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The fact that every tower can be reduced to 1 since the 1st move makes that tower not choosable again. So then, if the height of towers is 1 since the beggining makes P2 to win by default as P1 cannot reduce any tower and he goes first.
if the number of towers is pair then P2 will win as P1 will reduce one of them to 1 and in the next turn P2 will do the same, then no more chances for P1.
If the number of towers is odd, then P1 will win as he will have the last turn.
Why is that all players have to reduce the towers to 1 everytime they play? There is also an option to reduce the towers to a number (y) that can evenly divide the tower height (x)...
This had me confused a bit as well but I figured it out.
When the number of towers is even, P2 can just copy P1's move, not matter what it is and will always win, it not necesary to lower the tower to 1 in the first move.
When the number of towers is odd, if P1 lowers one tower to 1 that tower is out of the game and we are now in a even tower situation with the players reversed, so from now on P1 just has to copy P2 to win the game.
It was my initial thought as well but that assumes P1 decides to be greedy and reduce the T1 (of 2 towers) by 5 i.e. P1 takes 2nd option of their example:
*There are 2 towers, each 6 units tall. Player 1 has a choice of two moves:
- remove 3 pieces from a tower to leave 3 as 6 modulo 3 = 0
- remove 5 pieces to leave 1 *
However if P1 reduce T1 by only 3 and P2 is greedy and reduce T2 by 5. P1 gets to reduce T1 by 2. P2 can't make a move and loses. Greed != Win :)
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Tower Breakers
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The fact that every tower can be reduced to 1 since the 1st move makes that tower not choosable again. So then, if the height of towers is 1 since the beggining makes P2 to win by default as P1 cannot reduce any tower and he goes first. if the number of towers is pair then P2 will win as P1 will reduce one of them to 1 and in the next turn P2 will do the same, then no more chances for P1. If the number of towers is odd, then P1 will win as he will have the last turn.
Why is that all players have to reduce the towers to 1 everytime they play? There is also an option to reduce the towers to a number (y) that can evenly divide the tower height (x)...
This had me confused a bit as well but I figured it out. When the number of towers is even, P2 can just copy P1's move, not matter what it is and will always win, it not necesary to lower the tower to 1 in the first move. When the number of towers is odd, if P1 lowers one tower to 1 that tower is out of the game and we are now in a even tower situation with the players reversed, so from now on P1 just has to copy P2 to win the game.
It was my initial thought as well but that assumes P1 decides to be greedy and reduce the T1 (of 2 towers) by 5 i.e. P1 takes 2nd option of their example: *There are 2 towers, each 6 units tall. Player 1 has a choice of two moves: - remove 3 pieces from a tower to leave 3 as 6 modulo 3 = 0 - remove 5 pieces to leave 1 *
However if P1 reduce T1 by only 3 and P2 is greedy and reduce T2 by 5. P1 gets to reduce T1 by 2. P2 can't make a move and loses. Greed != Win :)