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)

I started with the obvious of 6,5,4,3,2,1 and by working my way from right to left I could get 17 gems in the cup (leaving me with 0,1,2,0,1,0).

Then I thought, what if instead of having 1 in the rightmost hole I started with 14 which is enough to right round the board and drop the last gem in the cup. This led me to 5,4,3,2,1,14 which gives 2 gems in the cup on the first move and then a second move starting with 6,5,4,3,2,1 - ie my total was now 19.

Extending this further I got to 3,2,1,16,13,10 which puts 23 gems in the cup. (ie a 4th move starting with 6,5,4,3,2,1 and 6 gems already in the cup). This strategy can't be extended further as there are only 48 gems in total.

Posted by fwaff on 2003-01-21 22:24:37