SOLUTION MANUAL
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS IS IS IS
1 - Nim and Combinatorial Games
IS IS IS IS IS
2 - Congestion Games
IS IS IS
3 - Games in Strategic Form
IS IS IS IS IS
4 - Game Trees with Perfect Information
IS IS IS IS IS IS
5 - Expected Utility
IS IS IS
6 - Mixed Equilibrium
IS IS IS
7 - Brouwer’s Fixed-Point Theorem
IS IS IS IS
8 - Zero-Sum Games
IS IS IS
9 - Geometry of Equilibria in Bimatrix Games
IS IS IS IS IS IS IS
10 - Game Trees with Imperfect Information
IS IS IS IS IS IS
11 - Bargaining
IS IS
12 - Correlated Equilibrium
IS IS IS
2
,Game Theory Basics IS IS
Solutions to Exercises I S I S
© Bernhard von Stengel 2022
I S IS IS IS
Solution to Exercise 1.1 IS IS IS
(a) Let ≤ be defined by (1.7). To show that ≤ is transitive, consider x, y, z with x ≤ y and y ≤ z. If x = y t
IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS
hen x ≤ z, and if y = z then also x ≤ z. So the only case left is x < y and y < z, which implies x < z
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
because < is transitive, and hence x ≤ z. IS IS IS IS IS IS IS IS
Clearly, ≤ is reflexive because x = x and therefore x ≤ x.
IS IS IS IS IS IS IS IS IS IS IS IS
To show that ≤is antisymmetric, consider x and y with x y and y ≤ x. If we had
IS IS
≤ x ≠ y then x <
ISISISISIS IS IS IS IS IS IS IS ISISISISIS IS IS ISISISISIS IS IS IS IS IS IS IS IS IS IS
y and y < x, and by transitivity x < x which contradicts (1.38). Hence x = y, as required. This sh
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
ows that ≤ is a partial order.
IS IS IS IS IS IS
Finally, we show (1.6), so we have to show that x < y implies x y and x ≠ y ≤
IS and vice versa. Let x <
IS IS IS IS IS IS IS IS IS IS IS IS IS ISISIS IS IS IS IS IS IS IS IS IS IS
y, which implies x y by (1.7). If we had x = y then
IS IS
≤ x < x, contradicting (1.38), so we also have x ≠ y.
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
Conversely, x y and x ≠ y imply by (1.7)x < y or x = y where the ≤
IS second case is excluded, hence
IS ISISI S IS IS IS IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS
x < y, as required.
IS IS IS IS
(b) Consider a partial order and≤assume (1.6) as a definition of <. To show that < is transitive, sup
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
pose x < y, that is, x y and x ≠ y, and y < z,≤that is, y z and y ≠ z. Because is transitive,
IS IS IS IS
≤ x z. If we IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS ISISISIS IS IS ISISISIS IS IS IS
had x = z then
≤ x y and y x
IS and
≤ hence
IS x = y by antisymmetry
IS
≤ of
IS , which
≤ contradicts x ≠ y, so IS ISISISISIS IS IS ISISISISIS IS IS IS IS IS IS IS IS ISISISIS IS IS IS IS IS IS I
we have x z and x ≠ z, that is,x < z by (1.6), as required.
≤ ≤
S IS IS ISISISI S IS IS IS IS IS IS IS I S IS IS IS IS IS
Also, < is irreflexive, because x < x would by definition mean x x and x ≠ x,≤but the latter is not
IS IS IS IS IS IS IS IS IS IS IS IS ISISIS IS IS IS IS IS IS IS IS IS IS
true.
Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, given
IS IS IS IS IS IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS
that < is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x ≠ y and then by defin
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
ition x < y. Hence, x ≤ y implies x < y or x = y. Conversely, suppose x < y or x = y. If x < y th
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS I S I S IS IS IS IS I S IS I S I S IS
en x ≤ y by (1.6), and if x = y then x ≤ y because ≤ is reflexive. This completes the proof.
IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
Solution to Exercise 1.2 IS IS IS
(a) In analysing the games of three Nim heaps where one heap has size one, we first lookat some e
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
xamples, and then use mathematical induction to prove what we conjecture to be the losing position
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
s. A losing position is one where every move is to a winning position, because then the opponen
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
t will win. The point of this exercise is to formulate a precise statement to be proved, and then t
IS IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
o prove it. IS IS
First, if there are only two heaps recall that they are losing if and only if the heaps are of equal
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
size. If they are of unequal size, then the winning move is to reduce thelarger heap so that both
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS I
heaps have equal size.
S IS IS IS
3
, Consider three heaps of sizes 1, m, n, where 1 m n.≤We ≤ observe the following: 1, 1, m is win
IS IS IS IS IS IS IS IS IS ISISISISIS ISISISISIS IS IS IS IS IS IS IS IS IS
ning, by moving to 1, 1, 0. Similarly, 1, m, m is winning, by moving to 0, m, m. Next, 1, 2, 3 is lo
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
sing (observed earlier in the lecture), and hence 1, 2, n for n 4 is winning. 1, 3, n is winning for
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
any n 3 by moving to 1, 3, 2. For 1, 4, 5, reducing any heap produces a winning position, so thi
≥ ≥
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
s is losing.
IS IS
The general pattern for the losing positions thus seems to be: 1, m, m 1, for even+numbers m.
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
This includes also the case m = 0, which we can take as the base case foran induction. We now pr
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
oceed to prove this formally. IS IS IS IS
First we show that if the positions of the form 1, m, n with m n are losing
IS IS
≤ when m is even an IS IS IS IS IS IS IS IS IS IS IS IS ISISISISISIS IS IS IS IS IS IS IS
d n = m 1, then these are+the only losing positions because any other position 1, m, n with m
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS I S IS I S I S
n is winning. Namely, if m≤ = n then a winning move from1, m, m is to 0, m, m, so we can assum
I S IS I S IS IS I S IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
e m < n. If m is even then n > m 1 (otherwise we would be in the position 1, m, m 1) and so
+
IS IS IS I S IS IS IS IS IS IS IS I S I S IS IS IS IS IS IS IS IS IS IS I S I S IS IS I
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
+ +
S IS IS IS IS IS IS IS I S I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
position 1, m 1, m (this would also be a winning move from 1, m, m so there the winning move is
– −
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
not unique). IS
Second, we show that any move from 1, m, m + 1 with even m is to a winning position,using as ind
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
uctive hypothesis that 1, mJ, mJ + 1 for even mJ and mJ < m is a losing position. The move to 0, m,
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
m + 1 produces a winning position with counter-
IS IS IS IS IS IS IS IS IS
move to 0, m, m. A move to 1, mJ, m + 1 for mJ < m is to a winning position with the counter-
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
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 winni
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
ng position with counter-
IS IS IS
move to 0, m, m. A move to 1, m, mJ with mJ < m is also to a winning position with the counter-
IS IS IS IS IS IS IS IS IS IS IS I S IS I S IS IS IS IS IS IS IS IS IS
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 becau
IS IS IS IS IS IS IS I S IS IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS I S IS IS IS
se m is even). This concludes the induction proof.
IS IS IS IS IS IS IS IS
+ +
This result is in agreement with the theorem on Nim heap sizes represented as sums of powers of 2:
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
1 m ∗n is+∗ losing 0
I S
+∗if and only if, except for 2 , the powers of 2 making upm and n come in pairs
I S I S I S I S IS
I S
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
. So these must be the same powers of 2, except for 1 = 20, which occurs in only m or n, where we ha
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
ve assumed that n is the larger number, so 1 appearsin the representation of n: We have m = 2
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS I S IS IS IS IS
a 2b 2c
ISISISISISIS ISISISISISIS
+ + + ··· ··· ≥ I S IS IS IS IS IS
for a > b > c > 1,so
+ + + · · · + +
IS I S IS I S IS I S ISISISISISISISI S IS IS
m is even, and, with the same a, b, c, . . ., n = 2
IS IS
a 2b 2 c IS 1 = m 1. Then IS IS IS IS IS IS IS IS IS IS I S IS
I S I S I S I S I S I S
IS IS IS IS
IS IS ISISISI S I S
1 m
∗ + ∗ + ∗ ≡∗
I
n 0.S
The following
ISISISISISI S
IS
is an example using the bit representation
ISISISISI S
IS
where ISISISISISI S
IS
I S IS IS IS IS IS IS IS IS IS
m = 12 (which determines the bit pattern 1100, which of course depends on m):
IS IS IS IS IS IS IS IS IS IS IS IS IS IS
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-
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
sum of the binary representations 01, 10, 11 is 00. Examples show that any other position iswi
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
nning. The three numbers are n, n 1, n +2. If n+is even then reducing the heap of size n 2 to 1 c
IS IS IS IS IS IS I S IS I S I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
reates the position n, n 1, 1 which is losing as shown in (a). If n is odd, then n 1 is even and
+ +
IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS I S IS IS IS IS
n 2 = n 1 1 so by the same argument, a winning move is to reduce the Nim heap of size n t
+ +
ISISIS IS I S ISISIS ISISIS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
o 1 (which only works if n > 1).
( + )+
IS IS IS IS IS IS IS IS
I S I S IS
4
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS IS IS IS
1 - Nim and Combinatorial Games
IS IS IS IS IS
2 - Congestion Games
IS IS IS
3 - Games in Strategic Form
IS IS IS IS IS
4 - Game Trees with Perfect Information
IS IS IS IS IS IS
5 - Expected Utility
IS IS IS
6 - Mixed Equilibrium
IS IS IS
7 - Brouwer’s Fixed-Point Theorem
IS IS IS IS
8 - Zero-Sum Games
IS IS IS
9 - Geometry of Equilibria in Bimatrix Games
IS IS IS IS IS IS IS
10 - Game Trees with Imperfect Information
IS IS IS IS IS IS
11 - Bargaining
IS IS
12 - Correlated Equilibrium
IS IS IS
2
,Game Theory Basics IS IS
Solutions to Exercises I S I S
© Bernhard von Stengel 2022
I S IS IS IS
Solution to Exercise 1.1 IS IS IS
(a) Let ≤ be defined by (1.7). To show that ≤ is transitive, consider x, y, z with x ≤ y and y ≤ z. If x = y t
IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS
hen x ≤ z, and if y = z then also x ≤ z. So the only case left is x < y and y < z, which implies x < z
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
because < is transitive, and hence x ≤ z. IS IS IS IS IS IS IS IS
Clearly, ≤ is reflexive because x = x and therefore x ≤ x.
IS IS IS IS IS IS IS IS IS IS IS IS
To show that ≤is antisymmetric, consider x and y with x y and y ≤ x. If we had
IS IS
≤ x ≠ y then x <
ISISISISIS IS IS IS IS IS IS IS ISISISISIS IS IS ISISISISIS IS IS IS IS IS IS IS IS IS IS
y and y < x, and by transitivity x < x which contradicts (1.38). Hence x = y, as required. This sh
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
ows that ≤ is a partial order.
IS IS IS IS IS IS
Finally, we show (1.6), so we have to show that x < y implies x y and x ≠ y ≤
IS and vice versa. Let x <
IS IS IS IS IS IS IS IS IS IS IS IS IS ISISIS IS IS IS IS IS IS IS IS IS IS
y, which implies x y by (1.7). If we had x = y then
IS IS
≤ x < x, contradicting (1.38), so we also have x ≠ y.
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
Conversely, x y and x ≠ y imply by (1.7)x < y or x = y where the ≤
IS second case is excluded, hence
IS ISISI S IS IS IS IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS
x < y, as required.
IS IS IS IS
(b) Consider a partial order and≤assume (1.6) as a definition of <. To show that < is transitive, sup
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
pose x < y, that is, x y and x ≠ y, and y < z,≤that is, y z and y ≠ z. Because is transitive,
IS IS IS IS
≤ x z. If we IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS ISISISIS IS IS ISISISIS IS IS IS
had x = z then
≤ x y and y x
IS and
≤ hence
IS x = y by antisymmetry
IS
≤ of
IS , which
≤ contradicts x ≠ y, so IS ISISISISIS IS IS ISISISISIS IS IS IS IS IS IS IS IS ISISISIS IS IS IS IS IS IS I
we have x z and x ≠ z, that is,x < z by (1.6), as required.
≤ ≤
S IS IS ISISISI S IS IS IS IS IS IS IS I S IS IS IS IS IS
Also, < is irreflexive, because x < x would by definition mean x x and x ≠ x,≤but the latter is not
IS IS IS IS IS IS IS IS IS IS IS IS ISISIS IS IS IS IS IS IS IS IS IS IS
true.
Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, given
IS IS IS IS IS IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS
that < is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x ≠ y and then by defin
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
ition x < y. Hence, x ≤ y implies x < y or x = y. Conversely, suppose x < y or x = y. If x < y th
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS I S I S IS IS IS IS I S IS I S I S IS
en x ≤ y by (1.6), and if x = y then x ≤ y because ≤ is reflexive. This completes the proof.
IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
Solution to Exercise 1.2 IS IS IS
(a) In analysing the games of three Nim heaps where one heap has size one, we first lookat some e
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
xamples, and then use mathematical induction to prove what we conjecture to be the losing position
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
s. A losing position is one where every move is to a winning position, because then the opponen
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
t will win. The point of this exercise is to formulate a precise statement to be proved, and then t
IS IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
o prove it. IS IS
First, if there are only two heaps recall that they are losing if and only if the heaps are of equal
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
size. If they are of unequal size, then the winning move is to reduce thelarger heap so that both
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS I
heaps have equal size.
S IS IS IS
3
, Consider three heaps of sizes 1, m, n, where 1 m n.≤We ≤ observe the following: 1, 1, m is win
IS IS IS IS IS IS IS IS IS ISISISISIS ISISISISIS IS IS IS IS IS IS IS IS IS
ning, by moving to 1, 1, 0. Similarly, 1, m, m is winning, by moving to 0, m, m. Next, 1, 2, 3 is lo
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
sing (observed earlier in the lecture), and hence 1, 2, n for n 4 is winning. 1, 3, n is winning for
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
any n 3 by moving to 1, 3, 2. For 1, 4, 5, reducing any heap produces a winning position, so thi
≥ ≥
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
s is losing.
IS IS
The general pattern for the losing positions thus seems to be: 1, m, m 1, for even+numbers m.
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
This includes also the case m = 0, which we can take as the base case foran induction. We now pr
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
oceed to prove this formally. IS IS IS IS
First we show that if the positions of the form 1, m, n with m n are losing
IS IS
≤ when m is even an IS IS IS IS IS IS IS IS IS IS IS IS ISISISISISIS IS IS IS IS IS IS IS
d n = m 1, then these are+the only losing positions because any other position 1, m, n with m
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS I S IS I S I S
n is winning. Namely, if m≤ = n then a winning move from1, m, m is to 0, m, m, so we can assum
I S IS I S IS IS I S IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
e m < n. If m is even then n > m 1 (otherwise we would be in the position 1, m, m 1) and so
+
IS IS IS I S IS IS IS IS IS IS IS I S I S IS IS IS IS IS IS IS IS IS IS I S I S IS IS I
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
+ +
S IS IS IS IS IS IS IS I S I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
position 1, m 1, m (this would also be a winning move from 1, m, m so there the winning move is
– −
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
not unique). IS
Second, we show that any move from 1, m, m + 1 with even m is to a winning position,using as ind
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
uctive hypothesis that 1, mJ, mJ + 1 for even mJ and mJ < m is a losing position. The move to 0, m,
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
m + 1 produces a winning position with counter-
IS IS IS IS IS IS IS IS IS
move to 0, m, m. A move to 1, mJ, m + 1 for mJ < m is to a winning position with the counter-
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
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 winni
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
ng position with counter-
IS IS IS
move to 0, m, m. A move to 1, m, mJ with mJ < m is also to a winning position with the counter-
IS IS IS IS IS IS IS IS IS IS IS I S IS I S IS IS IS IS IS IS IS IS IS
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 becau
IS IS IS IS IS IS IS I S IS IS IS IS IS IS I S IS IS IS IS IS IS IS IS IS I S IS IS IS
se m is even). This concludes the induction proof.
IS IS IS IS IS IS IS IS
+ +
This result is in agreement with the theorem on Nim heap sizes represented as sums of powers of 2:
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
1 m ∗n is+∗ losing 0
I S
+∗if and only if, except for 2 , the powers of 2 making upm and n come in pairs
I S I S I S I S IS
I S
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
. So these must be the same powers of 2, except for 1 = 20, which occurs in only m or n, where we ha
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
ve assumed that n is the larger number, so 1 appearsin the representation of n: We have m = 2
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS I S IS IS IS IS
a 2b 2c
ISISISISISIS ISISISISISIS
+ + + ··· ··· ≥ I S IS IS IS IS IS
for a > b > c > 1,so
+ + + · · · + +
IS I S IS I S IS I S ISISISISISISISI S IS IS
m is even, and, with the same a, b, c, . . ., n = 2
IS IS
a 2b 2 c IS 1 = m 1. Then IS IS IS IS IS IS IS IS IS IS I S IS
I S I S I S I S I S I S
IS IS IS IS
IS IS ISISISI S I S
1 m
∗ + ∗ + ∗ ≡∗
I
n 0.S
The following
ISISISISISI S
IS
is an example using the bit representation
ISISISISI S
IS
where ISISISISISI S
IS
I S IS IS IS IS IS IS IS IS IS
m = 12 (which determines the bit pattern 1100, which of course depends on m):
IS IS IS IS IS IS IS IS IS IS IS IS IS IS
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-
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
sum of the binary representations 01, 10, 11 is 00. Examples show that any other position iswi
IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
nning. The three numbers are n, n 1, n +2. If n+is even then reducing the heap of size n 2 to 1 c
IS IS IS IS IS IS I S IS I S I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
reates the position n, n 1, 1 which is losing as shown in (a). If n is odd, then n 1 is even and
+ +
IS IS IS IS I S IS IS IS IS IS IS IS IS IS IS IS IS IS IS I S IS IS IS IS
n 2 = n 1 1 so by the same argument, a winning move is to reduce the Nim heap of size n t
+ +
ISISIS IS I S ISISIS ISISIS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS IS
o 1 (which only works if n > 1).
( + )+
IS IS IS IS IS IS IS IS
I S I S IS
4
