 The idea of doing an addition with Turing machines when using unary representation is to shift the leftmost number n one square to the right. This is achieved by erasing the leftmost 1 of \(n +1\) (this is done in state \(q_1\)) and then setting the 0 between \(n+1\) and \(m+1\) to 1 (state \(q_2\)). We then have \(n + m + 2\) and so we still need to erase one additional 1. This is done by erasing the leftmost 1 (states \(q_3\) and \(q_4\)).  Figure 4  shows this computation for \(3 + 4\).