Scientific Sermon: Computing Winning Sets From Winning Strategies

In our last post of the series, we have seen how to obtain winning strategies in polytime if all you can do is compute winning sets (without strategies) in polynomial time. In this post, we’ll try to solve a related problem: given a pair of strategies for both players that are guaranteed to be winning on the winning sets of the game, can we extract these winning sets in polynomial time? The answer is, as we will see, yes.

In order to simplify the problem, let $G$ be a parity game and $\sigma$ be a player 0 strategy. We can define everything accordingly for player 1.
Define the player 0 $\sigma$-winning set as follows:
W_0(G,\sigma) = \{v \in V \mid \text{all $\sigma$-conforming plays starting in $v$ are won by pl~0}\}

It is easy to see that $W_0(G,\sigma) \subseteq W_0(G)$, and $W_0(G,\sigma) = W_0(G)$ iff $\sigma$ is a winning strategy on $W_0(G)$.

Hence, in order to solve our problem, it suffices to compute $W_0(G,\sigma)$.

Now we simplify our problem even more by considering the so-called strategy subgame. Given a game $G=(V,V_0,V_1,E,\Omega)$ and a player 0 strategy $\sigma$, the strategy subgame $G|_\sigma = (V,V_0,V_1,E’,\Omega)$ restricts the set of edges as follows:
E’ = \{(v,w) \in E \mid v \in V_0 \Rightarrow \sigma(v) = w\}
In other words all edges of player 1 remain in the game while all player 0 edges chosen by $\sigma$ are removed from the game.

It is not hard to see that $W_0(G,\sigma) = W_0(G|_\sigma)$, because the game on the right-hand side still includes the strategy $\sigma$. So we have a reduced our original problem to computing $W_0(G|_\sigma)$.

Why is that of any help? If you look closely, you’ll see that $W_0(G|_\sigma)$ is a so-called one-player game, i.e. a game in which at most one of the two players (in this case at most player 1) has more than one strategy. So our remaining task can be phrased as follows: How do we solve one-player parity games in polynomial time?

There are a bunch of algorithms that solve one-player parity games in polytime, but let us consider an algorithm very similar to the recursive algorithm that we have described earlier. You can also take a look at the Wikipedia page for an outline of the algorithm.

Our algorithm is recursive in the number of different priorities. If we only have one priority, we can immediately see who wins the whole game. If we have more than one priority, let $p$ be the largest one.

  1. Case $p$ is odd: Let $P$ be the set of all nodes with priority $p$. We compute the $1$-attractor $U$ of $P$. It is not too hard to see that $U \subseteq W_1$ with the attractor strategy of player 1 on $U$ as winning strategy. We remove $U$ from the game and continue with the rest.
  2. Case $p$ is even: Let $P$ be the set of all nodes with priority $p$. We compute the $0$-attractor $U$ of $P$ and remove it from the game, resulting in a game $G’$. We solve the game $G’$ by recursion and obtain winning sets $W_0’$ and $W_1’$. We compute the $1$-attractor $A$ in $G$ of $W_1’$ and remove it from the game as it is clearly won by player 1. This leaves us with $U \setminus A$ and $W_0′ \setminus A$.
    We now claim that $W_0′ \setminus A$ is won by player 0. Assume it’s not. Then, there must be a player 1 escape going through $A \setminus W_0′ \cup U$ reaching $W_1’$. But since $A$ is an attractor and the game is a one-player game, this is impossible. Let $X$ be the player 0 attractor of $W_0′ \setminus A$. Hence, we can remove $X$ as a save win for player 0 from the game.
    We are left with $U \setminus (A \cup X)$ with which we continue.

As an exercise, you can check that the algorithm runs in polytime.

In the next post of the series, we consider an assumption that you are allowed to make if you’re trying to find a polytime algorithm for solving parity games: you can assume that one of the two players, say player 0, wins the whole game. All you need to do, is to compute her winning strategy.

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