Redrawing a diode circuit in a clearer way
Probably better drawn this way:







simulate this circuit &ndash; Schematic created using CircuitLab

All currents (positive polarity) should flow from the top of the sheet towards the bottom. All signals should flow from the left side (inputs) of the sheet to the right side (outputs) of the sheet. The above schematic arranges all the diodes so that they point, correctly, from top towards bottom of the sheet.

The arrangement above doesn't say anything about which voltages are more positive or more negative. But if the diodes are active, then it is at least suggestive (as discussed at the bottom, below.) Which is the best you can hope for.

The circuit tells you immediately that \$V_O\$ is an output of interest. There is no particular need to show the (+) and (-) terminals for \$V_O\$, since it is already implied.



Assuming a fixed voltage drop for a diode, the following can be immediately determined from a casual examination now:


To be active, the path from \$V_1\$ to \$V_O\$ requires that it must be at least one diode drop above \$V_O\$. Otherwise, the path is inactive.
To be active, the path from \$V_2\$ to \$V_O\$ requires that it must be at least one diode drop below \$V_O\$. Otherwise, the path is inactive.
To be active, the path from \$V_3\$ to \$V_O\$ requires that it must be at least one diode drop below \$V_O\$. Otherwise, the path is inactive.
In all cases, there is a path from \$V_O\$ to ground via \$R_4\$.