Design a finite state machine (by its transition diagram) that performs the serial addition of three arbitrary binary numbers.
For a finite state machine for two binary numbers I got it, but for 3 numbers I'm considering an input set like
$\{000, 001, 010, 011, 100, 101, 110, 111\}$
I am considering two states, a carry state and a non-carry state, however I think I need more states for this problem. Could someone provide me with a solution?
You are right, three states are needed for adding three binary integers.
First, write the integers in reverse binary and potentially add zeros to make the three representations the same length. For example, if $n_1 = 38$, $n_2 = 53$ and $n_3 = 55$, write $n_1$ as $011001$ (since $2 + 4 + 32 = 38$), $n_2$ as $101011$ (since $1 + 4 + 16 + 32 = 53$) and $n_3$ as $111011$ (since $1 + 2 + 4 + 16 + 32 = 55$).
Next, write the three representations one below the other \begin{align*} &011001\\ &101011\\ &111011 \end{align*} and count the number of 1 bits in each column to obtain the following word on the alphabet $\color{blue}A = \{\color{blue}0, \color{blue}1, \color{blue}2, \color{blue}3\}$ $$ \color{blue}2 \color{blue}2 \color{blue}3 \color{blue}0 \color{blue}2 \color{blue}3 $$ The sequential automaton which realises the addition is shown below. It has the input alphabet $\color{blue}A$ and the output alphabet $\{0,1\}$. The vertical bars are separators. Thus $\color{blue}3 \mid \color{red}1$ means that on input $\color{blue}3$ the output is $1$. The states $0$, $1$ and $2$ encode the current carries. Note the final outputs $\color{green}\varepsilon$, $\color{green}1$ and $\color{green}0\color{green}1$ attached to the states $0$, $1$ and $2$, respectively.
In our example, the output is $\color{red}0\color{red}1\color{red}0\color{red}0\color{red}1\color{red}0\color{green}0\color{green}1$, which is the code of $146 = 2 + 16 + 128.$ And indeed, $38 + 53 + 55 = 146$!
See also this answer for adding two binary numbers.