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In the previous two blogs, we see that a nominal game (i.e. player who makes the last move wins) can be defined as G=(G

_{L}|G

_{R}) where G

_{L}(

_{R}resp.) is the set of possible configurations after L (R resp.) makes the first move.

Indeed, we can add two nominal games G=(G

_{L}|G

_{R}) and G'=(G'

_{L}|G'

_{R}) together and form a new nominal game. Symbolically, it is

G+G'=(G

_{L}+G', G+G'

_{L}| G

_{R}+G', G+G'

_{R})

where the sum is defined recursively.

What does it mean exactly? Indeed, we simply put the two games side by side, and at each step, the player can make a move on either game.

What is the effect on the games?

Suppose both G and G' are games that the second player wins.

If the first player moves on G, then the second player moves on G; if the first player moves on G', then the second player also moves on G'. The second player is still the second winner on both G and G' and therefore he wins on both games. In other words, G+G' is still a second player's game.

Suppose both G and G' are games that L (R resp.) wins.

It is trivial that L (R resp.) is still the winner.

Suppose both G and G' are games that the first player wins.

It is a bit more complicated. Consider the following two games:

- G(
*n*) consists of a board with*n*pieces on it. Two players in term takes a piece from the board. The player who takes the last piece wins. - G'(
*n*) consists of a board with*n*piece on it. Two players in term can take arbitrary number (possibly all) of pieces from the board. The player who takes the last piece wins.

*n*) is a first player's game if

*n*is odd but a second player's game if

*n*is even, and G'(

*n*) is always a first player's game-the first player can take everything at the first move.

G(1)+G(1)=G(2), so the sum of two first player's games can be a second player's game.

G(1)+G'(2) is, however, a first player's game. At the very first move, the first player will take away a piece from G'(2) and the new configuration is G(2) with himself as the second player. Therefore, he wins.

Can you figure out the remaining four cases?