Two players play the following game. There are 2 piles of stones. One pile consists of 4 stones and the other one consists of 6 stones. Players take turns one by one. The amount of stones is unlimited. During the turn a player can add 4 stones into one pile or multiply a number of stones by 2 in one pile. The player wins if there are more than 26 stones in either pile after his turn. Who will win in this game - the first player or the second player? What choices should the winning player make to secure the win?
It's like that -> 4,6 -> 4,12 -> 4,24 -> 4,28 (win) or 4,6 -> 8,6 -> 16,6 -> 32,6 (win).
I need a table that shows all the winning combinations. (I'm actually bad at maths, it's just that our school teachers thinks that we should be able to solve complex advanced problems, but I can't solve this one no matter how much I try.)
The answer must look like this (the captions are in Russian, so don't mind them): https://i.stack.imgur.com/rrC9d.jpg
The basic framework for thinking about combinatorial games is the P-position.
A P-position is a game state where the player whose turn it is will lose. An example of a P-position in your game is when both piles have 13 stones - the current player cannot win, but after their turn the next player is guaranteed a win.
The most important fact about P-positions is that, if you start from a P-position you cannot reach another, and if you are not at a P-position, then it's always possible to reach one with your move, thus guaranteeing a win.
As I mentioned above, $(13,13)$ is a P-position, as is $(m,n)$ whenever $m\geq 13$ or $n \geq 13$. Now you can work backward through the tree: any position that can "see" such an $(m,n)$ is not a P-position. By "see", I mean there exists a move that takes you to the P-position. Mark them all as such. Continue to work backward. Here's the first couple of steps to get you started:
Mark all $(m,n)$ that can see $(13,13)$ - these are not P-positions.
Now you go backward through the rest of the squares, marking them either 'P' or not P. Make sure that when you mark a square, you've already marked all the squares it can see! Then the rule is easy: If it can see a P-position, then it isn't one, and if it can't, then it is.
Now every square can see exactly the following squares: the four immediately to its right, the four immediately above it, the one at twice its horizontal position and the one at twice its vertical position. For example, since $(12,12)$ can't see a P-position, it is one. Intuitively: whatever my opponent does with $(12,12)$, I'll be able to either win on my turn or leave them a $(13,13)$, i.e. a P-position. So mark $(12,12)$ as a P-position, and then mark all the squares that can see it as not P-positions:
Continue in this way, only marking a square when you have already marked all the squares it can see. The answer to your question: if the starting position is a P-position, then the first player will lose. If it is not, then the first player will win (by making a move that puts the game in a P-position).