RE: Paradox, or, Breaking the mind of logic
Original Message: - From: John D. Giorgis [EMAIL PROTECTED] Date: Thu, 12 Oct 2006 21:43:28 -0400 To: brin-l@mccmedia.com Subject: RE: Paradox, or, Breaking the mind of logic Based on the explanations of Dan and Alberto, let me give this another crack: * >Case I > >If N(blue) = 0, then every native exists in: > > >State 1: > -Sees only red dots > -Doesn't know if N(blue) = 0 or N(blue)=1 > -Doesn't know if everyone else is existing in State 1 or State 2 > >This case obviously doesn't apply to the given example. > > >* > >Case II > >If N(blue) = 1, then: > >State 1: One native > -Sees only red dots > -Doesn't know if N(blue) = 0 or N(blue) =1 > -Doesn't know if everyone else is existing in State 1 or State 2 > >State 2: All other natives > -See one blue dot, > -Don't know if N(blue)= 1 or N(blue) = 2 > -Don't know if native with blue dot is in State 1 or State 2 > -Don't know if All other natives are in State 2 or State 3 > >In this case, the anthropologist imparts information to the one native >in State 1 that N(blue) = 1. That native commits suicide on the first >day, everyone else goes on the second day. > > > >Case III > >If N(blue) = 2 > >State 1: (No Natives) > -Sees only red dots > >State 2: Two Natives > -See one blue dot, > -Don't know if N(blue)= 1 or N(blue) = 2 > -Don't know if native with blue dot is in State 1 or State 2 > -Don't know if all other natives are in State 2 or State 3 > >State 3: All Other Natives > -See two blue dots > -Don't know if N(blue) = 2 or N(blue) = 3 > -Don't know if the two natives with blue dots are in State 2 or State 3 > -Don't know if all other natives are in State 3 or State 4 > >In this case, the anthropologist imparts second-order knowledge. The >two natives in State 2 don't know if the other one is blissfully living >in State 1 - believing that it is one happy island of red dots.When >the other one does not commit suicide on the 1sst day, they then realize >that it is because the other sees one blue dot - their own. They die >that night, and the rest die the next night. > > >** > >Case IV > > >If N(blue) = 3 > >State 1: (No Natives) > -Sees only red dots > >State 2: (No Natives) > -Sees one blue dot >State 3: Three Natives > -See two blue dots > -Don't know if N(blue) = 2 or N(blue) = 3 > -Don't know if the two natives with blue dots are in State 2 or State 3 > -Don't know if all other natives are in State 3 or State 4 >State 4: All Other Natives > -See three blue dots > -Don't know if N(blue) = 3 or N(blue) = 4 > -Don't know if the three natives with blue dots are in State 3 or >State 4 > -Don't know if all other natives are in State 4 or State 5 >I'm still not sure what the anthropologist imparts in this situation. Using your termonology, third order knowledge. You have shown, in Case III, that if there are two people with blue dots, then they will each deduce that on day two. Gor knows that too. >Every native knows that every other already sees at least one blue dot. >Thinking about the case of the three natives in State 3 - let's call >them Gor, Kull, and Tar as Alberto suggested. >Day 1: >Gor thinks: If I am red, then Kull and Tar each see one blue dot Kull >and Tar don't know if I see one blue dot or two blue dots. >Gor thinks: If I am blue, the Kull and Tar each see two blue dots. >Kull and Tar don't know if I see one blue dot or two blue dots. >In both cases, the anthropologist imparts no new information to anyone, >so no one commits suicide. >Day 2: >Gor thinks: Neither Kull nor Tar kicked the bucket last night. That is >because they each saw at least one blue dot - just as the anthropologist >said. But I knew *yesterday* that they each saw one blue dot. How is >today any different? Moreover, Kull and Tar see that I am still >around, so I must see at least one blue dot - but they knew that >yesterday too. Again, how is today any different? Day 3: Gor thinks, Kull and Tar did not both kick the bucket last night. That means neither one of them deduced that they were in state 2, one of two natives with blue dots. If they each only saw one blue dot, then they would have deduced that by yesterday. Thus, I must have a blue dotand kill myself along with Kull and Tar tonight. Dan M. ___ http://www.mccmedia.com/mailman/listinfo/brin-l mail2web - Check your email from the web at http://mail2web.com/ . ___ http://www.mccmedia.com/mailman/listinfo/brin-l
RE: Paradox, or, Breaking the mind of logic
Based on the explanations of Dan and Alberto, let me give this another crack: * Case I If N(blue) = 0, then every native exists in: State 1: -Sees only red dots -Doesn't know if N(blue) = 0 or N(blue)=1 -Doesn't know if everyone else is existing in State 1 or State 2 This case obviously doesn't apply to the given example. * Case II If N(blue) = 1, then: State 1: One native -Sees only red dots -Doesn't know if N(blue) = 0 or N(blue) =1 -Doesn't know if everyone else is existing in State 1 or State 2 State 2: All other natives -See one blue dot, -Don't know if N(blue)= 1 or N(blue) = 2 -Don't know if native with blue dot is in State 1 or State 2 -Don't know if All other natives are in State 2 or State 3 In this case, the anthropologist imparts information to the one native in State 1 that N(blue) = 1. That native commits suicide on the first day, everyone else goes on the second day. Case III If N(blue) = 2 State 1: (No Natives) -Sees only red dots State 2: Two Natives -See one blue dot, -Don't know if N(blue)= 1 or N(blue) = 2 -Don't know if native with blue dot is in State 1 or State 2 -Don't know if all other natives are in State 2 or State 3 State 3: All Other Natives -See two blue dots -Don't know if N(blue) = 2 or N(blue) = 3 -Don't know if the two natives with blue dots are in State 2 or State 3 -Don't know if all other natives are in State 3 or State 4 In this case, the anthropologist imparts second-order knowledge. The two natives in State 2 don't know if the other one is blissfully living in State 1 - believing that it is one happy island of red dots.When the other one does not commit suicide on the 1sst day, they then realize that it is because the other sees one blue dot - their own. They die that night, and the rest die the next night. ** Case IV If N(blue) = 3 State 1: (No Natives) -Sees only red dots State 2: (No Natives) -Sees one blue dot State 3: Three Natives -See two blue dots -Don't know if N(blue) = 2 or N(blue) = 3 -Don't know if the two natives with blue dots are in State 2 or State 3 -Don't know if all other natives are in State 3 or State 4 State 4: All Other Natives -See three blue dots -Don't know if N(blue) = 3 or N(blue) = 4 -Don't know if the three natives with blue dots are in State 3 or State 4 -Don't know if all other natives are in State 4 or State 5 I'm still not sure what the anthropologist imparts in this situation. Every native knows that every other already sees at least one blue dot. Thinking about the case of the three natives in State 3 - let's call them Gor, Kull, and Tar as Alberto suggested. Day 1: Gor thinks: If I am red, then Kull and Tar each see one blue dot Kull and Tar don't know if I see one blue dot or two blue dots. Gor thinks: If I am blue, the Kull and Tar each see two blue dots. Kull and Tar don't know if I see one blue dot or two blue dots. In both cases, the anthropologist imparts no new information to anyone, so no one commits suicide. Day 2: Gor thinks: Neither Kull nor Tar kicked the bucket last night. That is because they each saw at least one blue dot - just as the anthropologist said. But I knew *yesterday* that they each saw one blue dot. How is today any different? Moreover, Kull and Tar see that I am still around, so I must see at least one blue dot - but they knew that yesterday too. Again, how is today any different? JDG ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
On 12 Oct 2006, at 4:33PM, jdiebremse wrote: Case I If N(blue) = 0, then every native exists in: State 1: Sees only red dots, but doesn't know if N(blue) = 0 or N (blue) =1 This case obviously doesn't apply to the given example. ** **\ * Case II If N(blue) = 1, then: State 1: One native sees only red dots, but doesn't know if N (blue) = 0 or N(blue) =1 State 2: All other natives see one blue dot, but don't know if N (blue) = 1 or N(blue) = 2 Moreover, in this case All Natives know that each and every Native knows that each of them is either in State 1 or State 2. In this case, the anthropologist imparts information to the one native in State 1, causing the cascade. ** **\ * Case III If N(blue) = 2 State 2: Two natives see one blue dot, but don't know if N(blue) = 1 or N(blue) = 2 State 3: All other natives see two blue dots, but don't know if N (blue) = 2 or N(blue) = 3 Moreover, in this case All Natives know that each and every Native knows that each of them is either in State 2 or State 3. But they don't. The two blue-dot natives don't know if the other blue dot sees one blue dot or *none*. In this case, the anthropologist doesn't impart any information to anyone. Everyone knows that N(blue) >= 1.So, presuming that the island existed in a steady state before the anthropologist's arrival, then her arrival with the announcement that N(blue) >= 1 has no effect. It tells each of the two blue-dots that the other blue-dot now has enough information to kill himself if he is the only blue-dot. When neither does, both can deduce they are blue-dots the next day. All the reds follow the day after. Possibly Maru -- William T Goodall Mail : [EMAIL PROTECTED] Web : http://www.wtgab.demon.co.uk Blog : http://radio.weblogs.com/0111221/ If you listen to a UNIX shell, can you hear the C? ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
On 13/10/2006, at 1:33 AM, jdiebremse wrote: In this case, the anthropologist doesn't impart any information to anyone. Everyone knows that N(blue) >= 1.So, presuming that the island existed in a steady state before the anthropologist's arrival, then her arrival with the announcement that N(blue) >= 1 has no effect. Am I just missing something here? Yes. You've got a blue dot... Charlie ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
JDG wrote: > > Maybe I'm exhibiting my ignorance here, but if N(blue) = 4 then all the > natives *know* that there is *not* "only one blue-dotted native" before > the anthropologist even arrives. > The problem is that you are not assuming that _all_ natives are omniintelligent. > > *But*, if N(blue) > 2, then *every* native starts out with: > > Given: ~A > > Thus, the arrival of this anthropologist can't impart any additional > information, because the first step of the induction leads to a > conclusion that the natives have already reached anyways. > He does. > Maybe it will help to lay out different cases. It seems clear that if > a native can identify the value of N(blue), then mass suicide becomes > inevitable. > > > > Case I > > If N(blue) = 0, then every native exists in: > State 1: Sees only red dots, but doesn't know if N(blue) = 0 or N(blue) > =1 > > This case obviously doesn't apply to the given example. > Ok. In case, the anthropologist would say something like "why everybody has a red dot?" and mass suicide would happen in the first night. > \ > * > > Case II > > If N(blue) = 1, then: > > State 1: One native sees only red dots, but doesn't know if N(blue) = 0 > or N(blue) =1 > > State 2: All other natives see one blue dot, but don't know if N(blue) > = 1 or N(blue) = 2 > > Moreover, in this case All Natives know that each and every Native knows > that each of them is either in State 1 or State 2. > > In this case, the anthropologist imparts information to the one native > in State 1, causing the cascade. > Yes. > \ > * > > > Case III > > If N(blue) = 2 > > State 2: Two natives see one blue dot, but don't know if N(blue) = 1 or > N(blue) = 2 > > State 3: All other natives see two blue dots, but don't know if N(blue) > = 2 or N(blue) = 3 > > > Moreover, in this case All Natives know that each and every Native knows > that each of them is either in State 2 or State 3. > > In this case, the anthropologist doesn't impart any information to > anyone. Everyone knows that N(blue) >= 1.So, presuming that the > island existed in a steady state before the anthropologist's arrival, > then her arrival with the announcement that N(blue) >= 1 has no effect. > > Am I just missing something here? > Yes. Let's work it out in detail. Imagine that Gor and Kull are the two blue-dotted natives. When the anthropologist claims "I see a blue dotted native", Gor will reason this way: (G-B): I am blue dotted. Then, the anthropologist gave no new information to Kull. So, Kull will not know his color, and he will not suicide on the first night. (G-R): I am red dotted. Then, the anthropologist gave a new information to Kull - and he will know that he is blue dotted. So, he will suicide on the first night, poor bastard. On the first night, Kull does not suicide. Then, Gor will know that he is blue dotted, and he will suicide on the second night. // Case IV: if (N(blue) = 3) Now we have Gor, Kull and Tar with blue dots. And _Tar_, being omniintelligent, will reason like this: (T-R): Suppose I am red dotted. But Gor is very intelligent, and he will reason as he did in Case III: if Kull does not suicide on the first night, then Gor must suicide on the second night. Gor does not suicide on the second night. So, Tar deduces that he is blue dotted, and suicides on the third night. // Case V: if (N(blue) = 4) Now we have Gor, Kull, Tar and Zira with blue dots. Zira will reason like this: (Z-R): Suppose I am red dotted. Since Tar is very intelligent, he will reason as he did in Case IV: if Kull does not suicide on the first night and Gor does not suicide on the second night, then Tar must suicide on the third night. Tar does not suicide on the third night. So, Zira deduces that she is blue-dotted, and suicides on the fourth night. Can you see that the induction process can only be started when the anthropologist gives the _absolute_ information that there is at least one blue dotted native? Alberto Monteiro ___ http://www.mccmedia.com/mailman/listinfo/brin-l
RE: Paradox, or, Breaking the mind of logic
> -Original Message- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On > Behalf Of jdiebremse > Sent: Thursday, October 12, 2006 10:33 AM > To: Killer Bs Discussion > Subject: Re: Paradox, or, Breaking the mind of logic > > > > --- In [EMAIL PROTECTED], "Alberto Monteiro" <[EMAIL PROTECTED]> wrote: > > > But how does this work for N(blue) = 4? > > > > > The key point is that the natives are omniintelligent and > > know that all other natives are also omniintelligent. > > > > > The initial state is that each native has two cases: > > > > > > 1) There are three blue-dot natives, and each blue dot native sees > > > two blue dot natives. > > > > > > 2) There are four blue-dot natives, including himself, and each blue > > > dot native sees three blue dot natives. > > > > > > In this case, I don't see how the naturalist provides any additional > > > information. In the initial state, every native knows that every > other > > > native knows that there is at least one blue dot. > > > > > He does. Because of the omniintelligence hypothesis, each native > > can reason like this: > > > > (a) If there is only one blue dotted native, then, seeing that > > everybody else is red dotted, this native will commit ritual > > suicide in the first night. > > > Maybe I'm exhibiting my ignorance here, but if N(blue) = 4 then all the > natives *know* that there is *not* "only one blue-dotted native" before > the anthropologist even arrives. That's certainly true. But, what they don't know is what the other native can deduce. After the anthropologist makes his statement, even though it is a statement that every native knows to be true, the natives can surmise what the actions of other natives would be. Considering the case where there are 4 natives with a blue dot. The 4 with a blue dot know that there are either 3 natives with a blue dot or 4 natives with a blue dot. The rest know that either there are 4 natives with a blue dot or there are 5 natives with a blue dot. OK, that much is simple. Now, comes the harder partthe deductions made by the natives that have a blue dot about the knowledge of the other natives with a blue dot. The ones with a blue dot know that there are either 3 or 4 natives with a blue dot. But, they don't know what the other natives with a blue dot know. If they have a red dot, then there are only 3 natives with a blue dot, each of which think there are either 3 natives with a blue dot or two natives with a blue dot...depending on whether they have a blue dot themselves. If they themselves have a blue dot, then each of the other natives think there are either 3 or 4 natives with a blue dot. At this point, let me arbitrarily name the natives with the blue dots native A, B, C and D. Native A knows that natives B, C, & D will think there are 3 or 4 with a blue dot if A has a blue dot, and think there are 2 or three natives with a blue dot if A has a red dot. Now, native A does some deductions about the understanding of native B concerning the understanding of natives C & D if native A has a red dot. In this case, native B knows that, if he has a red dot, then native C thinks there is either one or two natives with a blue dot, depending on his dot. After D lives past the first midnight, he knows that the number must be 2, requiring him to have a blue dot. He would, if B had a red dot, kill himself (along with D who makes similar deductions) the second night. When this doesn't happen, if native A had a red dot, native B would know that he had a blue dot, and B, C, & D would kill themselves the third night. Since this doesn't happen, A, B, C, & D accurately deduce that they all have blue dots, and kill themselves on the 4th night. The key here is not what the natives know. In this case, it's a 4-fold regression...it's what the natives know about what the other natives know about what the other natives know about what the other natives know. It would not take an outsider to start the chainany certain statement that allows each and every native to start the regression of knowledge of knowledge of knowledge would suffice. Does that help? Dan M. ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
--- In [EMAIL PROTECTED], "Alberto Monteiro" <[EMAIL PROTECTED]> wrote: > > But how does this work for N(blue) = 4? > > > The key point is that the natives are omniintelligent and > know that all other natives are also omniintelligent. > > > The initial state is that each native has two cases: > > > > 1) There are three blue-dot natives, and each blue dot native sees > > two blue dot natives. > > > > 2) There are four blue-dot natives, including himself, and each blue > > dot native sees three blue dot natives. > > > > In this case, I don't see how the naturalist provides any additional > > information. In the initial state, every native knows that every other > > native knows that there is at least one blue dot. > > > He does. Because of the omniintelligence hypothesis, each native > can reason like this: > > (a) If there is only one blue dotted native, then, seeing that > everybody else is red dotted, this native will commit ritual > suicide in the first night. Maybe I'm exhibiting my ignorance here, but if N(blue) = 4 then all the natives *know* that there is *not* "only one blue-dotted native" before the anthropologist even arrives. > Induction Hypothesis: > > (b) Suppose that there are (N+1) blue dotted natives. Then, each > of these natives, noticing that the other (N) blue dotted natives > didn't commit suicide in the N-th night, will commit ritual > suicide in the (N+1)-th night > > The naturalist provides information because he starts the process, > by forcing step (a) of the induction. If I understand this correctly, here's how the "Induction Hypothesis" works, starting with step (a). Let A = There is one, and only one, blue dot native. Let B = One Native commites ritual suicide on the first night. The induction seems to be that: Given: If A then B. Given: ~B Then: ~A *But*, if N(blue) > 2, then *every* native starts out with: Given: ~A Thus, the arrival of this anthropologist can't impart any additional information, because the first step of the induction leads to a conclusion that the natives have already reached anyways. Maybe it will help to lay out different cases. It seems clear that if a native can identify the value of N(blue), then mass suicide becomes inevitable. Case I If N(blue) = 0, then every native exists in: State 1: Sees only red dots, but doesn't know if N(blue) = 0 or N(blue) =1 This case obviously doesn't apply to the given example. \ * Case II If N(blue) = 1, then: State 1: One native sees only red dots, but doesn't know if N(blue) = 0 or N(blue) =1 State 2: All other natives see one blue dot, but don't know if N(blue) = 1 or N(blue) = 2 Moreover, in this case All Natives know that each and every Native knows that each of them is either in State 1 or State 2. In this case, the anthropologist imparts information to the one native in State 1, causing the cascade. \ * Case III If N(blue) = 2 State 2: Two natives see one blue dot, but don't know if N(blue) = 1 or N(blue) = 2 State 3: All other natives see two blue dots, but don't know if N(blue) = 2 or N(blue) = 3 Moreover, in this case All Natives know that each and every Native knows that each of them is either in State 2 or State 3. In this case, the anthropologist doesn't impart any information to anyone. Everyone knows that N(blue) >= 1.So, presuming that the island existed in a steady state before the anthropologist's arrival, then her arrival with the announcement that N(blue) >= 1 has no effect. Am I just missing something here? JDG ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
- Original Message - From: "Ronn!Blankenship" <[EMAIL PROTECTED]> To: "Killer Bs Discussion" Sent: Wednesday, October 11, 2006 4:28 PM Subject: Re: Paradox, or, Breaking the mind of logic > At 03:32 PM Wednesday 10/11/2006, maru dubshinki wrote: >>On 10/11/06, Alberto Monteiro <[EMAIL PROTECTED]> wrote: >>. >>>He does. Because of the omniintelligence hypothesis, each native >>>can reason like this: >>> >>>(a) If there is only one blue dotted native, then, seeing that >>>everybody else is red dotted, this native will commit ritual >>>suicide in the first night. >>> >>>Induction Hypothesis: >>> >>>(b) Suppose that there are (N+1) blue dotted natives. Then, each >>>of these natives, noticing that the other (N) blue dotted natives >>>didn't commit suicide in the N-th night, will commit ritual >>>suicide in the (N+1)-th night >>> >>>The naturalist provides information because he starts the process, >>>by forcing step (a) of the induction. >>> >>>Alberto Monteiro >> >>This is basically my conclusion as well. I put it differently, >>though: >>with the outsider's pronouncement, each native can know reason about >>the beliefs of each of the others. While it is true that the >>outsider >>provides no new information about the physical situation to each >>blue >>dot, each blue dot now knows something new: that each native *must* >>believe there to be at least one blue dot, because the stranger told >>them all so, where before each blue dot could believe that they >>themself were red and the other blue dot ignorant of their status. >>With this forcing of belief, the induction argument becomes >>operative. > > > I take it this is your alternative explanation of what really > happened on Easter Island? > > On Easter Island an evil naturalist proclaimed publicly that until he had come to this island he had never seen anyone with a green dot. xponent Evil Con Carne Maru rob ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
At 03:32 PM Wednesday 10/11/2006, maru dubshinki wrote: On 10/11/06, Alberto Monteiro <[EMAIL PROTECTED]> wrote: . He does. Because of the omniintelligence hypothesis, each native can reason like this: (a) If there is only one blue dotted native, then, seeing that everybody else is red dotted, this native will commit ritual suicide in the first night. Induction Hypothesis: (b) Suppose that there are (N+1) blue dotted natives. Then, each of these natives, noticing that the other (N) blue dotted natives didn't commit suicide in the N-th night, will commit ritual suicide in the (N+1)-th night The naturalist provides information because he starts the process, by forcing step (a) of the induction. Alberto Monteiro This is basically my conclusion as well. I put it differently, though: with the outsider's pronouncement, each native can know reason about the beliefs of each of the others. While it is true that the outsider provides no new information about the physical situation to each blue dot, each blue dot now knows something new: that each native *must* believe there to be at least one blue dot, because the stranger told them all so, where before each blue dot could believe that they themself were red and the other blue dot ignorant of their status. With this forcing of belief, the induction argument becomes operative. I take it this is your alternative explanation of what really happened on Easter Island? -- Ronn! :) ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
On 10/11/06, Alberto Monteiro <[EMAIL PROTECTED]> wrote: . He does. Because of the omniintelligence hypothesis, each native can reason like this: (a) If there is only one blue dotted native, then, seeing that everybody else is red dotted, this native will commit ritual suicide in the first night. Induction Hypothesis: (b) Suppose that there are (N+1) blue dotted natives. Then, each of these natives, noticing that the other (N) blue dotted natives didn't commit suicide in the N-th night, will commit ritual suicide in the (N+1)-th night The naturalist provides information because he starts the process, by forcing step (a) of the induction. Alberto Monteiro This is basically my conclusion as well. I put it differently, though: with the outsider's pronouncement, each native can know reason about the beliefs of each of the others. While it is true that the outsider provides no new information about the physical situation to each blue dot, each blue dot now knows something new: that each native *must* believe there to be at least one blue dot, because the stranger told them all so, where before each blue dot could believe that they themself were red and the other blue dot ignorant of their status. With this forcing of belief, the induction argument becomes operative. ~maru ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
On 10/11/06, David Hobby <[EMAIL PROTECTED]> wrote: jdiebremse wrote: ... > But how does this work for N(blue) = 4? > > The initial state is that each native has two cases: > > 1) There are three blue-dot natives, and each blue dot native sees two > blue dot natives. > > 2) There are four blue-dot natives, including himself, and each blue dot > native sees three blue dot natives. > > In this case, I don't see how the naturalist provides any additional > information. In the initial state, every native knows that every other > native knows that there is at least one blue dot. > > JDG JDG-- Maru's original post didn't say this, but the puzzle has an additional assumption: All the natives are expert logicians, they all know that all are, they all know that everybody knows that all the natives are expert logicians, etc. Without this, nothing happens even for only two blues, as each would say, "So, maybe the other guy sees only reds but is dumb." True, true, but remember this is a logic problem, after all. If we wanted to specify all the assumptions, we'd get into silliness like "there exists an objective reality" or "each native will succeed in killing themselves should they try". The role of the outsider is to make it clear to everybody that any situation with only one blue leads to suicide. Of course when N = 4 everybody knows there are blues, but this is different. But isn't the case of only one blue already clear without the outsider? All the natives would eventually conclude that there could not be just three blues, since each of the three would only see two, and eventually wonder why those two hadn't killed themselves, finally concluding that the reason was that each of the two actually saw two blues, since the one thinking all this was the third blue. Etc! ---David And this reasoning stands for all N equal to or greater than 3? ~maru Mmm... sicilian... ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
jdiebremse wrote: ... But how does this work for N(blue) = 4? The initial state is that each native has two cases: 1) There are three blue-dot natives, and each blue dot native sees two blue dot natives. 2) There are four blue-dot natives, including himself, and each blue dot native sees three blue dot natives. In this case, I don't see how the naturalist provides any additional information. In the initial state, every native knows that every other native knows that there is at least one blue dot. JDG JDG-- Maru's original post didn't say this, but the puzzle has an additional assumption: All the natives are expert logicians, they all know that all are, they all know that everybody knows that all the natives are expert logicians, etc. Without this, nothing happens even for only two blues, as each would say, "So, maybe the other guy sees only reds but is dumb." The role of the outsider is to make it clear to everybody that any situation with only one blue leads to suicide. Of course when N = 4 everybody knows there are blues, but this is different. All the natives would eventually conclude that there could not be just three blues, since each of the three would only see two, and eventually wonder why those two hadn't killed themselves, finally concluding that the reason was that each of the two actually saw two blues, since the one thinking all this was the third blue. Etc! ---David Never argue with a Sicilian when death is on the line, Maru. ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
At 10:41 PM Tuesday 10/10/2006, maru dubshinki wrote: A while ago on #Wikipedia, I fell into a discussion with a fellow editor. He posed me a question about the following riddle: Suppose there is an island with a number of natives on it. Each native has either a red or a blue spot on their forehead. But they are not allowed to indicate to each other or otherwise divine in any direct observational fashion what the color of their particular spot might be. One of the iron-clad customs of these indigenous persons is that any native who deduces the color of their spot through logic must kill themselves that midnight. Now, suppose further that of all the natives there, only two have blue spots and all the rest have red spots. A outsider comes along (perhaps he is an ignorant ethnographer), and truthfully mentions to the natives that "At least one of you has a blue dot on your forehead." What will happen to the natives, and how long will it take? The smart ones get on eBay and order some moist towelettes. Or if that is insufficient, some good opaque coverup makeup. They ask for next-day express delivery. Reprogramming The Simulation [Kobayshi] Maru -- Ronn! :) ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
JDG wrote: > > But how does this work for N(blue) = 4? > The key point is that the natives are omniintelligent and know that all other natives are also omniintelligent. > The initial state is that each native has two cases: > > 1) There are three blue-dot natives, and each blue dot native sees > two blue dot natives. > > 2) There are four blue-dot natives, including himself, and each blue > dot native sees three blue dot natives. > > In this case, I don't see how the naturalist provides any additional > information. In the initial state, every native knows that every other > native knows that there is at least one blue dot. > He does. Because of the omniintelligence hypothesis, each native can reason like this: (a) If there is only one blue dotted native, then, seeing that everybody else is red dotted, this native will commit ritual suicide in the first night. Induction Hypothesis: (b) Suppose that there are (N+1) blue dotted natives. Then, each of these natives, noticing that the other (N) blue dotted natives didn't commit suicide in the N-th night, will commit ritual suicide in the (N+1)-th night The naturalist provides information because he starts the process, by forcing step (a) of the induction. Alberto Monteiro ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
--- In [EMAIL PROTECTED], "Bryon Daly" <[EMAIL PROTECTED]> wrote: > > On 10/10/06, maru dubshinki [EMAIL PROTECTED] wrote: > > > Now, the stranger appears to be absolutely useless, but nevertheless, > > removed from the picture the whole thing breaks down in the case where > > N = 2. What is the use of the useless stranger? > > > The key here as I see it is that prior to the stranger's announcement, each > of the blue-dot natives thinks that either: > 1) he is red-dot and there is only the one blue-dot native, who in turn sees > all red-dot natives > 2) he is a blue-dot, and the other blue-dot also sees one blue-dot native. But how does this work for N(blue) = 4? The initial state is that each native has two cases: 1) There are three blue-dot natives, and each blue dot native sees two blue dot natives. 2) There are four blue-dot natives, including himself, and each blue dot native sees three blue dot natives. In this case, I don't see how the naturalist provides any additional information. In the initial state, every native knows that every other native knows that there is at least one blue dot. JDG ___ http://www.mccmedia.com/mailman/listinfo/brin-l
Re: Paradox, or, Breaking the mind of logic
On 10/10/06, maru dubshinki <[EMAIL PROTECTED]> wrote: Now, the stranger appears to be absolutely useless, but nevertheless, removed from the picture the whole thing breaks down in the case where N = 2. What is the use of the useless stranger? The key here as I see it is that prior to the stranger's announcement, each of the blue-dot natives thinks that either: 1) he is red-dot and there is only the one blue-dot native, who in turn sees all red-dot natives 2) he is a blue-dot, and the other blue-dot also sees one blue-dot native. Similarly, the red dot natives see two blues, but can't be sure about themselves: they each think they could be blue or red and don't know for sure either way. So, for all the natives, seeing everyone else's color doesn't tell them anything about their own, without any extra information becoming available. So in the initial state, both the blue-dot natives cannot distinguish between cases 1 & 2 and do not act, and each red-dotters doesn't know for sure if there are two or three blue dotters (the two he sees, plus potentiall himself). But, once the stranger blabs, all the natives, particularly the blue-dot ones, knows that the blue-dot native he sees now has enough information to act, if he sees all red-dots as in case 1 above. If there was only 1 blue-dot, he would have seen every one else with reds and known he must be the blue and killed himself that first night. When everyone is still alive on the second day, both blue dotters know that case 1 above cannot be true, so case 2 must be correct, and thue they kill themselves that night. That is, assuming they all took the time to work out the logic and didn't just say "yeah, we know" and blow it off. Funny, if both blue dotters cheat on night 2 and didn't kill themselves, all the honest red dotters would assume they were a third blue-dotter, and kill themselves on night 3. -Bryon ___ http://www.mccmedia.com/mailman/listinfo/brin-l
RE: Paradox, or, Breaking the mind of logic
> -Original Message- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On > Behalf Of maru dubshinki > Sent: Tuesday, October 10, 2006 10:41 PM > To: Killer Bs Discussion > Subject: Paradox, or, Breaking the mind of logic > > A while ago on #Wikipedia, I fell into a discussion with a fellow > editor. He posed me a question about the following riddle: > > Suppose there is an island with a number of natives on it. Each native > has either a red or a blue spot on their forehead. But they are not > allowed to indicate to each other or otherwise divine in any direct > observational fashion what the color of their particular spot might > be. One of the iron-clad customs of these indigenous persons is that > any native who deduces the color of their spot through logic must kill > themselves that midnight. > > Now, suppose further that of all the natives there, only two have blue > spots and all the rest have red spots. A outsider comes along (perhaps > he is an ignorant ethnographer), and truthfully mentions to the > natives that "At least one of you has a blue dot on your forehead." > > What will happen to the natives, and how long will it take? My understanding, off the top of my head, is that the natives will be fine as long as they are not all together. After that, the first midnight, nothing will happen. Then, if the two people with blue dots meet after that midnight, they will kill themselves the following midnight. If the natives discover this before there are additional native, the rest will kill themselves by the following midnight. Dan M. > (Answer will be supplied below to further the cause of discussing the > question) > > > > > > spoiler > > Are you really sure you don't want to figure it out yourself? > > If you need a hint, you can always Google the problem. It's a classic > logic problem you know. > > > > > Well, if you're sure. > > Begin spoilers > > > > All the natives will eventually kill themselves. The precise number of > days is something like N+1 days. > The reason is that if either blue dot looks around, and sees the other > blue dot, they should believe that they are probably red (since most > natives are red), and thus there is only one blue dot, the other guy. > So they believe that the other blue dot will kill themselves that day. > Now, he won't (because there is another blue dot, and he is using the > same reasoning - to him, it's the *other* blue dot who should be > killing himself). On the second day, nobody will be dead, and so the > second blue dot must conclude that the reason for this is because they > themself are the second blue dot. Both will kill themselves. This same > reasoning can be generalized (mathematical induction?) for all numbers > 2 and above, since if 3 blue dots, they will wait to day 3 before all > the blues are dead, and so forth. If there is only one blue dot, then > they will kill themselves immediately, since there is at least one > blue dot, and they know that everybody except themself is red, thus > they must be the blue dot. > > Anyway, once the blues have killed themselves off, all the reds will > immediately commit suicide: they followed the blues' reasoning after > all, and know that all the blues are dead, which means that they are > red, and so since they know, they must kill themselves. > So that's that. Necessary background is over and done with. > > The problem that fellow editor posed me (once I'd solved the original > riddle) was this: in the case of 2 blue dots and a bunch of reds, each > blue dot *already* knows what the original stranger told them, that > there was at least one blue dot. They can see the other blue dot! So > it's quite obvious to them that there is at least one blue dot, and so > the stranger tells them absolutely nothing new - they already knew > what he told me, after all. What is not quite as obvious is that if > the stranger is removed, the 2 blues will exist in stasis; in the > first round, each will be waiting for the other to do something (their > situations remember are absolutely symmetrical), and so *every* round > they will be waiting for the other to do something, and of course > that means they never do anything. The reds are irrelevant since they > don't affect matters until the blues are gone. > > Now, the stranger appears to be absolutely useless, but nevertheless, > removed from the picture the whole thing breaks down in the case where > N = 2. What is the use of the useless stranger? > > ~maru > We put thirty spokes together and call it a wheel; > But it is on the space where there is nothing > That the usefu
Paradox, or, Breaking the mind of logic
A while ago on #Wikipedia, I fell into a discussion with a fellow editor. He posed me a question about the following riddle: Suppose there is an island with a number of natives on it. Each native has either a red or a blue spot on their forehead. But they are not allowed to indicate to each other or otherwise divine in any direct observational fashion what the color of their particular spot might be. One of the iron-clad customs of these indigenous persons is that any native who deduces the color of their spot through logic must kill themselves that midnight. Now, suppose further that of all the natives there, only two have blue spots and all the rest have red spots. A outsider comes along (perhaps he is an ignorant ethnographer), and truthfully mentions to the natives that "At least one of you has a blue dot on your forehead." What will happen to the natives, and how long will it take? (Answer will be supplied below to further the cause of discussing the question) spoiler Are you really sure you don't want to figure it out yourself? If you need a hint, you can always Google the problem. It's a classic logic problem you know. Well, if you're sure. Begin spoilers All the natives will eventually kill themselves. The precise number of days is something like N+1 days. The reason is that if either blue dot looks around, and sees the other blue dot, they should believe that they are probably red (since most natives are red), and thus there is only one blue dot, the other guy. So they believe that the other blue dot will kill themselves that day. Now, he won't (because there is another blue dot, and he is using the same reasoning - to him, it's the *other* blue dot who should be killing himself). On the second day, nobody will be dead, and so the second blue dot must conclude that the reason for this is because they themself are the second blue dot. Both will kill themselves. This same reasoning can be generalized (mathematical induction?) for all numbers 2 and above, since if 3 blue dots, they will wait to day 3 before all the blues are dead, and so forth. If there is only one blue dot, then they will kill themselves immediately, since there is at least one blue dot, and they know that everybody except themself is red, thus they must be the blue dot. Anyway, once the blues have killed themselves off, all the reds will immediately commit suicide: they followed the blues' reasoning after all, and know that all the blues are dead, which means that they are red, and so since they know, they must kill themselves. So that's that. Necessary background is over and done with. The problem that fellow editor posed me (once I'd solved the original riddle) was this: in the case of 2 blue dots and a bunch of reds, each blue dot *already* knows what the original stranger told them, that there was at least one blue dot. They can see the other blue dot! So it's quite obvious to them that there is at least one blue dot, and so the stranger tells them absolutely nothing new - they already knew what he told me, after all. What is not quite as obvious is that if the stranger is removed, the 2 blues will exist in stasis; in the first round, each will be waiting for the other to do something (their situations remember are absolutely symmetrical), and so *every* round they will be waiting for the other to do something, and of course that means they never do anything. The reds are irrelevant since they don't affect matters until the blues are gone. Now, the stranger appears to be absolutely useless, but nevertheless, removed from the picture the whole thing breaks down in the case where N = 2. What is the use of the useless stranger? ~maru We put thirty spokes together and call it a wheel; But it is on the space where there is nothing That the usefulness of the wheel depends. We turn clay to make a vessel; But it is on the space where there is nothing That the usefulness of the vessel depends. ___ http://www.mccmedia.com/mailman/listinfo/brin-l
