A Novel Approach to Solving Goal-Achieving Problems for Board Games

TL;DR

The paper tackles solving goal-achieving puzzles in board games, focusing on Go life-and-death and Hex connectivity. It introduces Relevance Zone-Based Search (RZS), which uses Relevance Zones (RZs) and Consistent-Replay (CR) conditions to prune outside-zone moves and replay winning strategies, and Faster to Life (FTL) to bias AlphaZero-like training toward faster wins. Across 7x7 and 19x19 Go tsumego benchmarks, RZS combined with FTL solves all 20 seven-by-seven problems and 68 of 106 19x19 problems, outperforming prior solvers that rely on hand-tuned heuristics. The approach is claimed to be generic, extendable to other goal-achieving problems and games such as Hex and Slither, by providing a principled framework for reduction of search and reuse of proven search trees. Overall, the work offers a cohesive method that blends zone-based reasoning with deep-search techniques to efficiently certify goal-oriented outcomes in complex board games.

Abstract

Goal-achieving problems are puzzles that set up a specific situation with a clear objective. An example that is well-studied is the category of life-and-death (L&D) problems for Go, which helps players hone their skill of identifying region safety. Many previous methods like lambda search try null moves first, then derive so-called relevance zones (RZs), outside of which the opponent does not need to search. This paper first proposes a novel RZ-based approach, called the RZ-Based Search (RZS), to solving L&D problems for Go. RZS tries moves before determining whether they are null moves post-hoc. This means we do not need to rely on null move heuristics, resulting in a more elegant algorithm, so that it can also be seamlessly incorporated into AlphaZero's super-human level play in our solver. To repurpose AlphaZero for solving, we also propose a new training method called Faster to Life (FTL), which modifies AlphaZero to entice it to win more quickly. We use RZS and FTL to solve L&D problems on Go, namely solving 68 among 106 problems from a professional L&D book while a previous program solves 11 only. Finally, we discuss that the approach is generic in the sense that RZS is applicable to solving many other goal-achieving problems for board games.

Figures (11)

• Figure 1: An example for Go. For position $p_a$, if Black plays at 1 as in $p_{b'}$, $p_{c'}$, and $p_{d'}$, White replies to win at 2 as in $p_b$, $p_c$, and $p_d$ respectively. Since positions $p_b$, $p_c$, and $p_d$ satisfy UCA, we obtain RZs as the shaded cells, denoted by $z_b$, $z_c$, and $z_d$ respectively. For their previous positions $p_{b'}$, $p_{c'}$, and $p_{d'}$, White wins by playing 2 and it is not hard to see that their corresponding RZs are also $z_b$, $z_c$, and $z_d$ respectively. Furthermore, consider $p_a$. For all moves outside $z_a$, denoting the shaded zone of $p_a$, White can simply play at E2 to win. Thus, $p_a$ wins. Note that G1 is illegal for Black since suicide is prohibited. In addition, $z_a$ is its RZ, a union of $z_b$, $z_c$, and $z_d$, since White follows the same strategy as above to win for all Black moves outside of it.
• Figure 2: Solution trees.
• Figure 3: An illustration of achieving connection between C3 to the bottom side in Hex. For Black B2 in (b), White replies at C2, then wins by either B1 or C1. The RZ is the shaded cells in (b), since the goal can be achieved regardless of any moves outside of the RZ. Similar to the example in Figure \ref{['fig:benson_examples']}, B2 is viewed as a null move, and the must-play region for Black is reduced to include only the unoccupied cells, C2, B1, and C1. Next, if Black plays at C2 as in (d), White replies at B2 and wins due to symmetry to (b). Again, since Black C2 is a null move, the RZ is the shaded cells as in (d) and the must-play region is reduced to B1 only, from the intersection of the two RZs in (b) and (d). Now, we only need to search Black B1, in which case White replies at D2 for a win as in (e), since C3 is guaranteed to be connected to D2, and D2 is guaranteed to be connected to the lower side independently. Since all unoccupied cells in the must-play region have been searched, the goal can be achieved, and the RZ for the position in (a) is the shaded area in (f), which is the union of the above RZs, denoted by $z_f$. Namely, C3 can be connected to the lower side, regardless of any moves or changes outside of $z_f$.
• Figure 4: A 7x7 kill-all Go example. First, Black tries to prevent White from achieving UCA by playing at, say, D2 as in (b). In this case, White replies at F4 in order to achieve UCA by making a second region at G3 (the first is at G1). If Black counters with F5, then White achieves UCA with G4, where the set of shaded cells is its RZ, denoted by $z_b$. Since F5 is outside of $z_b$, it is a null-move, so the must-play region includes the two legal moves G3 and G4 that Black can try. For G3, it is trivial to win at G4. Black now considers playing at G4 instead, leading to the position in (c). In this variation, White plays at F5 in response to Black's move at G4. Again, if Black counters with F6, then White achieves UCA at G5. This search continues with Black being forced to consider G5, and again White at F6 forces Black to play at G6 as in (d). Finally, White plays at F7 to achieve UCA. Thus, for the Black move at D2, we can conclude that White wins and the RZ $z_e$ is the union of the RZs as in (e). Interestingly, since D2 is outside of $z_e$, it is a null move and all of Black's moves outside $z_e$ can be disregarded. The must-play region is therefore the unoccupied cells in $z_e$. Now, let Black choose to play F4 (inside $z_e$) as in (f). Similarly, White wins with an RZ $z_f$ as in (f) due to symmetry to (e). Since we can disregard Black moves outside $z_e$ and $z_f$, the must-play region becomes the intersection of the two RZs as shown in (g), which contains no legal moves for Black. Thus we have proved White wins in (a), with large reductions to the search space. Note that the corresponding RZ for the position in (a) is the union of the two RZs, shaded in (h).
• Figure 5: A dilation case for $\pi_{\square}$ (White) with capturing blocks. For the position $p_a$ in (a), White is to play and wins as follows. White plays at F1 to capture four black stones, leading to the position $p_b$. Then, as in (c), for Black's move at G3, White replies G2 to achieve UCA and obtains an RZ $z_c$ as shaded. Since G3 is outside of the RZ, it is a null move, which reduces the must-play region to G1 and G2 only. It is trivial for White to reply at G2 for Black's move at G1. For Black's move at G2, White replies at G3 to achieve UCA and obtains an RZ $z_d$ as shaded in (d). Thus, we conclude that White wins for the position $p_b$ (Black to play). The union of RZs of the children of $p_b$ is $z_b$, as shaded in (b). This union satisfies the CR conditions, and thus is a valid RZ. Now, let us examine $p_a$ with White to play. Let $z_u$ be the union of F1 (the move that was searched) and $z_b$, which is still $z_b$. Unfortunately, in this case, $z_u$ can not serve as an RZ for $p_a$. As a counter-example, consider $p_e$, which shares the same pattern within $z_u$ as $p_a$. If $z_u$ is an RZ for $p_a$, for any position that shares the same pattern, say, $p_e$, White should be able to replay their winning strategy at F1. This is clearly not the case in $p_e$. In order to replay the winning strategy (guaranteed by satisfying all the CR conditions), we need to dilate $z_u$ to $z_a$.

Theorems & Definitions (2)

• Lemma 1
• proof