Of course a few of these mathematical puzzles assume that the human beings involved are all robotic, and capable of remembering a randomly ordered list 100 long. The blue dot red dot problem being the worst offender because the theorem doesn't hold up of you assume anyone in the village has the wrong internal state.
Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
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Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
#102Earlier quoted context omitted.
I'm mad at myself for not managing to figure this out myself. I guess i made more assumptions about the limitations than there were in the description of the problem.
Same. It was not clear that multiple padlocks can be added to the box.
It took me way too long for this reason. I got it instantly once I realized I had made that limitation up.
I feel like there is some sort of lesson, here.
Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
#103Ah, I love these kind of puzzles! Here's another one, similiar to the first one (Names in Boxes). Apologies for any incorrections in advance. There are 100 prisoners. At random times one prisoner is chosen uniformly at random and led into a room with a single lamp. The prisoner can choose to switch it on or off or leave it as the last visiting prisoner left it. Apart from the state of the lamp he must leave the room…
Let F be the only person who is permitted to turn the light oFF. The N be every body else---they will be turning the lights oN. F starts with a counter at 0. If the light is on when F gets to the room, they turn the light off, and they increment their counter. If the light is off when F gets to the room, do nothing. Each N starts with a counter at 2. When any N enters the room, if the light is on, do nothing, but if…
This is indeed the standard solution. There is also a more difficult version: Every prisoner is required to have the same strategy (so you cannot pick a unique person F). (Every thread runs the same program, and threads do not have access to a unique thread identifier.)
Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
#104Earlier quoted context omitted.
How are you sending the model safely?
I presume email, and that GP misspoke when they said "model" and meant a file that the key could be modeled and printed from. That being said, that solution definitely skirts the intention of the puzzle, but would certainly work given the restrictions the puzzle did list.
model as in 3D model
Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
#105Earlier quoted context omitted.
Let F be the only person who is permitted to turn the light oFF. The N be every body else---they will be turning the lights oN. F starts with a counter at 0. If the light is on when F gets to the room, they turn the light off, and they increment their counter. If the light is off when F gets to the room, do nothing. Each N starts with a counter at 2. When any N enters the room, if the light is on, do nothing, but if…
This is indeed the standard solution. There is also a more difficult version: Every prisoner is required to have the same strategy (so you cannot pick a unique person F). (Every thread runs the same program, and threads do not have access to a unique thread identifier.)
Ugh. This gives me a headache.
Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
#106Earlier quoted context omitted.
It's because you start with the box labeled (via the initial random labeling) with your name. If you cycle back to it then it means that you found your name on a piece of paper, since the next box you open is always the one matching the piece of paper in the last box. So it is impossible to start in a cycle that doesn't include your name.
It assumes, they can secretly assign and remember a 100 items random ordering and then execute it perfectly. Further, it assumes they can decide which orientation is the start vs end of the line. Thus, it's not actually possible, but 'in theory' it seems to work. EX: If I know your going to order based on which side is closet to the entry door nob when the door is closed. Well nothing says they all enter from the sam…
I must be missing something. Couldn't one instead assume that they have a private notebook and a pen? And that they can tell left from right? Seems more reasonable.
Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
#107Ah, I love these kind of puzzles! Here's another one, similiar to the first one (Names in Boxes). Apologies for any incorrections in advance. There are 100 prisoners. At random times one prisoner is chosen uniformly at random and led into a room with a single lamp. The prisoner can choose to switch it on or off or leave it as the last visiting prisoner left it. Apart from the state of the lamp he must leave the room…
Good one! A naive strategy where they never die, and hopefully stay in prison for slightly less than a few billion years: One prisoner is the light-counter (Mr C). Others follow a simple pattern - if the light is off, and they have never turned it on yet, turn it on. If it is already on, or they have already ever turned it on, do nothing. Mr C enters the room, if the light is on, remembers it, and turns it off. When…
You have to be careful, because if the light starts out on, Mr C could be fooled when only 98 other people have entered the room.
Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
#108SPOILER 1 Names in boxes I don't understand how this works. The answer says it works to a certain percentage if there are no cycles longer than 50. But even if chance has it that there are two cycles of length 50. Then it seems the chance would be very large that one of the 100 prisoners would en up in the "wrong" loop and thus not find their name?
It's because you start with the box labeled (via the initial random labeling) with your name. If you cycle back to it then it means that you found your name on a piece of paper, since the next box you open is always the one matching the piece of paper in the last box. So it is impossible to start in a cycle that doesn't include your name.
I don't understand: the solution text claims that some permutations are guaranteed to work every time[1]. But you still have to initially land in one of "your" cycles, right?
[1] "If it happens that the permutation has no cycles of length greater than 50, this process will work every time and the prisoners will be spared."
Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
#109[deleted]
Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
#110SPOILER 1 Names in boxes I don't understand how this works. The answer says it works to a certain percentage if there are no cycles longer than 50. But even if chance has it that there are two cycles of length 50. Then it seems the chance would be very large that one of the 100 prisoners would en up in the "wrong" loop and thus not find their name?
I coded a simulation here if that helps: https://gist.github.com/bjornedstrom/971574557b6f3179db08350...
The first two times I ran it I got 0.298 and 0.286. Out of 20 attempts 5 were below 30%.