Ideal diode circuit techniques
The first thing I usually do when facing a problem like this is re-arrange it!







simulate this circuit &ndash; Schematic created using CircuitLab

This re-arrangement should make it abundantly obvious that the nodes enclosed by the red rectangular box are pulled towards voltages that are more positive than the nodes enclosed by the blue rectangular box.

This immediately suggests the idea that all the ideal diodes may be ON. Whether or not they actually are will depend on if you can make the currents sum up correctly. In this particular case, it's best to just start with the assumption here and see if the details work out.

Given that these are ideal diodes (which I take in this case to mean that they have zero voltage drop across them when ON, regardless of current) and taking the assumption for now that all the diodes are ON, the nodes enclosed by the red rectangular box are at the same potential as the nodes enclosed by the blue rectangular box. See note to the right side pointing this out. If all our assumptions are correct about the diodes being ON, then this voltage is determined by the grounded node at the left side.

Note that in the end it does turn out that the currents all balance out. I had earlier pointed out in a comment (made when I was nearly asleep last night and barely conscious) that the ground node would supply 5 parts out of 10. You can easily see how this divides out because I've also added the currents I was imagining last night -- nicely spelled out for you.



The above worked out nice because all the currents were in concert with each other. Let's keep the topology and just randomly change the resistor values so that the concert becomes cacophonous, rather than so cutely worked out.







simulate this circuit

The nodal equations look like the following:

$$\begin{align*}
\frac{V_1}{R_1} &amp;= \frac{-8\:\textrm{V}}{R_1} + I_{D_1} + I_{D_2}\\\\
\frac{V_2}{R_2} + I_{D_2} + I_{D_3} &amp;= \frac{8\:\textrm{V}}{R_2}\\\\
\frac{V_3}{R_3} &amp;= \frac{-8\:\textrm{V}}{R_3} + I_{D_3} + I_{D_4}\\\\
\frac{V_4}{R_4} + I_{D_4} &amp;= \frac{8\:\textrm{V}}{R_4}
\end{align*}$$

Where we know the following:

if D1 is ON then V1 = V0 else ID1 = 0
if D2 is ON then V1 = V2 else ID2 = 0
if D3 is ON then V2 = V3 else ID3 = 0
if D4 is ON then V3 = V4 else ID4 = 0
... and in all cases ID1 &gt;= 0, ID2 &gt;= 0, ID3 &gt;= 0, and ID4 &gt;= 0


So just start walking through from the left. Assume \$D_1\$ is ON, but all the rest are off. Then from the first equation (the others don't yet apply) and our rules, we have:

$$\begin{align*}
V_1 = V_0 &amp;= 0\:\textrm{V}\\\\
I_{D_2} &amp;= 0\:\textrm{A}\\\\
\frac{V_1}{R_1} &amp;= \frac{-8\:\textrm{V}}{R_1} + I_{D_1} + I_{D_2}\\\\
\therefore\quad I_{D_1}&amp;=\frac{V_1+8\:\textrm{V}}{R_1}= 0.\overline{36}\:\textrm{mA}
\end{align*}$$

Since \$I_{D_1}\ge 0\$, our assumptions are good so far.

Now, change the assumption about \$D_2\$ and turn it ON. Then from the first two equations (the others don't yet apply) and our rules, we have:

$$\begin{align*}
V_2 = V_1 = V_0 &amp;= 0\:\textrm{V}\\\\
I_{D_3} &amp;= 0\:\textrm{A}\\\\
\frac{V_1}{R_1} &amp;= \frac{-8\:\textrm{V}}{R_1} + I_{D_1} + I_{D_2}\\\\
\frac{V_2}{R_2} + I_{D_2} + I_{D_3} &amp;= \frac{8\:\textrm{V}}{R_2}\\\\
\therefore\quad I_{D_1} &amp;= \frac{8\:\textrm{V}}{R_1}-\frac{8\:\textrm{V}}{R_2} = -2.\overline{30}\:\textrm{mA}\\\\
I_{D_2} &amp;= \frac{8\:\textrm{V}}{R_2} = 2.\overline{6}\:\textrm{mA}
\end{align*}$$ 

It's clear now that \$I_{D_1}\$ violates our requirement of being ON, so it must be off. Solving for that case yields:

$$\begin{align*}
V_2 &amp;= V_1\\\\
I_{D_1} = I_{D_3} &amp;= 0\:\textrm{A}\\\\
\frac{V_1}{R_1} &amp;= \frac{-8\:\textrm{V}}{R_1} + I_{D_1} + I_{D_2}\\\\
\frac{V_2}{R_2} + I_{D_2} + I_{D_3} &amp;= \frac{8\:\textrm{V}}{R_2}\\\\
\therefore\quad V_1 = V_2 &amp;= 8\:\textrm{V}\cdot\frac{R_1-R_2}{R_1+R_2} = 6.08\:\textrm{V}\\\\
I_{D_2} &amp;= \frac{16\:\textrm{V}}{R_1+R_2} = 640\:\mu\textrm{A}
\end{align*}$$ 

Since \$I_{D_2}\$ is positive, we can now say that if \$D_1\$ is OFF and \$D_2\$ is ON then we have satisfied things up to this point and can move on to the idea of now turning on \$D_3\$. As a double-check, you can see that \$V_1=V_2\$ is quite a bit more positive than the ground reference on the other side of \$D_1\$, so this should also confirm for you that \$D_1\$ really is OFF -- up to this point, anyway.

I'll stop at this point, since by now you have a recipe for continuing on your own. At least you can see that you don't have to make an exhaustive table of all 16 permutations. But.. there still is work to be done, too.



By incrementally solving the problem you can work out the right answer without having to exhaustively test all permutations. It merely requires a logical application of rules, one step at a time, allowing the pieces to fall where they may, as you move forward. You may need to walk backwards on your determined states if, upon adding the next stage some prior condition reverses itself. (As happened in this new case I "randomized" up, where the prior assumed state of \$D_1\$ had to be changed when the 2nd state was added.) When it does happen, change it and recheck the results, and if that satisfies things then advance again by adding in the following stage.