Practical Key exchange for Internet In section 3.2 (page 10) of Vikram Singh's paper A practical Key Exchange for the internet using Lattice Cryptography, he gives the number of elements in each set for odd $q$. However, the results do not match my calculation of the number of elements in each set for odd $q$.

For example, given an odd prime $q=17$, you can see that calculated result does not match. Can anyone explain this discrepancy?

$\textbf{Randomized Rounding Procedure:}$

Let $q=17$. Thus $q\bmod 4 = 17\bmod 4 = 1 \implies q \equiv 1 \pmod4$

Now,

$$I_0 = \mathbb{Z}_q \cap \left[0,\tfrac{q}4\right) = \mathbb{Z}_q \cap \left[0,\tfrac{17}4\right) = \left[0,4\right)$$
$$I^{\prime}_1 = \mathbb{Z}_q \cap \left[\tfrac{q}4,\tfrac{q}2\right) = \left[\tfrac{17}4,\tfrac{17}2\right) = \left[4,8\right)$$
$$I^{\prime}_0 = \mathbb{Z}_q \cap \left[\tfrac{q}2,\tfrac{3q}4\right) = \left[\tfrac{17}2,\tfrac{51}4\right) = \left[8,12\right)$$
$$I_1 = \mathbb{Z}_q \cap \left[\tfrac{3q}4,q\right) = \left[\tfrac{51}4,17\right) = \left[12,17\right)$$

For odd $q$ the number of elements in each set is:

$$\left(\begin{array}{cc}\#I_1 &amp; \#I_0 \\ \#I^{\prime}_0 &amp; \#I^{\prime}_1\end{array}\right) = \left(\begin{array}{cc}\tfrac{q-1}4 &amp; \tfrac{q+3}4 \\ \tfrac{q-1}4 &amp; \tfrac{q-1}4\end{array}\right) \textrm{ for } q \equiv 1\pmod4$$
$$= \left(\begin{array}{cc}\tfrac{16}4 &amp; \tfrac{20}4 \\ \tfrac{16}4 &amp; \tfrac{16}4\end{array}\right)$$
$$= \left(\begin{array}{cc}4 &amp; 5 \\ 4 &amp; 4\end{array}\right)$$

But $I_0 = \left[0,4\right) = \left\{0,1,2,3\right\}$ (No. of elements is not 5), and $I_1 = \left[12,17\right) = \left\{12,13,14,15,16\right\}$ (No. of elements is not 4)

So, the number of elements in $I_0$ is not 5. Why?
$4 &lt; 17/4 = 4.25$, so $4 \in [0, 17/4) = [0, 4.25)$.  Hence $I_0 = \{0,1,2,3,4\}$, not $\{0,1,2,3\}$.