This week's book giveaway is in the OCAJP 8 forum. We're giving away four copies of OCA Java SE 8 Programmer I Study Guide and have Edward Finegan & Robert Liguori on-line! See this thread for details.
I was thinking about this after the thread here. It the best tradition of PD threads, this will be vastly under specified and you'll have to make it up as you go.
The Problem: imagine a large (2x2x2 meter) cube made from wood with wire mesh sides. Add (but don't overfill) some chickens. Each day the chickens are moved to a new grazing area by tipping the cube onto one of its sides. We also have this additional criteria:
the chickens poop on the wire, so all sides should be used before reusing a particular side as the base for poopin' on.
the grass should be given a chance to regrow, so a patch of grass shouldn't be reused immediately.
What I'm expecting is either a solution that starts back where it started 6 days later, or a solution which defines an 'elegant repeatable pattern', whatever that means.
This may not be as interesting as I imagine, but hopefully it will stir some brain cells.
I do not think it is possible to end up where you started in a single cycle of 6(+1) without using the same side twice. As far as patterns, my favorite (so far) is a 21 day cycle: __EF__ _CDGH_ AB__IJ TS__LK _RQNM_ __PO__
You could put a gazebo in the middle. [ December 10, 2004: Message edited by: Marc Peabody ]
I don't think you can either using the rule ~ tip the cube on one side. That is because the gazebo has 6 sides and therefore 60 degrees angles (360/6) on the inside of each intersection. The outside angle of each intersection is 120 degrees (180 - 60). However, when the cube is placed on an edge of the gazebo and tipped over, it will make a 90 degree angle.
I was trying to thnk of a way to weight the use of each side and then use those results to chose the next rotation. But I quickly am face with a tie and don't know which side to turn next. :roll:
Sonny, I don't follow what you are saying about the gazebo.
If you look at my 'map' that I drew of cube placings you will see that there is space left in the middle, as the cycle make a bit of a donut shape. The sides/angles of that object does not make a difference.
The problem posted assumes the cube's sides to be 2x2 meters each. That would make the whole in the middle 4x4 meters. As long as the gazebo is not more than 4 meters wide it will fit. You might have to walk through a bit of chicken poop to get to it though.
Sonny, you seem to be saying that the gazebo is a hexagon, and the space in the middle of Marc's map is a square, therefore the gazebo won't fit. Marc however never said that the gazebo had to fill up all of the available square. So, you can put a gazebo in the middle, as long as it's smaller than the square space available.
"I'm not back." - Bill Harding, Twister
Joined: Jan 30, 2000
Can you describe you map?
Ummm, he's already posted it above. Here's another version, using a monospace font and '+' to indicate the corners of the squares.
[ January 07, 2005: Message edited by: Jim Yingst ]
Each letter represents a position for the cage and it will remain at each position for the span of one day. The cube starts in position 'A' and goes through the alphabet A,B,C,D... After reaching 'T' it will return to 'A'. [ January 10, 2005: Message edited by: Marc Peabody ]