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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 ...


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