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Marco and Leonardo decided to play a game on a checkerbard of 4×4 squares. The board is initially filled with two-sided identical coins.

The game notes that these two players play turns alternatively where the both players are permitted to flip the coins on any 2×2 size square or a line of 1×4 fom side A to side B .

The first person who finishes the game with all 4×4 board pieces turned to the B side wins.

enter image description here

Question:

Prove that player 2 is always bound to win

Rules :

  • After some members' interventions i imposed some rules and reduced the board size to make the problem within any one's grasp .

  • any player cannot repeat his last or his opponent's last move

  • and moreover , any player is always prioritized to take action of biggest number still available of A sided coins ( for example a set containing 3 non flipped coins is a priori to be played than any set having 2 or 1 non flipped coins)

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  • $\begingroup$ Do both players only flip coins from A to B, so no coin is ever flipped from B to A? Or do they flip every coin in the region, whatever its state? Please work on Rule 1. I find it impossible to understand. No ... can't is a double negative, reinverse??? I think it needs revision from when the two players were flipping the same shape rectangles. $\endgroup$
    – Ross Millikan
    Commented Mar 17, 2015 at 21:56
  • $\begingroup$ yes it can be done fom B to A . 2 . if four squares are flipped over , no one can turn them back until one square at least is played with other squares , i made this rule to avoid leoll2's particular case $\endgroup$
    – Abr001am
    Commented Mar 17, 2015 at 22:06
  • $\begingroup$ So if I am player 1 I can flip a1,a2,b1,b2 on one turn, then I can't flip them back next turn (unless player 2 flips at least one of them), but I can flip a2,a3,b2,b3 on the next turn and a1,a2,b1,b2 on the one after that. $\endgroup$
    – Ross Millikan
    Commented Mar 17, 2015 at 22:10
  • $\begingroup$ the content is adjusted to a limited game solution $\endgroup$
    – Abr001am
    Commented Mar 21, 2015 at 10:35

2 Answers 2

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First of all, it's impossible for $player_1$ to win, assuming that $player_2$ is smart enough. Indeed, $player_2$ can simply repeat $player_1$'s move, locking the game in an infinite (boring) loop. As well, if $player_2$ had a strategy to win, $player_1$ could apply the aforementioned strategy, replicating $player_2$'s last move. Shortly, this game can be forced to be endless by both players.

By the way, I'm the authentic Leonardo of the game; the one who submitted the original puzzle (Marco) is actually a friend of mine.

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  • $\begingroup$ i tried to make rules to avoid that humdrum loops but its boring anyway :/ $\endgroup$
    – Abr001am
    Commented Mar 17, 2015 at 12:04
  • $\begingroup$ fixed.............. $\endgroup$
    – Abr001am
    Commented Mar 21, 2015 at 14:49
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If we consider a circle as a 2×2 size move and a line as 1×4 move , all possible game-ends would be one of the following (symmetry considered)

enter image description here

As the number of tracings is always even , the game always ends when player 2 performs his last turn.

And as no one can play a recently played move , or touch a majority flipped coins , we guarantee no humdrumic loop repeating . game has an end after my last improvement .

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