For the +ve half cycle of input Middle branch has ideal diode in forward bias and Zener diode in reverse bias Thus, we get constant 5 V output due to Zener breakdown. For this situation component between $\mathrm{x} \& \mathrm{y}$ should act as open circuit. For -ve half cycle of input Middle branch has ideal diode in reverse bias, which makes it open branch without any current. for this situation component between $x \& y$ should give a constant -5 V output due to Zener breakdown. (Note : The input side must contain some resistance else $\mathrm{V}_{\text {in }}$ is equal to $\mathrm{V}_{\text {out }}$ for every cases)