# Gate constraints

At this stage, for each $$i \in \{1,\dots, n\}$$, we need to transform the computation of each gate to a unified form as follows: $$q^O_i \cdot x_{c_i} + q^L_i \cdot x_{a_i} + q^R_i \cdot x_{b_i} + q^M_i \cdot (x_{a_i} \cdot x_{b_i}) + q^C_i = 0$$ where $$q_i^O, q_i^L, q_i^R, q_i^M, q_i^C$$ are selectors uniquely determined by the corresponding gate. In particular,

• For addition gate, $$(q_i^O, q_i^L, q_i^R, q_i^M, q_i^C) = (-1, 1, 1, 0, 0)$$ since $$(-1) \cdot x_{c_i} + 1 \cdot x_{a_i} + 1 \cdot x_{b_i} + 0 \cdot (x_{a_i} \cdot x_{b_i}) + 0 = 0$$ is equivalent to $$x_{c_i} = x_{a_i} + x_{b_i}$$.

• For multiplication gate, $$(q_i^O, q_i^L, q_i^R, q_i^M, q_i^C) = (-1, 0, 0, 1, 0)$$ since $$(-1) \cdot x_{c_i} + 0 \cdot x_{a_i} + 0 \cdot x_{b_i} + 1 \cdot (x_{a_i} \cdot x_{b_i}) + 0 = 0$$ is equivalent to $$x_{c_i} = x_{a_i} \cdot x_{b_i}$$.

• For gate of addition with constant, $$(q_i^O, q_i^L, q_i^R, q_i^M, q_i^C) = (-1, 1, 0, 0, c)$$ since $$(-1) \cdot x_{c_i} + 1 \cdot x_{a_i} + 0 \cdot x_{b_i} + 0 \cdot (x_{a_i} \cdot x_{b_i}) + c = 0$$ is equivalent to $$x_{c_i} = x_{a_i} + c$$.

• For gate of multiplication with constant, $$(q_i^O, q_i^L, q_i^R, q_i^M, q_i^C) = (-1, c, 0, 0, 0)$$ since $$(-1) \cdot x_{c_i} + c \cdot x_{a_i} + 0 \cdot x_{b_i} + 0 \cdot (x_{a_i} \cdot x_{b_i}) + 0 = 0$$ is equivalent to $$x_{c_i} = x_{a_i} \cdot c$$.

We now take a look at the example achieved above, i.e., $$\begin{cases} x_{c_1} = x_{a_1} \cdot x_{b_1},\\ x_{c_2} = x_{a_2} \cdot x_{b_2},\\ x_{c_3} = x_{a_3} \cdot 3,\\ x_{c_4} = x_{a_4} + x_{b_4},\\ x_{c_5} = x_{a_5} + x_{b_5}, \\ x_{c_6} = x_{a_6} + 5. \end{cases}$$ In this example, we can transform the above system of equation into the unified form as follows: $$\begin{cases} (-1) \cdot x_{c_1} + 0 \cdot x_{a_1} + 0 \cdot x_{b_1} + 1 \cdot (x_{a_1} \cdot x_{b_1}) + 0 = 0,\\ (-1) \cdot x_{c_2} + 0 \cdot x_{a_2} + 0 \cdot x_{b_2} + 1 \cdot (x_{a_2} \cdot x_{b_2}) + 0 = 0,\\ (-1) \cdot x_{c_3} + 3 \cdot x_{a_3} + 0 \cdot x_{b_3} + 0 \cdot (x_{a_3} \cdot x_{b_3}) + 0 = 0,\\ (-1) \cdot x_{c_4} + 1 \cdot x_{a_4} + 1 \cdot x_{b_4} + 0 \cdot (x_{a_4} \cdot x_{b_4}) + 0 = 0,\\ (-1) \cdot x_{c_5} + 1 \cdot x_{a_5} + 1 \cdot x_{b_5} + 0 \cdot (x_{a_5} \cdot x_{b_5}) + 0 = 0,\\ (-1) \cdot x_{c_6} + 1 \cdot x_{a_6} + 0 \cdot x_{b_6} + 0 \cdot (x_{a_6} \cdot x_{b_6}) + 5 = 0. \end{cases}$$