*Part 1*

Here is the MATLAB code for solving the spin glass problem:

% set the number of spins in the glass N = 100; % random energies from a gaussian distribution with mean 0 and variance 1 J = randn(1, N); % print out the minimum possible energy -sum(abs(J)) % construct a symmetric energy matrix A D = diag(J); D1 = [D(:,end) D(:,1:end-1)]; D2 = [D(end,:); D(1:end-1,:)]; A = 0.5 * (D1 + D2); % allocate a symmetrix matrix for the spins Z = sdpvar(N); % constrain the spin matrix to be psd and have 1's on the diagonal F = set(Z > 0) + set(diag(Z) == 1); % solve the semidefinite program solvesdp(F, trace(A*Z)); % print out the energy obtained trace(A * double(Z))On the first run, this technique found a configuration with the minimal energy, as indicated by the following ouput:

ans = -78.2133 ans = -78.2133Simulated annealing, though solved much faster, does not reach as good a solution. Here are the energies it finds for three cooling rates:

Rate | Energy |

0.1 | -64.1778 |

0.01 | -66.7326 |

0.001 | -73.2866 |

*Part 2*

For this part, I did the same as above, except I wrote this ugly nested for loop to obtain the energy matrix A:

A = zeros(N); for i = [1:N] col1 = mod(i - 1, S) + 1; row1 = (i - col1) / S + 1; down = mod(row1, S) + 1; right = mod(col1, S) + 1; up = mod(row1 + S - 2, S) + 1; left = mod(col1 + S - 2, S) + 1; for j = [i + 1:N] col2 = mod(j - 1, S) + 1; row2 = (j - col2) / S + 1; if ((row1 == row2 && (col2 == right || col2 == left)) || (col1 == col2 && (row2 == down || row2 == up))) r = 0.5 * randn(1,1); A(i, j) = r; A(j, i) = r; end end endI kept running out of memory on 10x10 matrices, so I did 9x9 instead. On one trial, the minimum possible energy was -128.6243 and the SDP relaxation found a minimal energy of -107.5701. For the same energy matrix, simulated annealing did almost as well at the lowest cooling rate:

Rate | Energy |

0.1 | -81.1302 |

0.01 | -88.8959 |

0.001 | -94.6923 |

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