Given a $ 4\times 4$ squares table. How many ways that we can fill the table with $ \{0,1\}$ such that two neighbor squares (have one common side) have product which is equal to $ 0$?
To solve the problem of filling a \(4 \times 4\) table with \(\{0,1\}\) such that two neighboring squares (sharing a common side) have a product equal to \(0\), we need to ensure that no two adjacent squares both contain \(1\). This means that each \(1\) must be surrounded by \(0\)s.

1. **Define the Problem in Terms of Columns:**
   Let's denote the columns of the \(4 \times 4\) table as \(C_1, C_2, C_3, C_4\). Each column can be represented as a 4-tuple of \(\{0,1\}\).

2. **Constraints on Columns:**
   For any column \(C_i\), the entries must be such that no two adjacent entries are both \(1\). This means each column must be a valid configuration where no two \(1\)s are adjacent.

3. **Valid Column Configurations:**
   We need to count the number of valid configurations for a single column. A column can be represented as a 4-tuple \((a, b, c, d)\) where \(a, b, c, d \in \{0, 1\}\) and no two adjacent entries are both \(1\).

   The valid configurations for a single column are:
   - (0, 0, 0, 0)
   - (0, 0, 0, 1)
   - (0, 0, 1, 0)
   - (0, 1, 0, 0)
   - (1, 0, 0, 0)
   - (0, 0, 1, 1)
   - (0, 1, 0, 1)
   - (0, 1, 1, 0)
   - (1, 0, 0, 1)
   - (1, 0, 1, 0)
   - (1, 1, 0, 0)

   There are 11 valid configurations for a single column.

4. **Transition Between Columns:**
   The next step is to determine how these columns can transition from one to the next while maintaining the constraint that no two adjacent squares (in the same row) both contain \(1\).

   For example, if the first column is \((0, 0, 0, 0)\), the second column can be any of the 11 valid configurations. If the first column is \((0, 0, 0, 1)\), the second column must have \(0\) in the last position to avoid having two \(1\)s adjacent in the same row.

5. **Recursive Formula:**
   Let \(a_n\) be the number of ways to fill the \(n\)-th column such that the constraints are satisfied. The recursive formula for \(a_{n+1}\) can be derived based on the valid transitions from the \(n\)-th column to the \((n+1)\)-th column.

   The recursive formula is:
   \[
   a_{n+1} = a_n + 2b_n + 2c_n + 2d_n + e_n
   \]
   where \(a_n, b_n, c_n, d_n, e_n\) represent the counts of specific valid configurations in the \(n\)-th column.

6. **Sequence and Final Count:**
   The sequence of valid configurations for the \(4 \times 4\) table is given by the OEIS sequence [A051737](http://www.research.att.com/~njas/sequences/A051737). According to this sequence, the number of ways to fill the \(4 \times 4\) table with the given constraints is 511.

The final answer is \(\boxed{511}\)