# Please do not post
# Omar Aceval Garcia, due 2/02/2025 :) Assignment 3

# Quantum Problem 3.1
# Assuming that L is a Hermitian operator (matrix) on any finite dim space, we first argue that it has an eigenvector.
# This is because det(L - lamba Id) = 0, where Id is the n x n identity matrix, is an n-degree complex polynomial, hence it has a root.
# Since L - lambda Id has determinant zero, by the Rank-Nullity theorem it must have a vector v where (L - lambda Id)v = 0.
# So Lv = lambda v, and so L has an eigenvector. Now the argument is straightforward:
# Let w_1, v_2, ... , v_n be an orthonormal basis for the space, where w_1 is an eigenvector. 
# For any i > 1, we argue (in bra-ket notation) that Lv_i is orthogonal to W_1.
# By assumption <w_1 | v_i> = 0 = <v_i | w_1>, and L|w_1> = lambda |w_1>, and <v|L corresponds to L|v> since L is Hermitian.
# So it follows that <w_1 | Lv_i > = <w_1L | v_i> = lambda <w_1 | v_i> = 0
# Similarly <v_iL | w_1> = 0. 
# So we have just shown that L is a Hermitian operator on the subspace v_2, v_3, ... , v_n. Hence it contains an eigenvector...
# So the proper inductive argument guarantees that we can assume w_1, w_2, ... , w_n is an orthonormal basis of eigenvectors.

# Threshhold for connectivity
# For n = 10, 20, 30 and a sample size of K=200 I will guess the threshold for connectivity by using p = 1/10, 2/10, etc 
# and creating a matrix of probabilities of being connected.

# Estimates the probability (for large K) that a graph on n edges with prob p of having each edge is connected
PC:=proc(n,p,K) local i:evalf(coeff(add(IsCo(RG(n,p)),i=1..K),true,1)/K):end:
ThreshGuess:= proc(K) local i,j:
return [seq([seq(PC(10*i, j/10, K) , j = 1..9 )]  , i=1..3)]:
end:
# Running this returned the matrix of probabilities:
# [
# [0.005000000000, 0.2650000000, 0.5900000000, 0.8650000000, 0.9750000000, 1., 1., 1., 1.], 
# [0.05500000000, 0.6800000000, 0.9700000000, 1., 1., 1., 1., 1., 1.], 
# [0.2250000000, 0.9600000000, 1., 1., 1., 1., 1., 1., 1.]
# ]
# Which suggests that as n goes higher, the threshold of connectivity quickly falls. 
# If I was guessing what p value corresponded to having a 50% of connectivity, based on this data I would say:
# A little under/around .3 for n=10, maybe between .15 and .2 for n=20, and around 0.05 for n=30.