The problem is that your opponent can place a marble anywhere and removes any ring at the edge. This results in many different situations. (27 alternative places to place the ring, and 16 alternative rings to remove gives a total of 27*16=432 different possiblities.) So let's look at the possibilities in general. You can offer up to five marbles and remove six rings in order to isolate two white marbles without loosing the initiative. You can either isolate the two white marbles on the same side or one on each side. This works well, in fact I still haven't found a case that breaks the theory. Let's suppose you've used the red strategy. Here's a few examples on what to do: Your opponent places any marble on f2 or g2 and removes any ring but a4, d3, e1 and f1: Note that in this case you isolate one white marble on each side of the board. Your opponent | You | 15. | Nf2/g2,nn | 16. | Ng2/f2,b5 | 17. | xg2Ne2 | 18. | Wd3,b4 | 19. | xd3Nf1 | 20. | Ne1,d2 | 20. | xf1Nd1 | 21. | Wa4,e1 x Wa4, Wd1 |
Your opponent places any marble on d6 or e6 and removes any ring but a4, b5, c6, d7 and f5: Note that in this case you isolate the two white marbles on the same side of the board. Your opponent | You | 15. | Nd6/e6,nn | 16. | Ne6/d6,b4 | 17. | xe6Nb5 | 18. | Nd6,c4 | 19. | xb5Nf6 | 20. | Wf5,g4 | 20. | xf5Nd7 | 21. | Nc6,c5 | 22. | xd7Nb5 | 23. | Wa4,c6 x Wa4, Wb5 |
Your opponent places any marble on e5 or f5 and removes a4: Note that in this case you isolate the two white marbles on the same side of the board even though your opponent removed the uttermost corner on that side. Your opponent | You | 15. | Ne5/f5,a4 | 16. | Wf5/e5,c4 | 17. | xf5Nd5 | 18. | Ne5,d4 | 19. | xd5Nf5 | 20. | Ng4,d5 | 20. | xg4Ne6 | 21. | Wf5,g4 | 22. | xf5Nd7 | 23. | Nc6,c5 | 24. | xd7Nb5 | 25. | Wb4,c6 x Wb4, Wb5 |
Have fun! Karina |