# OK to post homework
# Vikrant, Feb 13 2022, Assignment 7

# ================================================================================
# 0. Code that has been given.
# ================================================================================

read("C7.txt"):

# ================================================================================
# 1. Valentine.
# ================================================================================

(*

randomize(296337491):
with(plots):
RoseCardioid:= t->t^2*(1 - sin(t)):
layers:= 10:
R:= 
    {
        plot(RoseCardioid(t),t=0..layers*Pi+Pi/2,coords=polar,axes=none,color="Red"), 
        plot(RoseCardioid(t),t=-layers*Pi+Pi/2..0,coords=polar,axes=none,color="Crimson")
    }:
c:= ["RosyBrown","Pink"]:
T:= {textplot([990,-1550,"I love Bernoulli(1/2)!"],rotation=Pi/4)}:
g:= 10:
for i from 0 to 3 do
    s:= 30+20*i:
    t:= [seq([RoseCardioid(Pi*t/g)-t,Pi*t/g,2-LoadedCoin(1/2),rotation=Pi*t/g+Pi/2,font=["times","roman",s/5]],t=s+3..s+g)]:
    T:= T union {textplot(t,coords=polar,color=c[LoadedCoin(1/2)])}:
    s:= -30-20*i:
    t:= [seq([RoseCardioid(Pi*t/g)+t,Pi*t/g,2-LoadedCoin(1/2),rotation=Pi*t/g+Pi/2,font=["times","roman",-s/5]],t=s..s+g-3)]:
    T:= T union {textplot(t,coords=polar,color=c[LoadedCoin(1/2)])}:
od:
display(R union T);

*)

# ================================================================================
# 2. TypeCounts22.
# ================================================================================

TypeCounts22:=
    proc(K1,K2)
        local G,i:
        local Gs:= [seq(RG(2,2,K1),i=1..K2)]:
        evalf(
            [
                nops([seq(`if`(nops(NE(G))=1,1,NULL),G in Gs)]),
                nops([seq(`if`(nops(NE(G))=2 and nops(MNE22(G))-nops(NE(G))=0,1,NULL),G in Gs)]),
                nops([seq(`if`(nops(NE(G))=0 and nops(MNE22(G))=1,1,NULL),G in Gs)]),
                nops([seq(`if`(nops(NE(G))=2 and nops(MNE22(G))-nops(NE(G))=1,1,NULL),G in Gs)])
            ]/K2 
        ):
    end:

(*

randomize(1783538750):
for i from 1 to 5 do
    print(TypeCounts22(10^8,10^4)):
od:

# Experiment 1
# [0.7512000000, 0., 0.1258000000, 0.1230000000]
# Experiment 2
# [0.7445000000, 0., 0.1243000000, 0.1312000000]
# Experiment 3
# [0.7503000000, 0., 0.1249000000, 0.1248000000]
# Experiment 4
# [0.7517000000, 0., 0.1227000000, 0.1256000000]
# Experiment 5
# [0.7480000000, 0., 0.1280000000, 0.1240000000]

*)

# ================================================================================
# 3. Why two pure Nash Equilibria implies the existence of a strictly mixed one.
# ================================================================================

# Just exhaust all 16 configurations of Best Response Correspondences and we will find exactly two with two Pure Nash Equilibria, and both of them have a strictly Mixed Nash Equilibrium.

# ================================================================================
# 4. Limiting values of TypeCounts22 experiments.
# ================================================================================

# [3/4, 0, 1/8, 1/8].

# For the first, exactly 12 of the 16 configurations of Best Response Correspondences have exactly one Pure Nash equilibrium.
# For the second, none of the 16 configurations of Best Response Correspondences have exactly two Pure Nash equilibria and no strictly Mixed Nash Equilibria.
# For the third, exactly 2 of the 16 configurations of Best Response Correspondences have no Pure Nash equilibria and exactly one strictly Mixed Nash Equilibrium.
# For the fourth, exactly 2 of the 16 configurations of Best Response Correspondences have exactly two Pure Nash equilibria and exactly one strictly Mixed Nash Equilibrium.

# ================================================================================
# 5. PayOffGG.
# ================================================================================

PayOffGG:=
    proc(G,p1,p2)
        local i,j:
        if add(p1)<>1 or add(p2)<>1 or nops(G)<>nops(p1) or nops(G[1])<>nops(p2) then
            return(FAIL):
        fi:
        expand(add(add(p1[i]*p2[j]*G[i][j],i=1..nops(p1)),j=1..nops(p2))):
    end: