Even variable map: For 4 variables (even), we have XOR and XNOR compliment of each other and can be represented in K-maps as follow: For XNOR gate we have 2n/2 number of min terms with output as 1 (i.e. we have even number of 0s)
For XOR gate we have 2n/2 number of min terms with output as 1 (i.e. we have odd number of 1s)

From the K-maps also we can we that both the function are compliment of each other. Where we have 1 in one K-map, we have the 0 for the corresponding square in other K-map and vice versa.Odd variable map: For 3 variables (odd), we have XOR and XNOR equal to each other and can be represented in K-maps as follow:
And equation is While the compliment of the above is represented by
The above map can be represented by either which is compliment of
