SOLUTION MANUAL aa aa
Game Theory Basics 1st Edition
aa aa aa aa aa aa
By Bernhard von Stengel. Chapters 1 -
aa aa aa aa aa aa aa aa
12 aa
aa aa
aa 1 aa
,TABLE OF CONTENTS 3d 3d 3d aa
aa
aa
1 - Nim and Combinatorial Games
3d 3d 3d 3d 3d aa
aa
2 - Congestion Games
3d 3d 3d aa
aa
3 - Games in Strategic Form
3d 3d 3d 3d 3d aa
aa
4 - Game Trees with Perfect Information
3d 3d 3d 3d 3d 3d aa
aa
5 - Expected Utility
3d 3d 3d aa
aa
6 - Mixed Equilibrium
3d 3d 3d aa
aa
7 - Brouwer’s Fixed-Point Theorem
3d 3d 3d 3d aa
aa
8 - Zero-Sum Games
3d 3d 3d aa
aa
9 - Geometry of Equilibria in Bimatrix Games
3d 3d 3d 3d 3d 3d 3d aa
aa
10 - Game Trees with Imperfect Information
3d 3d 3d 3d 3d 3d aa
aa
11 - Bargaining 3d 3d aa
aa
12 - Correlated Equilibrium
3d 3d 3d aa
aa aa
Game Theory Basics 3d 3d aa
Solutions to Exercises aa 3d aa 3d aa aa
© Bernhard von Stengel 2022 aa
3 d aa 3d 3d 3d
Solution to Exercise 1.1 aa 3d 3d 3d
(a) aaLet ≤ be defined aa by (1.7). To show that ≤ is transitive, consider x, y, z with x
3d 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa ≤ y and 3d 3d
y ≤ z. If x
aa
3d 3d 3d 3d 3d 3d
= y then x ≤ z, and if y = z then also x ≤ z. So the only case left is x < y and aa
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa y 3 aa d aa < 3 aa d
z, which implie aa 3d 3d
s x aa < aa z because < is transitive, and hence x aa ≤ z. aa
3d 3 d aa 3 d aa 3d 3d 3d 3d 3d 3d 3d 3 d aa 3d
Clearly, ≤ is reflexive because x = x and therefore x ≤ x. aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
To show that ≤is 3d 3d 3d3d3d3d3d aa 3d antisymmetric, consider x and y with x y and y≤ 3d 3d 3d 3d 3d 3d 3d3d3d3d3d 3d 3d 3d3d3d3d3d aa x. If we had≤ 3d 3d 3d
aa 3d x ≠ y then x aa< y and y < x, and by transitivity x < x which contradicts (1.38). Hence x
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 aa d aa = 3 aa
d
aa y, as required.
3d 3d 3 aa d aa Th
is shows that ≤ is a partial order. aa
3d 3d 3d 3d 3d 3d 3d
Finally, we show (1.6), so we have to show that x < y implies x y and x ≠≤ y
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3d3d 3d 3d 3d 3d 3
aa and vice versa. Let d 3d 3d 3d 3
aa 2 aa
, aa d x < y, which implies x y by (1.7). If we had x = y then x < x, contradicting (1.38), so we also ha
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
v aa≤ aa
e x ≠ y. aa aaConversely, x aa aay and x ≠ y imply by (1.7) x
3d 3d 3d 3 d 3d 3d3d3 d 3d 3d 3d 3d 3d 3d 3d d 33 aa d aa < 3 aa d aa y or x3d 3d 3 aa d aa = 3 aa
d
aay where the second case is exclud
3d 3d 3d 3d 3d 3d
ed, hence x
3d 3d 3 aa d aa < 3 aa d aa y, as required. aa
3d 3d ≤ aa
(b) aaConsider a partial order and≤ assume (1.6) as a definition of <. To show that < is transitive, su
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
ppose x < y, that is, x y and x ≠ y, and y <≤ z aa, that is, y z and y ≠ z. Because is transitive,≤
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3 d3d3 d 3d
aa x z. If 3d 3d3d3d3d 3d 3d
aawe had x = z then x y and y x and hence x = y by antisymmetry of , which contradicts aa
3d 3d 3d 3d 3d 3d 3d3d3d3d3d 3d 3d 3d3d3d3d3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3d3d3d 3d 3d 3 d
aax aa aa≠ aa
≤ ≤ ≤ 3
≤ d aa aa aa aa
3 aa aad y, so we have
3d 3d 3d 3 aa aa d x 3d3d3d3 aa aa d z and aa aax
3d 3 d 3 aa aad ≠ 3 aa aa d z, that is, x 3d 3d 3 d3 aa aa d < 3 aa aa d z by (1.6), as required. aa
3d 3d 3d 3d
≤ aa ≤ aa
Also, aa< aaisirreflexive, aabecause aax aa< aax aawouldbydefinitionmeanx aax aaandx aa≠≤x aa,but aathe
aalatteris
not true. aa
3d aa3d aa 3d aa3d aa3d aa3d aa 3d aa 3d aa 3d aa3d aa3d aa3d aa3d3d3d aa 3d aa 3d aa3d aa 3d aa3d aa3d aa3d aa3d aa3d aa 3d
Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, gi
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
ven that < is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x ≠ y and then b
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
y definition x < y. Hence, x ≤ y implies x < y or x = y. Conversely, suppose x < y or x = y. If
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa 3 d aa 3d 3d 3d 3d 3 d aa 3d
x < y then x
3 d aa ≤ y by (1.6), and if x = y then x aa ≤ y because ≤ is reflexive. aa
3 d aa 3d 3d 3 d aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa 3d 3d 3d 3d 3d 3 d
This completes the proof. aa
aa 3d 3d 3d aa
Solution to Exercise 1.2 aa 3d 3d 3d
(a) aaIn aa analysing aa aa the aa aa games aa of aa three aa aa Nim aa heaps aa where aa aa one aa aa heap aa
3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d
has aa aa size aa aa one, aa aa we aa first aa aa look at
aa 3 d aa 3 some aa
d aa 3 d aa 3 d aa 3 d aa 3d 3d 3d
examples, and aa then use aa mathematical induction to aa prove what we conjecture to 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa be the losing positi 3d 3d 3d 3d
ons. A losing position is one where every move is to a winning position, because
3d 3d aa then the 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa oppo 3d
nent will win. The point of this exercise aa is to
3d 3d 3 d aa 3d 3d 3d 3d 3d 3d
aa formulate a precise statement to be proved, and t
3d 3d 3d 3d 3d 3d 3d 3d 3d
hen to prove it. aa 3d 3d 3d
First, if there are only two heaps recall that they are losing if and only if the heaps are of equa
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
l size. aa If they are of unequal size, then the winning move is to reduce the larger
3d 3 d aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d
heap so that bo 3d 3d 3d 3d
th heaps have equal size. aa 3d 3d 3d 3d
Consider three heaps of sizes 1, m, n, where 1 m n.≤ We aa observe≤ aa
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3d3d3d3d 3d3d3d3d3d 3d 3d
the following: 1, 1, m is win 3d 3d 3d 3d 3d 3d 3d
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
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
sing (observed earlier in the lecture), and hence 1, 2, n for n 4 is winning. 1, 3, n is winning for
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
any n 3 by moving to 1, 3, 2. For 1, 4, 5, reducing any aaheap produces a winning position, so thi
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aas is≥ losing. aa aa aa
3d 3d
aa 3 aa
, The general pattern for the losing positions thus seems to be: 1, m, m 1, for even+ numbers m.
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3
aa d aa This includes also the case m = 0, which we can take as the base case for an
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa induction.
3d 3 aa d
aa We now p 3d 3d
roceed to prove this formally. aa 3d 3d 3d 3d
First we show that if the positions of the form 1, m, n with m n are losing≤
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3d3d3d3d3d 3d 3d
aa when m is even an
3d 3d 3d 3d 3d
aad n = m 1, then these are the only losing positions because any other position 1, m, n
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 aa d
aa with m 3d 3 aa d aa 3 aa
+
aan aais winning. aaNamely, aa if m
d 3d 3d 3d 3d 3d 3 d aa =n 3d 3 d aa then a winning aa move3d 3d 3d
aa 3d from1 , m, m is to aa 0, m, m, aa so aa we can assu
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
me m < n. If m is even then ≤ n > m 3d 3d 3d 3d aa 3d 3d 3d 3d 3 d 3d 3d 3 d 3 d
aa 1 (otherwise we aa would be in the aa position 1, m, m
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d 3d aa 1) and aa 3d 3d
so aa the aa winning aa move is to aa 1, m, m
3d 3d 3d 3d 3d 3d 3d 3d 3 d 3 d
aa 1. If m is odd then the aa winning aa move aa is to aa 1, +m, m 1, aa the aa same
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
as position 1, m 1, m (this would aa also aa be aa
3d 3d 3d 3d 3d 3d 3d 3d 3d aa 3 aa d aa a aa +winning
3 d aa 3 aa d aa move 3 aa d aa from 3 aa d aa 1, m, m 3d 3d 3
aa d aa so aa 3 aa d aa there aa the aa 3d aa 3 aad aa winning+ 3 aa d aa move aa 3 aa d aa is aa not 3 d aa 3 aa d aa unique). aa – aa − aa
Second, we show that any move aa from 1, m, m + 1 with aa even m is to aa a winning aa position, using
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3d 3 d
as ind 3d 3d
uctive hypothesis that 1, mJ, mJ + 1 for even mJ and mJ < m is a losing position. The move to 0,
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
m, m + 1 produces a winning position with counter-3d 3d 3d 3d 3d 3d 3d 3d 3d
move to 0, m, m. aa A move aa to aa 1, mJ, m + 1 for mJ < m is to aa a winning aa position with the
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa counter- 3d
move to 1, mJ, mJ + 1 if mJ is even and to 1, mJ, mJ − 1 if mJ is odd. A move to 3d aa 1, m, m is to3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa a winnin 3d 3d
g position with counter- 3d 3d 3d
move to aa 0, m, m. aa A move to aa 1, m, mJ with 3d mJ < m is also aa to
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa 3d 3 d aa 3d 3d 3d
aa a winning aa position with aa the aa counter-
3d 3d 3d 3d 3d 3d
move to aa 1, mJ − aa 1, mJ if mJ is odd, and aa to aa 1, mJ 1, mJ if mJ is even (in which aa case mJ
3d 3d 3d 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3 d
1 < m becaus
aa 3d 3d 3d
e m is even). aa This concludes the aa induction proof. aa
3d 3d 3d 3d 3d 3d 3d 3d
This result is in agreement+ aa with the aa theorem on Nim heap sizes represented+ aa as sums of
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa powers of
3d 3d 3d
2: aa aa1 aam d is if and aa only if,
3 d 3 d 3 d 3d 3 3 3 3 3
aa except for 20,
3d aa the aa powers of 2 making aa upm aaandn aacomeinpai
3 3 3d 3 3 3 3 3 d
+∗ aa 3 d d d d d d d d d d d d 3 3d 3d 3d 3d 3d
rs.Sothesemustbe aa thesame aa powersof2,except aa for1 aa =20,which aa occursinonly
aamorn,wherewe
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3
have aa assumed that n is the aa larger number, so aa 1 appears in the
d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 aad aa 3 aa d aa representation aa of 3 d 3 aa d aa n: 3 d
aa We aa have aa m aa
3 d aa 3 d aa 3 d
=2a 2b 2c 3d3d3d3d3d3d 3d3d3d3d3d3d
3 aa d
aa 4 aa
Game Theory Basics 1st Edition
aa aa aa aa aa aa
By Bernhard von Stengel. Chapters 1 -
aa aa aa aa aa aa aa aa
12 aa
aa aa
aa 1 aa
,TABLE OF CONTENTS 3d 3d 3d aa
aa
aa
1 - Nim and Combinatorial Games
3d 3d 3d 3d 3d aa
aa
2 - Congestion Games
3d 3d 3d aa
aa
3 - Games in Strategic Form
3d 3d 3d 3d 3d aa
aa
4 - Game Trees with Perfect Information
3d 3d 3d 3d 3d 3d aa
aa
5 - Expected Utility
3d 3d 3d aa
aa
6 - Mixed Equilibrium
3d 3d 3d aa
aa
7 - Brouwer’s Fixed-Point Theorem
3d 3d 3d 3d aa
aa
8 - Zero-Sum Games
3d 3d 3d aa
aa
9 - Geometry of Equilibria in Bimatrix Games
3d 3d 3d 3d 3d 3d 3d aa
aa
10 - Game Trees with Imperfect Information
3d 3d 3d 3d 3d 3d aa
aa
11 - Bargaining 3d 3d aa
aa
12 - Correlated Equilibrium
3d 3d 3d aa
aa aa
Game Theory Basics 3d 3d aa
Solutions to Exercises aa 3d aa 3d aa aa
© Bernhard von Stengel 2022 aa
3 d aa 3d 3d 3d
Solution to Exercise 1.1 aa 3d 3d 3d
(a) aaLet ≤ be defined aa by (1.7). To show that ≤ is transitive, consider x, y, z with x
3d 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa ≤ y and 3d 3d
y ≤ z. If x
aa
3d 3d 3d 3d 3d 3d
= y then x ≤ z, and if y = z then also x ≤ z. So the only case left is x < y and aa
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa y 3 aa d aa < 3 aa d
z, which implie aa 3d 3d
s x aa < aa z because < is transitive, and hence x aa ≤ z. aa
3d 3 d aa 3 d aa 3d 3d 3d 3d 3d 3d 3d 3 d aa 3d
Clearly, ≤ is reflexive because x = x and therefore x ≤ x. aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
To show that ≤is 3d 3d 3d3d3d3d3d aa 3d antisymmetric, consider x and y with x y and y≤ 3d 3d 3d 3d 3d 3d 3d3d3d3d3d 3d 3d 3d3d3d3d3d aa x. If we had≤ 3d 3d 3d
aa 3d x ≠ y then x aa< y and y < x, and by transitivity x < x which contradicts (1.38). Hence x
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 aa d aa = 3 aa
d
aa y, as required.
3d 3d 3 aa d aa Th
is shows that ≤ is a partial order. aa
3d 3d 3d 3d 3d 3d 3d
Finally, we show (1.6), so we have to show that x < y implies x y and x ≠≤ y
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3d3d 3d 3d 3d 3d 3
aa and vice versa. Let d 3d 3d 3d 3
aa 2 aa
, aa d x < y, which implies x y by (1.7). If we had x = y then x < x, contradicting (1.38), so we also ha
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
v aa≤ aa
e x ≠ y. aa aaConversely, x aa aay and x ≠ y imply by (1.7) x
3d 3d 3d 3 d 3d 3d3d3 d 3d 3d 3d 3d 3d 3d 3d d 33 aa d aa < 3 aa d aa y or x3d 3d 3 aa d aa = 3 aa
d
aay where the second case is exclud
3d 3d 3d 3d 3d 3d
ed, hence x
3d 3d 3 aa d aa < 3 aa d aa y, as required. aa
3d 3d ≤ aa
(b) aaConsider a partial order and≤ assume (1.6) as a definition of <. To show that < is transitive, su
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
ppose x < y, that is, x y and x ≠ y, and y <≤ z aa, that is, y z and y ≠ z. Because is transitive,≤
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3 d3d3 d 3d
aa x z. If 3d 3d3d3d3d 3d 3d
aawe had x = z then x y and y x and hence x = y by antisymmetry of , which contradicts aa
3d 3d 3d 3d 3d 3d 3d3d3d3d3d 3d 3d 3d3d3d3d3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3d3d3d 3d 3d 3 d
aax aa aa≠ aa
≤ ≤ ≤ 3
≤ d aa aa aa aa
3 aa aad y, so we have
3d 3d 3d 3 aa aa d x 3d3d3d3 aa aa d z and aa aax
3d 3 d 3 aa aad ≠ 3 aa aa d z, that is, x 3d 3d 3 d3 aa aa d < 3 aa aa d z by (1.6), as required. aa
3d 3d 3d 3d
≤ aa ≤ aa
Also, aa< aaisirreflexive, aabecause aax aa< aax aawouldbydefinitionmeanx aax aaandx aa≠≤x aa,but aathe
aalatteris
not true. aa
3d aa3d aa 3d aa3d aa3d aa3d aa 3d aa 3d aa 3d aa3d aa3d aa3d aa3d3d3d aa 3d aa 3d aa3d aa 3d aa3d aa3d aa3d aa3d aa3d aa 3d
Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice versa, gi
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
ven that < is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x ≠ y and then b
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
y definition x < y. Hence, x ≤ y implies x < y or x = y. Conversely, suppose x < y or x = y. If
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa 3 d aa 3d 3d 3d 3d 3 d aa 3d
x < y then x
3 d aa ≤ y by (1.6), and if x = y then x aa ≤ y because ≤ is reflexive. aa
3 d aa 3d 3d 3 d aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa 3d 3d 3d 3d 3d 3 d
This completes the proof. aa
aa 3d 3d 3d aa
Solution to Exercise 1.2 aa 3d 3d 3d
(a) aaIn aa analysing aa aa the aa aa games aa of aa three aa aa Nim aa heaps aa where aa aa one aa aa heap aa
3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d aa 3 d
has aa aa size aa aa one, aa aa we aa first aa aa look at
aa 3 d aa 3 some aa
d aa 3 d aa 3 d aa 3 d aa 3d 3d 3d
examples, and aa then use aa mathematical induction to aa prove what we conjecture to 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa be the losing positi 3d 3d 3d 3d
ons. A losing position is one where every move is to a winning position, because
3d 3d aa then the 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa oppo 3d
nent will win. The point of this exercise aa is to
3d 3d 3 d aa 3d 3d 3d 3d 3d 3d
aa formulate a precise statement to be proved, and t
3d 3d 3d 3d 3d 3d 3d 3d 3d
hen to prove it. aa 3d 3d 3d
First, if there are only two heaps recall that they are losing if and only if the heaps are of equa
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
l size. aa If they are of unequal size, then the winning move is to reduce the larger
3d 3 d aa 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d
heap so that bo 3d 3d 3d 3d
th heaps have equal size. aa 3d 3d 3d 3d
Consider three heaps of sizes 1, m, n, where 1 m n.≤ We aa observe≤ aa
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3d3d3d3d 3d3d3d3d3d 3d 3d
the following: 1, 1, m is win 3d 3d 3d 3d 3d 3d 3d
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
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
sing (observed earlier in the lecture), and hence 1, 2, n for n 4 is winning. 1, 3, n is winning for
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
any n 3 by moving to 1, 3, 2. For 1, 4, 5, reducing any aaheap produces a winning position, so thi
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aas is≥ losing. aa aa aa
3d 3d
aa 3 aa
, The general pattern for the losing positions thus seems to be: 1, m, m 1, for even+ numbers m.
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3
aa d aa This includes also the case m = 0, which we can take as the base case for an
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa induction.
3d 3 aa d
aa We now p 3d 3d
roceed to prove this formally. aa 3d 3d 3d 3d
First we show that if the positions of the form 1, m, n with m n are losing≤
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d3d3d3d3d3d 3d 3d
aa when m is even an
3d 3d 3d 3d 3d
aad n = m 1, then these are the only losing positions because any other position 1, m, n
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 aa d
aa with m 3d 3 aa d aa 3 aa
+
aan aais winning. aaNamely, aa if m
d 3d 3d 3d 3d 3d 3 d aa =n 3d 3 d aa then a winning aa move3d 3d 3d
aa 3d from1 , m, m is to aa 0, m, m, aa so aa we can assu
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
me m < n. If m is even then ≤ n > m 3d 3d 3d 3d aa 3d 3d 3d 3d 3 d 3d 3d 3 d 3 d
aa 1 (otherwise we aa would be in the aa position 1, m, m
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d 3d aa 1) and aa 3d 3d
so aa the aa winning aa move is to aa 1, m, m
3d 3d 3d 3d 3d 3d 3d 3d 3 d 3 d
aa 1. If m is odd then the aa winning aa move aa is to aa 1, +m, m 1, aa the aa same
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
as position 1, m 1, m (this would aa also aa be aa
3d 3d 3d 3d 3d 3d 3d 3d 3d aa 3 aa d aa a aa +winning
3 d aa 3 aa d aa move 3 aa d aa from 3 aa d aa 1, m, m 3d 3d 3
aa d aa so aa 3 aa d aa there aa the aa 3d aa 3 aad aa winning+ 3 aa d aa move aa 3 aa d aa is aa not 3 d aa 3 aa d aa unique). aa – aa − aa
Second, we show that any move aa from 1, m, m + 1 with aa even m is to aa a winning aa position, using
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3d 3 d
as ind 3d 3d
uctive hypothesis that 1, mJ, mJ + 1 for even mJ and mJ < m is a losing position. The move to 0,
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
m, m + 1 produces a winning position with counter-3d 3d 3d 3d 3d 3d 3d 3d 3d
move to 0, m, m. aa A move aa to aa 1, mJ, m + 1 for mJ < m is to aa a winning aa position with the
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa counter- 3d
move to 1, mJ, mJ + 1 if mJ is even and to 1, mJ, mJ − 1 if mJ is odd. A move to 3d aa 1, m, m is to3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa a winnin 3d 3d
g position with counter- 3d 3d 3d
move to aa 0, m, m. aa A move to aa 1, m, mJ with 3d mJ < m is also aa to
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 d aa 3d 3 d aa 3d 3d 3d
aa a winning aa position with aa the aa counter-
3d 3d 3d 3d 3d 3d
move to aa 1, mJ − aa 1, mJ if mJ is odd, and aa to aa 1, mJ 1, mJ if mJ is even (in which aa case mJ
3d 3d 3d 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3d aa 3d 3d 3d 3d 3 d
1 < m becaus
aa 3d 3d 3d
e m is even). aa This concludes the aa induction proof. aa
3d 3d 3d 3d 3d 3d 3d 3d
This result is in agreement+ aa with the aa theorem on Nim heap sizes represented+ aa as sums of
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d
aa powers of
3d 3d 3d
2: aa aa1 aam d is if and aa only if,
3 d 3 d 3 d 3d 3 3 3 3 3
aa except for 20,
3d aa the aa powers of 2 making aa upm aaandn aacomeinpai
3 3 3d 3 3 3 3 3 d
+∗ aa 3 d d d d d d d d d d d d 3 3d 3d 3d 3d 3d
rs.Sothesemustbe aa thesame aa powersof2,except aa for1 aa =20,which aa occursinonly
aamorn,wherewe
3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3
have aa assumed that n is the aa larger number, so aa 1 appears in the
d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3d 3 aad aa 3 aa d aa representation aa of 3 d 3 aa d aa n: 3 d
aa We aa have aa m aa
3 d aa 3 d aa 3 d
=2a 2b 2c 3d3d3d3d3d3d 3d3d3d3d3d3d
3 aa d
aa 4 aa
