# OK to post homework
# Vikrant, Feb 20 2022, Assignment 8

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

# No C8.
read("C9.txt"):

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# 1. Read and understand Chapter 1 of Sigmund.
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Done.

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# 2. (Defect,Defect) is always the unique Nash Equilibrium of the Donation Game.
#    Tragedy of T=R+0.01, R=10000P, S=P-0.01.
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# There is exactly one pure Nash Equilibrium, namely (Defect,Defect); we can check this via enumeration.
# There can thus be no other equilibria as we have shown, in hw7 via an enumeration of Best Response Correspondences of two player games with two strategies each, that if there is a strictly mixed Nash Equilibrium, then there are either 0 or 2 pure Nash Equilibria.

# When T=10000P+0.01 > R=10000P > P=P > S=P-0.01 :
# The Nash Equilibrium gives a payoff of (P,P).
# Cooperation, on the other hand, gives a much higher payoff of (10000P,10000P).
# Any one player could give in to temptation for the miniscule gain of 0.01 payoff from that of the cooperative solution.
# The players are thus left to settle for a 10000 times worse payoff lest the other player is greedy over what amounts to no considerable gain.

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# 3. (Pay,Refuse), (Refuse,Pay) are Nash Equilibria for Snowdrift.
#    Moreover, there is another mixed Nash Equilibrium.
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(*

# (Pay,Refuse) yields the payoffs (S,T);
# Row deviating yields payoffs (P,P), which is undesirable to Row (S>P), and
# Col deviating yields payoffs (R,R), which is undesirable to Col (T>R).
# (Pay,Refuse) is thus a Nash Equilibrium.

# Similarly, (Refuse,Pay) is a Nash Equilibrium.

# Since there are two pure Nash Equilibria, we know that there must be one stricly mixed Nash Equilibrium.
# In terms of T,R,S,P, this equilibrium solution is given by:
assume(T>R and R>S and S>P):
MNE22([[[R,R],[S,T]],[[T,S],[P,P]]]);
T:= 'T':
R:= 'R':
S:= 'S':
P:= 'P':
# which is ((S-P)/(T-R+S-P),(S-P)/(T-R+S-P)).

randomize(1886169186):
for i from 1 to 5 do
    T:= rand():
    R:= rand() mod T:
    S:= rand() mod R:
    P:= rand() mod S:
    for sol in MNE22([[[R,R],[S,T]],[[T,S],[P,P]]]) do
        print((sol[1]=1 and sol[2]=0) or (sol[1]=0 and sol[2]=1) or (sol[1]=(S-P)/(T-R+S-P) and sol[2]=(S-P)/(T-R+S-P))):
    od:
od:

*)