Well, as a player who knows some basic game strategies (not so good at deploying them XD), I think I can explain the idea behind this play—assuming you already know the chain rules and other basics of the game.
A very common strategy for the second player in a 5×5 board is what I like to call the “1 or 3” strategy (just a made-up name XD). In the simplest form, imagine there’s a long chain in the middle, and there are two sections on opposite sides of the board that haven’t yet become chains or loops. Before reaching the endgame, the game forces the first player to commit to one of these sections. If the first player makes a move that turns one section into a long chain, the second player does the same with the other section. If the first player sacrifices some boxes to avoid turning a section into a long chain, the second player simply mirrors that choice.
In the end, this approach leaves either 1 chain or 3 chains, both of which are winning scenarios for the second player. To complete the basic idea, imagine both sections are identical in shape.
However, the strategy becomes more complex when the sections are not identical or when chains (or loops) can intersect. To give more context, let’s define a “value” for any area/section: the number of moves needed for that section to become a long chain. Some sections might need 2 moves, others 3 and etc. (It’s common practice to know how many moves are needed for certain areas to become a long chain, and also how many boxes can be sacrificed to prevent that from happening.) If the values of both sections are the same, and there is an odd number of chains in other parts of the board, it’s theoretically a win for the second player. Still, in many scenarios, simply having the chain count in your favor isn’t enough to secure a win. loops or earlier sacrifices can turn the tables against the “favored” player.
About the game you mentioned, here’s my opinion (not 100% sure it’s correct):
At move 10.b5, Shark visibly tried the “1 or 3” method right from the start, making it very difficult for the player to find the right moves. In that situation, with one section needing two moves to become a chain, Shark was effectively one move ahead in applying this method on the board. To align the position with what I explained earlier: from a large remaining area on the board, Shark needed to find a chain and an area requiring exactly two moves to become a chain. That’s tricky to do perfectly for a human. While I can’t say for sure who was winning at that moment, I believe Shark had the advantage (considering it’s a bot and better at making optimal moves in such a complex position) at move 10. The game continued, and despite playing well, the player couldn’t find a way out. Eventually, at move 19.a2, the small area at the bottom was brought closer to becoming a chain (just 1 more move needed to complete or break it). Shark responded with 20.d7. Now, if the player played b1 to make a long chain, Shark would win by playing a6, forcing two chains to connect and winning with 3 chains(and if the player tried c6, sacrificing a box to aim for a 4-chain win, Shark would still win by forcing more sacrifices). So, the player went for b7, creating 3 potential chains plus the small area at the bottom, starting a sacrificing game for Shark. Shark played c2, preventing that area from becoming a chain. The player then tried d11 to turn the upcoming chain into a loop, but Shark later sacrificed a box, turned it into a chain, and secured a 3-chain win. You might ask: What if, after Shark’s f3, the player sacrificed two boxes with f9 to make a loop and win the chain battle? Well, sacrificing boxes to create loops is usually very costly for the player because double-crossing the loop to access your chains means giving up 4 more boxes besides the one you gave up eariler. In this case, after f9, Shark’s h9 would still win.
Overally, playing against Shark (or any human who’s good at the “1 or 3” strategy) is extremely tough. As the first player, you must make sure you don’t allow the second player to increase the complexity of the board.
P.S: The “1 or 3” isn’t the only strong strategy for second player. Another common winning endgame for second is: 2 long chains, 2 loops, and 3 points from small chains, while the first player gets fewer than 3 points.
Hope this was helpful <3