#Nathan Fox
#Homework 19
#I give permission for this file to be posted online

##Read old files
read(`C19.txt`):

#Help procedure
Help:=proc():
print(` NashTest(n) `):
end:

##Problem 1
##BRA case
#We have [[R,S],[T,P]]*[[y],[1-y]] = [[R*y+S*(1-y)],[T*y+P*(1-y)]]
#So, the payoff is (R*y+S*(1-y))*x+(T*y+P*(1-y))*(1-x).  The
#coefficient on x is R*y+S*(1-y)-T*y-P*(1-y).  Hence, we will have
#1 if this quantity is positive, 0 if it is negative, and all x
#between 0 and 1 if it is zero.  Solving yields (R-S-T+P)*y>P-S.
#Hence, if R-S-T+P>0, our 1 condition is y>(P-S)/(R-S-T+P). If
#R-S-T+P<0, our 1 condition is y<(P-S)/(R-S-T+P).  If R-S-T+P=0,
#then the value does not depend on y; it is determined by the
#constant term.

##BRB case
#We have [x,1-x]*[[R,T],[S,P]] = [[R*x+S*(1-x)],[T*x+P*(1-x)]]
#So, the payoff is (R*x+S*(1-x))*y+(T*x+P*(1-x))*(1-y).  The
#coefficient on y is R*x+S*(1-x)-T*x-P*(1-x).  Hence, we will have
#1 if this quantity is positive, 0 if it is negative, and all y
#between 0 and 1 if it is zero.  Solving yields (R-S-T+P)*x>P-S.
#Hence, if R-S-T+P>0, our 1 condition is x>(P-S)/(R-S-T+P). If
#R-S-T+P<0, our 1 condition is x<(P-S)/(R-S-T+P).  If R-S-T+P=0,
#then the value does not depend on x; it is determined by the
#constant term.

##Problem 2
#If R>T, then [1,1] will be a Nash equilibrium, since both players
#will want to cooperate if the other does.
#If P>S, then [0,0] will be a Nash equilibrium, since both players
#will want to defect if the other does.
#If S>P and T>R, then [1,0] and [0,1] will be Nash equilibria,
#since A will want to cooperate if B defects, and B will want to
#defect if A cooperates, and vice versa
#If (T=R and S>=P) or (S=P and T>=R), then there will be no pure
#strategy.

##Problem 3
#n = number of iterations
NashTest:=proc(n) local i,R,S,T,P,rd,A,B,nash:

rd:=rand(-10..10):

for i from 1 to n do
    R:=rd():
    S:=rd():
    T:=rd():
    P:=rd():
    print(`R,S,T,P=`):
    print(R,S,T,P):
    if R > T then
        print(`[1,1] should be a Nash equilibrium`):
    fi:
    if S > P and T > R then
        print(`[1,0] should be a Nash equilibrium`):
        print(`[0,1] should be a Nash equilibrium`):
    fi:
    if P > S then
        print(`[0,0] should be a Nash equilibrium`):
    fi:
    if (T = R and S >= P) or (S = P and T >= R) then
        print(`There should be no pure strategy`):
    fi:
    A:=[[R,S],[T,P]]:
    B:=[[R,T],[S,P]]:
    nash:=false:
    #Test [1,1]
    if BRA(A,1)[1] = 1 and BRB(B,1)[1] = 1 then
        print(`[1,1] is a Nash equilibrium`):
        nash:=true:
    fi:
    #Test [1,0]
    if BRA(A,0)[1] = 1 and BRB(B,1)[1] = 0 then
        print(`[1,0] is a Nash equilibrium`):
        nash:=true:
    fi:
    #Test [0,1]
    if BRA(A,1)[1] = 0 and BRB(B,0)[1] = 1 then
        print(`[0,1] is a Nash equilibrium`):
        nash:=true:
    fi:
    #Test [0,0]
    if BRA(A,0)[1] = 0 and BRB(B,0)[1] = 0 then
        print(`[0,0] is a Nash equilibrium`):
        nash:=true:
    fi:
    if not nash then
        print(`No pure strategy`):
    fi:
    print(``):
od:

end:

#All the tests I ran confirmed my answer to Problem 2