81–90 of 220 posts
#2 is super interesting! I wonder if it generalizes to higher dimensions.
I don't quite follow the solution, can anyone explain a bit more? What's the motivation for considering large epsilon?
0 The limit for ε to infinity must be >= 0 and is ((a+b+c) - (a'+b'+c'))π, therefore a+b+c >= a'+b'+c'.
"large epsilon" is a standard way to express "consider the asymptotic behavior for epsilon to infinity and you will see what I mean".
Earlier quoted context omitted.
Alternative solution: send her a model of the key to 3d-print
How are you sending the model safely?
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…
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 awhile when I first heard it.
Thank god I could solve "Love in Kleptopia". Would have been embarrassing being a founder of a security company.
This, aside from the fact that the key on the hasp is not "inside a padlocked box" ...
One that I heard last week: There is an 8x8 checkerboard in a room with a coin placed on each square. Each coin is either facing heads or tails up, and the face is determined randomly. Before you can inspect the board, a "master" comes in, picks a square of interest, and must make a manipulation to the board by flipping one of the 64 coins. He then exits the room. You are now allowed to enter, and must read out which…
I think you're referring to this puzzle: http://datagenetics.com/blog/december12014/index.html The version you gave is missing information and so can't be solved as stated.
Earlier quoted context omitted.
You don't actually need to label them you just need all the prisoners to be able to memorize which name is associated with which box, Alice's box is the first on the left, Bob's the second etc... Then when Zach. Z. Z. Vanderwall opens "his box" and finds the name Bob he follows procedure. You need a minimal perfect hash function but it's a thing that can be done, especially because prisoners in these sort of conundru…
And they all do what you, The Nerd, say. Prisoners are known to be well-behaved and respect Nerds! /s
Me: "Here's a plan that will defeat the werewolf in any case, even if I am the werewolf"
Them: "That sounds complicated, therefore suspicious, let's hang him"
I think there's a simpler solution to boxes in boxes. Assume that the outer box is axis-aligned. Project both boxes onto the X axis, so they become one-dimensional. It's easy to check that the "projected perimeter" of the inner box is smaller. Repeat for axes Y and Z. From that and the triangle inequality, the result follows.
Earlier quoted context omitted.
Alternative solution: send her a model of the key to 3d-print
How are you sending the model safely?
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…
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 terminate).