Abalone is a 2-player board game. Here's a brief explanation from Wikipedia:

The board consists of 61 circular spaces arranged in a hexagon, five on a side. Each player has 14 marbles that rest in the spaces and are initially arranged as shown below.

For each move, a player moves a line of one, two or three marbles one space. The move can be either in-line (serial in respect to the line of marbles) or broadside (parallel to the line of marbles), as illustrated below.

Black opens with a broadside move White counters with an in-line move


There is only one set of marbles on the board with initial state as above picture (it doesn't matter which color you choose). At least how many moves does it takes to migrate all your marbles to your opponents initial space?

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    $\begingroup$ Isn't this question for boardgames.stackexchange.com ? $\endgroup$ – klm123 Oct 30 '14 at 9:42
  • $\begingroup$ that's right. I didn't know SE has boardgames branch :) $\endgroup$ – Rafe Oct 30 '14 at 9:44
  • $\begingroup$ I have asked this question there, but seems there is no activity there, like here. too many unanswered questions.... here's a link btw: boardgames.stackexchange.com/questions/20522/abalone-migration $\endgroup$ – Rafe Oct 30 '14 at 9:57
  • $\begingroup$ yes, it is hard decision where to ask such questions:) $\endgroup$ – klm123 Oct 30 '14 at 10:04
  • $\begingroup$ are oppenent marbles in place or there are only your marbles on the game? what happens if different rows merge? You still consider them separated or you have to move the as a single row from there on? $\endgroup$ – Mauro Sampietro Oct 30 '14 at 10:46

Solution in 30 moves.

enter image description here

The number at the top left of each board is the number of moves from the previous position.

The last step repeats all movement symmetrically backwards on the other side.

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    $\begingroup$ that was awesome! very good $\endgroup$ – Rafe Nov 4 '14 at 7:19

You cannot do it in any less than 30 moves.

If you were to move each marble individually, you'd need 88 moves to get them all to the other side.

  • 4 moves each for moving the row of 3 marbles to the corresponding position on the other side
  • 6 moves each for moving the row of 6 marbles to the corresponding position on the other side
  • 8 moves each for moving the row of 5 marbles to the corresponding position on the other side

If we can move a maximum of 3 marbles at a time, theoretically it would take us 30 moves to get them all there. 29 moves would move 3 marbles 1 space for a total of 87 spaces. One more move for the last marble would be sufficient.

So, the question is, can this be achieved?

  • $\begingroup$ I found an obvious way for 32. However, this implies a set of two marbles needing to be moved 8 times. If any lower then 32 (so 31 or 30 is achievable), one would need to figure out 'cheaper' methods to fill the top row, which I doubt there's many options for. $\endgroup$ – Tim Couwelier Oct 30 '14 at 11:26
  • $\begingroup$ @TimCouwelier Yes - saw your answer. I agree - I don't see many ways to keep moving units of three. $\endgroup$ – Trenin Oct 30 '14 at 11:30

31 Move Solution

While this is (possibly) not optimal as 30 is the theoretical lower limit...

The solution below uses only 31 moves...

The trick to it is really in the first 2 moves...

enter image description here

  • $\begingroup$ Given this beats my 32-move solution, I deleted mine. Neat find at the start. Could there be a way to move the 3's past the 2, so at the end you can reverse the trick from the start? $\endgroup$ – Tim Couwelier Oct 30 '14 at 15:47
  • $\begingroup$ I tried to find one... no luck yet, and I cant prove it to be impossible yet either... $\endgroup$ – RTL Oct 30 '14 at 19:08
  • $\begingroup$ For 30-move solution then we need to find a position which can be reached in 15 moves from either side, and exactly two moves moving two marbles (or one move moving one marble, but I don't think this move will be in the solution, since it would make it not symmetric). $\endgroup$ – justhalf Oct 31 '14 at 5:00
  • $\begingroup$ It doesn't need to be symmetric. This example isn't symmetric - you could either perform the trick at the beginning, or (if you look at the boards in reverse) at the end. $\endgroup$ – Trenin Oct 31 '14 at 14:34
  • $\begingroup$ I agree symmetry is not required, but if possible it would save a lot of the work... Also I agree that using this trick at the end would work however I don't have a way to get into this position in 28 moves (for the last 2 moves to be the mirror of the first). $\endgroup$ – RTL Oct 31 '14 at 14:39

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