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please help, its damn urgent .it is the trickiest of all!
 Imagine you are playing a new version of the game of chess, called x chess. the rules are same, with just one exception. You have to play 2k moves at a time, where k is any natural number. Prove that, black who plays second always has a non-losing strategy.

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You posed this question yesterday.  I am confused by the rules.  Tell me if I am understanding you
Lets choose k=4.  Then each side can play 2k=8 moves at a time.  White can start out with 
8)Any legal move at all (but don't move either Knight)
This places Black in double-check position (both Knights are checking the black king).  Since the Black King cannot get out of check on his 1st move (ie can't take both White Knights in one move), then Black has been checkmated and the game is over.
Therefore Black does NOT have a non-losing strategy.
Perhaps I'm missing something?
Perhaps you are implying that Black has 8 moves to 'get out of check' in which case both knights can be taken and the game can continue.  If that's the case, this is indeed a very difficult problem to prove.  I'll continue thinking about it.