151–160 of 220 posts
Thank god I could solve "Love in Kleptopia". Would have been embarrassing being a founder of a security company.
Edit: Ah, charlesdenaul already proposed that. You wouldn't need cryptography, as the problem doesn't state that the postal service also monitors the Internet.
Earlier quoted context omitted.
Can you expand on this last step? it's not obvious to me what you're applying the triangle inequality to and why it gives the desired result.
The length of a line segment can't be greater than the sum of its projections onto all three axes. Therefore the "perimeter" of the inner box can't be greater than the sum of its projections. But that sum is bounded from above by the "perimeter" of the outer box (which is equal to the sum of its own projections, because the outer box is axis-aligned).
What definition of "perimeter" are you using here?
I collect riddles like these! Here are a few of them. Numbered Hats A warden places a hat on the head of each of 100 prisoners, each with a random number from 1 to 100. There may be duplicates. Each prisoner can see everybody's hat but their own. Each prisoner then guesses their own number; if any guess correctly, they all go free. The group may not communicate in any way during the trial, but may strategize beforeha…
EDIT: Happy now? ;P
Earlier quoted context omitted.
>Actually that is incorrect as there is no guarantee that every prisoner has been in the room at least once over any amount of time. IIRC, in the original statement of the problem, there's also a stipulation that, for all prisoners, the king/chooser/whatever will visit them an infinite number of times (so at any time it must be true that each prisoner will be visited [again or for the first time] if the game doesn't…
So? It doesn't have to be true that each prisoner will be visited within any particular time frame.
Earlier quoted context omitted.
I think the puzzle as phrased is slightly incomplete. The stranger must communicate something _new_ each day to make the induction work. For instance, if the number of blues is 25, and the (merciful) stranger says that the number of blues is not prime every day, no one ever has enough information. And in fact, even that doesn't seem to be enough. The non-trivial part must be that at least one person knows more than t…
The clock is certainly necessary. But for information, I think its as easy as "somebody has blue eyes!" said once. If only one person has blue eyes, they see no one else with blue eyes so kill themselves. If two people have blue eyes, they see the other guy and think "maybe its only him". Next day, that guy is still alive; they now know the only remaining blue-eye is themselves; midnight comes and they both suicide.…
The 'dot-town suicides' is a more general version of a puzzle I know, "The Island with Blue-Eyed People". The solution is an induction, which is unusual in these kinds of problems.
I think the puzzle as phrased is slightly incomplete. The stranger must communicate something _new_ each day to make the induction work. For instance, if the number of blues is 25, and the (merciful) stranger says that the number of blues is not prime every day, no one ever has enough information. And in fact, even that doesn't seem to be enough. The non-trivial part must be that at least one person knows more than t…
I think that's what they meant by non-trivial. If the stranger tells them something they already know, that doesn't count.
You can assume that an individual person can count the blue dots they see, but they can't count their own dot. If they count 9 dots, then the true number must be either 9 or 10, so they already know the number can't be prime.
Ah, 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…
>The "easiest" solution would simply be to wait a few billion years Actually that is incorrect as there is no guarantee that every prisoner has been in the room at least once over any amount of time. Of course the probability will get higher and higher that everyone was in the room once if it was completely random but that probability will never be 100%. I know of two legitimate answers to this problem, it took me aw…
Thank god I could solve "Love in Kleptopia". Would have been embarrassing being a founder of a security company.
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.
This accounts for some high percentage of the failures in figuring out brain teasers :)
I collect riddles like these! Here are a few of them. Numbered Hats A warden places a hat on the head of each of 100 prisoners, each with a random number from 1 to 100. There may be duplicates. Each prisoner can see everybody's hat but their own. Each prisoner then guesses their own number; if any guess correctly, they all go free. The group may not communicate in any way during the trial, but may strategize beforeha…
Ah, the Colored Hats is awsome (Haven't tried the other ones, thanks for those)! It gets even more interesting by using three colors instead of two. The solution stays the same in principle, but [guvaxvat zbqhyb guerr vf n ybg yrff boivbhf guna jbexvat jvgu rira naq bqq.] (ROT13) EDIT: Happy now? ;P
Earlier quoted context omitted.
The length of a line segment can't be greater than the sum of its projections onto all three axes. Therefore the "perimeter" of the inner box can't be greater than the sum of its projections. But that sum is bounded from above by the "perimeter" of the outer box (which is equal to the sum of its own projections, because the outer box is axis-aligned).
> Therefore the "perimeter" of the inner box can't be greater than the sum of its projections. What definition of "perimeter" are you using here?