Finding ground state energy using numerical real space renormalization group

I want to find ground state energy (as well as wavefunction) for spinless $$tV$$ model using Real-Space Renormalization Group (RSRG) approximation. The $$tV$$ model is defined as $$H=H_t+H_{int}=-t\sum_{i=1}^N (c_i^\dagger c_{i+1}+c_{i+1}^\dagger c_i) + V\sum_{i=1}^N n_i n_{i+1}$$ where $$n_i$$ is number operator.

The RSRG works on hypothesis that the groundstate of a system is composed of low-energy states of the system's bipartitions. And an algorithm of RSRG is:

1. Construct Hamiltonian $$H_N$$ for exactly diagonalizable $$N$$ sites
2. Diagonalize $$H_N=\sum E_i|E_i\rangle\langle E_i|$$ where $$E_i$$ are eigenvalues in incrasing order
3. Apply a projector $$P$$ on $$H_N$$ to find space spanned by lowest $$m$$ eigenstates, $$P=\sum_{i=1}^m |E_i\rangle\langle E_i|$$
4. Calculate projected Hamiltonian $$\tilde{H}_N=P^\dagger H_N P$$
5. Construct Hamiltonian of size $$2N$$ by $$H_{2N}=\tilde{H}_N\otimes I + I\otimes \tilde{H}_N + \tilde{H}_{int}$$ where $$\tilde{H}_{int}=\tilde{A}_N\otimes \tilde{B}_N$$ and $$\tilde{A}_N(\tilde{B}_N)$$ are the projected operator acting on each bipartition $$\tilde{A}=P^\dagger A P(\tilde{B}=P^\dagger B P)$$
6. repeat step 2-5 until desired system size is acheived

My attempt

For simplicity, I considered a system at half-filling and used the binary basis to write $$H$$ for 4 sites, $$H_4$$. Using $$t=1, V=1$$, $$H_4$$ is

Step:1

$$H_4= \begin{bmatrix} 1 & -1 & 0 & 0 & 1 & 0 \\ -1 & 0 & -1 & -1 & 0 & 1 \\ 0 & -1 & 1 & 0 & -1 & 0 \\ 0 & -1 & 0 & 1 & -1 & 0 \\ 1 & 0 & -1 & -1 & 0 & -1 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ \end{bmatrix}$$

Step:2-3

It is easy to diagonalize $$H_4$$ and calculate $$P$$ for $$m=3$$ $$P= \begin{bmatrix} 0.69 & 0.24 & 0 & 0 & -0.24 & 0.31\\ 0.24 & 0.62 & 0.24 & 0.24 & 0 & -0.24\\ 0 & 0.24 & 0.19 & 0.19 & 0.24 & 0\\ 0 & 0.24 & 0.19 & 0.19 & 0.24 & 0\\ -0.24 & 0 & 0.24 & 0.24 & 0.62 & 0.24\\ 0.31 & -0.24 & 0 & 0 & 0.24 & 0.69 \end{bmatrix}$$

Step:4 is also straightforward.

Step:5

I am having trouble in calculating the $$\tilde{H}_{int}$$. What are $$A$$ and $$B$$ operators here? Are they just equal to $$H_{int}=diagonal[1,0,1,1,0,1]$$ i.e. $$\tilde{H}_{int}=[P^\dagger H_{int} P] \otimes [P^\dagger H_{int} P]$$

Another little confusion: After projecting out $$m$$ (which is 3 in my attempt) low-energy eigenstates, the size of Hamiltonian is still $$6\times 6$$ (it is not changed). If the size of Hamiltonian is not changed then how is RSRG efficient in saving the computational memory?