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Help Titus again in this sequel to Azada, or meet him for the first time in this stand-alone game. This even holds true for a 1 X infinity grid. 76.190.157.141 18:18, 21 June 2009 (UTC) at infinity*infinity the first player has wining strategy: if we mark the rows with 1,2,3,4 and columns with 1,2,3,4 the first step is (2,2). then if the opponent makes step (a,b ) you just make (b,a ). a or b will be 1 but this doesn't matter, after your turn the table goes simmetric, and after your secound step, the table goes finite, so after finitely many steps you win. for a*infinity i don't have the general solution, but the answer is: NO here is a (nontrivial) example: 2*infinity: with the same markings: a) If first player goes (1,x) the table is a finite rectangular table: 2nd have winning strategy. b) If first player goes (2,x) 2nd player will take (1,x+1) and so on: for every first player's step (a,b) 2nd player answers (a-1,b+1) or (a+1,b-1) [ only one of theese will exist in the table so that's a bijection for every field exept the (1,1) what is a loosing place ] - protos_drone 2:30, 01 Aug 2009 for a*infinity: a=1 -> first player has winning strategy. a=2 2nd player has winning strategy (like mentioned above) a>2 first player has winning strategy: he makes the field a 2*infinity, after that he wins with the above mentioned strategy. tha answer to the 2nd question: YES! consider an infinity*infinity table but only the following places are free to choose: (1,1),(1,2),(2,1),(2,2) ( 2X2 rectangle ) and the places (1,n) and (n,n) Tha table has only 1 infinitely long row/column: the first row, and any place is a loosing position because you can win the game by replacing (a,b) places to (a,2) if b>1 so you get the 2*infinity table, where the 2nd have wining strategy, and every step has the same result at the 2*infinity table and at the first mentioned table.( actualy you couldn't make it from the infinity*infinity table with regular steps, but the question doesn't included that it used to be :D ) Sorry, but I don't understand the last part.

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Nevil Maskelyne

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Sophie Blackall

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Hen" Fetsch

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Alberto Nietsnie

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