md65536

I haven't nearly the intellect required to make sense of all that, but the diagram reminds me of another, the second last here: http://scienceblogs.com/startswithabang/2011/10/why_we_think_theres_a_multiver.php It may have related ideas? I'd bet you have a better chance of detecting evidence of your "lines" than others have of finding real physical evidence of a multiverse, so I look forward to hearing about such discoveries!

I think we're in agreement but just splitting hairs. Yes, before the visitor's statement everyone already knows about the existence of a blue-eyed islander, but it isn't common knowledge. There's no need to assume that, because the OP stated, "if there is some way by which someone can deduce their eye-colour, they will do so instantly." As long as we assume it's possible, we don't need to consider how likely. The puzzle doesn't rely on any such educated guesses or assumptions about eye color, even "very probable" ones. No, they don't need to know what everyone else is using their logic to figure out. All they need to know is that the others are following the rules. If that's a "task at hand", then I agree with you. You could say that the islanders are trying to figure out if they have blue eyes. You could equally say they're trying to avoid it (because they don't want to die), but that they accidentally find out anyway. All that matters is that eventually, the blue will deduce their eye color and die. Yes, they do need common knowledge of this fact in order to deduce their own eye color. They do need common knowledge that everyone is following the rules, from which they can deduce that the blue will all leave on day N, from which they can deduce that N is greater than k after day k passes without the blue leaving, from which they can (and by the rules, must) deduce if they have blue eyes.

I imagine it would be similar to burning away the surface of anything... wood with a magnifying glass, or metal with a welding electrode, etc. What the damage looks like would depend on how and where the energy is focused. The Mars trilogy is amazing. It seemed very "realistic", like a visionary colonization of Mars is the story, rather than just a backdrop for a sci-fi story. It was filled with a lot of really colossal ideas.

With enough energy the area of Earth could be vaporized. I've heard of this before in fiction, perhaps in the Mars trilogy... http://en.wikipedia.org/wiki/Mars_trilogy I think they had satellites set up to purposefully vaporize strips of the Martian surface, as part of a terraforming effort, to increase the density of the Martian atmosphere.

Good point. I hadn't thought of that. I guess it's possible that John was an anonymous model in an art class or something like that. But then, it's possible the singer mingled with the audience at his debut, and was "in the audience" that night. It's possible the painter did a self-portrait, and that Joe is planning an autobiography. I can no longer find a satisfying solution.

I agree, but that's not enough. I don't think that John Cuthber understands the full meaning of common knowledge in logic. To quote the wikipedia link given 3 or so times already, "There is common knowledge of p in a group of agents G when all the agents in G know p, they all know that they know p, they all know that they all know that they know p, and so on ad infinitum." Every order of knowledge, from 1st to (k-1)th, is necessary in order for the knowledge to be common knowledge. The puzzle is an example of how every order can be important. In the puzzle, the islanders are explained to not have common knowledge of the existence of a blue eyed islander, until the outsider provides it. It shows how even with (k-2) orders of knowledge of what everyone knows, that (k-1)th order can make a difference. I think we'll have to have the OP weigh in on this. I disagree about the OP's intentions, because the given solution explains it like it's the standard blue-eyed islander puzzle, which relies on it not being common knowledge that there exists a blue-eyed islander. I'm pretty sure that is OP's intentions, because the whole point of the visitor's statement is that it provides the common knowledge. I believe that OP meant to give a fairly standard "blue-eyed islander" puzzle, but accidentally skipped some details, leaving it ambiguous. Yes, but the islanders are "hyper logical", and we can't assume that they would assume anything. It is definitely not a safe assumption that the existence of a blue-eyed islander is common knowledge before the visitor's statement. If you were on an island of 1000 where everyone around you has purple eyes, would you deduce that you have purple eyes? (And do you have purple eyes? So would you be wrong? Would it be flawed logic?) If you were in a forest where everyone around you was a tree, would you deduce that you were a tree, if there was no way to directly determine it? No, they don't need a common plan of what to do. All they need is the understanding that everyone else is acting "hyper logically". The islanders don't need a plan, to know that if day k passes and no one is gone, it is common knowledge that there are at least k blue. If I'm blue, and I see 99 blue, and all the blue go on day 99... BUT if it's allowed that they can kill themselves, then I don't know for sure if they all spontaneously decided to kill themselves, or if they suicided because they found out they were blue. The xkcd versions prevents this ambiguity in the rules, with "Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay." The wikipedia version currently has some ambiguity.

Find another post made by the answerer in other forums. If there are no other posts, wait for one.

I stand by my statement that no one (not islanders nor puzzle solver) needs to know how many possible eye colors there are and what those colors are, in order for the puzzle to work (ie. that all N blue leave on day N). Yes, but John Cuthber's statement says that the common knowledge given by a God is "perfect logic". In other words, the God sets the puzzle's rules, and makes those rules common knowledge. His example specifically excluded giving common knowledge of the existence of blue-eyed islanders. His point is that the puzzle is the same without that knowledge. His example is flawed. It is not the same as the original puzzle, and to claim so indicates a complete misunderstanding of the solution, and of the importance of the common knowledge given by the visitor in the original puzzle. Yes, but the start time is a part of the common knowledge. The start time is needed, but the start time alone (without the common knowledge of a blue-eyed islander) is not enough for anyone to deduce their eye color. That's true, I should have stuck to the topic and used OP's version. OP is incorrect. It must be common knowledge that the only two colors are brown and blue, in order that the brown also deduce their eye color. This is not stated in the puzzle. If the answer relies on assuming that things are common knowledge, then the point of the puzzle kind of falls apart. One might falsely assume all sorts of different common knowledge, as John Cuthber does in assuming that the existence of a blue eyed islander is already common knowledge, and that they could have deduced their eye color a year ago. Technically, OP's puzzle also omits the important information of the rule that "every islander knows the rules". We are told that everyone follows the religion, so we might assume that the islanders know that too. But to be fair, we really can't, because it is an assumption of common knowledge. Then the puzzle doesn't work, even for the blue-eyed islanders. I'd been using the xkcd version because it doesn't leave such ambiguities, but I should have stuck to OP's version. I will concede, that in OP's version, unless the islanders know something more that we don't know they do, nothing will happen. Ironically enough, in that case, the common knowledge of a blue-eyed islander, though essential, is not enough! And then the god-given common knowledge of "Oh by the way, everyone knows the rules" would start the countdown on the day of the new knowledge.

Wrong again. Sure an islander can try to think that, but if a blue assumes that their eyes are not blue they'll eventually be able to derive a contradiction. Thus they know it can't be that their eyes are not blue. Perhaps one way to ensure you stay is to kill all the blue-eyed islanders you see. I'm not sure if it would suffice to kill just one, but I think it would. Wrong again. Check the puzzle as written here: http://www.xkcd.com/blue_eyes.html It specifically states "assorted eye colors" and doesn't state what they are, or how many. The puzzle still works. Have you not been looking at any of the links that people provide? All your "questions" are answered in the links.

There is information in the topic's title that might imply an assumption, albeit a flimsy assumption. You'll have to post again to receive your second upvote!

No, the sudden acquisition of logic does not give them common knowledge (I mean the precise definition in logic) that there exist blue-eyed people. It is still possible to assume that someone assumes that someone assumes that someone assumes ... (repeated many times) that there are no blue-eyed people, and not deduce a contradiction. The puzzle doesn't rely only on an implicit consensus on a "start time", as if there is some secret shared plan that works as long as everyone chooses to follow it. It just relies on the fact that if everyone is acting perfectly logically (which is true by the rules), then if a blue-eyed islander assumes she is not blue-eyed, she will derive a contradiction on the N-1th day, and thus know that she has blue eyes. If the guru had only said "You can all start your logical voodoo now", no one would leave the island, because no one would be able to deduce their eye color. The guru's statement about blue eyes really is essential. Further, even if there are only blue and brown-eyed islanders, and all the blue leave, then (if it's not common knowledge that there were only blue or brown), the brown could stay indefinitely. For, if I see only brown eyes around me, but I still have no way to say for sure that I don't have green eyes, then I don't know my eye color. In other words, if I can assume without possible contradiction that I have green eyes, and I can assume without contradiction that I have brown eyes, then I don't know my eye color. All I know at that point is whether or not I have blue eyes.

I was going to come to your defence because I think Queen of Wands is being unfairly harsh... But looking at that... that's some of the worst math and/or logic that I've seen in a long time. Each man should pay 8.33. How do you get that they end up paying more after being reimbursed?! If they each put in 10 and ended up with 1 back, how do you figure that any one of the paid 9.33?! The nonsense of the original problem involves mixing addition and subtraction incorrectly, not division. It comes up with a nonsense result. You're applying double-nonsense to get the right answer the wrong way ("fudging"). If you want to do fractions, do this: The men each paid 8.33 and 1/3c for the room, and .66 and 2/3c for the "gratuity". The total paid is 27. 9 each. There is no misdirection involving fractions, because the whole problem is set up to be done with physical, whole dollars. I can't imagine what the +1 vote was for.

But they don't say that. If they did then it'd be common knowledge (in the case of just 1 blue existing). That everyone knows that everyone knows that everyone knows (and so on) that there are blue-eyed people is not known. The synchronization only happens when it becomes common knowledge. Yes they know first order knowledge and they know second order knowledge and they can deduce k'th-order knowledge up to k-1, which they can't deduce without the outsider's information. What happened a year before he arrived? What I meant was, why does everyone whose edits resulted in the current version of the wiki http://en.wikipedia....owledge_(logic) get it wrong? Edit: Sorry, I guess I tend to be a bit of an arse about some things. Perhaps I read your posts the wrong way -- when you say "the given solution must be incorrect," do you really only mean to say "I don't understand the given solution"? Cause then you're not intentionally saying that everyone who gives the correct solution is wrong.

This is a variation of the puzzle discussed in another thread http://www.sciencefo...rs-and-puzzles/. Some of this is taken from http://www.xkcd.com/blue_eyes.html, edited for this variation. A group of people with blue or brown eyes color live on an island. There are some unknown number N of blue. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every morning, a ferry stops at the island. Any islanders who had figured out the color of their own eyes the previous day must leave the island on that ferry, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph. An exception to the above, there is a guru with grey eyes who is allowed to speak once about eye color. It is common knowledge that everyone else has either blue or brown eyes, but not common knowledge that there are at least one of either. The guru says "I see someone with blue eyes." It is common knowledge that the islanders are all unerringly truthful, and that all use identical reasoning. Variation 1: It is common knowledge that any islander who knows for certain (barring any unpredictable events including new outsider knowledge) what day they will leave, must announce it the following day, and tell everyone else what day they're leaving. Variation 2: After the guru's announcement, on that day and each day afterward, every islander must announce to all, their best guess of how long they think they'll stay on the island. This is also common knowledge. What happens? Hint:

By the way I think my proof is flawed. If it worked as-is, I think you could replace "blue" with "brown" and conclude that it would work with any color, and it shouldn't. I think I need to add something like: B0. If a blue islander assumes there are 0 blue, they will leave today. Proof: This contradicts the guru's information. Then B3 fails in the case of N=1, because in that case A1 can't be supposed without contradiction. That would need to be fixed. For other colors B3 doesn't fail in the case of N=1. Another subtle flaw is that the proof relies on the islanders using the conclusions of the lemmas, in order to prove that they can deduce that information. So the proof should be rewritten from their perspective. That would include not giving any information that an islander doesn't know, including the actual number of blue. So the induction base case wouldn't use the number of blue. I was wrong; the islanders do have to consider the existence of much less than 100 blue---or at least the possibility that someone might think that someone might think that (repeat a bunch of times) that there might be no blue islanders.

I think this is totally off topic, but gladly. The prisoner assumes that a Friday execution is impossible because otherwise it wouldn't be a surprise. Once she's committed to that logic, she will be surprised if the execution happens on Friday. The prisoner assumes that the judge's statement is true. I.e. that the prisoner will be executed on some day and that it will be a surprise. The prisoner then uses logic to deduce that she won't be executed. Thus by contradiction, the judge's statement cannot be true. In which case, the prisoner has used terribly false logic. This is the opposite of the blue-eye puzzle, where everyone uses perfect logic! The proof by induction in the hanging paradox contradicts the assumption used to set up the induction. Therefore it is an invalid proof. There is no such contradiction in the blue-eyed islanders puzzle. I have failed. Wikipedia explains this better and more concisely than I have: http://en.wikipedia.org/wiki/Common_knowledge_(logic) Perhaps if you still don't get it after that, you can edit the wiki with your proof that everyone else is wrong.

I think that if every islander read the notice on the same day and every islander knew that everyone else read the note that day, that it would be the same effect. Dear lord, don't tell me that we now have to prove that proof by induction is a valid form of proof, before you'll accept the solution?! There is no paradox in this puzzle. I've written an induction proof that never relies on considering the existence of less than 100 blue-eyed islanders. Unfortunately it got far more complicated than I expected. The key bit of understanding is in B3 below, in understanding how an unknown false assumption about a known false assumption can be built up into a chain of assumptions about what others can assume about others. I was hoping to do an induction on the length of the chain of assumptions, but it didn't come to me. Unfortunately, this proof probably offers no help in understanding why it works, because that's lost in the complicated details. But I think that once you get the gist of why the solution works (using any proof or reasoning), then a proof like this can help nail down the intricate details. I'm sure that something a lot simpler is possible. Please let me know of any improvements you see! There are likely mistakes due to being braindead. ---- I'm using the XKCD version: A guru tells everyone she sees someone with blue eyes on day 1. Anyone who knows their eye color must leave the day they find out. Suppose there are M blue-eyed islanders. B1. If there are N blues, they will each see N-1 blues. Proof: It is clearly so. QED. B2. Given N blues, if they can't assume without contradiction that there are N-1 blues, then they will leave today. Proof: Suppose there are N blues. Suppose any one of them assumes that there are N-1 blues and derives a contradiction. Then (since their logic is flawless), there must not be N-1 blues. But by B1 they see N-1 blues, and they know there can only be that many or more, so they know their own eyes must be blue, and by the rules will leave today. QED. B3. If any blue person P sees N blue and assumes without any possible contradiction that there are N blue, and none of the blue leave today, then P can assume without contradiction that the N observed blue can assume without contradiction that there are N-1 blue. Proof: Suppose A1: that some blue person P sees and assumes that there are N blue, with no possible contradiction and that none of the blue leave today. Assume hypothesis B3 is false, and that P cannot assume without contradiction that the N observed blue can assume without contradiction that there are N-1 blue. Then P knows that the N observed blue can't assume without contradiction that there are N-1 blue. If there were N blue, then P knows by B2 they will leave today. Since they don't leave today, P knows that there are not N blue. Therefore P cannot assume without contradiction that there are N blue. This contradicts A1. Thus B3 by contradiction. QED. B4(n): On day n with no one having left yet, every blue knows that there are at least n blue-eyed islanders. B5(n): On day n with no one having left yet, every blue knows that every blue knows there are at least n blue-eyed islanders. Proof by induction: Base case: Everybody sees at least M-1 islanders so everyone knows there are at least M-1 islanders. So clearly every blue knows there are at least 1 blue on day 1, and that every blue knows it. Inductive case: Assume B4(k) and B5(k) holds for some k. On day k with no one having left, every blue knows there are at least k blue, and every blue knows that every blue knows it. Now say on day k+1 no one leaves that day. Assume B4(k+1) is false. Assume that on day k+1 with no one left yet, that not every blue knows that there are at least k+1 blue. Then it is possible for some blue person P to assume that there are k blue without contradiction. If P sees k-1 blue, then since P knows there are at least k blue, P would know she is blue and leave today; since that doesn't happen P must see k blue. By B3, P can assume without contradiction that the k observed blue can assume without contradiction that there are k-1 blue. But since by B5(k) P knows that the blue know there are at least k blue, they cannot assume without contradiction that there are only k-1 blue. Thus by contradiction B4(k+1) must be true. Since each blue has flawless logic, each is able to do the above proof of B4(k+1), and know that on day k+1 with no one having left yet, every blue knows that there are at least n blue-eyed islanders. Thus B5(k+1). By induction, on day M every blue knows that there are at least M blue. By B2, they will leave today.

Edit: By the way the first statement is incorrect. It's not even true that they'd be doomed if they knew that everyone knew about someone with blue eyes. They'd be doomed if they knew that everyone knew that everyone knew that someone had blue eyes. --- No wait, I don't think that's right. I think they're only doomed if they know that a long list of assumptions about assumptions, about 100 long, is wrong. Okay so put yourself in the position of someone on the island. You know the information about the birthmark and that everyone else knows it and you know everyone has perfect logic. Do you (or will you) know if your eyes are blue or not? There is a proof by induction on the XKCD link I think, that proves the answer to the original version of the problem is correct. You can't disprove it by showing the absurdity of an absurd variation on it. You'd have to at the very least show a mistake in the induction proof. If you do an induction on the problem size of your variation, it's not going to work like it does for the original problem. Suppose there was only one person with blue eyes. Do you think she would know she has blue eyes, after being told about birthmarks? If it doesn't work for 1 person, it's not going to work for 100. In the original problem, you can do an easy induction on the problem size, i.e. the number of blue-eyed people, but you don't accept that because it doesn't have any intuitive connection to the case where there are 100 blue-eyeds. But it doesn't matter because it works whether or not you understand it and whether or not it makes intuitive sense. Unfortunately for the islanders they all have perfect logic. I think you can also do a much harder induction, without ever considering a case where there are less than 100 blue, on the number of blue eyed people might be assumed to be seen by the other blue eyed people. I could try to do this if you want. I was once told that you don't truly understand something until you can explain it. I believe I understand this problem intuitively now, but it's so twisted that my grasp is frail. I think I'll try an intuitive proof sometime... (But if I fail it only shows that I don't intuitively understand the solution well enough, not that it's incorrect.) Here's an intuitive hint though: If I assume I'm brown, then I think I know something that all the blue-eyed people don't know: If I'm brown, and I see 99 blue, I know there are 99 blue. I also know that all the blue think there might be 1 less. But if I'm blue, that assumption would be incorrect. Either way, I realize that all the blues can think the same way as me. If there are 99 blue, they might assume that there are 98, and they might assume that those 98 think there are 97 and that the 98 assume that the 97 that are assumed to be seen in turn assume there are 96. This works despite my knowledge that all those assumptions are wrong. I also know that the 98 other blue (in any order) don't assume there are only 97, but I also know that the 99 blue I see don't know that! So perhaps I think I know everything about everyone else's assumptions and whether they're right or wrong. What I don't know is whether or not MY assumption is wrong. (I know that everyone else knows whether my assumption is wrong---I don't know anything about my own eye color but everyone else does---but that doesn't even matter).

Well I think the original conjecture is interesting and logical and something that I hadn't considered before. However I'd be more satisfied with a statement something along the lines of: If there exists sufficiently advanced extraterrestrial life, and conjecturing that such life would become autotrophic, and assuming that this would likely entail the use of chlorophyll for photosynthesis, it is likely that some of that life would be green. I think this better captures the multiplicity and enormity of the assumptions that go into the idea.

I think that both First↔Cause ♀ and ydoaPs are correct. It is a trick. I think that's an acceptable answer. It is also a trick question, and it has no answer. It's not a sensible question. Any sensible answer would answer something sensible, which changes the question being asked. It's easy to do that. It might be in the form "What happened to the physical $30 that the men brought in?"---The hotel has 25, the bellboy 2, and the men each have 1. Yes, there is a definite answer, but that answer doesn't directly answer the nonsense question "What happened to the 1$[...]?" So everybody's correct! (Except in saying that the others are wrong.) So I guess everyone's wrong for saying others are wrong, so I guess that makes me wrong... I'd assumed they were photos of her??? Like a tag-line or signature; superfluous but not out of place.

From the hotel. If you're adding up amounts paid, each of the men paid 10, and the hotel paid 5 to the bellboy. As per your previous post, I wish for you to supply the answer! I thought ydoaPs' link sufficed. Thanks. ---- Here's a variation on that one. The men each paid $10. The hotel gives the bellboy a $5 bill and he pockets it, because he thinks that cutting it in 3 pieces would be pointless. 10 + 10 + 10 + 5 = $35. The next day he goes to a candy store and uses it to buy $1 of candy. The shopkeeper gives him $4 change. So now on top of the $35, the shopkeeper now has a $5 bill and the bellboy gets $4 in coins. 35 + 5 + 4 = $44. Where does the money keep coming from?

If they all knew that everyone shared the same dream, then the logic would also work. All the blues would leave/die on day 100. Edit: AND only if they all knew that everyone else knew that everyone shared the same dream. Yes! And I know that. And you know that. But I don't know whether you know that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that every other knows that. Roughly speaking (I've underestimated the number of blues here). I said no one dies first. You're saying someone dies first. So maybe you tell me. Don't worry, you'll get this eventually. It doesn't matter. He doesn't know the order, he doesn't care. He can make the same assumption about any of the blues that he sees, an assumption that will prove to be false if that blue (i.e. any blue, i.e. all 99 blues that he sees) are still around after the 99th day. All 100 go on day 100. The ship leaving holds all 100. The guillotine accommodates all 100 at once. They can all go simultaneously. If they happen to go in some order, the order doesn't matter. Edit: Perhaps there is a misconception that the blues suicide when they see others suicide. That's not the case. They suicide because they see the 99th day go by where they expected (or at least hoped for) all the blues to kill themselves. They know their eyes are blue because of the lack of anyone killing themselves first.

If the islanders are acting on dream information then they're not acting on perfect logic. The solution requires that every blue acts with flawless logic AND, CRUCIALLY, also assumes that every other blue will act with flawless logic. Edit: Imagine a case where there are 1 or 2 people with blue eyes and you dream that a guru says "someone has blue eyes". Do you know if you have blue eyes? If there are 5 blues and I'm one of them, I can assume that [there are only 4 blues which can assume that (there are 3 blues which assume that there are 2 blues which assume there is 1 blue which assumes there might be no blues)]. If someone tells everyone that there is at least 1 blue, then I know that MY assumption is wrong (though I already knew that everyone else's assumptions listed here were wrong), and that no one can assume that anyone can assume that anyone can assume that anyone can assume that there might be no blues. If I'm brown, I'll find out that my assumption was right when the 4 blues derive their contradiction on the 4th day. Otherwise if the blues don't leave then, I'll realize my assumption (including that I'm not blue) is wrong, and I'll leave the next day. No one dies first, second, etc. All the blues are symmetrical. They all realize they're blue on the same day. The solution is correct. The weird part is, I'll never assume that anyone actually assumes that there are no blue-eyed islanders. But I can assume that someone else assumes that [long list of assumptions here] that someone else can! And eventually they'll realize that their assumption is wrong. If they don't, then I'll realize my assumption is wrong. Another way to look at it is this: Say I'm one of 100 blue. I see 99 blue. I know there are 99 or 100 blue. I know the browns see either 99 or 100 blue. But the browns may think they're blue. The browns may think there are 101 blue. I know the blues see either 98 or 99 blue. The blues may think there are 98 blue! If there are 99 blue (there aren't, but I don't know that yet), then the blues may, like me, think that the other blues might see only 97 blue. I know that assumption's wrong, but if my assumption's right (it's not), then the other blues will assume that the blues that they see will think that the blues that they see will see only 96 blue, and so on. I know that no one sees only 96, and I know that the blues I see know that no one sees 96, but I don't know whether the blues that I see know that no one else knows whether anyone sees 96. And so forth. Yet another way to state this: I know that nothing's going to happen until day 99 or day 100, but I don't know if perhaps the blues I see think something will happen on day 98 or 99. If I'm a blue, then all the other blues are dealing with the exact same logic problem as I am. But if I'm brown, then all the blues I see are dealing with the problem less one blue. I know this, and I know that the blues are thinking the same way as me. So they may think that the other blues are dealing with one less blue. And so forth. ------------ Here's a variation. Same rules as the original. 100 people with blue eyes. The guru/visitor says to the crowd "Someone on this island has blue eyes." Then Goofuth enters the room and says (truthfully), "What was that??? I was in the bathroom, I missed that!" You look at Goofuth's eyes, and are relieved to know that no one will have to leave the island. What color are Goofuth's eyes? (Not sure if this one works as I think it does...)

The abstract looks like an introduction and doesn't summarize any results. I found this helpful: http://www.lightbluetouchpaper.org/2007/03/14/how-not-to-write-an-abstract/ It looks like you're multiplying velocities by the Lorentz factor. Is that intentional or a misunderstanding of special relativity? Sorry if it's a stupid question -- I'm no expert and I didn't read the paper, only skimmed it a bit -- but that stands out as a potential problem.

There's some info here: http://en.wikipedia.org/wiki/Global_Consciousness_Project -- doesn't seem so "sold" on the idea as the videos. The videos present the topic in a very biased way. I don't think they're reliable sources. I read about the Princeton Eggs years ago and found it fascinating. I think there is probably something to it, something interesting that should be explainable in the future with better physics understanding.