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The Flying Dutchman is a ghost ship legend from the seventeenth century. Legend says that this ship was never able to make port, but doomed to sail the seven seas forever. In Mur, a player can execute a move which is called a 'Flying Dutchman'. It is a seemingly absurd move which leaves a ship in a position to be marked (trapped). However, similar to the 'poisoned pawn' position in chess, if the opponent dares to trap the abandoned ship (therefore sending the ship back to port), the consequence is a winning game position in favour of the player who played the absurd move! Now usually a Flying Dutchman is a single: meaning that the trap score is usually 2-1 and the player with the higher score sees that by allowing his opponent to trap one of his ships the resulting game position will allow him to get that last trap he needs to win; the term 'Double Dutch' is applied in the case where two traps are planned after a point is given over. However, it is possible, during a 0-0 or 0-1 score to play an absurd move and allow your opponent to trap one of your ships so that the resulting game position is so favourable that it results in you winning the entire game! Such a play is called a 'Triple Dutch'. Kindly watch the following short video:



The trap score for this game so far is 0-0. It is Black to play. Black is in danger of having his c1 ship trapped by White. It seems, then, that the best play would be for Black to move 1. b3*a-h X (Black using his b3 piece to knock White's piece on a3 to intersection h3) and after White is forced to play the withdrawal 2. Xh3 to then move his c1 piece to b1, d1 or 0 so that it is safe and cannot be trapped.


However, the best move is a seemingly absurd move which completely ignores White's threat on Black's c1 game piece. The best move is 1. h2-f!! (Black moves his h2 game piece to intersection f2). White assumes that Black has blundered and immediately traps Black's c1 piece with 2. +b1 X (White places his off-board ship on b1 and acquires his first trap for this game). Black is forced to withdraw with 3. Xc1. It is now White's turn. White sees that since Black has withdrawn a game piece, Black may trap the kraken and win if he plays 5. +d2 XXX. White must prevent this by occupying the d2 intersection. White has only two options. Let us call them option A and option B. Option A is to move his c2 ship to d2. 




White decides to play 4. c2-d and so now it is Black's turn. Black is in a position to force a series of moves resulting in three traps on White. Black begins his assault by placing his recently withdrawn ship on c2 with 5. +c2; this placement is an attack on d2 and a threat to game by marking (trapping) the kraken. To prevent Black from marking the kraken, White has no choice but to promote his d2 piece to an alpha and the only way to do this is to move his d1 piece to h3. This move is recorded as 6. d1-h3. Black now traps White's a3 piece by knocking it into his eye on f3 with 7. b3*a-f X and acquires his first mark (trap) of the game. White is forced to withdraw with 8. Xf3. Now it is Black's turn again. Black now acquires his second trap of the game with 9. h1-2 XX and White is forced to withdraw with 10. Xh3. It is now Black's turn again. Black now threatens White's g2 ship by moving his e1 alpha piece to the center of the board with 11. e1*0-a1 which threatens 13. 0-g1 XXX game. White, then, is forced to play 12. g2-g1. It is Black's turn again. Black now places White's a2 piece in a pocket by moving his c2 piece to b2; this is recorded as 13. c2-b. Now nothing can prevent Black from acquiring his third trap on a2 to win the game. Even if White were to play 14. g1-h1 then Black would create a mixed traps position with 15. 0*a1-0 and White would be forced to withdraw with 16. Xa2.


Let us go back now and look at option B:



In option B White decides to play 4. d1-d2 (an offside move since a count of three may only be reached if the player counts off the edge to acquire his third count). It is Black's turn. Solution: Black plays 5. e3-d! threatening 7. e1*2-3 . . . 9. g3-f XXX and if White plays 6. g2-h then 7. e1*K-3, 8. a3-f, 9. d3*K-h3 X, 10. Xf3, and 11. b3-a is a winning position since White's a2 piece is blocked. White then, having no choice, plays 6. c2*K-h (using his c2 piece to knock the kraken to h2). Black pushes the kraken to h3 with 7. h1*K-3. White, needing to promote a3 so as to prevent Black from gaming on h3, is forced to play 8. a2-c. Now, by re-entering his off-board piece,  Black seals a winning position with 9. +a2 (if White's a3 moves or knocks the kraken, Black can win); Black can now play 11. g3*h-c X followed by 12. Xa3 and next 13. d3*K-a XXX (Black uses his d3 piece to knock the c3 kraken to Black's eye on a3). Note: Should White play, in response to 9. +a2, 10. 0*c2-3 then Black marks with 11. a2*c-aX. After White withdraws with 12. Xc3, Black games with 13. g3*h-cXXX.


In conclusion, this small extract from the billions of possible game positions is certainly intriguing. If the material on a Mur board consists of 7 Black, 6 White and the kraken, the resulting number of game position possibilities are 13,385,572,200. Add exactly 13,385,572,200 more for the inverse (6 Black, 7 White and the kraken). If there were 7 Black, 7 White and the kraken, then the number of possible game positions reaches 21,034,470,600. If the material is 6 Black and 5 White and one kraken then you're looking at 3,603,807,900 more positions and another 3,603,807,900 for the inverse of that. Basically there are over 54 billion possible ways to arrange the game pieces on a Mur board; with each position there are nine different possible trap scores. Within the hundreds of billions of possible game scenarios, how many of them are triple dutch positions?

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