EventHorizon

Reasoning:

Some maximum permutations/combinations Some code (yay for global variables.... yeah... i was lazy) code also attached. main.cpp Example usage

So people don't need to register for an account at mathhelpboards.com to see those images...

Looks like a variation of the Siamese method for constructing n-odd magic squares.

How about...

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.

My thoughts and solutions:

You got it. Well done. I'll post my solutions later.

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 5-8 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.

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.

Perhaps this question will help move things along...

Can anyone find the groundhog using less than 5 hunters?

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.