https://mirror.codeforces.com/contest/1349/problem/D might be helpful

Adding one more condition $$$E(x, y) = E(y, x)$$$. Next if we try finding values of $$$E(1, n)$$$ for different values of $$$n$$$, we would get $$$E(1, n) = 2^n - 1$$$.

Next we have $$$E(1, n) = 1 + \frac{n}{n+1} E(2, n-1)$$$ and so on, we can calculate recursively. I haven't simplified the general expression for $$$E(x, y)$$$ but I think we might get some $$$O(1)$$$ formula.

I did the same thing in the original problem and observed that $$$E(x, y) = xy$$$. Also your recurrence is wrong. It should be $$$E(x, y) = 1 + \frac{x}{x+y} E(x-1, y+1) + \frac{y}{x+y} E(x+1, y-1)$$$

For simplicity, I'll denote their (constant) sum as $$$s$$$ and then the recurrence you mentioned as:

In terms of computation, as you've mentioned the immediate approach is dp but since the states form cycles, it doesn't quite work. However the equalities are linear in the values $$$f(k)$$$ so the next idea should be to solve these linear equations, in time $$$O(s^3)$$$.

Now, I can't quite find a closed form, but I can give an $$$O(s)$$$ algorithm to compute $$$f$$$: first we can reorder the equality in the following way:

This is a recurrence on the adjacent differences, with conditions $$$f(0) = f(s) = 0$$$. If we denote $$$f(1) = c$$$, all the values of $$$f$$$ are then a function of $$$c$$$, where by the condition $$$f(s) = 0$$$ the answer is unique. Specifically, if my calculations are right:

And then it must hold that $$$f(s) = 0$$$, but this looks disgusting. As for the algorithm, we can define the homogeneous version of the equation above:

and with the single condition $$$g(0) = 0$$$, it's easy to find one solution: choose $$$g(1) = 1$$$, and then:

Observe that for any $$$f$$$ which is a solution to the non-homogeneous equation, with the condition $$$f(0) = 0$$$ (but without $$$f(s) = 0$$$), and for any constant $$$c$$$, the function $$$f + cg$$$ is also a solution. So we can find any single $$$f$$$ and then shift it using $$$g$$$ so that $$$f(s) + cg(s) = 0$$$.

To find a single $$$f$$$, just choose $$$f(1) = 0$$$ and then compute the adjacent differences in $$$O(s)$$$, and then prefix sum.

Usually, when your $$$dp$$$ has "cycles" but they form linear relationships, you can solve it using Gaussian Elimination. However, this has complexity $$$O(n^3)$$$, which is pretty sad. However, for some special forms of equation systems (in particular, ones with a "small lookahead", where dependencies are only a tiny bit into the future), you can do some neat trick. I'll exemplify here for this problem, but you can think of how to generalize it.

Let's denote $$$n = x+y$$$ and $$$E_i = E(i, n - i)$$$. We know that $$$E_0 = E_{n} = 0$$$ and that $$$E_i = 1 + \frac{i}{n} E_{i-1} + \frac{n-i}{n} E_{i+1}$$$. This is, as you said, a "cyclic dp". However, let's do the following trick:

First, denote $$$X = E_1$$$ (undeterminate). Then, we can use equation $$$E_1 = 1 + \frac{1}{n} E_0 + \frac{n-1}{n} E_2$$$ to solve for $$$E_2$$$ (as a function of $$$X$$$): $$$E_2 = X \frac{n}{n - 1} - 1 \frac{n}{n-1}$$$, if I'm not mistaken. Consequently, we can use equation $$$E_2 = ...$$$ to solve for $$$E_3$$$, etc. You can see that doing all this would yield equations of form $$$E_i = a_i X + b_i$$$ (linear functions in $$$X$$$), for all $$$i$$$.

Nevertheless, doing all these steps you'll obtain that $$$E_{n} = a_n X + b_n$$$. However, at this point you can use the last "extra" equation $$$E_n = 0$$$ to finally solve for $$$X$$$.

This approach is $$$O(n)$$$ if coded, but you can see that the equation of $$$a_n$$$ and $$$b_n$$$ comes from a product of $$$2 \times 2$$$ matrices, and if that product can be computed analytically, the approach may be solvable in faster complexity. I'll try a bit on paper and see if this is the case.