Feature vector construction

We obtained $u_{ij}$ for each $i$-th tile and $j$-th. Let the number of bins is $b$ and the number of tiles is $t$. Then our feature vector $\mathbf{\mu}$ of image $I$ is:

$$\mathbf{\mu}=(\mu_1, \mu_2, \dots, \mu_{b*t})$$

where $\mu_k$ is computed as:

$$\mu_k = u_{{\lfloor k/b \rfloor},{k \pmod {b}}}$$

The dimension of vector $\mathbf{\mu}$ depends on number of tiles $t$ and number of bins $b$ of the SIFT descriptor and is $b*t$.