It is trickier however to get the phase \( (-1)^l \).
One possibility is as follows
- Let \( S_1 \) be a word that represents the $1-$bit to be removed and all others set to zero.
In the previous example \( S_1=|1000000000000000\rangle \)
- Define \( S_2 \) as the similar word that represents the bit to be added, that is in our case
\( S_2=|0000100000000000\rangle \).
- Compute then \( S=S_1-S_2 \), which here becomes
$$
S=|0111000000000000\rangle
$$
- Perform then the logical AND operation of \( S \) with the word containing
$$
\Phi_{0,3,6,10,13} = |1001001000100100\rangle,
$$
which results in \( |0001000000000000\rangle \). Counting the number of $1-$bits gives the phase. Here you need however an algorithm for bitcounting. Several efficient ones available.