Two players in turn paint the sides of an n-gon. The first one can paint the side that borders either zero or two colored sides, the second – the side that borders one painted side. The player who can not make a move loses. At what n can the second player win, no matter how the first player plays?
On a table there are 2022 cards with the numbers 1, 2, 3, …, 2022. Two players take one card in turn. After all the cards are taken, the winner is the one who has a greater last digit of the sum of the numbers on the cards taken. Find out which of the players can always win regardless of the opponent’s strategy, and also explain how he should go about playing.
Two people are playing. The first player writes out numbers from left to right, randomly alternating between 0 and 1, until there are 2021 numbers in total. Each time after the first one writes out the next digit, the second switches two numbers from the already written row $($ when only one digit is written, the second misses its move $)$. Is the second player always able to ensure that, after his last move, the arrangement of the numbers is symmetrical relative to the middle number?
A game takes place on a squared 9 × 9 piece of checkered paper. Two players play in turns. The first player puts crosses in empty cells, its partner puts noughts. When all the cells are filled, the number of rows and columns in which there are more crosses than zeros is counted, and is denoted by the number K, and the number of rows and columns in which there are more zeros than crosses is denoted by the number H $($ 18 rows in total $)$. The difference B = K – H is considered the winnings of the player who goes first. Find a value of B such that
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1$)$ the first player can secure a win of no less than B, no matter how the second player played;
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2$)$ the second player can always make it so that the first player will receive no more than B, no matter how he plays.
Solution 1
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One of the possible strategies of the first player: with his first move he places a cross at the center of the board; then for every move of the second player in any cell, it responds with a move in a cell symmetric with respect to the center. This is possible, since symmetry is violated each time by the second player.
In the final position in each pair of cells, symmetrical with respect to the center, there is a cross and a nought. Therefore, in the central row and in the central column, there are more crosses than zeros; and each row $($ column $)$ that does not pass through the center corresponds to a symmetrical row $($ column $)$, with one of them having more zeros than crosses and one of them having fewer zeros than crosses. Thus, with this strategy, the first player’s win is equal to 2.
Now we indicate the strategy of the second player. If he has the opportunity to occupy a cell that is a symmetrical $($ relative to the center $)$ cell, which has just been occupied by the first player then it does so. Otherwise, he makes any move. With such a strategy, after the turn of the second player, the set of occupied cells is either symmetrical $($ until the first has occupied the center $)$, or differs from the symmetrical one by exactly one cell. In this case, in each pair of occupied symmetrical cells there will be one cross and one zero. With his last move, the first player will have to create the position described in the previous paragraph, and his winnings are again 2.
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Solution 2
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We cover the board with tiles as shown in the figure. The strategy of the first: take the first lower left corner, and then go to the same domino tile on the board, where the second one went before it.
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The strategy of the second: if possible, take a cell in the same part of the board, where the first player went first; if it is impossible – any move.
If at least one of the players adheres to the strategy indicated for him, then in the end in the lower left corner there will be a cross, and in each of the other parts of the board there will be an equal number of noughts and crosses $($ for reasons similar to those stated in the first solution: after each move of the second player in all filled dominoes there will stand a cross and a nought, and, perhaps, in one there is just a nought $)$.
Consider the two upper rows. In total, they contain 9 crosses and 9 noughts $($ since they are divided into 9 “dominoes” $)$. Therefore, in one of them there are more crosses, in the other – more zeroes. A “Draw” is also achieved in the third and fourth, …, seventh and eighth rows from the top, and in the bottom rows the first “wins” $($ it has 5 crosses and 4 noughts $)$. The situation is similar in the columns $($ pairs of neighboring columns, starting counting from the right, also consist of 9 “dominos” $)$.
B = 2.
Two play tic-tac-toe on a $10 × 10$ board according to the following rules. First they fill the whole board with noughts and crosses, putting them in turn $($ the first player puts crosses, their partner – noughts $)$. Then two numbers are counted: $K$ is the number of five consecutively standing crosses and $H$ is the number of five consecutively standing zeros. $($ Five, standing horizontally, vertically and parallel to the diagonal are counted, if there are six crosses in a row, this gives two fives, if there are seven, then three, etc.$)$. The number $K-H$ is considered to be the winnings of the first player $($ the losses of the second $)$.
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a$)$ Does the first player have a winning strategy?
b$)$ Does the first player have a non-losing strategy?
There is a chocolate bar with five longitudinal and eight transverse grooves, along which it can be broken $($ in total into 9 * 6 = 54 squares $)$. Two players take part, in turns. A player in his turn breaks off the chocolate bar a strip of width 1 and eats it. Another player who plays in his turn does the same with the part that is left, etc. The one who breaks a strip of width 2 into two strips of width 1 eats one of them, and the other is eaten by his partner. Prove that the first player can act in such a way that he will get at least 6 more chocolate squares than the second player.
Two players in turn increase a natural number in such a way that at each increase the difference between the new and old values of the number is greater than zero, but less than the old value. The initial value of the number is 2. The winner is the one who can create the number 1987. Who wins with the correct strategy: the first player or his partner?
A cat tries to catch a mouse in labyrinths A, B, and C. The cat walks first, beginning with the node marked with the letter “K”. Then the mouse $($ from the node “M”$)$ moves, then again the cat moves, etc. From any node the cat and mouse go to any adjacent node. If at some point the cat and mouse are in the same node, then the cat eats the mouse.
Can the cat catch the mouse in each of the cases A, B, C?
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A B C
In the cases A, C it is sufficient for the mouse to walk to a node centrally symmetric to the node on which the cat is located each time.
$($ For case A, it’s still simpler. $)$ If we color the nodes of the labyrinth in a chess coloring format, we see that the cat, after its turn, is always in a node with an opposite color to that of the node of the mouse, so it can not eat the mouse with any sequence of moves.
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In the case B we also consider the chess coloring of the nodes $($ see Fig. $)$. First, the cat goes to node M. If the mouse during this time “passes” on the diagonal segment, the cat catches it with the next move.
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Otherwise, the cat itself goes diagonally and returns to M. If the mouse is not yet eaten, at that moment it is standing on a black node and must go to a white one. But there are only four white nodes, which have not yet been visited by the cat. Two of them are angular, if the mouse gets into one of them, the cat “clamps” it in this corner and eats it on the next move. Otherwise, the cat makes a move towards the mouse, and the mouse is forced to retreat into the black corner, where the cat will catch it in the next turn.
In the labyrinths A and B it can not, but in C it can.
There are three piles of rocks: in the first pile there are 10 rocks, 15 in the second pile and 20 in the third pile. In this game (with two players), in one turn a player is allowed to divide one of the piles into two smaller piles. The loser is the one who cannot make a move. Which player would be the winner?
A White Rook pursues a black bishop on a board of $3 \times 1969$ cells $($they walk in turn according to the usual rules$)$. How should the rook play to take the bishop? White makes the first move.
Two players play on a square field of size 99 × 99, which has been split onto cells of size 1 × 1. The first player places a cross on the center of the field; After this, the second player can place a zero on any of the eight cells surrounding the cross of the first player. After that, the first puts a cross onto any cell of the field next to one of those already occupied, etc. The first player wins if he can put a cross on any corner cell. Prove that with any strategy of the second player the first can always win.
There is a system of equations
$* x + * y + * z = 0$,
$* x + * y + * z = 0$,
$* x + * y + * z = 0$.
Two people alternately enter a number instead of a star. Prove that the player that goes first can always ensure that the system has a non-zero solution.
A rectangular chocolate bar size 5 × 10 is divided by vertical and horizontal division lines into 50 square pieces. Two players are playing the following game. The one who starts breaks the chocolate bar along some division line into two rectangular pieces and puts the resulting pieces on the table. Then players take turns doing the same operation: each time the player whose turn it is at the moment breaks one of the parts into two parts. The one who is the first to break off a square slice 1\times 1 (without division lines) a) loses; b) wins. Which of the players can secure a win: the one who starts or the other one?
Hannah and Emma have three coins. On different sides of one coin there are scissors and paper, on the sides of another coin – a rock and scissors, on the sides of the third – paper and a rock. Scissors defeat paper, paper defeats rock and rock wins against scissors. First, Hannah chooses a coin, then Emma, then they throw their coins and see who wins $($if the same image appears on both, then it’s a draw$)$. They do this many times. Is it possible for Emma to choose a coin so that the probability of her winning is higher than that of Hannah?
A chequered strip of $1 \times N$ is given. Two players play the game. The first player puts a cross into one of the free cells on his turn, and subsequently the second player puts a nought in another one of the cells. It is not allowed for there to be two crosses or two noughts in two neighbouring cells. The player who is unable to make a move loses.
Which of the players can always win $($no matter how their opponent played$)$?
The rook stands on the square a1 of a chessboard. For a move, you can move it by any number of cells to the right or up. The one who puts the rook on the h8 square will win. Who wins with the right strategy?
Given a board (divided into squares) of the size: a) 10×12, b) 9×10, c) 9×11, consider the game with two players where: in one turn a player is allowed to cross out any row or any column if there is at least one square not crossed out. The loser is the one who cannot make a move. Is there a winning strategy for one of the players?
On a plane there are 100 sheep-points and one wolf-point. In one move, the wolf moves by no more than 1, after which one of the sheep moves by a distance of no more than 1, after that the wolf again moves, etc. At any initial location of the points, will a wolf be able to catch one of the sheep?
Prove that in a game of noughts and crosses on a 3×3 grid, if the first player uses the right strategy then the second player cannot win.
Two grandmasters in turn put rooks on a chessboard $($one turn – one rook$)$ so that they cannot capture each other. The person who cannot put a rook on the chessboard loses. Who will win with the game – the first or second grandmaster?
