and spin-1 operators in the $i$th cell. $J_1$ characterizes the NN exchange coupling and $J_2$ the NNN coupling. We consider the antiferromagnetic interaction by taking $J_1,J_2>0$. $N$ is the total number of spins and here we choose it be even. Also we adopt the periodic boundary condition.

Four-spin model
---------------

### Eigenenergy and ground-state entanglement

The model with four spins is the simplest model with NNN interactions. We first solve the eigenvalue problem of this model. The key step is to write the four-spin Hamiltonian in the following form, $$\begin{aligned}
\label{H4}
H&=&\frac{1}{2}\big\{J_1{\bf S}^2+(2J_2-J_1)[({\bf s}_1+{\bf
s}_2)^2+({\bf S}_1+{\bf S}_2)^2]\nonumber\\
&-&2J_2({\bf s}_1^2+{\bf s}_2^2+{\bf S}_1^2+{\bf S}_2^2 )\big\},\end{aligned}$$ where ${\bf S}={\bf S}_1+{\bf S}_2+{\bf s}_1+{\bf s}_2$ denotes the total spin. From the above form, by angular momentum coupling theory, one can readily obtain all eigenvalues of the system as $$\begin{aligned}
E_1&=&J_2n(n+1)-\frac{11}{2}J_2,\nonumber\\
E_2&=&nJ_1+n(n+1)J_2-\frac{7}{2}J_2,\nonumber\\
E_3&=&-J_1+n(n+1)J_2-\frac{7}{2}J_2,\nonumber\\
E_4&=&-(n+1)J_1+n(n+1)J_2-\frac{7}{2}J_2, \label{E}\end{aligned}$$ where parameter $n=0,1,2$ in expressions $E_1$ and $E_2$ and $n=1,2$ in expressions $E_3$ and $E_4$, respectively. Then, from Eq. (\[E\]), we may find the ground-state energy as $$\label{Egs}
E_\text{GS}=\left\{
\begin{array}{ll}
-3J_1+\frac{5}{2}J_2 &\; \text{when}\;\;\; 0\leq J_2<J_1/4, \\
-2J_1-\frac{3}{2}J_2 &\; \text{when}\;\;\; J_1/4<J_2< J_1/2, \\
-\frac{11}{2}J_2     &\;\text{when} \;\;\;J_2>J_1/2,
\end{array}
\right.$$ Clearly, there are two level-crossing points, which will greatly affect behaviors of ground-state entanglement.

