# Please do not post homework
# Omar Aceval Garcia, 2/02/2025, Assignment 2

# Quantum Problem 2.1
# |l> = 1/sqrt(2) (|u> - |d>) and |r> = 1/sqrt(2) (|u> + |d>), which makes:
# <r|l> = 1/2 (<u|u> - <d|d> + <d|u> - <u|d>) = 1/2 ( 1 - 1 + 0 - 0) = 0
# <l|r> = 1/2 (<d|d> - <u|u> + <u|d> - <d|u>) = 1/2 ( 1 - 1 + 0 - 0) = 0

# Quantum Problem 2.2
# |i> = 1/sqrt(2) (|u> + i|d>) and |o> = 1/sqrt(2) (|u> - i|d>)
# Of course these are not unique in that you can do a phase change by multiplying by e^{i*theta} for any real theta to either 
# <i|o> = 1/2 (<u|u> + i<u|d> + i <d|u> + i^2 <d|d> ) = 1/2 (1 + 0 + 0 - 1) = 0
#	Similarly <o|i> = 0
# <i|r> = 1/2 (<u|u> + <u|d> - i <d|u> - i <d|d> ) = 1/2 (1 + 0 - i - 0) = 1/2 (1-i)
# 	Similarly <r|i> = 1/2 (1+i), hence since (1-i^2) = 2 we have <r|i><i|r> = 1/2
# All other calculations are similar to the above two lines (some easier since they involve |u> and |d> instead of |l> and |r>)

# Quantum Problem 2.3
# If |i> = alpha|u> + beta|d> and |o> = gamma|u> + delta|d> then clearly:
# <u|i> = alpha<u|u> + beta<u|d> = alpha
# <i|u> = alpha*<u|u> + beta*<d|u> = alpha*
# Hence the condition <u|i><i|u> = 1/2 says |alpha|^2 = alpha(alpha*)= 1/2
#	Similar for beta, gamma, delta. This is part (a)
# <r|i> = 1/sqrt(2) (<u| + <d|)(alpha|u> + beta |d>) = 1/sqrt(2) (alpha + beta)
# <i|r> = 1/sqrt(2) (alpha*<u| + beta*<d|)(|u> + |d>)= 1/sqrt(2) (alpha + beta)*
# Hence <i|r><r|i> = 1/2 (|alpha|^2 + |beta|^2 + alpha(beta*) + beta(alpha*))
# Since the above equaling 1/2 is a requirement, by (a) we must have that the final piece is 0 = alpha(beta*) + beta(alpha*)
# Note that alpha(beta*) is the conjugate of beta(alpha*) and hence since their sum is 0, alpha(beta*) is purely imaginary
# We conclude that alpha and beta cannot both be real. The same reasoning applies to gamma and delta.


with(Statistics):

# TotCliques problem
TotClique := proc(G,k):
return (nops(Cliques(G, k)) + nops(Cliques(Comp(G), k))):
end: 

# CheckRamsey problem
# Makes a random graph on 18 vertices and checks its number of k-cliques/anticliques. Outputs the minimum after doing so K times.
# This does NOT check if it accidentally generated the same graph 2 or more times. That's possible but unnecessary in my opinion
CheckRamsey := proc(k,K) local i, m:
m := 1000000:
i := 1:
while (m <> 0) and (i <= K) do
  m:= min(m, TotClique(RG(18, 1/2), k)):
  i++:
od:
return m:
end:

# Estimating avg cliques
EstimateAverageClique := proc(n,p,k,K) local i,S:
i := 1:
S := seq(nops(Cliques(RG(n, p), k)) ,i =1..K):
return Mean([S]):
end:
# When trying out EstimateAverageClique(10, 1/2, 4, 10000) I got 3.375