How can we numerically simulate a Schramm-Loewner evolution?

Question

I would like numerically simulate the curve generated by a Schramm–Loewner evolution, but there are two issues which I'm not able to resolve.

Firs of all, let me recapulate the following result of the Loewner equation:

If $(\xi_t)_{t\ge0}$ is a continuous process, then

1. There is a uniqe $\zeta(z)\in(0,\infty]$ and a unique continuous $g(z):[0,\zeta(t))\to\mathbb C$ with $$g_t(z)\ne\zeta_t\tag1$$ and $$g_t(z)=z+\int_0^t\frac2{g_s(z)-\xi_s}\:{\rm d}s\tag2$$ for all $t\in[0,\zeta(z))$. If $\zeta(z)<\infty$, then $$g_t(z)-\xi_t\xrightarrow{t\to\zeta(z)}0\tag3.$$
2. If $\zeta(\xi_0):=0$, then $$C_t:=\{z\in\mathbb C:t<\zeta(z)\}$$ is open and $$g_t:C_t\to\mathbb C$$ is holomorphic for all $t\ge0$.

Let $\mathbb H:=\{z\in\mathbb C:\Im(z)>0\}$, $$K_t:=\{z\in\mathbb H:t\ge\zeta(z)\}$$ and $$H_t:=\mathbb H\setminus K_t.$$ We restric $\zeta$ to $\mathbb H$ and $g_t$ to $H_t$.

Now, if I got this right, the curve we are considering is $$\gamma_t:=g_t^{-1}(\xi_t)$$ (up to which $t$ is this defined and why does the inverse exist at all?) and my guess is that we can show $$K_t=\{\gamma_s:s\in[0,t]\}\tag4$$ (but how?).

The Schramm-Loewner evolution with parameter $\kappa\ge0$ is giving by running this for $$\xi_t:=\sqrt\kappa B_t\;\;\;\text{for }t\ge0,$$ where $(B_t)_{t\ge0}$ is a standard Brownian motion.

Question 1: How do I run this numerically? I mean, I can solve a system like $(2)$ using the Euler method and I clearly can simulate a Brownian motion. But how do I numerically compute the inverse occuring in the definition of the curve $t\mapsto\gamma_t$? (Or is this the wrong approach?)

Question 2: I actually don't want to obtain a curve on $\mathbb H$, but a curve on $[0,1)^2$. Can I achieve this somehow? The curve should also be "space-filling" (which can be shown for the curve on $\mathbb H$, whenever $\kappa\ge8$).