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 Posted: Sun Sep 09, 2007 3:27 pm
jestingrabbit zz1000zz The answer based upon noting odd/even number of hats does *not* work. The original puzzle stated: Quote: So they line everyone up in such a way that each captive can only see the captive in front of them. This clearly states the captive can only see one hat, that of the person directly in front of him/her. Assuming the captives had time to discuss a plan before this happened, there is a simple solution. When guessing the color of one's hat, all the captive can make some signal of the next person's hat. An example would be to say, "Uh" before you guess if the next hat is black, and nothing if it is white. Right you are. I do think the version where each captive can see everyone in front of them is more interesting though. I would still use the same method, as it is by far the simplest.
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Posted: Sun Sep 09, 2007 7:58 pm
Dave the lost Dewdew jestingrabbit Dewdew Nobody's solved my riddle yet (my maths professor told us this one) So I'll post it here Quote: Okay, so there are ten people on a boat out at sea, when pirates come along and take them captive. Now the pirates aren't your ordinary pirates who just kill everyone, they like to give everyone a chance to live. So they line everyone up in such a way that each captive can only see the captive in front of them. The pirates than place either a black or a white hat on each captives head. They have no idea what the colour of the hat on their head is, only the colour of the person standing in front of them. Then, starting from the captive at the back of the line, they ask each captive to guess the colour of their hat. If they guess right they are allowed to live, if they get it wrong they have to walk the plank. The question is what can the captives do to guarantee that at least 9 out of ten of them would survive? This is one I like. I'll white the answer. The person at the back of the queue, who speaks first, can tell whether there are an even or an odd number of black hats. If there is an even number, they say black, and if there is an odd number they say white. Everyone else then knows whether there are an odd or even number of black hats on the remaining nine heads. If everyone behind them, except the last person in the queue, has answered right, they know how many black hats were behind them, and they know how many are in front. With this info they know the color of their hat.
So at most one person dies.Nice puzzle, seen it before though. Your answer does work. My answer was different, though I was never told the answer I figured one out myself. My answer is based one the fact that there are actually four different statements each person can make. I would be interested to hear your answer. okay my answer in white. Because there are only two possible colour for the person's hat they have four possible statements "My hat is white" "My hat is not white", "my hat is black" "My hat is not black" (This is based on logic which says if it is not true it must be false. So everyone can see the colour of the hat in front of them. The first person makes a statement with the colour of the person in front of thems hat. E.g if the person in front is wearing a white hat they could say "my hat is white" or "my hat is not white". They have a 50% chance of dyeing.
The next person now knows they are wearing a white hat, so they will ever say "My hat is white" if the person in front of them's hat is white or "My hat is not black" if the person in front of thems hat is black. and so forth.
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Posted: Mon Sep 10, 2007 10:56 pm
...?
By that reasoning each person could say, "My hat is not pink" and live.
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Posted: Tue Sep 11, 2007 8:01 am
Layra-chan Riddle stolen from a friend: You have n 3-state switches, and a 3-state light bulb (say, white, red, and off). The switches are rigged up so that changing the states of all of the switches will change the state of the light bulb. Show that one switch controls the light bulb regardless of the state of the other switch. Do you mean regardless of the state of the other switches? Hmm... I'm trying to build a little model of this but I'm not sure if I fully understand the setup. Would it look something like this for example?  Where n = 4 ?
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Posted: Tue Sep 11, 2007 4:27 pm
zz1000zz ...? By that reasoning each person could say, "My hat is not pink" and live. ' No they couldn't as pink is not an option, By saying "My hat is not white" they are eliminating the option the their hat is white, which means that their hat has to be black. If they said "My hat is not pink" they are eliminatin the option that their hat is not pink, but this was never an option. The option was always between black and white. if you want to be precise they could say "My hat is not white, therefor it must be black"
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Posted: Tue Sep 11, 2007 6:58 pm
Dewdew zz1000zz ...? By that reasoning each person could say, "My hat is not pink" and live. ' No they couldn't as pink is not an option, By saying "My hat is not white" they are eliminating the option the their hat is white, which means that their hat has to be black. If they said "My hat is not pink" they are eliminatin the option that their hat is not pink, but this was never an option. The option was always between black and white. if you want to be precise they could say "My hat is not white, therefor it must be black" Which is effectively the same method as mine.
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Posted: Tue Sep 11, 2007 11:19 pm
Morberticus Layra-chan Riddle stolen from a friend: You have n 3-state switches, and a 3-state light bulb (say, white, red, and off). The switches are rigged up so that changing the states of all of the switches will change the state of the light bulb. Show that one switch controls the light bulb regardless of the state of the other switch. Do you mean regardless of the state of the other switches? Hmm... I'm trying to build a little model of this but I'm not sure if I fully understand the setup. Would it look something like this for example? Where n = 4 ? Right, switches. So flipping all of the switches to b would make the light change to, say, blue, and flipping all of the switches to c would make the light change to, say, green, and flipping some of the switches to b and the rest to c would make the light change color, and so on. It ends up that only one switch needs to be toggled to change the light, but of course the trick is to prove that.
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Posted: Tue Sep 11, 2007 11:25 pm
zz1000zz Dewdew zz1000zz ...? By that reasoning each person could say, "My hat is not pink" and live. ' No they couldn't as pink is not an option, By saying "My hat is not white" they are eliminating the option the their hat is white, which means that their hat has to be black. If they said "My hat is not pink" they are eliminatin the option that their hat is not pink, but this was never an option. The option was always between black and white. if you want to be precise they could say "My hat is not white, therefore it must be black" Which is effectively the same method as mine. The entire problem here is that linguistically, there are many options, but semantically there are only two relevant options, as far as the individual is concerned. For instance, one could make the four choices: "Black" "My hat is black" "White" and "My hat is white" and still impart the necessary information. One could in fact say: "My hat is white and the hat in front of me is black" or variations thereof. The puzzle only becomes interesting when the linguistic options are limited to the semantic options; otherwise it's more an exercise in English more than one of logic.
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Posted: Wed Sep 12, 2007 7:08 am
Layra-chan Morberticus Layra-chan Riddle stolen from a friend: You have n 3-state switches, and a 3-state light bulb (say, white, red, and off). The switches are rigged up so that changing the states of all of the switches will change the state of the light bulb. Show that one switch controls the light bulb regardless of the state of the other switch. Do you mean regardless of the state of the other switches? Hmm... I'm trying to build a little model of this but I'm not sure if I fully understand the setup. Would it look something like this for example? Where n = 4 ? Right, switches. So flipping all of the switches to b would make the light change to, say, blue, and flipping all of the switches to c would make the light change to, say, green, and flipping some of the switches to b and the rest to c would make the light change color, and so on. It ends up that only one switch needs to be toggled to change the light, but of course the trick is to prove that. Last question (hopefully). So you mean only one switch is actually connected? And if any of the other switches were similarly connected, then the condition [changing the states of all of the switches will change the state of the light bulb] could not be met? Or do you mean if we have this condition [changing the states of all of the switches will change the state of the light bulb] then we can configure the switches in such a way that only one needs to be turned to change the bulb?
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Posted: Wed Sep 12, 2007 10:16 pm
Layra-chan zz1000zz Dewdew zz1000zz ...? By that reasoning each person could say, "My hat is not pink" and live. ' No they couldn't as pink is not an option, By saying "My hat is not white" they are eliminating the option the their hat is white, which means that their hat has to be black. If they said "My hat is not pink" they are eliminatin the option that their hat is not pink, but this was never an option. The option was always between black and white. if you want to be precise they could say "My hat is not white, therefore it must be black" Which is effectively the same method as mine. The entire problem here is that linguistically, there are many options, but semantically there are only two relevant options, as far as the individual is concerned. For instance, one could make the four choices: "Black" "My hat is black" "White" and "My hat is white" and still impart the necessary information. One could in fact say: "My hat is white and the hat in front of me is black" or variations thereof. The puzzle only becomes interesting when the linguistic options are limited to the semantic options; otherwise it's more an exercise in English more than one of logic. I agree wholeheartedly, but unless you reword the problem, there is no logical method that would work.
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Posted: Thu Sep 13, 2007 1:23 am
Morberticus Layra-chan Morberticus Layra-chan Riddle stolen from a friend: You have n 3-state switches, and a 3-state light bulb (say, white, red, and off). The switches are rigged up so that changing the states of all of the switches will change the state of the light bulb. Show that one switch controls the light bulb regardless of the state of the other switch. Do you mean regardless of the state of the other switches? Hmm... I'm trying to build a little model of this but I'm not sure if I fully understand the setup. Would it look something like this for example? Where n = 4 ? Right, switches. So flipping all of the switches to b would make the light change to, say, blue, and flipping all of the switches to c would make the light change to, say, green, and flipping some of the switches to b and the rest to c would make the light change color, and so on. It ends up that only one switch needs to be toggled to change the light, but of course the trick is to prove that. Last question (hopefully). So you mean only one switch is actually connected? And if any of the other switches were similarly connected, then the condition [changing the states of all of the switches will change the state of the light bulb] could not be met? Or do you mean if we have this condition [changing the states of all of the switches will change the state of the light bulb] then we can configure the switches in such a way that only one needs to be turned to change the bulb? The system acts as if there is only one switch connected to the light.
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