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SOLUTION MANUAL Game Theory Basics 1st Edition By Bernhard von Stengel. Chapters 1 - 12
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SOLUTION MANUAL Game Theory Basics 1st Edition By Bernhard von Stengel. Chapters 1 - 12
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SOLUTION MANUALGame Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS 6 6 6
1 - Nim and Combinatorial Games
6 6 6 6 6
2 - Congestion Games
6 6 6
3 - Games in Strategic Form
6 6 6 6 6
4 - Game Trees with Perfect Information
6 6 6 6 6 6
5 - Expected Utility
6 6 6
6 - Mixed Equilibrium
6 6 6
7 - Brouwer’s Fixed-Point Theorem
6 6 6 6
8 - Zero-Sum Games
6 6 6
9 - Geometry of Equilibria in Bimatrix Games
6 6 6 6 6 6 6
10 - Game Trees with Imperfect Information
6 6 6 6 6 6
11 - Bargaining
6 6
12 - Correlated Equilibrium
6 6 6
2
,Game Theory Basics
6 6
Solutions to Exercises
6 6
©6 Bernhard6von6Stengel62022
Solution6to6Exercise61.1
(a) Let6≤6be6defined6by6(1.7).6 To6show6that6≤6is6transitive,6consider6x,6y,6z6with6x6 ≤6y6and6y6≤6z.6If6x6=6y6t
hen6x6≤6z,6and6if6y6=6z6then6also6x6≤6z.6So6the6only6case6left6is6x6<6y6and6 y6 <6 z,6which6implies6x6 <6 z6
because6<6is6transitive,6and6hence6x6 ≤6z.
Clearly,6≤6is6reflexive6because6x6=6x6and6therefore6x6 ≤6x.
To6show6that66666≤is6antisymmetric,6consider6x6and6y6with6x66666y6and6y≤
66666x.6If6we6had6x6≠6y6then6x6<
≤
6y6and6y6<6x,6and6by6transitivity6x6<6x6which6contradicts6(1.38).6Hence6x6 =6 y,6as6required.6 This6sh
ows6that6≤6is6a6partial6order.
Finally,6we6show6(1.6),6so6we6have6to6show6that6x6<6y6implies6x666y6and6x6≠6y6≤and6vice6versa.6Let6x6
<6y,6which6implies6x6y6by6(1.7).6If6we6had6x6=6y6then ≤ 6x6<6x,6contradicting6(1.38),6so6we6also6have6x6
≠6y.6 Conversely,6x666 y6and6x6≠6y6imply6by6(1.7)6x6 <6 y6or6 x6 =6 y6where6the
≤ 6second6case6is6excluded,6he
nce6 x6 <6 y,6as6required.
(b) Consider6a6partial6order6and≤ 6assume6(1.6)6as6a6definition6of6<.6To6show6that6<6is6transitive,6sup
pose6x6<6y,6that6is,6x6y6and6x6≠6y,6and6y6<6z,≤6that6is,6y6z6and6y6≠6z.6Because6666is6transitive,
≤ 6x6666z.6If6we
6had6x6=6z6then6x66666y6and6y66666x6and6hence6x6=6y6by6antisymmetry6of6666,6which6contradicts6 x6 ≠6 y,6s
≤ ≤ ≤ ≤
o6we6have6 x6666z6 and6 x6 ≠6 z,6that6is,6x6 <6 z6by6(1.6),6as6required.
≤ ≤
Also,6<6is6irreflexive,6because6x6<6x6would6by6definition6mean6x666x6and6x6≠6x,≤6but6the6latter6is6not
6true.
Finally,6we6show6(1.7),6so6we6have6to6show6that6x6 ≤6y6implies6x6<6y6or6x6=6y6and6vice6versa,6given6
that6<6is6defined6by6(1.6).6Let6x6≤6y.6Then6if6x6=6y,6we6are6done,6otherwise6x6≠6y6and6then6by6defin
ition6x6<6y.6Hence,6x6≤6y6implies6x6<6y6or6x6=6y.6Conversely,6suppose6x6 <6 y6or6x6=6y.6 If6x6 <6 y6then6
x6 ≤6y6by6(1.6),6and6if6x6=6y6then6x6 ≤6 y6because6≤6is6reflexive.6 This6completes6the6proof.
Solution6to6Exercise61.2
(a) In6 analysing6 the6 games6 of6 three6 Nim6 heaps6 where6 one6 heap6 has6 size6 one,6 we6 first6 look6at6some6exa
mples,6and6then6use6mathematical6induction6to6prove6what6we6conjecture6to6be6the6losing6positions.6
A6losing6position6is6one6where6every6move6is6to6a6winning6position,6because6then6the6opponent6
will6win.6 The6point6of6this6exercise6is6to6formulate6a6precise6statement6to6be6proved,6and6then6to6
prove6it.
First,6if6there6are6only6two6heaps6recall6that6they6are6losing6if6and6only6if6the6heaps6are6of6equal6
size.6 If6they6are6of6unequal6size,6then6the6winning6move6is6to6reduce6the6larger6heap6so6that6both
6heaps6have6equal6size.
3
, Consider6three6heaps6of6sizes61,6m,6n,6where6166666m66666n.
≤6We≤6observe6the6following:61,61,6m6is6wi
nning,6by6moving6to61,61,60.6Similarly,61,6m,6m6is6winning,6by6moving6to60,6m,6m.6Next,61,62,636is6
losing6(observed6earlier6in6the6lecture),6and6hence61,62,6n6for6n646is6winning.61,63,6n6is6winning6f
or6any6n636by6moving6to61,63,62.6For61,64,65,6reducing6any6heap6produces6a6winning6position,6so6
≥ ≥
this6is6losing.
The6general6pattern6for6the6losing6positions6thus6seems6to6be:61,6m,6m61,6for6even+ 6numbers6m.6
This6includes6also6the6case6m6=60,6which6we6can6take6as6the6base6case6for6an6induction.6 We6now6
proceed6to6prove6this6formally.
First6we6show6that6if6the6positions6of6the6form61,6m,6n6with6m666666n6are6losing ≤ 6when6m6is6even6an
d6n6=6m61,6then6these6are+6the6only6losing6positions6because6any6other6position61,6m,6n6 with6m6 6 n
6 is6winning.6 Namely,6if6m6 =6n6 then6a6winning6move6from61,6 m,6 m6is6to60,6 m,6 m,6so6we6can6assume6m6
≤
<6n.6 If6m6is6even6then6n6>6m6 6 16(otherwise6we6would6be6in6the6position61,6m,6m6 6 1)6and6so6the6win
+
ning6move6is6to61,6m,6m6 6 1.6If6m6is6odd6then6the6winning6move6is6to61,6m,6m61,6the6same6as6position6
+ +
1,6m61,6m6(this6would6 also6 be6 a6 winning6 move6 from6 1,6m,6m6 so6 there6 the6 winning6 move6 is6 not6 uniqu
e). – −
Second,6we6show6that6any6move6from61,6m,6m6+616with6even6m6is6to6a6winning6position,6using6as6ind
uctive6hypothesis6that61,6mJ,6mJ6+616for6even6mJ6and6mJ6<6m6is6a6losing6position.6The6move6to60,6m
,6m6+616produces6a6winning6position6with6counter-
move6to60,6m,6m.6A6move6to61,6mJ,6m6+616for6mJ6<6m6is6to6a6winning6position6with6the6counter-
move6to61,6mJ,6mJ6+616if6mJ6is6even6and6to61,6mJ,6mJ6−616if6mJ6is6odd.6A6move6to61,6m,6m6is6to6a6winni
ng6position6with6counter-
move6to60,6m,6m.6A6move6to61,6m,6mJ6with6 mJ6<6 m6is6also6to6a6winning6position6with6the6counter-
move6to61,6mJ6−61,6mJ6if6 mJ6is6odd,6and6to61,6mJ6 1,6mJ6if6mJ6is6even6(in6which6case6mJ6 16<6m6because6
m6is6even).6This6concludes6the6induction6proof.
+ +
This6result6is6in6agreement6with6the6theorem6on6Nim6heap6sizes6represented6as6sums6of6powers6of62:
0
6 16 6 m6 6 n6is6losing6if6and6only6if,6except6for62 ,6the6powers6of626making6up6m6and6n6come6in6pairs.6S
∗6 +∗ +∗
o6these6must6be6the6same6powers6of62,6except6for616=620,6which6occurs6in6only6m6or6n,6where6we6have
6assumed6that6n6is6the6larger6number,6so616appears6in6 the6 representation6 of6 n:6 We6 have6 m6 =6 2
a66666
6 b666666 c
2 2
+ + +6 ·6 ·6 · ·6 ·6 ·6 ≥
for6 a6 >6 b6 >6 c6 >66666666 1,6so6 m6
+ + +6 ·6 ·6 ·6 + +
is6 even,6 and,6 with6 the6 same6 a,6b,6c,6.6.6.,6 n6 =6 2a6 6 6 2b6 6 6 2c 16 =6 m6666 1.6 Then
∗16 666666
+6 ∗m66666
+6 ∗n666666
≡60.
∗ 6 The6 following6 is6 an6example6 using6 the6 bit6 representation6 where
m6 =6126(which6determines6the6bit6pattern61100,6which6of6course6depends6on6m):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) We6use6(a).6Clearly,61,62,636is6losing6as6shown6in6(1.2),6and6because6the6Nim-
sum6of6the6binary6representations601,610,6116is600.6Examples6show6that6any6other6position6is6wi
nning.6The6three6numbers6are6n,6n6 1,6n6 6 2.+6If6n6is6+
even6then6reducing6the6heap6of6size6n626to616cre
ates6the6position6n,6n6 1,616which6is6losing6as6shown6in6(a).6If6n6is6odd,6then6n6 16is6even6and6n666
+ +
26=6 n666166616so6by6the6same6argument,6a6winning6move6is6to6reduce6the6Nim6heap6of6size6n6to616
+ + (6 +6 )6+
(which6only6works6if6n6 >61).
4
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