6

Yes it is possible, but it may not be as efficient as the other methods listed in that PDF file. In fact, I'm still not sure there's any problem for which QUBO is the best way to solve it (see this: What are some real-world applications of QUBO?, and if interested in this area you may also find this one interesting: Are there any real-world problems where ...


4

The log-likelihood function $L$ is continuous while QUBO is discrete. If you really wanted to formulate it as a QUBO, this is how I would do it. In summary, the steps would be as follows: $L \to \text{ Discrete-}L \to \text{ Binary-} L \to \text{Higher-Order-Binary-} L \to \text{ QUBO } L$ $L \to \text{ Discrete-}L$ This step would simply be restricting ...


3

Luckily in this case, the exponential can be treated in a way very similar to how we're already familiar. If we use the example that I chose in my answer to your recent question, we would have, where your $\vec{r}$ is just $(s_1,s_2)$: $$\tag{1} Z=\sum_{s_1,s_2} e^{\left(\mathcal{J}s_1s_2 + hs_1\right)}. $$ If $\mathcal{J}=2.5$ and $h=3$, then we would have: ...


3

\begin{align}\text{diff}^2&=c^2+4\left(\left(\sum s_jx_j\right)^2-c\sum s_jx_j\right)\\&=c^2+4\left(\sum s_j^2x_j^2+\sum_{\rm cyc}s_ks_\ell x_kx_\ell-c\sum s_jx_j\right)\\&=c^2+4\left(\sum x_j\boldsymbol{s_j(s_j-c)}x_j+\sum_{\rm cyc}x_k\boldsymbol{s_ks_\ell}x_\ell\right)\\ &=c^2+4x^\top Qx\end{align}


3

According to the equality constraint (the equal to 4 one), at least two of the $x_{?}$ are 1. Therefore, the slack for the 1st constraint is at most 3. According to the equality constraint, the worst case is $x_1$, $x_4$, $x_5$ are 1 while other xs are 0. The slack is 6 in that case for the 3rd constraint. Alternatively, we can approximate the bound of the ...


3

Your bisection code is returning a tuple of [a,b] but in your main function you are only retrieving a, which should be causing the type error. There's also another bug in your bracket_minimum function - it can potentially get stuck in an infinite loop if your return condition is never satisfied. Ideally you want to have max iterations and return no solution ...


2

Considering that all your plots will be for the same problem you don't need to normalise by using relative error, so you can use either representation as long as you use it consistently for everything. The thing I would consider foremost is the readability of the charts, so I would suggest to plot both and see which representation is better scaled to ...


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