wrote:You asked me the exact same thing before. I then offered three posts, none of which got a response from you. You later told me you didn't understand them. I haven't simplified the logic, so I can't help you there.

Those posts were just rambling. And as I said, general claims about hypotheticals are not a demonstration.


Where did you demonstrate false steps in James reasoning? And did James concur that he stood corrected? I must have missed all that, but given James compliant attitude now perhaps it did happen.

I demonstrated the false steps in his reasoning in that post of mine in which I made some phrases blue and some red. James need not concur: he has not shown how his reasoning is impervious to my critique.

Yeah.... whatever you say I guess... lol.
It took you two years and a very intensive private workshop to see the logic to value ontology. Before that time it had been "just rambling". This is a bit predictable.
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Re: Math Fun

Postby Silhouette » Sat May 04, 2013 7:54 pm

Flannel Jesus wrote:I personally doubt that there's any single proof which covers both.

Personally I think it's ridiculous to say that your demonstration of "why it is unsolvable" is invalid as soon as the squares are coloured differently to a chessboard (with 2 opposite corners removed). As you say, if all squares are the same colour, they can of course still be treated as though they were coloured like a chessboard.

This talk of a generalised solution is interesting. Here's a start, perhaps (not sure if it really needs to be tabbed anymore):
To advance the observation that, starting along any given side, because each side has an odd number of squares, at least one 2x1 rectangular tile would have to protrude into the next row/colomn, and that this would necessarily also give the next row/colomn (at most) 7 empty spaces like the previous one:

It is important to note that ONLY an odd number of 2x1 rectangular tiles can ever protrude into the next row/colomn, from the previous one.
E.g. if all 2x1 rects protrude into the next row/colomn, there are 7 of them. If all but 1 protude into the next row/colomn, there are 5 that do protrude (a 2x1 rect turned 90 degrees from its protruding state obviously covers 2 squares), if all but 2 protrude, there are 3 that do protrude, and if all but 3 protrude, there is only 1 that protrudes.
As already noted, this pattern will continue all the way to the other side of the board.

Now consider a property of the 8x8 board (with 2 opposite corners removed): it has ROTATIONAL symmetry when empty of 2x1 rects.
This means that if we rotate any 4x8 half (with 1 corner missing), it will map over onto the other half.

So now imagine we only advance this "odd number of protruding rects" pattern only half way. In any given combination, there will be an odd number of protruding rects, when filled so that there are no gaps, only protrusions.

Consider now that an odd number of protruding 2x1 rects from one half CANNOT lock into the other half because of this rotational symmetry, which requires that each protrusion MUST have a matching protrusion, an equal number of squares from the middle, which could be removed to let the one half lock into the other. Rotational symmetry requires that (on an even x even square board) there must be an even number of protrusions from either half. There can only be an odd number.

Writing this, I realise that each half doesn't have to be symmetrical with the other, but somehow I can intuit that this isn't actually a proper problem to this starting point.

As for the "don't remove two squares that would have the same colour if coloured like a chessboard, remove two that would have opposite colours" thing:
A simple algorithm that resembles "laying bricks", with 2x1 rects parallel to the base of the board whenever possible (only "upright" when necessary), and completing one layer before advancing to the one on top, will solve any 8x8 grid with 2 squares removed that would be of opposite colour if coloured like a chessboard.

I dunno how to explain that in proof form, but it just works.

Fixed Cross wrote:You asked me the exact same thing before. I then offered three posts, none of which got a response from you. You later told me you didn't understand them. I haven't simplified the logic, so I can't help you there.

Might I suggest you try explaining them again, but in an improved way?

Perhaps if they were not clear enough to merit a meaningful criticism the first time around, you ought to re-write it in a clearer way. I vaguely remember not understanding what you were getting at either, though perhaps at least offering the criticism that you cannot judge "what a smaller number of islanders would do if there were that smaller number of them" whilst artificially attributing knowledge to them that which only a larger number of islanders would know. I seem to remember that inconsistency in method applying to at least some of what you have offered.
Last edited by Silhouette on Sat May 04, 2013 8:02 pm, edited 2 times in total.
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Re: Math Fun

Postby Fixed Cross » Sat May 04, 2013 8:00 pm

I am very used to people coming around to my ideas a couple of years after I present them. Almost without exception, such ideas are at first arrogantly dismissed (without proper argument) by the people who later come to embrace them, even though by then they've often forgotten where they got them. So until someone responds to what my points rationally and to the point (I am hoping Carleas will in a promised PM) I am not quite exactly prompted to think that they've been understood.
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Re: Math Fun

Postby Sauwelios » Sat May 04, 2013 8:17 pm

Fixed Cross wrote:
Sauwelios wrote:
Fixed Cross wrote:You asked me the exact same thing before. I then offered three posts, none of which got a response from you. You later told me you didn't understand them. I haven't simplified the logic, so I can't help you there.

Those posts were just rambling. And as I said, general claims about hypotheticals are not a demonstration.


Where did you demonstrate false steps in James reasoning? And did James concur that he stood corrected? I must have missed all that, but given James compliant attitude now perhaps it did happen.

I demonstrated the false steps in his reasoning in that post of mine in which I made some phrases blue and some red. James need not concur: he has not shown how his reasoning is impervious to my critique.

Yeah.... whatever you say I guess... lol.
It took you two years and a very intensive private workshop to see the logic to value ontology. Before that time it had been "just rambling". This is a bit predictable.

Your posts were just rambling—i.e., all over the place. And indeed, why must you refer to three posts?... Kindly give a systematic account of your reasoning, in one post. You can copy/paste from those three posts if you think it's functional (I doubt it).

By the way, I still don't think value ontology is the revolutionary innovation you so arrogantly proclaimed it was.
"Someone may object that the successful revolt against the universal and homogeneous state could have no other effect than that the identical historical process which has led from the primitive horde to the final state will be repeated. But would such a repetition of the process--a new lease of life for man's humanity--not be preferable to the indefinite continuation of the inhuman end? Do we not enjoy every spring although we know the cycle of the seasons, although we know that winter will come again?" (Leo Strauss, "Restatement on Xenophon's Hiero".)
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Re: Math Fun

Postby Fixed Cross » Sat May 04, 2013 8:27 pm

Silhouette wrote:
Flannel Jesus wrote:I personally doubt that there's any single proof which covers both.

Personally I think it's ridiculous to say that your demonstration of "why it is unsolvable" is invalid as soon as the squares are coloured differently to a chessboard (with 2 opposite corners removed). As you say, if all squares are the same colour, they can of course still be treated as though they were coloured like a chessboard.

I think so too. It's a great explanation making use of a valid example-form. I very much enjoy this sort of insight.

Fixed Cross wrote:You asked me the exact same thing before. I then offered three posts, none of which got a response from you. You later told me you didn't understand them. I haven't simplified the logic, so I can't help you there.

Might I suggest you try explaining them again, but in an improved way?

Perhaps if they were not clear enough to merit a meaningful criticism the first time around, you ought to re-write it in a clearer way. I vaguely remember not understanding what you were getting at either, though perhaps at least offering the criticism that you cannot judge "what a smaller number of islanders would do if there were that smaller number of them" whilst artificially attributing knowledge to them that which only a larger number of islanders would know. I seem to remember that inconsistency in method applying to at least some of what you have offered.


Yeah.... I think you're just way too lazy on this one. As I said I'm used to people taking a lot of time coming around to my logic, which admittedly gets very involved and counterintuitive.
The fact that y'all copied the canonical and began defending it at a point where it wasn't even being attacked, but all failed, even refused to demonstrate the epistemic necessity of the guru (which is the point you're relying on), has gradually led me to think perhaps you didn't really penetrate the logic of that solution.

I did penetrate that, first by thinking that the hypothesis "if there was only one and the guru said she said one" was false. And it still is. But that doesn't matter, as it is meant to be false. Its falseness is its epistemic status. Start there, do you understand what I mean when I say that?

This is a puzzle about epistemology, about the status of knowledge in epistemic agents. On the surface, the guru is a great tool. But the structure is not as simple as the three of you make it out to be.
By the by, as I've shown (and as Sauwelios even admitted when I told him "humbly" (the key to teaching) the explicit text doesn't even allow for the canonical solution, as all people see each other at all times, disallowing for the whole "in the morning they were still there" scenario. But that's just a minor flaw, should not distract from the major falseness of the first hypothesis.
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Re: Math Fun

Postby Fixed Cross » Sat May 04, 2013 8:31 pm

Sauwelios wrote:Your posts were just rambling—i.e., all over the place. And indeed, why must you refer to three posts?... Kindly give a systematic account of your reasoning, in one post. You can copy/paste from those three posts if you think it's functional (I doubt it).

By the way, I still don't think value ontology is the revolutionary innovation you so arrogantly proclaimed it was.

I was right then, you've contributed exactly nothing. And indeed, to nothing, no one need concur.
The strong do what they can, the weak accept what they must.
- Thucydides
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Re: Math Fun

Postby Fixed Cross » Sat May 04, 2013 8:34 pm

Seriously, all I get from you is "I don't understand, and that's your fault".
But it's not only my fault - I make things too complex for you, you want things to be more simple than I can make them.
The strong do what they can, the weak accept what they must.
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Re: Math Fun

Postby Sauwelios » Sat May 04, 2013 8:37 pm

Fixed Cross wrote:y'all copied the canonical and began defending it at a point where it wasn't even being attacked, but all failed, even refused to demonstrate the epistemic necessity of the guru (which is the point you're relying on)

This is just an ungrounded insinuation. Are you in fact trolling here?


I did penetrate that, first by thinking that the hypothesis "if there was only one and the guru said she sa[w] one" was false. And it still is.

The "if" part of an "if/then" statement (jargon?) cannot be true or false, as it's not a complete statement... Anyway, your reasoning seems to go like this:

1. The one, blue-eyed person does not know his own eye colour.
2. The guru tells him that she sees a blue-eyed person.
3. The one, blue-eyed person, supposing that the guru is to be believed, knows he has blue eyes.

Now according to you, it seems, the 3rd step somehow retroactively cancels the 1st. This is nonsense.


By the by, as I've shown (and as Sauwelios even admitted when I told him "humbly" (the key to teaching) the explicit text doesn't even allow for the canonical solution, as all people see each other at all times, disallowing for the whole "in the morning they were still there" scenario.

I didn't "admit" that, it's just a truism.
"Someone may object that the successful revolt against the universal and homogeneous state could have no other effect than that the identical historical process which has led from the primitive horde to the final state will be repeated. But would such a repetition of the process--a new lease of life for man's humanity--not be preferable to the indefinite continuation of the inhuman end? Do we not enjoy every spring although we know the cycle of the seasons, although we know that winter will come again?" (Leo Strauss, "Restatement on Xenophon's Hiero".)
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Re: Math Fun

Postby Sauwelios » Sat May 04, 2013 8:38 pm

Fixed Cross wrote:Seriously, all I get from you is "I don't understand, and that's your fault".
But it's not only my fault - I make things too complex for you, you want things to be more simple than I can make them.

Perhaps it's because you don't think in words, that you can't put your thoughts into words very well.
"Someone may object that the successful revolt against the universal and homogeneous state could have no other effect than that the identical historical process which has led from the primitive horde to the final state will be repeated. But would such a repetition of the process--a new lease of life for man's humanity--not be preferable to the indefinite continuation of the inhuman end? Do we not enjoy every spring although we know the cycle of the seasons, although we know that winter will come again?" (Leo Strauss, "Restatement on Xenophon's Hiero".)
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Re: Math Fun

Postby Silhouette » Sat May 04, 2013 8:55 pm

Fixed Cross wrote:Yeah.... I think you're just way too lazy on this one. As I said I'm used to people taking a lot of time coming around to my logic, which admittedly gets very involved and counterintuitive.

Yeah.... I think you're just getting carried away on a cloud of self-believed but undemonstrated superiority on this one.

The fact that you're attacking a strawman just leads me to think you didn't really penetrate the logic of my/our solutions.

See how when I emulate how you speak, it sounds annoying? Yeah, that's how you're sounding when you speak. I recommend you follow your own advice and switch back to humble mode if you at least expect your theory on teaching attitudes to be valid - else it becomes apparent that you don't (and perhaps never did) intend to teach anything, just say stuff and "proclaim" it's correct, but just too deep for us to get (that JSS-style tactic won't fly I'm afraid).

Please start by explaining why the Guru "saying she sees one" is incompatible with the thought experiment of "what one would do if there was only one".

As for the "all people see each other at all times" thing - this is not incompatible with anyone leaving or staying on the island (perhaps the ferry doesn't take them out of view, and they might even see perfectly well at night and never sleep for all we know). The whole "see them in the morning" thing is perfectly fine when taken as it was intended, as "seen on the island even after the ferry had been and gone, picking up anyone who knew their eye colour while it was there at midnight - as in the puzzle". Even this attempt at a minor point seems to hold no water.

And please, to both you and Sau, stop squabbling. It really doesn't lend yourself any of the credibility either of you are fighting for here.
Last edited by Silhouette on Sat May 04, 2013 8:57 pm, edited 1 time in total.
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