# Tag Info

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Call the number of days until expiration the "freshness" of the cookie. A cookie with freshness $1$ is here (and can be eaten) today, but will be gone tomorrow; a cookie with freshness $n+1$ will have freshness $n$ tomorrow. If all the cookies' freshnesses have the same parity, then this will persist for the entire game: one player (Eve) will only see ...

23

Methods Starting from the base url http://www.cross-tables.com/annotated.php?a=1 I used a combination of Python's urllib, multiprocessing and BeautifulSoup to extract the first 10000 games. The games were parsed and turned into numpy 15x15 Boolean matrices. The matrices were then turned into graphs making an edge if two adjacent cells in the matrix were ...

19

In 3-dimensional chess, it is possible to force checkmate starting with a finite number of rooks. As this fact still appears to be open, I'll post a method of forcing checkmate with 96 rooks, even though it should be clear that this is not optimal. You can remove some of the rooks from the method I'll give below, but I am aiming for a simple explanation of ...

10

There are several standard texts. You may want to start with a short essay, "The History of Combinatorial Game Theory", by Richard J. Nowakowski. The paper lists the main texts you may want to consult. For a more exhaustive list of references, see "Combinatorial games: Selected bibliography with a succinct GourmetIntroduction", by Aviezri Fraenkel. Thomas ...

8

It turns out that the maximum possible diameter uses the full hundred tiles! My friend Carl illustrated this in 2008 in a game he made up, here: http://www.cross-tables.com/annotated.php?u=2493#0# . Thanks for the question!

8

This is known as Wythoff's game. The losing positions are $(\lfloor n \varphi\rfloor, \lfloor n \varphi^2\rfloor)$ where $\varphi$ is the golden ratio. An interesting way of presenting this game is to have the players move a Queen on a chessboard, which you can find at cut-the-knot here.

7

There is some discussion of this topic at the very end of the second edition (2000) On Numbers and Games by Conway. He describes work by himself, Simon Norton, and Martin Kruskal to define integration. According to the description, it looked good for a while, producing workable logarithm function in terms of the integral of $x^{-1}$, but then got stuck, and ...

7

Here is an example of a game where the second player (B) has a winning strategy. It is more comfortable to describe it using multisets, this should lead to no confusion. Our base set has $3\times a_i$ and $6\times b_i$ for $i=0,1,2$, thus a total of $27$ elements. The winning sets are (with indices mod $3$) the following. ($3\times a_i);\ (3\times a_i,\ ... 7 Here's one option of a type you didn't suggest. Take any symmetric game. Modify the game as follows: After the first player's first move, give the second player the option to continue the game as usual, or to switch places, taking the first player's position and giving the (former) first player the next move as second player. Assuming that ties are ... 6 You can do it generically. Treat the case of the multiplier separately. So there's 11C5 (eleven choose 5) ways with no multiplier. You want the average. It's just 5 times the average of those eleven numbers. Then there's 11C4 ways with the multiplier. Let the multiplier be k. The average for the other 4 will be 4 times the average of the eleven. So, if A's ... 6 Astoundingly enough, this has already been studied. And I'm almost embarrassed to say that I'm familiar with the result. I used to freecell a lot. And FYI, 11982 is the impossible Frecell game. But I recommend entering in games -1, -2, -3, etc too. So here are some stats from some studies of freecell. Firstly, the depth of the aces, i.e. how many cards ... 6 The minimum number of switches required is$100$. This puzzle can be turned into a linear algebra problem over${\Bbb Z}/2{\Bbb Z}$by letting the vector$v\in({\Bbb Z}/2{\Bbb Z})^{100}$record which switches are pressed ($1$if pressed,$0$if not), the vector$w\in({\Bbb Z}/2{\Bbb Z})^{100}$record which lights are initially on ($1$if on,$0$if off), ... 6 This PDF might be a useful starting point. It also mentions a book Fair Game: How to Play Impartial Combinatorial Games, by Richard K. Guy. I’ve not seen it, but it was published by COMAP; the combination of author and publisher suggests that it might be well worth looking into. (There is of course the massive Winning Ways, by Berlekamp, Conway, & Guy, ... 6 This is a most intriguing problem and took many, many, many thinking hours to solve, I even wrote a script to better understand it. It is possible for the sinners (players) to win every time. The number of players playing is irrelevant and the number of numbers they choose per turn is irrelevant as long as the players get to choose more numbers than the ... 6 The Sprague-Grundy theory should kill this problem pretty dead; it solves any game of a large class that includes this game. Without getting into the theoretical details of the theory: You can tabulate the "value" of each position, where the "value" is a non-negative integer A position with no legal moves has value 0. A position with legal moves to one or ... 6 For a$6 \times 3$board, it's impossible. Consider the board below, and the squares, marked with an$\times$. $$\begin{matrix} \cdot & \cdot & * & \cdot & \cdot & \cdot \\ \times & \cdot & \cdot & \cdot & \times & \cdot \\ \cdot & \cdot & * & \cdot & \cdot & \cdot \\ \end{matrix}$$ The only ... 6 So, to interpret your question, you have two different procedures for filling out a table of numbers which has a definite top and left edge, but continues indefinitely down and to the right. First is Don't fill in any cell before all the positions directly above and directly to the left of it are filled. Write in each cell the smallest nonnegative ... 6 In short, because — as emphasized by the last phrase of your bolded passage — having extra pieces on the board in Tic-Tac-Toe is never bad. It's not just 'having more squares' vs. 'having fewer squares'; it's that if position A is exactly position B with an extra X on it, then position A is always at least as good for player X as position B is. ... 5 Are you including$1$as a reasonable move? If so, this game ("Prime Nim" created by CE Shannon) has been solved. If not, the article above has Martin Gardner claiming an analysis of the strategy would be very difficult. 5 Every ordinal number$\alpha$has a canonical copy as a surreal number$\hat\alpha$, whose left set has as members$\hat\beta$for every$\beta\lt\alpha$, and whose right set is empty. Succinctly, $$\hat\alpha=\{\ \{\hat\beta\mathrel{:} \beta\lt\alpha\ \}\mathrel{|} \}.$$ One can prove by induction that$\alpha\mapsto\hat\alpha$is isomorphism of the ... 5 There is unlikely to be a simple formula for general$N$and$M$though there will be bounds. The standard version has$N=4$pegs drawn from$M=6$colours, with repeats allowed. Koyama, K. and Lai, T. W. "An Optimal Mastermind Strategy." J. Recr. Math. 25, 251-256, 1993 provides an algorithm with an average of 4.340 guesses and a maximum of 6, and ... 5 You can show by induction:$N=1$is a losing position as there is no move.$N>1$and even is a winning position: you can subtract an odd proper divisor such as$1$to reach an odd number$N>1$and odd is a losing position: you must subtract an odd proper divisor to reach an even number 5 Assign point$(x,y)$the weight$2^{-(x+y)}$. Then the weight of$(x,y)$equals the weight of$(x+1,y)$plus the weight of$(x,y+1)$, so the total weight of the occupied points never changes. At the beginning, with a single coin at the origin, the total weight is just$1$. The points outside that square have total weight$15/16$, which is less than$1$, so ... 5 Each of the 16 pawns can move at most 6 times and there are 30 captures possible. Therefore$(16\cdot6+30)\cdot 50=6300$is a rough upper bound (for example, not all pawns can make it to the opposite line without sometimes capturing - which would mean that sometimes pawn move and capture occur together). 5 Going to higher dimension will not help; it makes the problem worse. The Hales-Jewett theorem implies that for any fixed$p$and$n$, the first player can win the$p$-in-a-row game on the$n^d$board for all sufficiently large$d$. So adding dimensions makes these games less fair, not more fair. For example: The first player obviously wins 2-in-a-row ... 4 This is a cute little problem. The insight here is to look for an invariant that is preserved by the allowed moves. In this case, the right way to approach it is to think of the original coin at the origin as having a fixed mass of 1. The rule for replacing the coin with two others can be thought of as breaking the coin in half, to be replaced by two coins ... 4 You're correct that this is a nim variant. More generally, all impartial games (those where both players have the same moves available to them) of this sort fall under the broad category of the Sprage-Grundy theorem which says that their values are equivalent to the value of some nim-pile. But for determining who's to win from a given 'single-position' ... 4 The first player can win. Here is a winning strategy (someone else may have a more elegant explanation). Winning Strategy Say the person who goes first is called$F$, and the person who goes last is called$L$. If a person removes one pebble from an odd pile, that person wins the corresponding sequence of removing one pebble at a time. This makes it easy ... 4 One can show that the second player to play wins if and only if the starting number of stones is exactly a power of two. Proof: Let$(n, k)$denote the position where$n$stones remain and$k$stones are the maximum allowed to be taken. If$k\geq n$the next player to play wins. Otherwise, we show that the first player wins if and only if$k\$ is at least ...

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