Think of a Chinese Game Board, and how you can make jumps. Now think of actual checkers where you capture pieces. How many marbles can you place without a capture possible?

There's one slight rule to make sure this is not easy (without this rule you could just fill every slot):

You cannot place marbles in the outer rim, but you can, when the game starts, capture in that direction. AKA: A marble can jump over another onto the outer rim, but you cannot place it there at the start.

Also, it's important to note that you only use the big part, the outer triangles are unusable.

Note: Any jump can capture, as long as you jump the same distance on the other side as the current distance, and there are no obstructions. Colour do not matter.

  • $\begingroup$ On a Chinese game board or a vanilla checkers game board? $\endgroup$ – John Meacham Oct 26 '14 at 10:05
  • $\begingroup$ Chinese checkers board. $\endgroup$ – warspyking Oct 26 '14 at 13:55
  • $\begingroup$ Can pieces jump over any piece where jumping twice the distance is unblocked, or only adjacent pieces? $\endgroup$ – Joe Z. Oct 26 '14 at 17:04
  • $\begingroup$ @Joe Z. Any symmetrical jump. $\endgroup$ – warspyking Oct 26 '14 at 17:07
  • $\begingroup$ Symmetrical jumps count stuff like x-oo-* as well, though (where x jumps to *). $\endgroup$ – Joe Z. Oct 26 '14 at 17:09

You can put maximum 13 marbles.


The first thing to note is that you cannot have adjacent pieces. If you have a line of 2 or more adjacent pieces, it has to stop somewhere and it is not a border because the rim is empty. That means the next-to last in a line can always capture the last one.

Without any 2 pieces adjacent, the smallest distance between pieces is as in the solution above. This means that arrangement is the densest possible. The only chance of improvement is a better use of the borders. But the borders are quite well filled already.

  • $\begingroup$ I'll give some more time for others to be 13 if it's possible. $\endgroup$ – warspyking Oct 26 '14 at 23:20
  • 1
    $\begingroup$ Notably, this solution is a maximal independence set of the graph formed by the center of the Chinese checkers board. As Florian F notes in their answer, if any two pieces were adjacent to each other, they could capture each other. $\endgroup$ – Kevin Oct 29 '14 at 5:31

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