It is interesting that nobody could write the B part of the question during the exam.

How to solve this problem？B part is too difficult

Cant find the solution anywhere so…bump this

falantrng wrote: Baklavas with nuts are laid out on the table in a row at the Nowruz celebration. Kosa and Kechel saw this and decided to play a game. Kosa eats one baklava from either the beginning or the end of the row in each move. Kechel either doesn't touch anything in each move or chooses the baklava he wants and just eats the nut on it. They agree that the first Kosa will start the game and make $20$ moves in each step, and the Kechel will only make $1$ move in each step. If the last baklava eaten by the Kosa is a nut, he wins the game. It is given that the number of baklavas is a multiple of $20.$ $A)$ If the number of baklavas is $400,$ prove that Kosa will win the game regardless of which strategy Kechel chooses. $B)$ Is it always true that no matter how many baklavas there are and what strategy Kechel chooses, Kosa will always win the game? For $B$ I don’t think so…. Kechel can eat every single one and Kosa will lose the game

Bump....

B) Kechel has a winning strategy if the number of initial baklavas is large enought. Let $n=20$ and $k$ the initial number of baklavas. Replace the baklavas with white and black balls (initialy every ball is white, Kosa every move delete $n$ balls, Kechel every move change a ball from white to black and Kechel's gool is that the finale taken ball is black). Suppose that Kosa can delete the ball has he want such that the Total number of delete balls after $t$ moves is at most $nt+t_0$ were $t_0$ is a fixed costant. We prove that Kechel has a winning strategy for large enought $k$ by induction on $n$. Base case is left to the reader. Suppose the statment is true for $n-1$. Kechel start coloring a ball "in the middle" then every next move he color the ball such that the distance between two consecutive black balls is exactly $n$. He stops when there are $x$ black balls $b_1,...b_x$ and before $b_1$ and after $b_x$ there are at most $n$ white balls. Notice that Kechel can garanties $x$ arbitrarly large if he choose $k$ large enought. Because in every move Kosa eat at most $1$ black ball and given that he start coloring in the middle in the first $\dfrac{k}{4n}$ Kosa cannot eat any black ball. Now we color every black ball of purple. From now after every number $t$ of moves Kosa has eat at least $t-2$ purple balls. Now Kechel can play like there are no purple balls and use the strategy for $n-1$ because from now Kosa eat $n-1$, $n$ or $n-2$ not purple balls at every move such that the number of times he have eaten $n$ ball is at most $2$ plus the number of times he have eaten $n-2$ not purple balls.