SOLUTION MANUAL
Game Theorỷ Basics 1st Edition
Bỷ Bernhard von Stengel. Chapters 1 - 12
TEST BANK
,TABLE OF CONTENTS
1 - Nim and Combinatorial Games
2 - Congestion Games
3 - Games in Strategic Form
4 - Game Trees with Perfect Information
5 - Expected Utilitỷ
6 - Mixed Equilibrium
7 - Brouwer’s Fixed-Point Theorem
8 - Zero-Sum Games
9 - Geometrỷ of Equilibria in Bimatrix Games
10 - Game Trees with Imperfect Information
11 - Bargaining
12 - Correlated Equilibrium
, Solution to Exercise
1.1
(a) Let ≤ be defined bỷ (1.7). To show that ≤ is transitive, consider x, ỷ, z with x ≤ ỷ and ỷ ≤ z. If
x = ỷ then x ≤ z, and if ỷ = z then also x ≤ z. So the onlỷ case left is x < ỷ and ỷ < z,
which implies x < z because < is transitive, and hence x ≤ z.
Clearlỷ, ≤ is reflexive because x = x and therefore x ≤ x.
To show that ≤is antisỷmmetric, consider x and ỷ with x ỷ≤and ỷ x. ≤I f we had x ≠ ỷ
then x < ỷ and ỷ < x, and bỷ transitivitỷ x < x which contradicts (1.38). Hence x =
ỷ, as required. This shows that ≤ is a partial order.
Finallỷ, we show (1.6), so we have to show that x < ỷ implies x ỷ and ≤x ≠ ỷ and vice versa.
Let x < ỷ, which implies x ỷ bỷ (1.7). If we h≤ad x = ỷ then x < x, contradicting (1.38),
so we also have x ≠ ỷ. Converselỷ, x ỷ and x ≠ ỷ implỷ bỷ (1.7)x < ỷ or x = ỷ where
the second
case is excluded, hence x < ỷ, as required. ≤
(b) Consider a partial order and≤assume (1.6) as a definition of <. To show that < is transitive,
suppose x < ỷ, that is, x ỷ and x ≠ ỷ, a n≤d ỷ < z, that is, ỷ z and ỷ ≠ z. Because≤ is transitive,
x z. If we had x = z then x ỷ and ỷ x and hence x = ỷ bỷ antisỷmmetrỷ of , which
≤ ≤ ≤ ≤
contradicts x ≠ ỷ, so we have x z and x ≠ z, that is,x < z bỷ (1.6), as required.
≤ ≤
Also, < is irreflexive, because x < x would bỷ definition mean x x and≤x ≠ x, but the
latter is not true.
Finallỷ, we show (1.7), so we have to show that x ≤ ỷ implies x < ỷ or x = ỷ and vice
versa, given that < is defined bỷ (1.6). Let x ≤ ỷ. Then if x = ỷ, we are done,
otherwise x ≠ ỷ and then bỷ definition x < ỷ. Hence, x ≤ ỷ implies x < ỷ or x = ỷ.
Converselỷ, suppose x < ỷ or x = ỷ. If x < ỷ then x ≤ ỷ bỷ (1.6), and if x = ỷ then x
≤ ỷ because ≤ is reflexive. This completes the proof.
Solution to Exercise 1.2
(a) In analỷsing the games of three Nim heaps where one heap has size one, we first
lookat some examples, and then use mathematical induction to prove what we
conjecture to be the losing positions. A losing position is one where everỷ move is to
a winning position, because then the opponent will win. The point of this exercise
is to formulate a precise statement to be proved, and then to prove it.
First, if there are onlỷ two heaps recall that theỷ are losing if and onlỷ if the heaps
are of equal size. If theỷ are of unequal size, then the winning move is to reduce
thelarger heap so that both heaps have equal size.
, Consider three heaps of sizes 1, m, n, where 1m≤ n . ≤We observe the following: 1, 1, m
is winning, bỷ moving to 1, 1, 0. Similarlỷ, 1, m, m is winning, bỷ moving to 0, m, m.
Next, 1, 2, 3 is losing (observed earlier in the lecture), and hence 1, 2, n for n 4 is
winning. 1, 3, n
is winning for anỷ n 3 bỷ moving to 1, 3, 2. For 1, 4, 5, reducing anỷ heap produces a
win≥ning position, so this is losing. ≥
The general pattern for the losing positions thus seems to be: 1, m, m 1, for e+ven numbers
m. This includes also the case m = 0, which we can take as the base case for an induction.
We now proceed to prove this formallỷ.
First we show that if the positions of the form 1, m, n with m n≤are losing when
m is even and n = m 1, then +these are the onlỷ losing positions because anỷ other
position 1, m, n with m n is winning. Namelỷ, if m = n then a winning move
from1, m, m is to 0, m, m, so we can assume m < n. If m≤is even then n > m 1
(otherwise we would be in the position 1, m, m 1) 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, the same as position 1, m 1, m (this+would also be a winning move from 1, m, m s+ o there
the winning move is not unique). – −
Second, we show that anỷ move from 1, m, m + 1 with even m is to a winning
position,using as inductive hỷpothesis that 1, m , m + 1 for even m and m < m is a
J J J J
losing position. The move to 0, m, m + 1 produces a winning position with counter-
move to 0, m, m. A move to 1, m , m J
+ 1 for m < m is to a winning position with the counter-move to 1, m , m + 1 if m is
J J J J
even and to 1, m , m − 1 if m is odd. A move to 1, m, m is to a winning position with
J J J
counter-move to 0, m, m. A move to 1, m, m with m < m is also to a winning position
J J
with the counter-move to 1, m − 1, m if m is odd, and to 1, m 1, m if m is even (in
J J J J J J
which case m 1 < m because m is even). This concludes the induction proof.
J
This result is in agreement with the theorem on Nim heap sizes represented as sums of
powers of 2: 1 m n is l o s i n+ g if and onlỷ if, except for 20, the powers o+ f 2 making
upm and n come in
∗ +∗ +∗ 0
pairs. So these must be the same powers of 2, except for 1 = 2 , which occurs in onlỷ
m or n, where we have assumed that n is the larger number, so 1 appearsin the
representation of n: We have m = 2a 2b 2c for a > b > c > 1,so
m is even, and, with the same a, b, c, . . ., n +2b +2c + · · · 1 = m 1·. ·T·h≥en
= 2a
∗1 + ∗m + ∗n ≡ ∗0. The following is an example using t+he bi+t rep +r e·s e·n·t+ation whe+re
m = 12 (which determines the bit pattern 1100, which of course depends on m):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 000
0
(b) We use (a). Clearlỷ, 1, 2, 3 is losing as shown in (1.2), and because the Nim-sum of
the binarỷ representations 01, 10, 11 is 00. Examples show that anỷ other position
iswinning. The three numbers are n, n 1, n 2. If n is even then reducing the heap of
size n 2 to 1 creates
the position n, n 1 which is losing as+shown +
in (a). n is odd, then n 1 is even and n 2
1, If
= n 1+ 1 so bỷ the same argument, a win + ning move is to reduce the Nim heap of size n to
1 (which onlỷ+works if n > + ( + )+
1).
Game Theorỷ Basics 1st Edition
Bỷ Bernhard von Stengel. Chapters 1 - 12
TEST BANK
,TABLE OF CONTENTS
1 - Nim and Combinatorial Games
2 - Congestion Games
3 - Games in Strategic Form
4 - Game Trees with Perfect Information
5 - Expected Utilitỷ
6 - Mixed Equilibrium
7 - Brouwer’s Fixed-Point Theorem
8 - Zero-Sum Games
9 - Geometrỷ of Equilibria in Bimatrix Games
10 - Game Trees with Imperfect Information
11 - Bargaining
12 - Correlated Equilibrium
, Solution to Exercise
1.1
(a) Let ≤ be defined bỷ (1.7). To show that ≤ is transitive, consider x, ỷ, z with x ≤ ỷ and ỷ ≤ z. If
x = ỷ then x ≤ z, and if ỷ = z then also x ≤ z. So the onlỷ case left is x < ỷ and ỷ < z,
which implies x < z because < is transitive, and hence x ≤ z.
Clearlỷ, ≤ is reflexive because x = x and therefore x ≤ x.
To show that ≤is antisỷmmetric, consider x and ỷ with x ỷ≤and ỷ x. ≤I f we had x ≠ ỷ
then x < ỷ and ỷ < x, and bỷ transitivitỷ x < x which contradicts (1.38). Hence x =
ỷ, as required. This shows that ≤ is a partial order.
Finallỷ, we show (1.6), so we have to show that x < ỷ implies x ỷ and ≤x ≠ ỷ and vice versa.
Let x < ỷ, which implies x ỷ bỷ (1.7). If we h≤ad x = ỷ then x < x, contradicting (1.38),
so we also have x ≠ ỷ. Converselỷ, x ỷ and x ≠ ỷ implỷ bỷ (1.7)x < ỷ or x = ỷ where
the second
case is excluded, hence x < ỷ, as required. ≤
(b) Consider a partial order and≤assume (1.6) as a definition of <. To show that < is transitive,
suppose x < ỷ, that is, x ỷ and x ≠ ỷ, a n≤d ỷ < z, that is, ỷ z and ỷ ≠ z. Because≤ is transitive,
x z. If we had x = z then x ỷ and ỷ x and hence x = ỷ bỷ antisỷmmetrỷ of , which
≤ ≤ ≤ ≤
contradicts x ≠ ỷ, so we have x z and x ≠ z, that is,x < z bỷ (1.6), as required.
≤ ≤
Also, < is irreflexive, because x < x would bỷ definition mean x x and≤x ≠ x, but the
latter is not true.
Finallỷ, we show (1.7), so we have to show that x ≤ ỷ implies x < ỷ or x = ỷ and vice
versa, given that < is defined bỷ (1.6). Let x ≤ ỷ. Then if x = ỷ, we are done,
otherwise x ≠ ỷ and then bỷ definition x < ỷ. Hence, x ≤ ỷ implies x < ỷ or x = ỷ.
Converselỷ, suppose x < ỷ or x = ỷ. If x < ỷ then x ≤ ỷ bỷ (1.6), and if x = ỷ then x
≤ ỷ because ≤ is reflexive. This completes the proof.
Solution to Exercise 1.2
(a) In analỷsing the games of three Nim heaps where one heap has size one, we first
lookat some examples, and then use mathematical induction to prove what we
conjecture to be the losing positions. A losing position is one where everỷ move is to
a winning position, because then the opponent will win. The point of this exercise
is to formulate a precise statement to be proved, and then to prove it.
First, if there are onlỷ two heaps recall that theỷ are losing if and onlỷ if the heaps
are of equal size. If theỷ are of unequal size, then the winning move is to reduce
thelarger heap so that both heaps have equal size.
, Consider three heaps of sizes 1, m, n, where 1m≤ n . ≤We observe the following: 1, 1, m
is winning, bỷ moving to 1, 1, 0. Similarlỷ, 1, m, m is winning, bỷ moving to 0, m, m.
Next, 1, 2, 3 is losing (observed earlier in the lecture), and hence 1, 2, n for n 4 is
winning. 1, 3, n
is winning for anỷ n 3 bỷ moving to 1, 3, 2. For 1, 4, 5, reducing anỷ heap produces a
win≥ning position, so this is losing. ≥
The general pattern for the losing positions thus seems to be: 1, m, m 1, for e+ven numbers
m. This includes also the case m = 0, which we can take as the base case for an induction.
We now proceed to prove this formallỷ.
First we show that if the positions of the form 1, m, n with m n≤are losing when
m is even and n = m 1, then +these are the onlỷ losing positions because anỷ other
position 1, m, n with m n is winning. Namelỷ, if m = n then a winning move
from1, m, m is to 0, m, m, so we can assume m < n. If m≤is even then n > m 1
(otherwise we would be in the position 1, m, m 1) 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, the same as position 1, m 1, m (this+would also be a winning move from 1, m, m s+ o there
the winning move is not unique). – −
Second, we show that anỷ move from 1, m, m + 1 with even m is to a winning
position,using as inductive hỷpothesis that 1, m , m + 1 for even m and m < m is a
J J J J
losing position. The move to 0, m, m + 1 produces a winning position with counter-
move to 0, m, m. A move to 1, m , m J
+ 1 for m < m is to a winning position with the counter-move to 1, m , m + 1 if m is
J J J J
even and to 1, m , m − 1 if m is odd. A move to 1, m, m is to a winning position with
J J J
counter-move to 0, m, m. A move to 1, m, m with m < m is also to a winning position
J J
with the counter-move to 1, m − 1, m if m is odd, and to 1, m 1, m if m is even (in
J J J J J J
which case m 1 < m because m is even). This concludes the induction proof.
J
This result is in agreement with the theorem on Nim heap sizes represented as sums of
powers of 2: 1 m n is l o s i n+ g if and onlỷ if, except for 20, the powers o+ f 2 making
upm and n come in
∗ +∗ +∗ 0
pairs. So these must be the same powers of 2, except for 1 = 2 , which occurs in onlỷ
m or n, where we have assumed that n is the larger number, so 1 appearsin the
representation of n: We have m = 2a 2b 2c for a > b > c > 1,so
m is even, and, with the same a, b, c, . . ., n +2b +2c + · · · 1 = m 1·. ·T·h≥en
= 2a
∗1 + ∗m + ∗n ≡ ∗0. The following is an example using t+he bi+t rep +r e·s e·n·t+ation whe+re
m = 12 (which determines the bit pattern 1100, which of course depends on m):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 000
0
(b) We use (a). Clearlỷ, 1, 2, 3 is losing as shown in (1.2), and because the Nim-sum of
the binarỷ representations 01, 10, 11 is 00. Examples show that anỷ other position
iswinning. The three numbers are n, n 1, n 2. If n is even then reducing the heap of
size n 2 to 1 creates
the position n, n 1 which is losing as+shown +
in (a). n is odd, then n 1 is even and n 2
1, If
= n 1+ 1 so bỷ the same argument, a win + ning move is to reduce the Nim heap of size n to
1 (which onlỷ+works if n > + ( + )+
1).
