Ring Signatures: Why is the combination function easy to solve? I'm reading Rivest's How To Leak A Secret and am on the part where they describe their signature scheme ("Generating a ring signature").

Why is computing $C_{k, v}(y_1, ..., y_r) = v$ easy? The computation seems to be "linear" or single-threaded, i.e. of the form $E_k(y_r \oplus E_k(y_{r-1} \oplus E_k(...E_k(y_1 \oplus v))))$. So let's say $r_i=s$ -- we require the value of $y_i$ to proceed with the computation (i.e. we can't continue to encrypt if we don't know its value), and so wouldn't we just be stuck trying all values of $y_i$ until we hit one that results in $v$ when the computation finishes?

Wikipedia also says that $C_{k, v}$ is trivially invertible given $r - 1$ parameters. I'm blanking out - why is this the case?

I did look at this answer but I didn't understand the line 


  Start again at 12 o'clock and calculate going counter-clockwise until you arrive at 6 o'clock. 


How can you "go counter-clockwise" when all the computations counter-clockwise have inputs dependent on all previous computation? All of the XORs have the previous computation as input.



Sorry if I've made any gaffs, this is my first *Exchange question.
I think I might have a potential solution, anyone care to confirm this? Let's take an example with 5 public keys (including ours). Let's assume our $s = 3$.

Let's borrow from that previous Stack Overflow post. Let's define a left var as 

$$l = E(E(E(v \oplus y_1) \oplus y_2) \oplus y_3 )$$

and we know that

$$ z = r = E(E(l \oplus y_4) \oplus y_5))$$

Now let's start decrypting $z$:

$$D(z) = E(l \oplus y_4) \oplus y_5$$
$$D(z) \oplus y_5 = E(l \oplus y_4)$$
$$D(D(z) \oplus y_5) = l \oplus y_4$$
$$D(D(z) \oplus y_5) \oplus y_4 = l$$

We've solved for $l$! Solving for $y_s = y_3$ is easy now:

$$l = E(E(E(v \oplus y_1) \oplus y_2) \oplus y_3 )$$

$$D(l) = E(E(v \oplus y_1) \oplus y_2) \oplus y_3$$

$$D(l) \oplus E(E(v \oplus y_1) \oplus y_2) = y_3$$

And then we just do $g^{-1}(y_3)=x_3$, giving us our full signature.