This is a 2-person game.

Let $\ P\ $ be any arbitrary projective space (of any dimension $\ \ge2$ and any cardinality, etc., but typically, let it be a finite plane over a field). Let $\ S_0\subseteq P\ $ be an arbitrary finite set (typically, $\ S_0=P\ $ or $\ S_0=P\setminus L\ $ or $\ S_0=P\setminus(L\cup Ł)\ $ where $\ L\ $ and $\ Ł\ $ are straight lines, and $\ P\ $ is a finite projective plane -- all three of these cases are mathematically equivalent).

The positions of the game are subsets $\ S_n\ $ of $\ S_0\ $ such that

• $\ S_n\subseteq S_{n-1}; $
• $\ |S_{n-1}\setminus S_n|\ \ge\ 3; $
• there is a straight line $\ L_n\ $ in $\ P\ $ such that $$ S_{n-1}\ \subseteq\ S_n\cup L, $$ for every $\ n\in\mathbb N.$

The winner (of the positive flavored game) is the player of the last legal move.

One may also consider the negative flavor when the last legal move causes a game loss.

Question  As a minimum, one would like to decide about the winning strategies for the projective planes over fields $\ \mathbb Z/7\ $ and $\ \mathbb Z/11,\ $ where $\ S_0:=P\ $ is the respective projective plane.

Some years ago, I wrote a program in Perl that solved the positive flavor for the projective plane $\ P\ $ over $\ \mathbb Z/5\ $, for arbitrary $\ S_0\subseteq P. $ I should be able to find that program (I hope) -- with my today's computer, possibly it should handle $\ \mathbb Z/7\ $ (or only after some improvements? -- hmm, I have a much more powerful computer these days than in the past but my ability went down the drain hence it is a tie).

This is a 2-person game.

Let $\ P\ $ be any arbitrary projective space (of any dimension $\ \ge2$ and any cardinality, etc., but typically, let it be a finite plane over a field). Let $\ S_0\subseteq P\ $ be an arbitrary finite set (typically, $\ S_0=P\ $ or $\ S_0=P\setminus L\ $ or $\ S_0=P\setminus(L\cup Ł)\ $ where $\ L\ $ and $\ Ł\ $ are straight lines, and $\ P\ $ is a finite projective plane -- all three of these cases are mathematically equivalent).

The positions of the game are subsets $\ S_n\ $ of $\ S_0\ $ such that there exists straight line $\ L_n\ $ such that

$$\ S_n\ := S_{n-1}\setminus L_n $$ and $$ S_{n-1}\cap L_n\ \ge 3. $$

The winner (of the positive flavored game) is the player of the last legal move.

One may also consider the negative flavor when the last legal move causes a game loss.

Question  As a minimum, one would like to decide about the winning strategies for the projective planes over fields $\ \mathbb Z/7\ $ and $\ \mathbb Z/11,\ $ where $\ S_0:=P\ $ is the respective projective plane.

Some years ago, I wrote a program in Perl that solved the positive flavor for the projective plane $\ P\ $ over $\ \mathbb Z/5\ $, for arbitrary $\ S_0\subseteq P. $ I should be able to find that program (I hope) -- with my today's computer, possibly it should handle $\ \mathbb Z/7\ $ (or only after some improvements? -- hmm, I have a much more powerful computer these days than in the past but my ability went down the drain hence it is a tie).

This is a 2-person game.

Let $\ P\ $ be any arbitrary projective space (of any dimension $\ \ge2$ and any cardinality, etc., but typically, let it be a finite plane over a field). Let $\ S_0\subseteq P\ $ be an arbitrary finite set (Typically, $\ S_0=P\ $ or $\ S_0=P\setminus L\ $ or $\ S_0=P\setminus(L\cup Ł)\ $ where $\ L\ $ and $\ Ł\ $ are straight lines, and $\ P\ $ is a finite projective plane -- all three of these cases are mathematically equivalent).

The positions of the game are subsets $\ S_n\ $ of $\ S_0\ $ such that

• $\ S_n\subseteq S_{n-1}; $
• $\ |S_{n-1}\setminus S_n|\ \ge\ 3; $
• there is a straight line $\ L_n\ $ in $\ P\ $ such that $$ S_{n-1}\ \subseteq\ S_n\cup L, $$ for every $\ n\in\mathbb N.$

The winner (of the positive flavored game) is the player of the last legal move.

One may also consider the negative flavor when the last legal move causes a game loss.

Question  As a minimum, one would like to decide about the winning strategies for the projective planes over fields $\ \mathbb Z/7\ $ and $\ \mathbb Z/11,\ $ where $\ S_0:=P\ $ is the respective projective plane.

Some years ago, I wrote a program in Perl that solved the positive flavor for the projective plane $\ P\ $ over $\ \mathbb Z/5\ $, for arbitrary $\ S_0\subseteq P. $ I should be able to find that program (I hope) -- with my today's computer, possibly it should handle $\ \mathbb Z/7\ $ (or only after some improvements? -- hmm, I have a much more powerful computer these days than in the past but my ability went down the drain hence it is a tie).

This is a 2-person game.

Let $\ P\ $ be any arbitrary projective space (of any dimension $\ \ge2$ and any cardinality, etc., but typically, let it be a finite plane over a field). Let $\ S_0\subseteq P\ $ be an arbitrary finite set (typically, $\ S_0=P\ $ or $\ S_0=P\setminus L\ $ or $\ S_0=P\setminus(L\cup Ł)\ $ where $\ L\ $ and $\ Ł\ $ are straight lines, and $\ P\ $ is a finite projective plane -- all three of these cases are mathematically equivalent).

The positions of the game are subsets $\ S_n\ $ of $\ S_0\ $ such that

• $\ S_n\subseteq S_{n-1}; $
• $\ |S_{n-1}\setminus S_n|\ \ge\ 3; $
• there is a straight line $\ L_n\ $ in $\ P\ $ such that $$ S_{n-1}\ \subseteq\ S_n\cup L, $$ for every $\ n\in\mathbb N.$

The winner (of the positive flavored game) is the player of the last legal move.

One may also consider the negative flavor when the last legal move causes a game loss.

Question  As a minimum, one would like to decide about the winning strategies for the projective planes over fields $\ \mathbb Z/7\ $ and $\ \mathbb Z/11,\ $ where $\ S_0:=P\ $ is the respective projective plane.

Some years ago, I wrote a program in Perl that solved the positive flavor for the projective plane $\ P\ $ over $\ \mathbb Z/5\ $, for arbitrary $\ S_0\subseteq P. $ I should be able to find that program (I hope) -- with my today's computer, possibly it should handle $\ \mathbb Z/7\ $ (or only after some improvements? -- hmm, I have a much more powerful computer these days than in the past but my ability went down the drain hence it is a tie).