SOLUTION MANUAL
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS DK DK DK
1 - Nim and Combinatorial Games
DK DK DK DK DK
2 - Congestion Games
DK DK DK
3 - Games in Strategic Form
DK DK DK DK DK
4 - Game Trees with Perfect Information
DK DK DK DK DK DK
5 - Expected Utility
DK DK DK
6 - Mixed Equilibrium
DK DK DK
7 - Brouwer’s Fixed-Point Theorem
DK DK DK DK
8 - Zero-Sum Games
DK DK DK
9 - Geometry of Equilibria in Bimatrix Games
DK DK DK DK DK DK DK
10 - Game Trees with Imperfect Information
DK DK DK DK DK DK
11 - Bargaining
DK DK
12 - Correlated Equilibrium
DK DK DK
2
,Game Theory Basics DK DK
Solutions to Exercises D K D K
© Bernhard von Stengel 2022
D K DK DK DK
Solution to Exercise 1.1 DK DK DK
(a) Let ≤ be defined by (1.7). To show that ≤ is transitive, consider x, y, z with x ≤ y and y ≤ z. If x =
DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK
y then x ≤ z, and if y = z then also x ≤ z. So the only case left is x < y and y < z, which implies x
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK
< z because < is transitive, and hence x ≤ z.
D K DK DK DK DK DK DK DK DK D K DK
Clearly, ≤ is reflexive because x = x and therefore x ≤ x.
DK DK DK DK DK DK DK DK DK DK DK DK
To show that ≤is antisymmetric, consider x and y with x y and ≤
DK DK y x. If we ≤had x ≠ y then x
DKD KDKD KDK DK DK DK DK DK DK DK DKDKDKDKD K DK DK D KDKDKD KDK DK DK DK DK DK DK DK DK
< y and y < x, and by transitivity x < x which contradicts (1.38). Hence x = y, as required. Thi
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K
s shows that ≤ is a partial order.
DK DK DK DK DK DK DK
Finally, we show (1.6), so we have to show that x < y implies x y and x ≠ ≤
DK DK y and vice versa. Let DK DK DK DK DK DK DK DK DK DK DK DK DKDKDK DK DK DK DK DK DK DK DK DK
x < y, which implies x y by (1.7). If we had x = y≤then x < x, contradicting (1.38), so we also have
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
x ≠ y. Conversely, x y and x ≠ y imply by (1.7)x < y or x = y where
DK DK DK
≤ the second case is excluded, DK DKDKD K DK DK DK DK DK DK DK D
K D K D K DK DK D K DK DK DK DK DK DK DK D
hence x < y, as required.
K DK D K D K DK DK
(b) Consider a partial order and≤assume (1.6) as a definition of <. To show that < is transitive, su
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
ppose x < y, that is, x y and x ≠ y, and y <≤z, that is, y z and y ≠ z. Because is transitive,
DK DK DK
≤
DK x z. If DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DKD KDK DK DK DK DKD KDKD K DK
we had x ≤
DK = z then x y and ≤
DK y x and hence x = y by antisymmetry
DK
≤ DK of ≤, which
DK contradicts
DK x DK DKD KDKDKD K DK DK DKDKDKD KDK DK DK DK DK DK DK DK DK DKD KDKD K DK DK DK D
≠ y, so we have x z and x ≠ z, that is,x < z by (1.6), as required.
≤ ≤
K DK DK DK DK DK DKD KDKD K DK DK D K DK DK DK D
K D K DK DK DK DK DK
Also, < is irreflexive, because x < x would by definition mean x x and x ≠ ≤x, but the latter is n
DK DK DK DK DK DK DK DK DK DK DK DK D KDKD K DK DK DK DK DK DK DK DK DK
ot true. DK
Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, give
DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK
n that < is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x ≠ y and then by d
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
efinition x < y. Hence, x ≤ y implies x < y or x = y. Conversely, suppose x < y or x = y. If x
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K D K DK DK DK DK D K DK D K
< y then x ≤ y by (1.6), and if x = y then x ≤ y because ≤ is reflexive. This completes the pro
D K DK DK D K DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK DK D K DK DK DK
of.
Solution to Exercise 1.2 DK DK DK
(a) In analysing the games of three Nim heaps where one heap has size one, we first lookat some
DK D K D K D K D K D K D K D K D K D K D K D K D K D K D K D K DK DK DK
examples, and then use mathematical induction to prove what we conjecture to be the losing positi
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
ons. A losing position is one where every move is to a winning position, because then the oppo
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
nent will win. The point of this exercise is to formulate a precise statement to be proved, and t
DK DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
hen to prove it. DK DK DK
First, if there are only two heaps recall that they are losing if and only if the heaps are of equa
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
l size. If they are of unequal size, then the winning move is to reduce thelarger heap so that bo
DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK D
K DK DK DK DK
th heaps have equal size.
DK DK DK DK
3
, Consider three heaps of sizes 1, m, n, where 1 m ≤n. We ≤ observe the following: 1, 1, m is w
DK DK DK DK DK DK DK DK DK D KDKDKD KDK DKD KDKDKD K DK DK DK DK DK DK DK DK DK
inning, by moving to 1, 1, 0. Similarly, 1, m, m is winning, by moving to 0, m, m. Next, 1, 2, 3 i
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
s losing (observed earlier in the lecture), and hence 1, 2, n for n 4 is winning. 1, 3, n is winnin
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
g for any n 3 by moving to 1, 3, 2. For 1, 4, 5, reducing any heap produces a winning position,
≥ ≥
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
so this is losing.
DK DK DK DK
The general pattern for the losing positions thus seems to be: 1, m, m 1, for even+numbers m.
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D
This includes also the case m = 0, which we can take as the base case foran induction. We now
K DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D
K DK D K DK DK
proceed to prove this formally. DK DK DK DK
First we show that if the positions of the form 1, m, n with m n are losing
DK DK
≤ when m is even a DK DK DK DK DK DK DK DK DK DK DK DK DKDKDKDKD KDK DK DK DK DK DK DK DK
nd n = m 1, then these are
+ the only losing positions because any other position 1, m, n with m
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K DK D
K n is winning. Namely, ≤
D K if m = n then a winning move from1, m, m is to 0, m, m, so we can as
D K DK D K DK DK D K DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
sume m < n. If m is even then n > m 1 (otherwise we would be in the position 1, m, m 1)
+
DK DK DK D K DK DK DK DK DK DK DK D K D K DK DK DK DK DK DK DK DK DK DK D K D K DK
and so the winning move is to 1, m, m 1. If m is odd then the winning move is to 1, m, m 1, th
+ be a winning move from 1, m, m so there the +winni
DK DK DK DK DK DK DK DK DK D K D K DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
e same as position 1, m 1, m (this would also
– −
DK DK DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK DK D K DK D K DK
ng move is not unique).
DK DK D K DK
Second, we show that any move from 1, m, m + 1 with even m is to a winning position,using as in
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
ductive hypothesis that 1, mJ, mJ + 1 for even mJ and mJ < m is a losing position. The move to 0
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
, m, m + 1 produces a winning position with counter-
DK DK DK DK DK DK DK DK DK DK
move to 0, m, m. A move to 1, mJ, m + 1 for mJ < m is to a winning position with the counter-
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
move to 1, mJ, mJ + 1 if mJ is even and to 1, mJ, mJ − 1 if mJ is odd. A move to 1, m, m is to a win
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
ning position with counter- DK DK DK
move to 0, m, m. A move to 1, m, mJ with mJ < m is also to a winning position with the counter-
DK DK DK DK DK DK DK DK DK DK DK D K DK D K DK DK DK DK DK DK DK DK DK
move to 1, mJ − 1, mJ if mJ is odd, and to 1, mJ 1, mJ if mJ is even (in which case mJ 1 < m bec
DK DK DK DK DK DK DK D K DK DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK D K DK DK DK
ause m is even). This concludes the induction proof.
DK DK DK DK DK DK DK DK
+ +
This result is in agreement with the theorem on Nim heap sizes represented as sums of powers of 2
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
0
: 1 D m ∗n +∗ is losing
K
+∗ if and only if, except for 2 , the powers of 2 making upm and n come in pa
D K D K D K D K
D K
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
irs. So these must be the same powers of 2, except for 1 = 20, which occurs in only m or n, where w
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
e have assumed that n is the larger number, so 1 appearsin the representation of n: We have m
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K DK DK
= 2a 2 b 2c DKDKDKDKDKD K DKD KDKDKD KD K
+ + + ··· ··· ≥
D K DK
for a > b > c > 1,so
D K DK DK DK DK DK
+ + + · · · + +
D K D K D K D K D K D K DKDKDKDKDKDKDKD K D
K D
a b c
Km is even, and, with the same a, b, c, . . ., n = 2
D K D K 2 2 1 = m 1. Then
D K DK D K D K D K DK DK DK DK DK D K D K D K
D K D K D K D K D K D K
D K DK DK D K
DK DK DKDKDKD K D K
1 m
∗ + ∗ + ∗ ≡∗D
n 0. K
The following
DKDKD KDKDKD K is an
DK
example using the bit representation
DKDKDKD KD
DK
where K DKDKDKD KDKD K
DK
D K DK DK DK DK DK DK DK DK DK
m = 12 (which determines the bit pattern 1100, which of course depends on m):
DK DK DK DK DK DK DK DK DK DK DK DK DK DK
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) We use (a). Clearly, 1, 2, 3 is losing as shown in (1.2), and because the Nim-
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
sum of the binary representations 01, 10, 11 is 00. Examples show that any other position is
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D
K
winning. The three numbers are n, n 1, n+ 2. If + n is even then reducing the heap of size n 2 t
DK DK DK DK DK DK D K DK D K D K DK DK DK DK DK DK DK DK DK DK DK DK DK
o 1 creates the position n, n 1, 1 which is losing as shown in (a). If n is odd, then n 1 is eve
+ +
DK DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K DK DK
n and n 2 = n 1 1 so by the same argument, a winning move is to reduce the Nim heap o
+ +
DK DK DKDKD K DK D K DKDKD K D KDK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
f size n to 1 (which only works if n > 1).
( + )+
DK DK DK DK DK DK DK DK DK DK DK
D K D K DK
4
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS DK DK DK
1 - Nim and Combinatorial Games
DK DK DK DK DK
2 - Congestion Games
DK DK DK
3 - Games in Strategic Form
DK DK DK DK DK
4 - Game Trees with Perfect Information
DK DK DK DK DK DK
5 - Expected Utility
DK DK DK
6 - Mixed Equilibrium
DK DK DK
7 - Brouwer’s Fixed-Point Theorem
DK DK DK DK
8 - Zero-Sum Games
DK DK DK
9 - Geometry of Equilibria in Bimatrix Games
DK DK DK DK DK DK DK
10 - Game Trees with Imperfect Information
DK DK DK DK DK DK
11 - Bargaining
DK DK
12 - Correlated Equilibrium
DK DK DK
2
,Game Theory Basics DK DK
Solutions to Exercises D K D K
© Bernhard von Stengel 2022
D K DK DK DK
Solution to Exercise 1.1 DK DK DK
(a) Let ≤ be defined by (1.7). To show that ≤ is transitive, consider x, y, z with x ≤ y and y ≤ z. If x =
DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK
y then x ≤ z, and if y = z then also x ≤ z. So the only case left is x < y and y < z, which implies x
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK
< z because < is transitive, and hence x ≤ z.
D K DK DK DK DK DK DK DK DK D K DK
Clearly, ≤ is reflexive because x = x and therefore x ≤ x.
DK DK DK DK DK DK DK DK DK DK DK DK
To show that ≤is antisymmetric, consider x and y with x y and ≤
DK DK y x. If we ≤had x ≠ y then x
DKD KDKD KDK DK DK DK DK DK DK DK DKDKDKDKD K DK DK D KDKDKD KDK DK DK DK DK DK DK DK DK
< y and y < x, and by transitivity x < x which contradicts (1.38). Hence x = y, as required. Thi
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K
s shows that ≤ is a partial order.
DK DK DK DK DK DK DK
Finally, we show (1.6), so we have to show that x < y implies x y and x ≠ ≤
DK DK y and vice versa. Let DK DK DK DK DK DK DK DK DK DK DK DK DKDKDK DK DK DK DK DK DK DK DK DK
x < y, which implies x y by (1.7). If we had x = y≤then x < x, contradicting (1.38), so we also have
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
x ≠ y. Conversely, x y and x ≠ y imply by (1.7)x < y or x = y where
DK DK DK
≤ the second case is excluded, DK DKDKD K DK DK DK DK DK DK DK D
K D K D K DK DK D K DK DK DK DK DK DK DK D
hence x < y, as required.
K DK D K D K DK DK
(b) Consider a partial order and≤assume (1.6) as a definition of <. To show that < is transitive, su
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
ppose x < y, that is, x y and x ≠ y, and y <≤z, that is, y z and y ≠ z. Because is transitive,
DK DK DK
≤
DK x z. If DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DKD KDK DK DK DK DKD KDKD K DK
we had x ≤
DK = z then x y and ≤
DK y x and hence x = y by antisymmetry
DK
≤ DK of ≤, which
DK contradicts
DK x DK DKD KDKDKD K DK DK DKDKDKD KDK DK DK DK DK DK DK DK DK DKD KDKD K DK DK DK D
≠ y, so we have x z and x ≠ z, that is,x < z by (1.6), as required.
≤ ≤
K DK DK DK DK DK DKD KDKD K DK DK D K DK DK DK D
K D K DK DK DK DK DK
Also, < is irreflexive, because x < x would by definition mean x x and x ≠ ≤x, but the latter is n
DK DK DK DK DK DK DK DK DK DK DK DK D KDKD K DK DK DK DK DK DK DK DK DK
ot true. DK
Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, give
DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK
n that < is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x ≠ y and then by d
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
efinition x < y. Hence, x ≤ y implies x < y or x = y. Conversely, suppose x < y or x = y. If x
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K D K DK DK DK DK D K DK D K
< y then x ≤ y by (1.6), and if x = y then x ≤ y because ≤ is reflexive. This completes the pro
D K DK DK D K DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK DK D K DK DK DK
of.
Solution to Exercise 1.2 DK DK DK
(a) In analysing the games of three Nim heaps where one heap has size one, we first lookat some
DK D K D K D K D K D K D K D K D K D K D K D K D K D K D K D K DK DK DK
examples, and then use mathematical induction to prove what we conjecture to be the losing positi
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
ons. A losing position is one where every move is to a winning position, because then the oppo
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
nent will win. The point of this exercise is to formulate a precise statement to be proved, and t
DK DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
hen to prove it. DK DK DK
First, if there are only two heaps recall that they are losing if and only if the heaps are of equa
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
l size. If they are of unequal size, then the winning move is to reduce thelarger heap so that bo
DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK D
K DK DK DK DK
th heaps have equal size.
DK DK DK DK
3
, Consider three heaps of sizes 1, m, n, where 1 m ≤n. We ≤ observe the following: 1, 1, m is w
DK DK DK DK DK DK DK DK DK D KDKDKD KDK DKD KDKDKD K DK DK DK DK DK DK DK DK DK
inning, by moving to 1, 1, 0. Similarly, 1, m, m is winning, by moving to 0, m, m. Next, 1, 2, 3 i
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
s losing (observed earlier in the lecture), and hence 1, 2, n for n 4 is winning. 1, 3, n is winnin
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
g for any n 3 by moving to 1, 3, 2. For 1, 4, 5, reducing any heap produces a winning position,
≥ ≥
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
so this is losing.
DK DK DK DK
The general pattern for the losing positions thus seems to be: 1, m, m 1, for even+numbers m.
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D
This includes also the case m = 0, which we can take as the base case foran induction. We now
K DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D
K DK D K DK DK
proceed to prove this formally. DK DK DK DK
First we show that if the positions of the form 1, m, n with m n are losing
DK DK
≤ when m is even a DK DK DK DK DK DK DK DK DK DK DK DK DKDKDKDKD KDK DK DK DK DK DK DK DK
nd n = m 1, then these are
+ the only losing positions because any other position 1, m, n with m
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K DK D
K n is winning. Namely, ≤
D K if m = n then a winning move from1, m, m is to 0, m, m, so we can as
D K DK D K DK DK D K DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
sume m < n. If m is even then n > m 1 (otherwise we would be in the position 1, m, m 1)
+
DK DK DK D K DK DK DK DK DK DK DK D K D K DK DK DK DK DK DK DK DK DK DK D K D K DK
and so the winning move is to 1, m, m 1. If m is odd then the winning move is to 1, m, m 1, th
+ be a winning move from 1, m, m so there the +winni
DK DK DK DK DK DK DK DK DK D K D K DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
e same as position 1, m 1, m (this would also
– −
DK DK DK DK DK DK DK DK DK DK DK DK D K DK DK DK DK DK D K DK D K DK
ng move is not unique).
DK DK D K DK
Second, we show that any move from 1, m, m + 1 with even m is to a winning position,using as in
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
ductive hypothesis that 1, mJ, mJ + 1 for even mJ and mJ < m is a losing position. The move to 0
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
, m, m + 1 produces a winning position with counter-
DK DK DK DK DK DK DK DK DK DK
move to 0, m, m. A move to 1, mJ, m + 1 for mJ < m is to a winning position with the counter-
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
move to 1, mJ, mJ + 1 if mJ is even and to 1, mJ, mJ − 1 if mJ is odd. A move to 1, m, m is to a win
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
ning position with counter- DK DK DK
move to 0, m, m. A move to 1, m, mJ with mJ < m is also to a winning position with the counter-
DK DK DK DK DK DK DK DK DK DK DK D K DK D K DK DK DK DK DK DK DK DK DK
move to 1, mJ − 1, mJ if mJ is odd, and to 1, mJ 1, mJ if mJ is even (in which case mJ 1 < m bec
DK DK DK DK DK DK DK D K DK DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK D K DK DK DK
ause m is even). This concludes the induction proof.
DK DK DK DK DK DK DK DK
+ +
This result is in agreement with the theorem on Nim heap sizes represented as sums of powers of 2
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
0
: 1 D m ∗n +∗ is losing
K
+∗ if and only if, except for 2 , the powers of 2 making upm and n come in pa
D K D K D K D K
D K
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
irs. So these must be the same powers of 2, except for 1 = 20, which occurs in only m or n, where w
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
e have assumed that n is the larger number, so 1 appearsin the representation of n: We have m
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K DK DK
= 2a 2 b 2c DKDKDKDKDKD K DKD KDKDKD KD K
+ + + ··· ··· ≥
D K DK
for a > b > c > 1,so
D K DK DK DK DK DK
+ + + · · · + +
D K D K D K D K D K D K DKDKDKDKDKDKDKD K D
K D
a b c
Km is even, and, with the same a, b, c, . . ., n = 2
D K D K 2 2 1 = m 1. Then
D K DK D K D K D K DK DK DK DK DK D K D K D K
D K D K D K D K D K D K
D K DK DK D K
DK DK DKDKDKD K D K
1 m
∗ + ∗ + ∗ ≡∗D
n 0. K
The following
DKDKD KDKDKD K is an
DK
example using the bit representation
DKDKDKD KD
DK
where K DKDKDKD KDKD K
DK
D K DK DK DK DK DK DK DK DK DK
m = 12 (which determines the bit pattern 1100, which of course depends on m):
DK DK DK DK DK DK DK DK DK DK DK DK DK DK
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) We use (a). Clearly, 1, 2, 3 is losing as shown in (1.2), and because the Nim-
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
sum of the binary representations 01, 10, 11 is 00. Examples show that any other position is
DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK D
K
winning. The three numbers are n, n 1, n+ 2. If + n is even then reducing the heap of size n 2 t
DK DK DK DK DK DK D K DK D K D K DK DK DK DK DK DK DK DK DK DK DK DK DK
o 1 creates the position n, n 1, 1 which is losing as shown in (a). If n is odd, then n 1 is eve
+ +
DK DK DK DK DK DK D K DK DK DK DK DK DK DK DK DK DK DK DK DK DK D K DK DK
n and n 2 = n 1 1 so by the same argument, a winning move is to reduce the Nim heap o
+ +
DK DK DKDKD K DK D K DKDKD K D KDK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK DK
f size n to 1 (which only works if n > 1).
( + )+
DK DK DK DK DK DK DK DK DK DK DK
D K D K DK
4
