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Stat 155 Fall 2019Midterm 1: SolutionElla Hiesmayr1.(a) This game is PBICG since the two players take turns moving, the rules are the same for eachplayer and the game ends in a finite number of moves no matter how it is played (the number ofsquares is an upper bound on the maximum number of moves from a position).(b) This game is not impartial because one player can only move the black stones and the other playercan only move the white stones.(c) This is not a combinatorial game since the player choose their action simultaneously.2. First, note that the game is impartial and progressively bounded, so all the positions are P- or N-position. If 6∈Pthen 4∈Nas there is a move 4→6. If 6∈Nthen 5∈P, as the only move from 5goes to 6. Then 4∈Nas there is a move 4→5.In both cases 4∈N, the first player wins.P.S.Note that this is the same argument as the one we used in chomp: here adding 1 to the initialposition is analogous to chopping the top-right cell.3. We can represent this game as a Nim-game with 5 piles, one corresponding to each coin. The numberof chips in each pile are the number of empty slots to the left of the coin (which corresponds to theposition of the coin). Moving a coin to the left by a certain amount of slots corresponds to reducing thepile corresponding to that coin by the same amount of chips. The game in the figure thus correspondsto a Nim-game with piles (2,3,4,4,5).To find out whether this is aPorNposition, we write the height of the piles in binary expansion andtake their Nim-sum. This gives2 = (0,1,0)3 = (0,1,1)4 = (1,0,0)4 = (1,0,0)5 = (1,0,1).The Nim-sum of these is (1,0,0), so this is anNposition and a winning move is to take away one ofthe piles of height 4.4. See p. I - 15 in Ferguson.