
Content Count
546 
Joined

Last visited

Days Won
5
Content Type
Profiles
Forums
Calendar
Gallery
Blogs
Everything posted by EventHorizon

Burning Rope Question (only one rope)
EventHorizon replied to matohak's question in New Logic/Math Puzzles
How about... 
Groundhog in a Hole  Yet Again
EventHorizon replied to EventHorizon's question in New Logic/Math Puzzles
While the hole that the groundhog starts in is unknown, it is one specific hole and its movements are as described. No matter what hole it starts in, given its described movement, the strategy given will eventually find it. You make it seem like the groundhog can teleport just because it's starting hole was unknown. Given your "about 17 holes outside" example, it would take just one or two (depending on parity) rounds of area clearing/expansion to get it. It does not matter that there are more holes outside the area, it will eventually be trapped and caught. 
Groundhog in a Hole  Yet Again
EventHorizon replied to EventHorizon's question in New Logic/Math Puzzles
My thoughts and solutions: 
Groundhog in a Hole  Yet Again
EventHorizon replied to EventHorizon's question in New Logic/Math Puzzles
You got it. Well done. I'll post my solutions later. 
Groundhog in a Hole  Yet Again
EventHorizon replied to EventHorizon's question in New Logic/Math Puzzles
Unfortunately, that does not work for the known parity case. Notice day 4 checks even holes, and day 5 also checks even holes. So days 58 are checking the wrong parity. Fixing this, the numbers change such that day 4k+x checks hole 4k+y, so the groundhog can escape by fleeing away from hole 0. 
Groundhog in a Hole  Yet Again
EventHorizon replied to EventHorizon's question in New Logic/Math Puzzles
There is no end to start from. Every hole has an infinite number of holes on both sides of it. And, yes, both you and the groundhog are immortal. Your friend, not so much. Luckily, you'll always be able to enlist a member of his posterity to help check one hole each day, so close enough. 
Groundhog in a Hole  Yet Again
EventHorizon replied to EventHorizon's question in New Logic/Math Puzzles
Perhaps this question will help move things along... 
Groundhog in a Hole  Yet Again
EventHorizon replied to EventHorizon's question in New Logic/Math Puzzles
Can anyone find the groundhog using less than 5 hunters? 
Groundhog in a Hole  Yet Again
EventHorizon replied to EventHorizon's question in New Logic/Math Puzzles
Negative infinity to positive infinity. You could pick any hole to be the origin. 
There is an infinite line of holes in an infinite field. You know there is a groundhog hiding in one of the holes. You can check a hole once a day. At night the groundhog will move one hole to the left or one hole to the right. Knowing you cannot find the groundhog yourself, you enlist a friend to help. Now you can check 2 holes a day. Can you find the groundhog? If so, how would you do it? Original: http://brainden.com/forum/topic/11943/ My Additions: http://brainden.com/forum/topic/12010/

I went and found the first time I encountered this puzzle: http://brainden.com/forum/topic/11943/ Here's my additional puzzles on the same theme: http://brainden.com/forum/topic/12010/ I just thought of an additional one... will post shortly :)

Beat me to it, kudos CaptainEd. I guessed the solution and came up with a proof while trying to get to sleep last night. The first step was inelegant, so I was going to work on it a bit before posting. This problem reminded me a lot of


I'd say that the minimum value of f is slightly less than... I'll play around with it a bit to see how much I can lower it. Another interesting addition might be, once the minimum f is found, to find the minimum travel distance (e.g., amount of gas) needed for some f a little higher than the minimum.

addendum: I didn't come up with this myself. Monty Hall was asked about the Monty Hall problem, and his response was such.


n factorial (represented as n!) is the product of all positive integers up to n. So the number you are looking for is 1,000,000! ("one million factorial")
 1 reply

 1

I'm sure I know what BMAD was saying. Yes, ignore the units / squared units difference. The problem doesn't involve a circle though. You need to find a function f(x) such that the area under the curve between two points is the same as the arc length between those two points along the function. Here the arc length is like the length of a string representing the function cut at the two points. How to find the arc length of a function: Restatement of problem using previous spoiler: One answer is simple, the other is less so.
