Analysis of a circuit with 6 diodes
Enough time has gone by. Your schematic is horribly drawn (perhaps by intention of its author.) Re-drawing it helps a great deal:







simulate this circuit &ndash; Schematic created using CircuitLab

Assuming ideal diodes with \$V_\text{FWD}=0\:\text{V}\$ and tentatively ignoring \$D_3\$ and \$D_4\$ for the moment, you can almost immediately see now that \$D_1\$ and \$D_2\$ are equally on and that \$V_y=0\:\text{V}\$ and also that \$D_6\$ will be on and pull \$V_z=1\:\text{V}\$ (\$D_5\$ will be off.)

Ignoring the resistors for a moment, you'd expect the lower of \$V_y\$ and \$V_z\$ to pull \$V_x\$ to its level. So ignoring the resistors, \$V_x=V_y=0\:\text{V}\$. Here, you also do NOT find any conflicts. In other words, this just means that \$D_4\$ is off and that \$D_3\$ is on. (And you'll note that \$R_1\$ can easily sink the current that \$R_2\$ can source.)

But suppose you take the other position; that \$V_x=V_z=1\:\text{V}\$? Then both \$D_3\$ and \$D_4\$ would be on, \$V_y=1\:\text{V}\$, and that \$D_1\$ and \$D_2\$ are off. But now this means that \$R_1\$ and \$R_3\$ can together sink far more current than \$R_2\$ can source, so this would pull \$V_x\$ well into the negative voltage region (but above \$-10\:\text{V}\$.) Both \$D_3\$ and \$D_4\$ would still be on, but \$V_x\$ and \$V_y\$ would also be well into the negative voltage region, too. But if so, \$D_1\$, \$D_2\$, \$D_5\$, and \$D_6\$ would all be on and would have \$V_\text{FWD}\gg 0\:\text{V}\$, which by definition cannot be the case.

So you don't need to go through all of the permutations to work it out.