SOLUTION MANUAL
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS QWE QWE QWE
1 - Nim and Combinatorial Games
QWE QWE QWE QWE QW E
2 - Congestion Games
QWE QWE QWE
3 - Games in Strategic Form
QWE QWE QWE QW E QWE
4 - Game Trees with Perfect Information
QWE QWE QWE QWE QWE QWE
5 - Expected Utility
QWE QWE QWE
6 - Mixed Equilibrium
QWE QWE QW E
7 - Brouwer’s Fixed-Point Theorem
QWE QWE QWE QW E
8 - Zero-Sum Games
QWE QWE QWE
9 - Geometry of Equilibria in Bimatrix Games
QWE QWE QWE QWE QWE QWE QWE
10 - Game Trees with Imperfect Information
QWE QWE QW E QWE QWE QW E
11 - Bargaining
QWE QWE
12 - Correlated Equilibrium
QWE QWE QWE
2
,Game Theory Basics QWE QWE
Solutions to Exercises Q W E Q W E
© Q W E Bernhard von Stengel 2022 QW E QW E QW E
Solution to Exercise 1.1 QWE QWE QWE
(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 then
QW E QW E QW E QW E QW E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E QW E QW E QW E QWE QWE QWE QWE QWE QW E QWE QWE
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 because < is
QWE QW E QW E QWE QW E QWE QWE QW E QWE QWE QWE QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE QW E QW E Q W E QW E QWE QWE QWE Q W E QW E QW E QW E QWE Q
transitive, and hence x ≤ z.
WE QWE QWE QWE Q W E QW E
Clearly, ≤isreflexivebecause x = x andtherefore x ≤ x. QWE E
QW QWE QWE QWE QW E QWE QWE QWE QWE QWE QWE
To show that is≤antisymmetric, consider x and y with x y and y x. If≤
QWE we had x ≠ y then
≤ x < y and y < QWE QWEQW EQW EQW EQW E QW E QWE QWE QWE QWE QW E QWE QWEQWEQWEQWEQWE QWE QWE QWEQWEQWEQWEQWE QWE QW E QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE QW E QWE
x, and by transitivity x < x which contradicts (1.38). Hence x = y, as required. This shows that ≤is a
QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE QW E Q W E Q W E QWE QWE Q W E QWE QWE QWE QW
E QWE QWE
partial order. QWE
Finally, we show (1.6), so we have to show that x < y implies x y and x ≠ y and vice
≤ versa. Let x < y, whi QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE QWE QW E QWE QWEQWE QW E QWE QWE QWE QWE QW E QW E QWE QW E QWE QWE QW E QWE
ch implies x y by (1.7). If we had x = y then x < x, contradicting
QWE
≤ (1.38),sowealsohave x ≠ y. Conversely, QWE QWE QW E QW E QW E QWE QW E QWE QW E QWE QWE QWE QWE QW E QW E E
QW E
QW E
QW E
QW E
QW QWE QWE QWE Q W E Q
x yand x ≠ yimplyby(1.7)x < y or x = y where the second case is excluded,
WE QWEQWEQ W E QWE
≤ hence x < y, as required.
QWE QWE QWE QWE E
QW E
QW E
QW Q W E Q W E QWE QWE Q W E Q W E QWE QWE QWE QWE QWE QWE QWE QWE Q W E Q W E QWE QWE
(b) Consider a partial order and assume ≤ (1.6) as a definition of <. To show that < is transitive, suppose QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE
x < y, that is, x y and x ≠ y, and y < z, that is, y≤z and y ≠ z. Because is transitive, x z. If we
QWE QW E QWE
≤ had x = z then QWE QW E QW E QWE QWE QWE QWE QWE QW E QWE QWE QW E QWE QWE QW E QWE QWE QW E QWE QW E QWEQWE QWE QW E QW E QWE QWEQWEQWE QWE QWE QWE QW E QWE QWE QWE QWE QWE
x y and y ≤x and hence x = y by≤antisymmetry of , which contradicts
QWEQWEQWEQWEQWE QWE
≤ x ≠ y,≤ so we have x z and x ≠ QWE QW EQW EQW EQW EQW E QWE QWE QWE QWE QW E QWE QW E QW E QWEQWE QWE Q W E QW E QW E Q W E Q W E Q W E QW E QW E QW E Q W E QW EQWEQWE Q W E QW E Q W E Q W E Q
z, that is,x < z by (1.6), as required.
≤ ≤
W E QW E QW E E
QW Q W E Q W E QWE QWE QWE QWE
Also, < is irreflexive, because x < x would by definition mean x x and x ≠ x, but≤the latter is not true.
QWE QW E QWE QWE QWE QWE QW E QW E QWE QWE QWE QWE QWEQWEQW E QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE
Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, given that QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE Q W E QW E QWE QWE QWE QWE QW E QW E QW E QWE QW E QW E QWE QWE QWE QWE
< is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x ≠ y and then by definition x < y.
QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QW E QWE QWE QW E QWE QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE
Hence, x ≤ y implies x < y or x = y. Conversely, suppose x < y or x = y. If x < y then x ≤ y by ( QWE QWE QWE QWE QWE QWE QWE QW E QW E QWE QWE QWE QWE QWE Q W E Q W E QW E QW E QW E QWE Q W E QWE Q W E Q W E QWE QWE Q W E QWE QWE QWE
1.6), and if x = y then x ≤ y because ≤is reflexive. This completes the proof.
QWE QWE QWE QW E QW E QWE QWE Q W E QWE QW E QWE E
QW QWE Q W E QWE QWE QWE
Solution to Exercise 1.2 QWE QWE QWE
(a) In analysing the games of three Nim heaps where one heap has size one, we first lookat some exa
Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E QWE QW E QW E
mples, and then use mathematical induction to prove what we conjecture to be the losing positions. A l
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QWE QWE QWE
osing position is one where every move is to a winning position, because then the opponent will win
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E QW E QW E QW E
. The point of this exercise is to formulate a precise statement to be proved, and then to prove it.
Q W E QW E QW E QW E QW E QW E QW E QW E QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
First, if there are only two heaps recall that they are losing if and only if the heaps are of equal size.
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE Q W E
If they are of unequal size, then the winning move is to reduce thelarger heap so that both heaps hav
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE E
QW QWE QWE QWE QWE QWE QWE
e equal size.
QWE QWE
3
, Consider three heaps of sizes 1, m, n, where 1 m n. We≤observe ≤ the following: 1, 1, m is winning, QWE QWE QWE QWE QWE QWE QWE QWE QWE QW EQWE QWE QWE QWE QWEQWEQWEQWEQWE QW E QWE QWE QWE QW E QWE QWE QWE QWE QWE
by moving to 1, 1, 0. Similarly, 1, m, m is winning, by moving to 0, m, m. Next, 1, 2, 3 is losing (observ
QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QW E QWE QW E QWE QWE QW E QWE QWE QWE QWE
ed earlier in the lecture), and hence 1, 2, n for n 4 is winning. 1, 3, n is winning for any n 3 by movin
QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E QWE QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
g to 1, 3, 2. For 1, 4, 5, reducing any heap produces a winning position, so this is losing.
≥ ≥
QWE QWE QW E QWE QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
The general pattern for the losing positions thus seems to be: 1, m, m 1, for even numbers
QWE
+ m. Thisi QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QWE QWE QW E QWE Q W E QWE
ncludesalsothecase m =0,whichwecantakeasthebasecaseforan induction. We now proceed to pr QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE QWE QW
E QWE QWE QWE E
QW QWE Q W E QWE QWE QWE QWE
ove this formally. QWE QWE
First we show that if the positions of the form 1, m, n with m n are losing when
QWE
≤ m is even and n = m QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QWEQWEQWEQWEQWEQWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
1, then these are the only losing
QWE
+ positions because any other position 1, m, n with m n is winn
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E Q W E QW E Q W E Q W E Q W E QW E
ing. Namely, if m = n then ≤ a winning move from1, m, m is to 0, m, m, so we can assume m < n. If m i
Q W E QW E QW E Q W E QW E Q W E QW E QW E QW E QW E QWE QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E QW E QW E
s even then n > m 1 (otherwise we would be in the position 1, m, m 1) and so the winning m
+
QW E QW E QW E QW E QW E Q W E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E Q W E QW E QW E QW E QW E QW E
ove 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 (
+ +
QW E QW E QW E QW E QW E Q W E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QW E QW E QW E QW E QW E QW E QWE QW E QW E
this would also be a winning move from 1,m,m so there the winning move is not unique).
– −
QW E Q W E Q W E Q W E Q W E Q W E Q W E Q W E QWE QWE Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E
Second, we show that any move from 1, m, m + 1 with even m is to a winning position,using as inductiv QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QWE QWE
e hypothesis that 1, mJ, mJ + 1 for even mJ and mJ < m is a losing position. The move to 0, m, m + 1 pro
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QW E QW E QWE QWE QWE QWE QW E QWE QWE QW E QW E
duces a winning position with counter- QWE QWE QWE QWE QWE
move to 0, m, m. A move to 1, mJ, m + 1 for mJ < m is to a winning position with the counter-
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E
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 winning posit
QWE QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E QW E QW E QW E QW E QW E
ion with counter- QW E QW E
move to 0, m, m. A move to 1, m, mJ with mJ < m is also to a winning position with the counter-
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E QW E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E
move to 1, mJ − 1, mJ if mJ is odd, and to 1, mJ 1, mJ if mJ is even (in which case mJ
QW E 1 < m because QW E QW E QW E QW E QW E QW E Q W E QW E QW E QW E QW E QW E QW E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E QW E QW E QW E QW E
m is even). This concludes the induction proof.
+ +
QW E QW E QW E QWE QWE QWE QWE
This result is in agreement with the theorem on Nim heap sizes represented as sums of powersof2:
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QWE Q W E
1 m Q n∗islosing
+∗ +∗
W ifandonlyif, exceptfor20, thepowersof2making upm and n come in pairs. So the
E Q W E Q W E Q W E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E QW E QW E QW E QW E
se must be the same powers of 2, except for 1 = 20, which occurs in only m or n, where we have assumed t
Q W E
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E
hat n is the larger number, so 1 appearsin the representation of n: We have m = 2a 2b 2c
QW E QW E QW E QW E QW E QW E QW E QW E QWE Q WE Q WE QWE Q W E Q W E Q WE Q W E Q W E Q W E
QWEQWEQWEQWEQWEQW E QWEQWEQWEQWEQWEQW E
for a > b > c > 1,so m
+ +c + ··· ··· ≥
Q W E Q W E Q W E Q W E Q W E Q W E QWEQWEQWEQWEQWEQW EQW EQ W E E
QW Q W E Q W
is even, and, with the same a,b, c,..., n = 2 a 2 b 2 1 = m 1. Then Q W E Q W E Q W E Q W E Q W E Q W E
Q W E QW E Q W E QW E QW E Q W E
+ + + ··· + +
E Q W E Q W E Q W E Q W E Q W E Q W E QWE QWE QWE QWE QWE Q W E Q W E Q W E Q W E Q W E QWEQWEQWE Q W E Q W E
∗1 +m∗ +n∗ 0. ≡ ∗ The following is an example using the bit representation where
Q W E Q W E Q W E Q W E
QWEQWEQW EQW EQW EQ W E QWEQWEQWEQWEQ W E QWEQWEQWEQW EQW EQ W E Q W E QWE Q WE QWE QWE QWE Q WE QWE QWE QW E
Q W E Q W E Q W E QWE
m = 12 (which determines the bit pattern 1100, which of course depends on m):
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
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-
QWE QWE QWE QW E QWE QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE
sum of the binary representations 01, 10, 11 is 00. Examples show that any other position iswinni
QWE QW E QWE QWE QWE QW E QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE E
QW
ng. The three numbers are n, n 1, n
QW E 2.+If n is even
+ then reducing the heap of size n 2 to 1 creates QW E QW E QW E QW E QW E Q W E QW E Q W E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QW E QW E QW E QWE QWE QWE
the position n, n 1, 1 which is losing as shown in (a). If n is odd, then n 1 is even and n 2 = n
+ +
QWE QWE QWE Q W E QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QW E QWE QWE Q W E QWE QWE QWE QWE QWEQWE QWE QWE Q W E QWEQWE QW E
1 1 so by the same argument, a winning move is to reduce the Nim heap of size n to 1 (which only w
+ +
QWEQWE QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE
orks if n > 1).
( + )+
QWE QW E Q W E QW E
Q W E Q W E QWE
4
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS QWE QWE QWE
1 - Nim and Combinatorial Games
QWE QWE QWE QWE QW E
2 - Congestion Games
QWE QWE QWE
3 - Games in Strategic Form
QWE QWE QWE QW E QWE
4 - Game Trees with Perfect Information
QWE QWE QWE QWE QWE QWE
5 - Expected Utility
QWE QWE QWE
6 - Mixed Equilibrium
QWE QWE QW E
7 - Brouwer’s Fixed-Point Theorem
QWE QWE QWE QW E
8 - Zero-Sum Games
QWE QWE QWE
9 - Geometry of Equilibria in Bimatrix Games
QWE QWE QWE QWE QWE QWE QWE
10 - Game Trees with Imperfect Information
QWE QWE QW E QWE QWE QW E
11 - Bargaining
QWE QWE
12 - Correlated Equilibrium
QWE QWE QWE
2
,Game Theory Basics QWE QWE
Solutions to Exercises Q W E Q W E
© Q W E Bernhard von Stengel 2022 QW E QW E QW E
Solution to Exercise 1.1 QWE QWE QWE
(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 then
QW E QW E QW E QW E QW E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E QW E QW E QW E QWE QWE QWE QWE QWE QW E QWE QWE
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 because < is
QWE QW E QW E QWE QW E QWE QWE QW E QWE QWE QWE QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE QW E QW E Q W E QW E QWE QWE QWE Q W E QW E QW E QW E QWE Q
transitive, and hence x ≤ z.
WE QWE QWE QWE Q W E QW E
Clearly, ≤isreflexivebecause x = x andtherefore x ≤ x. QWE E
QW QWE QWE QWE QW E QWE QWE QWE QWE QWE QWE
To show that is≤antisymmetric, consider x and y with x y and y x. If≤
QWE we had x ≠ y then
≤ x < y and y < QWE QWEQW EQW EQW EQW E QW E QWE QWE QWE QWE QW E QWE QWEQWEQWEQWEQWE QWE QWE QWEQWEQWEQWEQWE QWE QW E QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE QW E QWE
x, and by transitivity x < x which contradicts (1.38). Hence x = y, as required. This shows that ≤is a
QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE QW E Q W E Q W E QWE QWE Q W E QWE QWE QWE QW
E QWE QWE
partial order. QWE
Finally, we show (1.6), so we have to show that x < y implies x y and x ≠ y and vice
≤ versa. Let x < y, whi QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE QWE QW E QWE QWEQWE QW E QWE QWE QWE QWE QW E QW E QWE QW E QWE QWE QW E QWE
ch implies x y by (1.7). If we had x = y then x < x, contradicting
QWE
≤ (1.38),sowealsohave x ≠ y. Conversely, QWE QWE QW E QW E QW E QWE QW E QWE QW E QWE QWE QWE QWE QW E QW E E
QW E
QW E
QW E
QW E
QW QWE QWE QWE Q W E Q
x yand x ≠ yimplyby(1.7)x < y or x = y where the second case is excluded,
WE QWEQWEQ W E QWE
≤ hence x < y, as required.
QWE QWE QWE QWE E
QW E
QW E
QW Q W E Q W E QWE QWE Q W E Q W E QWE QWE QWE QWE QWE QWE QWE QWE Q W E Q W E QWE QWE
(b) Consider a partial order and assume ≤ (1.6) as a definition of <. To show that < is transitive, suppose QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE
x < y, that is, x y and x ≠ y, and y < z, that is, y≤z and y ≠ z. Because is transitive, x z. If we
QWE QW E QWE
≤ had x = z then QWE QW E QW E QWE QWE QWE QWE QWE QW E QWE QWE QW E QWE QWE QW E QWE QWE QW E QWE QW E QWEQWE QWE QW E QW E QWE QWEQWEQWE QWE QWE QWE QW E QWE QWE QWE QWE QWE
x y and y ≤x and hence x = y by≤antisymmetry of , which contradicts
QWEQWEQWEQWEQWE QWE
≤ x ≠ y,≤ so we have x z and x ≠ QWE QW EQW EQW EQW EQW E QWE QWE QWE QWE QW E QWE QW E QW E QWEQWE QWE Q W E QW E QW E Q W E Q W E Q W E QW E QW E QW E Q W E QW EQWEQWE Q W E QW E Q W E Q W E Q
z, that is,x < z by (1.6), as required.
≤ ≤
W E QW E QW E E
QW Q W E Q W E QWE QWE QWE QWE
Also, < is irreflexive, because x < x would by definition mean x x and x ≠ x, but≤the latter is not true.
QWE QW E QWE QWE QWE QWE QW E QW E QWE QWE QWE QWE QWEQWEQW E QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE
Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, given that QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE Q W E QW E QWE QWE QWE QWE QW E QW E QW E QWE QW E QW E QWE QWE QWE QWE
< is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x ≠ y and then by definition x < y.
QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QW E QWE QWE QW E QWE QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE
Hence, x ≤ y implies x < y or x = y. Conversely, suppose x < y or x = y. If x < y then x ≤ y by ( QWE QWE QWE QWE QWE QWE QWE QW E QW E QWE QWE QWE QWE QWE Q W E Q W E QW E QW E QW E QWE Q W E QWE Q W E Q W E QWE QWE Q W E QWE QWE QWE
1.6), and if x = y then x ≤ y because ≤is reflexive. This completes the proof.
QWE QWE QWE QW E QW E QWE QWE Q W E QWE QW E QWE E
QW QWE Q W E QWE QWE QWE
Solution to Exercise 1.2 QWE QWE QWE
(a) In analysing the games of three Nim heaps where one heap has size one, we first lookat some exa
Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E QWE QW E QW E
mples, and then use mathematical induction to prove what we conjecture to be the losing positions. A l
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QWE QWE QWE
osing position is one where every move is to a winning position, because then the opponent will win
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E QW E QW E QW E
. The point of this exercise is to formulate a precise statement to be proved, and then to prove it.
Q W E QW E QW E QW E QW E QW E QW E QW E QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
First, if there are only two heaps recall that they are losing if and only if the heaps are of equal size.
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE Q W E
If they are of unequal size, then the winning move is to reduce thelarger heap so that both heaps hav
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE E
QW QWE QWE QWE QWE QWE QWE
e equal size.
QWE QWE
3
, Consider three heaps of sizes 1, m, n, where 1 m n. We≤observe ≤ the following: 1, 1, m is winning, QWE QWE QWE QWE QWE QWE QWE QWE QWE QW EQWE QWE QWE QWE QWEQWEQWEQWEQWE QW E QWE QWE QWE QW E QWE QWE QWE QWE QWE
by moving to 1, 1, 0. Similarly, 1, m, m is winning, by moving to 0, m, m. Next, 1, 2, 3 is losing (observ
QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QW E QWE QW E QWE QWE QW E QWE QWE QWE QWE
ed earlier in the lecture), and hence 1, 2, n for n 4 is winning. 1, 3, n is winning for any n 3 by movin
QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E QWE QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
g to 1, 3, 2. For 1, 4, 5, reducing any heap produces a winning position, so this is losing.
≥ ≥
QWE QWE QW E QWE QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
The general pattern for the losing positions thus seems to be: 1, m, m 1, for even numbers
QWE
+ m. Thisi QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QWE QWE QW E QWE Q W E QWE
ncludesalsothecase m =0,whichwecantakeasthebasecaseforan induction. We now proceed to pr QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE QWE QW
E QWE QWE QWE E
QW QWE Q W E QWE QWE QWE QWE
ove this formally. QWE QWE
First we show that if the positions of the form 1, m, n with m n are losing when
QWE
≤ m is even and n = m QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QWEQWEQWEQWEQWEQWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
1, then these are the only losing
QWE
+ positions because any other position 1, m, n with m n is winn
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E Q W E QW E Q W E Q W E Q W E QW E
ing. Namely, if m = n then ≤ a winning move from1, m, m is to 0, m, m, so we can assume m < n. If m i
Q W E QW E QW E Q W E QW E Q W E QW E QW E QW E QW E QWE QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E QW E QW E
s even then n > m 1 (otherwise we would be in the position 1, m, m 1) and so the winning m
+
QW E QW E QW E QW E QW E Q W E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E Q W E QW E QW E QW E QW E QW E
ove 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 (
+ +
QW E QW E QW E QW E QW E Q W E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QW E QW E QW E QW E QW E QW E QWE QW E QW E
this would also be a winning move from 1,m,m so there the winning move is not unique).
– −
QW E Q W E Q W E Q W E Q W E Q W E Q W E Q W E QWE QWE Q W E Q W E Q W E Q W E Q W E Q W E Q W E Q W E
Second, we show that any move from 1, m, m + 1 with even m is to a winning position,using as inductiv QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QWE QWE
e hypothesis that 1, mJ, mJ + 1 for even mJ and mJ < m is a losing position. The move to 0, m, m + 1 pro
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QW E QW E QWE QWE QWE QWE QW E QWE QWE QW E QW E
duces a winning position with counter- QWE QWE QWE QWE QWE
move to 0, m, m. A move to 1, mJ, m + 1 for mJ < m is to a winning position with the counter-
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E
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 winning posit
QWE QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E QW E QW E QW E QW E QW E
ion with counter- QW E QW E
move to 0, m, m. A move to 1, m, mJ with mJ < m is also to a winning position with the counter-
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E QW E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E
move to 1, mJ − 1, mJ if mJ is odd, and to 1, mJ 1, mJ if mJ is even (in which case mJ
QW E 1 < m because QW E QW E QW E QW E QW E QW E Q W E QW E QW E QW E QW E QW E QW E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E Q W E QW E QW E QW E QW E
m is even). This concludes the induction proof.
+ +
QW E QW E QW E QWE QWE QWE QWE
This result is in agreement with the theorem on Nim heap sizes represented as sums of powersof2:
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QWE Q W E
1 m Q n∗islosing
+∗ +∗
W ifandonlyif, exceptfor20, thepowersof2making upm and n come in pairs. So the
E Q W E Q W E Q W E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QW E QW E QW E QW E QW E QW E
se must be the same powers of 2, except for 1 = 20, which occurs in only m or n, where we have assumed t
Q W E
QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E QW E
hat n is the larger number, so 1 appearsin the representation of n: We have m = 2a 2b 2c
QW E QW E QW E QW E QW E QW E QW E QW E QWE Q WE Q WE QWE Q W E Q W E Q WE Q W E Q W E Q W E
QWEQWEQWEQWEQWEQW E QWEQWEQWEQWEQWEQW E
for a > b > c > 1,so m
+ +c + ··· ··· ≥
Q W E Q W E Q W E Q W E Q W E Q W E QWEQWEQWEQWEQWEQW EQW EQ W E E
QW Q W E Q W
is even, and, with the same a,b, c,..., n = 2 a 2 b 2 1 = m 1. Then Q W E Q W E Q W E Q W E Q W E Q W E
Q W E QW E Q W E QW E QW E Q W E
+ + + ··· + +
E Q W E Q W E Q W E Q W E Q W E Q W E QWE QWE QWE QWE QWE Q W E Q W E Q W E Q W E Q W E QWEQWEQWE Q W E Q W E
∗1 +m∗ +n∗ 0. ≡ ∗ The following is an example using the bit representation where
Q W E Q W E Q W E Q W E
QWEQWEQW EQW EQW EQ W E QWEQWEQWEQWEQ W E QWEQWEQWEQW EQW EQ W E Q W E QWE Q WE QWE QWE QWE Q WE QWE QWE QW E
Q W E Q W E Q W E QWE
m = 12 (which determines the bit pattern 1100, which of course depends on m):
QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE
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-
QWE QWE QWE QW E QWE QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE QWE
sum of the binary representations 01, 10, 11 is 00. Examples show that any other position iswinni
QWE QW E QWE QWE QWE QW E QW E QWE QW E QWE QWE QWE QWE QWE QWE QWE E
QW
ng. The three numbers are n, n 1, n
QW E 2.+If n is even
+ then reducing the heap of size n 2 to 1 creates QW E QW E QW E QW E QW E Q W E QW E Q W E Q W E QW E QW E QW E QW E QW E QW E QW E QW E QW E QWE QW E QW E QW E QWE QWE QWE
the position n, n 1, 1 which is losing as shown in (a). If n is odd, then n 1 is even and n 2 = n
+ +
QWE QWE QWE Q W E QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QW E QWE QWE Q W E QWE QWE QWE QWE QWEQWE QWE QWE Q W E QWEQWE QW E
1 1 so by the same argument, a winning move is to reduce the Nim heap of size n to 1 (which only w
+ +
QWEQWE QW E QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QWE QW E QWE QWE QWE QWE QWE
orks if n > 1).
( + )+
QWE QW E Q W E QW E
Q W E Q W E QWE
4
