Well, dany's comment is right but someone can ask you to apply Thevenin simplification anyway.
I'll try to explain what I do by words (at this moment, I have not any media to draw the circuit).
Vs and the two R can be thought as forming a circuit connected to the D1-V1 and D2-V2 "branches".
So Vs-R-R is simplified by Thevenin as Vs/2 with a R/2 in series and connected to the two mentioned branches.
If you now draw the new circuit, the analysis is strongly simplified: you have not any node! during the two positive and negative waves of Vs.
- During positive values of Vs (over the V1 value), D1 is ON.
- During negative values of Vs (over the V2 value), D2 is ON.
Currents in (ideal) diodes are:
- D1: 2x(Vs-V1)/R
- D2: 2x(Vs-V2)/R.
Obviously, only when Vs>V1 and Vs>V2, respectively. 0A when Vs value is between the V2 and v1 values.