II. (50 points)
As shown in Figure 3, in a $7 \times 8$
rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected." Now, some of the 56 chess pieces are to be removed so that no 5 remaining chess pieces are in a straight line (horizontally, vertically, or diagonally) in sequence. What is the minimum number of chess pieces that need to be removed to meet this requirement? Explain your reasoning.
Second, at least 11 chess pieces must be taken out to possibly meet the requirement.
The reason is as follows:
If a square is in the $i$th row and the $j$th column, then this square is denoted as $(i, j)$.

Step 1 Proof: If any 10 chess pieces are taken, then the remaining chess pieces must have a five-in-a-row, i.e., five chess pieces are consecutively connected in a straight line (horizontally, vertically, or diagonally).
Proof by contradiction.
Assume that 10 chess pieces can be taken such that the remaining chess pieces do not have a five-in-a-row. As shown in Figure 8, one chess piece must be taken from each of the first five squares in every row, and one chess piece must be taken from each of the first five squares in the last three columns. Thus, the 10 taken chess pieces will not be distributed in the shaded area in the lower right corner of Figure 8. By symmetry, they will not be distributed in the shaded areas in the other corners.

The 1st and 2nd rows must each have 1 chess piece taken, and these can only be distributed in the squares $(1,4)$, $(1,5)$, $(2,4)$, and $(2,5)$.

Similarly, at least 2 chess pieces must be taken from the squares $(6,4)$, $(6,5)$, $(7,4)$, and $(7,5)$.

In the 1st, 2nd, and 3rd columns, at least 1 chess piece must be taken from each column, distributed in the squares $(3,1)$, $(3,2)$, $(3,3)$, $(4,1)$, $(4,2)$, $(4,3)$, $(5,1)$, $(5,2)$, and $(5,3)$.

Similarly, at least 3 chess pieces must be taken from the squares $(3,6)$, $(3,7)$, $(3,8)$, $(4,6)$, $(4,7)$, $(4,8)$, $(5,6)$, $(5,7)$, and $(5,8)$.

Thus, at least 10 chess pieces must be taken from these areas.
Therefore, no chess pieces can be taken from the central shaded area.
Since at most 2 of the 4 chess pieces (1), (2), (3), and (4) can be taken, there will inevitably be a five-in-a-row from the diagonal direction. This is a contradiction.
Step 2 Construction of a method:

Take a total of 11
chess pieces, and the
remaining chess pieces
will not have a five-in-a-row.
As shown in Figure 9, as long as the chess pieces in the marked positions are taken,
the remaining chess pieces will not
have a five-in-a-row.
In summary, at least 11 chess pieces must be taken to possibly meet the requirement.