Since the 10V battery and the diodes D1 and D2 are connected such that both the diodes are forward biased all the time. So this output of this circuit has got no relation with the input of the circuit.
simulate this circuit – Schematic created using CircuitLab
As far as output is concerned, the voltage across a short-circuited battery is the output here. Since the diodes are ideal, the battery 'burns out' and output becomes zero volts, in case of non-ideal battery. Talking about ideal case, you can not short circuit an ideal voltage.