by the original Hamiltonian $\tilde{H}(x_1,x_2,\cdots, x_{N})=H(q,p)$ and induced constraints $\tilde{G}_b(x_1,x_2,\cdots, x_{N})=0$ $(b=1, \cdots, N-2)$.*]{} Note that $\tilde{H}(x_1,x_2,\cdots, x_{N})$ is not uniquely determined, because of the freedom to add a term $\sum_b\tilde{\lambda}_b(x_1,x_2,\cdots, x_{N})\tilde{G}_b(x_1,x_2,\cdots, x_{N})$ to $\tilde{H}(x_1,x_2,\cdots, x_{N})$. Here $\tilde{\lambda}_b(x_1,x_2,\cdots, x_{N})$ are some functions.

### Examples {#Examples}

Here we present two simple examples to show how induced constraints are obtained for given multiplets.\
(a) $N=3$\
Consider composite variables, $$\begin{aligned}
x=\frac{1}{4}\left(q^2 - p^2\right)~,~~ 
y=\frac{1}{4}\left(q^2 + p^2\right)~,~~ z=\frac{1}{2}qp~,
\label{Ex1}\end{aligned}$$ which satisfy the following relations: $$\begin{aligned}
\{x, y\}_{\mbox{\tiny{PB}}} = z~,~~
 \{y, z\}_{\mbox{\tiny{PB}}} = x~,~~ \{z, x\}_{\mbox{\tiny{PB}}} = -y~.
\label{Ex1-PB}\end{aligned}$$ Then the conditions (Eq. (\[xyzG\])) become $$\begin{aligned}
\frac{\partial \tilde{G}}{\partial x} = x~,~~ 
\frac{\partial \tilde{G}}{\partial y} = -y~,~~ 
\frac{\partial \tilde{G}}{\partial z} = z~,
\label{Ex1-dG}\end{aligned}$$ and $\tilde{G}$ is obtained by $$\begin{aligned}
\tilde{G}= \frac{1}{2}\left(x^2 - y^2 + z^2\right)+C~,
\label{Ex1-G}\end{aligned}$$ where $C$ is a constant. Redefining $\tilde{G}$ as $\tilde{G}-C$, we obtain the induced constraint $\tilde{G}(x,y,z)=G(q,p)=0$.

  \
(b) $N=4$\
Consider variables including composite ones, $$\begin{aligned}
x_1=q~,~~ x_2=p~,~~ x_3=x_3(q, p)~,~~ x_4=x_4(q, p)~,
\label{Ex2}\end{aligned}$$ which satisfy the following relations: $$\begin{aligned}
&& \{x_1, x_2\}_{\mbox{\tiny{PB}}} 
= 1~,~~ \{x_1, x_3\}_{\mbox{\tiny{PB}}} = \frac{\partial x_3}{\partial p}~,~~ 
\{x_1, x_4\}_{\mbox{\tiny{PB}}} = \frac{\partial x_4}{\partial p}~,~~
\nonumber \\
&& \{x_2, x_3\}_{\mbox{\tiny{PB}}} = 
-\frac{\partial x_3}{\partial q}~,~~ 
\{x_2, x_4\}_{\mbox{\tiny{PB}}} = -\frac{\partial x_4}{\partial q}~,~~ 
\nonumber 