Analysis of diode biasing in the given bridge connection of four diodes
The voltages across the diodes will vary with the current through them. But they only vary about \$100\:\textrm{mV}\$ (different diode types may vary a little here) for each factor of 10 change in current. That's not much of a voltage change for quite a lot of change in current. So as a first order approximation for now, you can consider the voltage drops across the diodes as approximately fixed.

When the input is at \$0\:\textrm{V}\$, there will be about \$10\:\textrm{V}\$ across both \$10\:\textrm{k}\Omega\$ resistors to the two power supplies -- or, roughly \$1\:\textrm{mA}\$. This splits in two directions through the pairs of diodes. So that means about \$500\:\mu\textrm{A}\$. I suspect that there will be perhaps from \$650-700\:\textrm{mV}\$ across each diode, then. Let's call it \$680\:\textrm{mV}\$ just to pick a number.

If you assume that the diodes all have fixed voltages across them no matter what, then the output should follow \$V_s\$ the entire time. (\$V_s\$, plus \$D_1\$'s diode drop, but then minus \$D_2\$'s diode drop, again.) But there's a problem. As \$V_s\$ declines towards \$-10\:\textrm{V}\$, at some point it crosses over a voltage after which \$D_3\$ simply turns off. Below that point, all that you have is the \$10\:\textrm{k}\Omega\$ output resistor sourcing current from ground, through \$D_4\$ and into the negative rail resistor. This sets up a current that will have a magnitude of \$10\:\textrm{V}\$, less a diode drop, divided by the resistance sum of \$20\:\textrm{k}\Omega\$. That current creates a voltage drop across the negative rail \$10\:\textrm{k}\Omega\$ resistor so that the voltage at the cathode of \$D_4\$ and \$D_3\$ will be \$-10\:\textrm{V}+\frac{10\:\textrm{V}-680\:\textrm{mV}}{10\:\textrm{k}\Omega+10\:\textrm{k}\Omega}\cdot 10\:\textrm{k}\Omega\$ or about \$-5.34\:\textrm{V}\$. This means that \$V_s\$ will be \$-5.34\:\textrm{V}+680\:\textrm{mV}=-4.66\:\textrm{V}\$ at this point. So when \$V_s \lt-4.66\:\textrm{V}\$ nothing more happens at the output. So the output tracks \$V_s\$ to that point in the negative direction, but no further.

A similar argument also has it that when \$V_s \gt +4.66\:\textrm{V}\$, the output no longer tracks \$V_s\$.

So the answer is that:

$$\begin{align*}
\begin{array}{cc}
V_s &amp; V_{out} \\ \\ -10\:\textrm{V} \rightarrow -4.66\:\textrm{V} &amp; -4.66\:\textrm{V} \\ -4.66\:\textrm{V}\rightarrow +4.66\:\textrm{V} &amp; V_s \\ +4.66\:\textrm{V} \rightarrow +10\:\textrm{V}  &amp; +4.66\:\textrm{V}
\end{array}
\end{align*}$$

You could get a little silly about it and remember that there's a \$100\:\textrm{mV}\$ change for each factor of 10 current and imagine that as \$\vert V_s\vert\$ gets closer to that magnitude of \$4.66\:\textrm{V}\$ that the current gradually shuts off and that while that happens, the voltage drop for the diode also diminishes. But this might add half (remember here we divided by 2, earlier), or another \$50\:\textrm{mV}\$ of extra range. So I might round things to a magnitude of \$4.7\:\textrm{V}\$ perhaps. But this is nit-picking and I'd probably should have rounded it, anyway.

So let's go with:

$$\begin{align*}
\begin{array}{cc}
V_s &amp; V_{out} \\ \\ -10\:\textrm{V} \rightarrow -4.7\:\textrm{V} &amp; -4.7\:\textrm{V} \\ -4.7\:\textrm{V}\rightarrow +4.7\:\textrm{V} &amp; V_s \\ +4.7\:\textrm{V} \rightarrow +10\:\textrm{V}  &amp; +4.7\:\textrm{V}
\end{array}
\end{align*}$$

Added information and image taken from additional work by the OP:


  jonk has already answered it correctly. 
  
  So, I post my simulation output here
  
  (Cutin voltage for the diode was 0.4 V in the simulation)