【Question 12】  
This is a game played by two people. The two players take turns moving chips on a $20 \times 1$ rectangular strip. In each round, any one of the four chips can be moved any number of squares to the right. However, it cannot be placed on top of another chip or beyond another chip. Initially, the chips are positioned as shown in the figure. The winner is the last player to move a chip (after his move, the four chips exactly occupy the four rightmost squares of the strip, making it impossible to move any further). The first player should move $\qquad$ to the right by $\qquad$ squares to ensure a win.
Because the four chips in the final winning state are symmetrically distributed about the boundary between chip $B$ and chip $C$;
we can adopt a symmetric strategy to ensure that the intervals between the four chips (between $A$ and $B$, and between $C$ and $D$) remain symmetric throughout the movement; that is, the interval between $A$ and $B$ is equal to the interval between $C$ and $D$.
Winning Strategy (1): The first mover should move $A$ 2 spaces to the right. At this point, no matter how the opponent moves, the intervals will not be equal. On the next turn, the first mover just needs to ensure that the intervals are equal.
Winning Strategy (2): The first mover should move $D$ 2 spaces to the right, and proceed as in Winning Strategy (1).