In the game of mancala there is a board with 2 rows of 6 holes.(To make this easier imagine these rows travelling from left to right)

In each of these holes there are 4 gems. Also, At the end of each row is a special cup. (this cup is not included as one of the 6 holes) Everytime you drop a stone into this cup you guarantee yourself that gem.

In order to get gems into that cup you must pick up all the gems in any 1 hole and drop 1 gem in every hole ahead of it in order from left to right. If you drop it in your rightmost hole you then drop 1 in your cup and proceed dropping along your opponents cups from right to left.(still from your point of view.) For those who are still confused you are pretty much picking up all the gems in one hole and dropping them one at a time counter-clockwise in every hole and cup you come across.

If the last gem you drop lands in your cup you get to go again.

So the question I ask you is this: What is the maximum number of gems that can be on your side that can be sunk on 1 turn(including everytime you get to go again)? Also, how would they have to be set up?

(Still not clear on the rules? Play it here)

Hey, I don't see a solution to this problem using the criteria that you have to have no beads when you're done, so mine is as follows:

starting from 3/1/16/2/16/7, (44 beads), you can put 26 in your cup and end with 0, putting 18 in the opponents cup. The other 4 start between your cup and your opponents board.

Note: following the rules of mancala, as provided in the link, you would just need 47 beads in one of your opponents first five cups from the left, and one bead in your cup of value (6- # of opponents cup with beads) from the left. Then you could simply take all of the 47, and your 1, but moving forward 1 square.

(I'm a big mancala fan)

~Ryan

Posted by Ryan on 2003-05-07 07:00:44