These are homework assignments for HONORS 391A, a one-credit seminar on Conway's combinatorial game theory. Responses are to be handed in on paper in class on the Monday specified.
(1) In class we have made complete analyses of two games, Nim and Division Nim. We can now tell, for each position in either game, which player has a winning strategy and if so what their first move should be. Your assignment is to make a similar analysis for One-or-Two Nim, the variant of Nim where a move consists of taking either one or two stones from one pile. Given any starting position (a finite number of piles each with a finite number of stones), explain how to determine whether the first player has a winning strategy and if so what move she should make.
(2) At the end of class we defined the game of Red-Blue Hackenbush. In this game a position consists of a number of blue and red edges, each one connected to a "floor" by zero or more other edges. A Left move consists of erasing a blue edge, and a Right move consists of erasing a red edge. If an erasure causes another edge to no longer be connected to the floor, it is also erased.
Consider the position where there are five edges, two blue and three red, called B1, B2, R1, R2, and R3. B1, B2, and R3 are connected to the floor, R1 is connected only to B1, and R2 is connected only to B2. Who wins this game if Left moves first and both players play optimally? Who wins if Right moves first and both players play optimally?
(1) Today we defined addition of two games, and the additive inverse of a game. We also defined two games G and H to have the same value if the sum of G and H's inverse is a zero game. (Recall that a zero game is one in which the second player wins no matter whether Left or Right goes first.) Prove that this relation SV(G, H) is an equivalence relation, that is, that it is reflexive, symmetric, and transitive.
(2) Here are five Red-Blue Hackenbush positions:
(a) Points g (ground), a, b, c, d, e: blue edges g to b and g to d, red edges b to a, d to c, and g to e.
(b) Points g (ground), a, b, c: blue edges g to a and b to c, red edge a to b.
(c) Points g (ground), a, b, c: blue edge g to a, red edges a to b and b to c.
(d) Points g (ground), x, a, b, c: blue edge g to x, x to a, and x to b, red edges g to x, g to x again, x to c.
(e) Points g (ground), a, b, c: blue edges g to a and g to c, red edges a to b and b to c.
You proved in assignment 1 that (a) is a zero game. The numerical value of a single isolated red edge on the ground is -1, so the value of a blue edge with a red edge atop it must satisfy the equation 2x + 1 = 0 and so be 1/2.
It turns out that position (b) has value 3/4. To prove this we can put two of them together with a single isolated red edge (-1) and a red edge with a blue edge atop it (-1/2, as it is the inverse of the blue edge with a red atop it), and show that the resulting game is a zero game.
Can you find the values of (c), (d), and (e), and verify these values by constructing a zero game by adding these games to other games with known values?
Define an RBH position to be grassy if every edge is in a stack with one edge on the ground and any others above it forming a straight line. In class we defined a number to be a dyadic rational if it is of the form n/2k, where n is an integer and k is a natural number. We observed that a number is a dyadic rational if and only if it has a terminating binary representation.
(1) Give a method to determine the numerical value of any finite grassy RBH position, and show that it must be a dyadic rational. Explain why every dyadic rational is the value of some such position.
We also defined the Toads and Frogs game as follows. The game board is a row of cells, and a cell may hold either one toad or one frog. A legal move for Left is to have a toad either move one cell right into an empty cell, or jump to the right over one frog into an empty cell. A legal move for Right is move a frog one cell left into an empty cell, or to jump left over one frog into an empty cell. Of course if there are no legal moves on a player's turn, they lose.
(2) Investigate all the possible positions with four cells, one toad, and one frog. Find numerical values for as many positions as you can. You may find it useful to make composite games with more than one row, with four cells, one frog, and one toad on each row. Of course the value of such a game is the sum of the values of its rows.
In the last homework and in class we determined the numerical values of the 12 possible positions of a Toads-and-Frogs game with four cells, one toad, and one frog. Six positions had value 0, and the others were 2, 1, 1/2, -1/2, -1, and -2.
For this assignment, consider the same game with five cells, two toads, and two frogs. This time there are 30 possible positions, as there are five choices for the black cell and six for the relative position of the toads and frogs. I alleged here that all 30 positions are reachable from the start position TT_FF. That is not true -- three of them are not reachable.
It's easy to believe that the start position has value 0. We know that this position is its own additive inverse, as if we reverse it left-to-right and exchange toads and frogs, we get TT_FF back. But this does not mean that it must be 0! 0 is the only number that is its own additive inverse, but it is not the only game value with that property.
Consider a single non-empty pile in ordinary Nim. If we exchange the roles of Left and Right, we get the same game back, as the game is non-partisan. So this game is its own inverse, but it is not a zero game since the first player, not the second, always wins. The value of a Nim pile of one stone is called * or "star". The values of larger Nim piles are different, but each is also its own additive inverse.
We looked in class at evaluation of general games where all the options for both players were numbers. The Fundamental Theorem of Conway Numbers says that in such a game where Left's best option is a number x, Right's best option is a number y, and x < y, the value of the entire game is the "simplest" number that is strictly between x and y. This rule will get you a long way in evaluating Toads-and-Frogs positions. If there is just one option for each player and those two options have numerical values, we are doing a strange sort of arithmetic. The integers are defined by 1 = {0|}, 2 = {1|},... and -1 = {|0}, -2 = {|-1},..., and for integers m and n with m < n we have that {m|n} is the simplest number that is strictly between m and n. If there is an integer between m and n, {m|n} is the one of those integers that is closest to 0. If n = m + 1, then {m|n} = m + 1/2.
But if there is a Left option that is as good or better than any Right option, the conditions of the Theorem do not apply. An example of this is a size-1 Nim pile, where Left and Right each have one option, which is 0. We can define the game value {0|0} as *, and * is not a number.
If you complete the entire derivation, you will get some values of *, and some other values that are not numbers. Try to determine whether these game values are positive (Left always wins), negative (Right always wins), zero (second player always wins) or fuzzy (first player always wins). If a non-number value is positive or negative, how does it compare to other numbers?
We also talked about single-stalk RBH games that have infinitely many edges. If the edges are all blue, the value of the game is larger than any given positive integer, but is still a "number" in Conway's terminology. We call its value ω or "omega", after the first infinite ordinal in set theory. Ordinals include the natural numbers -- along with the axioms about 0 and successor, we add a new axiom that any set of ordinals has a least upper bound. The ordinal ω is the least upper bound of the set of naturals, and we can identify it with the game {0,1,2,3,...|} which is the value of the infinite stalk of blue RBH edges.
Consider an infinite RBH stalk with one blue edge at the bottom and the rest of its edges red. This game is positive, but less than 2-k for any positive integer k. We can call it 1/ω. Your first assignment is to determine which of 1/ω and "up" is greater, and prove your answer.
Our new game for the week is called Domineering. The game board is a set of squares, as on a chessboard. Each connected component of squares is a separate game, and these games can be added. A Left move is to place a 1 by 2 vertical domino on two of the squares, covering them and removing them from the game. A Right move is to place a 2 by one horizontal domino. Of course the first player who cannot move is the loser.
Try playing some games, like a 2 by 2 square, a 3 by 3 square, or an L-shaped figure like this:
O
O
OOO
Here are some more patterns to evaluate. All but one have values that we have seen -- the new one has to stay in curly-bracket notation.
O OO O O OOO
OO OO OOO OOOO O
O OOO
(1) In Grundy's Game, we have zero or more nonempty piles of stones as in Nim. A legal move is to divide any pile into two nonempty piles of different sizes. As usual, a player who cannot move loses. It turns out that every pile in Grundy's Game has a value equal to that of some Nim pile. Determine, for as many values of n as you can, the Nim pile equivalent to a Grundy pile of n stones. Explain your reasoning.
(2) Northcott's Game is played on an 8 by 8 chessboard, with one White piece and one Black piece on each row. You may move any piece of your own color to any vacant square on the same row, provided you do not jump over the opponent's piece on that row. The first player who cannot move is the loser. It is not obvious that this game satisfies the property that a game cannot go on forever. In fact two players who want the game to go on forever can make it do so. But from any position, one player has a winning strategy that ends the game in finite time. Can you show this for the 3 by 3 version of the game?
(3) Read pages 1-18 of the Knuth book, available for free download here.
(1) Investigate the variant of Grundy's game where a legal move is to divide a pile into non-empty piles, which may be equal in size. Determine the Nim stack equivalent to this game with n = 117, and prove your answer.
(2) Dawson's Chess is played on a chessboard with three rows and n columns. White and Black each have a row of pawns on the row closest to them. A legal move is to either move a pawn one square forward or to capture an enemy pawn by moving diagonally one square forward. Captures are mandatory: if a player has a capture available she must take it, and if she has more than one capture available she must choose one of them.
Despite appearances, this is an impartial game and thus by the Sprague-Grundy Theorem it is equivalent to some Nim-pile for each n. Determine these Nim-values for all n from 0 through 20, explaining how you are getting your answers.
(3) Read Surreal Numbers through page 54. Think about how the stone's definition of "adding numbers" relates to our definition of "adding games".
(1) Your main assignment is to choose groups and a topic for your final presentation, in consultation (probably email) with Dan and me. I know of three of the five groups -- Conor, Sara, George, Luke, and Andrew need to form a group of three and a group of two.
(2) Using the Conway definition of multiplication, compute the number (1/2)ω, where ω is the game {0,1,2,3,...|}.
(3) Let X be the game {0,1,2,3,...|ω, ω/2, ω/4, ω/8,...}, where "ω/8", for example, means (1/8) times ω. Show that by Conway's definition of multiplication, X2 = ω.
Optional Extra Credit for 18 November 2013:
(1) Compute the product of 1/2 and 1/2.
(2) Compute the product of 1/3 and 3/2, where 1/3 is defined to be {0, 1/4, 5/16, 21/64,...|1/2, 3/8, 11/32, 43/128/...}
(3) Compute the product of ω and ω, where ω is given above.
(4) Let X be the number {1|2, 3/2, 5/4, 9/8,...}. Compute the product of X and ω.
Last modified 17 November 2013