Two players are trying to figure out the binary code Two players guess code that is hidden in $2^n$ ($n\geq2$, n $\in$ $\Bbb N$) nodes that each contain either 0 or 1 which makes a binary code that consists of zeros and/or ones, for example, '0010' is binary code, that has $2^2$ nodes that consist only of 0's and 1's.
There are two players - first player and second player. The first player starts and guesses the node. The second player only knows if the node that was chosen by the first player contains 0 or 1 but the second player do not know which node the first player chose. There are two rules that have to be followed:
A) If a player chooses node and guesses right, he can choose three consecutive nodes that contain one of the nodes that he guessed this turn and send to other player those three consecutive nodes as partial information.
B) If a player chooses node and guesses wrong, he can choose two consecutive nodes that contain one of the nodes that he guessed this turn and send to other player those two consecutive nodes as partial information.
After first player followed rules A and B, it is second player's turn and he does the same: guesses the node, sends partial information by the rules A and B, but the difference is that the second player becomes the one that is sending information and the first player is the one who is receiving it. Then it is first player's turn again, then the second player's turn and so on until one of the players can tell whole code containing of $2^n$ nodes of zeros or ones, but the player is considered to know the code when he knows all the node values for sure.
It seems to me that if we could find out strategy for trivial n=2 or n=3 we could just use the mathematical induction and prove that strategy works for all natural numbers n. Also, for clarification, by default first and last node is not connected (the nodes are not in a closed circle) - it would make the strategy more difficult or even impossible to find.
I probably should make myself more clear about the task.
Let us say we have 4 nodes and the exact code is '0100'.
(I) First player guesses that 2nd node is '1'.
Public info: the first player guessed right about the '1'.
Private info for first player: 2nd node is '1'.
Public info: either 1st, 2nd or 3rd is the node containing '1'
(first player guessed right so he chose 3 consecutive nodes this way).
(II) Second player guesses that 1st node is '1'.
Public info: the second player guessed wrong about the '1'.
Private info for second player: 1st node is '0'.
Public info: either 1st or 2nd is the node containing '0'
(it is forced to reveal this info because 1st node is only near the 2nd)
(I) Since first player knew that 2nd is '1', so he already knows that 1st is '0' from the public info.
First player guesses that 4th node is '0'.
Public info: the first player guessed right about the '0'.
Private info for first player: 4th node is '0'.
Public info: either 2nd, 3rd or 4th is the node containing '0'.
(II) Second player knows that 1st one is '0', either 2nd or 3rd is '1', either 2nd, 3rd or 4th is '0'.
Basically, second player knows almost nothing.
Public info: the second player guessed wrong about the '1'.
Private info for second player: 3rd node is '0'.
Public info: either 2nd or 3rd is the node containing '0'.
(I) Since the first player knew that 2nd node contains '1', he knows then that 3rd contains '0'.
But he figures out the code that is '0100' and the first player wins.
Now, talking about the quantity of nodes. $2^n$ is the reason because of forward-backward induction idea that involves such numbers because we have to first assume that from every natural number n it follows for $2^n$ is true and then we have to prove that if for all n is true then for all n-1 is also true.
It also is related to binary codes, so why not stick to $2^n$.
 A: This is not an answer but an attempt to rewrite the task/rules in a simpler form. I think there are some translation issues. OP, if you agree this is useful, please feel free to copy/paste, and I'll delete later.

The rules are thus: Two players are attempting to discover a binary code of $2^n$ bits, unknown to either. Each turn, one player is active and the other is passive. A turn plays as follows:

*

*The active player chooses a bit, and guesses whether it is $0$ or $1$. The game system tells the active player whether their guess is correct or not.

*The active player then chooses several consecutive bits containing the guessed bit: they choose three if they guessed correctly, two otherwise.

*The game system then tells the passive player the true value of the guessed bit and the two or three bits selected by the active player.

*This ends the turn: the active player becomes passive and vice versa.

The end result of a turn, in terms of revealed information, is that the active player learns the true value of one bit, and the passive player learns that one of two (or three) consecutive bits has that value. The game ends when one player can determine the code correctly; that player wins.
As an example: the hidden code is $0100$. We will call the players X and Y, and label the bits $ABCD$.

*

*Turn $1$X: X guesses that bit $B$ is $1$, and is told this is correct. X chooses to reveal that one of the bits $A,B,C$ is $1$.

*Turn $1$Y: Y guesses that bit $A$ is $1$, and is told this is incorrect. Y has no choice but to reveal that one of the bits $A,B$ is $0$.

*X now knows the code is $01CD$. Y knows the code is $0BCD$, where either $B$ or $C$ is $1$.

*Turn $2$X: X guesses that bit $D$ is a $0$, and is told this is correct. X has no choice but to reveal that one of the bits $B,C,D$ is $0$.

*Turn $2$Y: Y, lacking information, guesses that bit $C$ is a $1$, and is told this is incorrect. Y chooses to reveal that one of the bits $B,C$ is $0$.

*Y now knows that the code is $010D$. But X now knows the entire code, and wins.


The question is: How can one develop a strategy for this game, for larger $n$?
