2D Mancala

When I first learned how to play Mancala, I was quite fascinated by the possibilities that opened up when your last stone fell into the home pit, giving you another move in the same turn. I quickly realized that a whole series of moves could be made, given the right layout of stones and judicious play (much to the chagrin of my girlfriend at the time).

Clearly, if you have one stone in the pit closest to the home pit (Pit 1), you can play it and get a free move. If you have one stone in Pit 1 and two in the next pit (Pit 2), then you could have two free moves, allowing you to clear three stones (by playing Pit 1, Pit 2, and then Pit 1 again, in that order). It thus seemed logical to me that one could clear six stones by having three in Pit 3, two in Pit 2 and one in Pit 1, and so on.

Alas, it is not that simple. The most stones that can be cleared with three pits is five (three stones in Pit 3 and one each in Pits 2 and 1). However, the patterns that arise from the question of, “What is the maximum number of stones that can be cleared in one turn (with free moves), using a given number of pits?” are quite interesting (to me, anyway; see my article about it here).

So I started thinking about changing the game. The standard Mancala board has six pits (on each side), which can be viewed as one row of six pits (duh). What might happen if there were more than one row? Say, two rows of pits, or 22 rows. Would it then be possible to clear a board with n stones in each pit of Column n? I kept the rules basically the same: in each move, harvest all the stones in one pit and sow one stone in each column to the right. You can sow a stone into any pit in that column, but you must sow one and only one stone per column. If the last sown stone lands in the Home pit, you get another move.

The cases for one and two columns of pits are easy: you only need one row to clear one stone from one pit or three stones from two pits (two in Pit 2 and one in Pit 1). At three columns, it becomes interesting. Two rows are needed to clear six stones in a row (three in each pit in Column 3, two in each pit in Column 2, and one in each pit in Column 1). Here’s one way to do it:
  1. From row 2, column 1 to home
  2. From row 2, column 2 to row 2, column 1 and to home
  3. From row 2, column 1 to home
  4. From row 1, column 1 to home
  5. From row 1, column 2 to row 1, column 1 and to home
  6. From row 1, column 1 to home
  7. From row 1, column 3 to row 1, column 2, to row 1 column 1, and to home
  8. From row 1, column 1 to home
  9. From row 2, column 3 to row 1, column 2, to row 1, column 1, and to home
  10. From row 1, column 1 to home
  11. From row 1, column 2 to row 1, column 1 and to home
  12. From row 1, column 1 to home
Four columns of pits (with four stones in each pit in Column 4 down to one each in Column 1) takes six rows, while five columns only takes four. The six-column case seems to be where my computational strategies are breaking down. So far, all I’ve found is that clearing six columns can’t be done with six or fewer rows. The number of possibilities is just so large that I haven’t yet found a way to efficiently handle them. But I haven’t given up!