1's complement addition of outer carry to the result Let's take for example this addition: 3 + (-1).

1 in binary is 001, and to obtain it's 1's complement counterpart we
flip the bits. So it is: 110.

3 in binary is 011.



011 + 110 = 1001
That first 1 which is in bold has to be added to the number formed by the last 3 bits as follows:
001 + 1 = 010 (2 in decimal).

Why do we do the last step, adding that outer carry? Which is the logic behind?
I only understand two's complement

Let's take two 3 bits numbers, X and Y

Complementing the bits is like calculating 111-X

(it is obvious as $\overline {b} = 1-b$, $\sum_{i=0}^n 2^i*(1-b_i) = \sum_{i=0}^n 2^i -  \sum_{i=0}^n 2^i*b_i$...)

The addition becomes 111-X+Y

If you add 1 you get 1000-X+Y. The leftmost carry is discarded.