Several years ago I found a method based on continuous optimizaton which can sometimes
find lower bound. It also applies to the SAT problem but it may not be exactly the same as the SAT solver method.

Given the number of vertices $n$, and positive
integers $k,l$, we can code all $2$-colorings of the edges of $K_n$
by (the upper triangular part of) a binary matrix $x_{ij}, 1\le i <
j \le n$, and write down the energy function

$$\phi_{kl}^{(n)}(x)= \sum_{1 \le i_1 <...<i_k \le n} \prod_{1 \le s<
t \le k} x_{i_s i_t} + \sum_{1 \le j_1<...<j_l \le n} \prod_{1 \le s
< t \le l}(1- x_{j_s j_t}),$$

which counts the total number of monochromatic $1$-$K_k$ or
$0$-$K_l$. We now interpolate and think of $\phi(x)$ as a polynomial
function of the ${ n \choose 2}$ real variables $x_{ij}$ which is
linear in each, and must therefore satisfies a strong mean-value
property :

**Lemma**. Let $f(x_1,...,x_n)$ be a real function which is linear in
each variable. Then for any axes-parallel rectangular box of any
dimension $k$, the value of $f$ at the center is the average value of
$f$ over the vertices of the box.

**Proof**. This is true for $k=1$ (i.e. for a line), since $f$ is linear. For a $k$-dimensional box, we just have to take the average over the straight line joining the centers of two opposite $k-1$-dimensional faces which equals the
average over the whole box.

Applying this to $\phi$, we see that if
$$\phi^{(n)}_{kl}(1/2,...,1/2)={ n \choose k } 2^{-{k \choose 2}}
+{n \choose l} 2^{-{l \choose 2}}<1,$$
we get a lower bound $R(k,l)>n$, which is the same bound as the original probabilistic counting method. We now observe that we can make the same conclusion
if we can find any $x$ inside the unit box $[0,1]^{n \choose 2}$,
with $\phi(x)<1$ because of the maximum principle: $\phi$ is
harmonic in any subset of the variables so any extremal value of
$\phi$ inside the box must already occur at some vertex. So instead
of doing brute force searching over the vertices discretely, we can
try to minimize $\phi$ over the whole box continuously by some form
of gradient descent algorithm.

Finding the minimum of a polynomial inside a box is however an NP-hard problem since the energy function which counts the number of
falsified clauses of a truth assignment of a boolean formula in
conjunctive normal form has exactly the same sum-product form as
$\phi$ and SAT is NP-complete.

Evaluating $\phi$ and its gradient is compute intensive but is
polynomial in $n$ for fixed $k$ and $l$ and is certainly doable for
the range of $R(5,5)$ up to $n=49$.

Maybe someone with more computing power and better optimization
algorithm can try this. The idea is that we don't look at all the
vertices just the good ones following the gradient. The problem is
of course that we cannot be sure the best we have found is the true minimum
unless it is of zero energy.

Evaluating $\phi$ at $x=(1/2)^m, m=\binom{n}2$ gives the same bound as the original probabilistic existence method. However our method is constructive. One can always change each $x_{ij}=1/2$ to either $0$ or $1$ without increasing $\phi$ since $\phi$ is linear in each variable. In $m$ steps we will arrive at a vertex $x_0$ with $\phi(x_0) \le \phi(x)<1$ and hence must be zero since $\phi$ is non-negative integral at the verticesr. So $x_0$ defines an extremal graph with no monochromatic. The same method works for 2-colorings of hypergraphs.