I am studying DES for my Cryptoexam .As far as I understood , the initial plaintext is permuted and then the left half is called L0 and Right half is called R0.
After that for each round the process is as follows
L(i) = R (i-1)
R(i) = L(i-1) XOR f(R(i-1), K(i))
This relation holds till the last round
i..e L(16) = R(15)
and R(16) = L(15) XOR f(R(15),K(16))
After the last round the text is interchanged into R(16)L(16) and the inverse permutation is applied.This is what Stinson Book says.
But When I read the same topic from Forouzan , he says that in the last round there is no swapper i..e L(16) != R(15) and R(16) = R(15) . I searched on net regarding this but didn't find anything useful. Is my understanding wrong?