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