Problem : Compute the position values for 0 through 16 for 1-row misere Nim where the legal moves are to take 1, 4, or 5 stones.

Problem : Compute the Grundy numbers for 0 through 16 for 1-row normal Nim where the legal moves are to take 1, 4, or 5 stones. Find a winning move (or state that there is none) for the 3-row version of this game when there are 3, 5, and 7 stones in the rows.

Problem : Find a winning move (or state that there is none) for the following Kayles positions:

• xxxx xx xxxx Take the group of two: xxxx .. xxxx
• xxxxxxx xxxxx xxx xxxx Take the middle from the group of 5: xxxxxxx xx.xx xxx xxxx
• xxx xxxx xxxxxxx no winning move

Problem : Consider a normal game played with $n$ rows of stones. On each turn, the current player can take any number of stones from any row, or may move any number of stones to the row directly above it (including to a row that has been emptied, not skipping over any row that has not been emptied, and not to a row that never existed).

• Explain why the state with 3 rows of one stone each cannot be viewed simply as the sum of three subgames.
  

  The rows are not independent games since stones can move between rows.

• Compute the Grundy numbers for all positions in this game reachable from an initial position with 3 rows of 1 stone each.

  Position (top/middle/bottom)	Moves	Grundy Numbers of Moves	Grundy Number
  1/1/1	{2/0/1, 1/2/0, 0/1/1, 1/0/1, 1/1/0}	{*0, *2, *0, *2, *0}	*1
  2/0/1	{2/1/0, 1/0/1, 0/0/1, 2/0/0}	{*1, *2, *2, *2}	*0
  1/2/0	{3/0/0, 2/1/0, 0/2/0, 1/1/0, 1/0/0}	{*3, *1, *1, *0, *1}	*2
  2/1/0	{3/0/0, 1/1/0, 0/1/0, 2/0/0}	{*3, *0, *2, *2}	*1
  0/1/1	{1/0/1, 0/2/0, 0/0/1, 0/1/0}	{*2, *1, *1, *2}	*0
  1/0/1	{1/1/0, 0/0/1, 1/0/0}	{*0, *1, *1}	*2
  1/1/0	{2/0/0, 0/1/0, 1/0/0}	{*2, *2, *1}	*0
  3/0/0	{2/0/0, 1/0/0, 0/0/0}	{*2, *1, *0}	*3
  0/2/0	{2/0/0, 1/1/0, 0/1/0, 0/0/0}	{*2, *0, *2, *0}	*1
  2/0/0	{1/0/0, 0/0/0}	{*1, *0}	*2
  0/0/1	{0/1/0, 0/0/0}	{*2, *0}	*1
  0/1/0	{1/0/0, 0/0/0}	{*1, *0}	*2
  1/0/0	{0/0/0}	{*0}	*1
  0/0/0	{}	{}	*0

• Compute a winning move in the summed game consisting of two independent copies of the original game (stones cannot move between copies), for the position with the 1st game in the inital position of one with 3 rows with one stone each, and the other game with just one stone left in the middle position (1/1/1 + 0/1/0), or determine that no winning move exists. Repeat for (1/1/1 + 0/2/0).
  1/1/1 + 0/1/0 is equivalent to *1 + *2. The Nim-sum (exclusive-or) of 1 and 2 is 3. We then need to move from *2 to *1 to make the new Nim-sum *1 + *1. One possible move is in the second game from 0/1/0 to 1/0/0.

  1/1/1 + 0/2/0 is equivalent to *1 + *1. The Nim-sum is 0. So there is no winning move.

• How many positions with 5 total stones are reachable from 0/0/6?

  The 5 stones can be in any of the three rows, so we can view the positions as multisets where each element in the multiset indicates the row a particular stone is in; for example 3/1/1 would be {1,1,1,2,3}. The number of size-5 multisets of {1,2,3} is ${7 \choose 5} = 21$.

Problem :

• Coino is a game played with $n$ coins. On each turn, the player chooses how many of the remaining coins to flip. The player earns one point for each head. If the flip results in at least as many tails as heads then loses the flipped coins for the rest of the game, but the player keeps the points for any heads on that turn. The game is over when the player is out of coins or reaches $k$ points. The player wins after reaching $k$ points and loses after running out of coins with $\lt k$ points. Write a program that computes the strategy for Coino that maximizes the probability of winning and determines that maximum probability.

  def prob(heads, total):
      return len(list(itertools.combinations([i for i in range(total)], heads))) / 2 ** total
  
  def solve_coino(s, g):
      # e[c][t] will be the probability of winning given that we
      # have c coins left and a total of t
      e = [[0.0] * (g + 1) for c in range(s + 1)]
      opt = [[0] * (g + 1) for c in range(s + 1)]
  
      # have already won when t >= g (we clip the columns at g)
      for c in range(s + 1):
          e[c][g] = 1.0
  
      # on every turn the number of coins goes down or the total goes up;
      # compute the rest of e in the opposite of that order
      # (increasing number of coins, decreasing total)
      for c in range(1, s + 1):
          for t in range(g - 1, -1, -1):
              # compute the value of each choice (the number to flip)
              values = [0.0]
              for f in range(1, c + 1):
                  # prob is a function that gives the probability of
                  # flipping exactly h heads out of f flipped coins
                  value = sum(prob(h, f) * e[c - (f if h <= (f - h) else 0)][min(t + h, g)] for h in range(0, f + 1))
                  values.append(value)
              # determine which choice has the max value
              e[c][t] = max(values)
              opt[c][t] = max(enumerate(values), key=lambda p: p[1])[0]
      return e, opt
  

• Suppose we modify Coino so that you get one do-over: at most once during the game you may discard the result of a flip and reflip any number of coins if the initial result was not to your liking. How would you modify the algorithm to find the optimal strategy and the expected winning percentage in this case?
  Let the goal be $g$ and the inital number of coins be $s$. Denote by $E(c,t,r)$ the expected winning percentage given that the game has reached the position with $c$ coins left, $t$ total heads, and $r$ reflips available. Then $E(c,t,r) = 1.0$ when $t >= g$, and $E(c,t,r) = 0.0$ when $c = 0$ and $t \lt g$. For $0 \lt c \lt s$ and $0 \le t \lt g$ we have $$E(c,t,0) = \max_{1 \le f \le c} \left(\sum_{0 \le h \le {f \over 2}}^f P(h {\rm \ heads\ out\ of\ } f)*E(c-f,t+h,0) + \sum_{{f \over 2} \lt h \le f} P(h {\rm \ heads\ out\ of\ } f)*E(c,t+h,0) \right)$$ as before, and $$E(c,t,1) = \max_{1 \le f \le c} \left(\sum_{0 \le h \le {f \over 2}}^f P(h {\rm \ heads\ out\ of\ } f)*\max(E(c,t,0), E(c-f,t+h,1)) + \sum_{{f \over 2} \lt h \le f} P(h {\rm \ heads\ out\ of\ } f)*\max(E(c,t,0),E(c,t+h,1)) \right)$$ where the inner maxes represent the option of taking the result of a flip or using the reflip.

  $OPT(c,t,r)$ would record the argmax of the outer max for each entry to record the optimal choices of how many coins to flip; $OPT(c,t,r,f,h)$ would store the argmax of the inner max to record, for each position $(c,t,r)$, choice ($f$), and outcome of that choice ($h$), whether the optimal choice was to reflip or not.

Problem : Consider a solitaire version of Can't Stop with the objective of minimizing the expected number of turns to end the game. Assuming you had the computational resources to handle the large state space, what algorithm would you use to compute the optimal strategy?

Problem : Consider a press-your-luck solitaire coin-tossing game in which the player flips two coins and scores one point for each head. After each flip, the player can bank the total points and start a new turn, or flip again, except that if any flip ends in two tails, the unbanked points are lost and a new turn starts. The game ends when the player gets to 5 banked points.

• Consider the policy where the player stops a turn when they have 3 points on the current turn, or when banking points ends the game. Set up the system of linear equations for the values of the states reachable under the policy.
  $\begin{array}{rcl} v_{00} & = & -1 + \frac{1}{4} v_{00} + \frac{1}{2}v_{01} + \frac{1}{4} v_{02} \\ v_{01} & = & \frac{1}{4} v_{00} + \frac{1}{2}v_{02} + \frac{1}{4} v_{03} \\ v_{02} & = & \frac{1}{4} v_{00} + \frac{1}{2}v_{03} + \frac{1}{4} v_{04} \\ v_{03} & = & v_{30} \\ v_{04} & = & v_{40} \\ v_{30} & = & -1 + \frac{1}{4} v_{30} + \frac{1}{2}v_{31} + \frac{1}{4} v_{32} \\ v_{31} & = & \frac{1}{4} v_{30} + \frac{1}{2}v_{32} + \frac{1}{4} v_{33} \\ v_{32} & = & v_{50} \\ v_{33} & = & v_{60} \\ v_{40} & = & \frac{1}{4} v_{40} + \frac{1}{2}v_{41} + \frac{1}{4} v_{42} \\ v_{41} & = & v_{50} \\ v_{42} & = & v_{60} \\ v_{50} & = & 0 \\ v_{60} & = & 0 \\ \end{array}$
• If $v_\pi(0,0) = -3\frac{8}{15}$, $v_\pi(0,4) = -\frac{4}{3}$, $v_\pi(0,5) = 0$, and $v_\pi(3,0) = -\frac{8}{5}$ after an iteration of policy iteration, what is $\pi_{new}(0,3)$?
  $q((0,3),flip) = \frac{1}{4} v_\pi(0,0) + \frac{1}{2}v_\pi(0,4) + \frac{1}{4}v_\pi(0,5) = -\frac{31}{20}$; $q((0, 3), stop) = v_\pi(3,0) = -\frac{8}{5}$. $-\frac{31}{20} \gt -\frac{8}{5}$, so $\pi_{new}(0, 3) = flip$.

Problem : Consider a constant-sum game with payoff matrix $$\left( \begin{array}{ccc} 0 & 2 & -1 \\ 1 & -2 & 0 \\ -2 & 0 & 2 \end{array} \right)$$.

• Find $v^-$ and $v^+$ for this game.
  $v^- = -1$, $v^+ = 1$
• Does this game have an equilibrium in pure strategies?
  No, because $v^- \ne v^+$.
• Compute the value of $E(X,Y)$ for $X=(0 {1 \over 2} {1 \over 2})$ and $Y=({2 \over 3} 0 {1 \over 3})$.
  $E(X, Y) = \frac{1}{2} \cdot \frac{2}{3} \cdot 1 + \frac{1}{2} \cdot \frac{1}{3} \cdot 0 + \frac{1}{2} \cdot \frac{2}{3} \cdot -2 + \frac{1}{2} \cdot \frac{1}{3} \cdot 2 = 0$
• Using Theorem 1.3.8c, verify that $X^*=({2 \over 5} {2 \over 5} {1 \over 5})$,$Y^*=({2 \over 5} {1 \over 5} {2 \over 5})$ is a saddle point for this game.
  $E(X^*, Y^*) = 0$. \begin{array}{rcccccl} E(X^*, 1) & = & \frac{2}{5} \cdot 0 + \frac{2}{5} \cdot 1 + \frac{1}{5} \cdot -2 & = & 0 & \ge & 0 \\ E(X^*, 2) & = & \frac{2}{5} \cdot 2 + \frac{2}{5} \cdot -2 + \frac{1}{5} \cdot 0 & = & 0 & \ge & 0 \\ E(X^*, 3) & = & \frac{2}{5} \cdot -1 + \frac{2}{5} \cdot 0 + \frac{1}{5} \cdot 2 & = & 0 & \ge & 0 \\ E(1, Y^*) & = & \frac{2}{5} \cdot 0 + \frac{1}{5} \cdot 2 + \frac{2}{5} \cdot -1 & = & 0 & \le & 0 \\ E(2, Y^*) & = & \frac{2}{5} \cdot 1 + \frac{1}{5} \cdot -2 + \frac{2}{5} \cdot 0 & = & 0 & \le & 0 \\ E(3, Y^*) & = & \frac{2}{5} \cdot -2 + \frac{1}{5} \cdot 0 + \frac{2}{5} \cdot 2 & = & 0 & \le & 0 \\ \end{array} All inequalities are true, so $X^*, Y^*$ is indeed a saddle point.
• Find player I's best response to $Y=({1 \over 3} {1 \over 3} {1 \over 3})$.
  \begin{array}{rcccl} E(1, Y) & = & \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot 2 + \frac{1}{3} \cdot -1 & = & \frac{1}{3} \\ E(2, Y) & = & \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot -2 + \frac{1}{3} \cdot 0 & = & -\frac{1}{3} \\ E(3, Y) & = & \frac{1}{3} \cdot -2 + \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot 2 & = & 0 \\ \end{array} The maximum value is $E(1, Y)$, so the best response is $(1\ 0\ 0)$.

Problem : Find a saddle point and its value for the constant-sum game with payoff matrix $$ \left( \begin{array}{cc} 3 & 2 \\ 0 & 4 \\ \end{array} \right) $$

Problem : Set up the linear program to find a saddle point for the constant-sum game with payoff matrix $$ \left( \begin{array}{ccc} 4 & 7 & 3 \\ 0 & 5 & 4 \\ 10 & 6 & 2 \\ \end{array} \right) $$