Stat 155 Fall 2019
Midterm 1: Solution
Ella Hiesmayr
1.
(a) This game is PBICG since the two players take turns moving, the rules are the same for each
player and the game ends in a finite number of moves no matter how it is played (the number of
squares 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 player
can 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
∈
P
then 4
∈
N
as there is a move 4
→
6. If 6
∈
N
then 5
∈
P
, as the only move from 5
goes to 6. Then 4
∈
N
as 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 initial
position 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 number
of chips in each pile are the number of empty slots to the left of the coin (which corresponds to the
position of the coin). Moving a coin to the left by a certain amount of slots corresponds to reducing the
pile corresponding to that coin by the same amount of chips. The game in the figure thus corresponds
to a Nim-game with piles (2
,
3
,
4
,
4
,
5).
To find out whether this is a
P
or
N
position, we write the height of the piles in binary expansion and
take their Nim-sum. This gives
2 = (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 an
N
position and a winning move is to take away one of
the piles of height 4.
4. See p. I - 15 in Ferguson.