It's A Big Digi World! #2

Observe that the top part and the bottom part are identical (swapping the fifth input and the sixth input doesn't change anything). Thus they give the same answer, and when fed to the final XOR gate, the result is 0 .

as both input in XOR circuit are equal, so for XOR circuit the equal inputs give the result 0.

0 will be the output

Both parts before the last XOR gate are giving the opposite output thus the solution for XOR gate by: X =AB'+A'B is 0. NOTE: Opposite input always answer in 0 for XOR gate'

obviously 0. Its easy

A three input NXOR will always input 1, those two 1's will feed to the final XOR and produce a 0. You can ignore the rest of the diagram.

Observe the pattern. The upper and the lower part is same.Hence the both input from lower and upper half to the last gate i.e. XOR gate is going to be same and it is the rule of XOR gate that whenever both the input are same then its output is 0.

The top and the bottom part of thid circuit is symmetrical, so the final two inputs for the XOR gate in the end would be the same. This implies that 0 must be the output.

0 ,1,0,1 |||
1,1||| 1,1,1,1||| 0,0||| 0

the way to reach the final outputis given by putting the outputs of each gates in their order.

i solved and found it

thanks you made me remember my faculty

As both da inputs of XOR gate iz 1, so da output will b 0.

outputs of both XNOR gates would be same and it comes out to be 1. XOR gate gate has expansion AB'+A'B =1 0+0 1 =0

The output at the first levels are 1 1 1 1 then at the second level 0 0 at third 0 1 0 1 ,at fourth 0 0 so Ex-or of 0 0 is 0 hence the answer is zero

0 is the answer

Solution is simple. Just how : 1x0=0=0 1x0=0=0=0=0 1x0=0=0=0=0 1x0=0=0