4. Place the natural numbers $1,2,3,4, \cdots, 2 n$ in any order on a circle. It is found that there are $a$ groups of three consecutive numbers that are all odd, $b$ groups where exactly two are odd, $c$ groups where exactly one is odd, and $d$ groups where none are odd. Then $\frac{b-c}{a-d}=$ $\qquad$ .
4. -3 .

Let the numbers on a circle be recorded in reverse order as $x_{1}, x_{2}$, $\cdots, x_{2 n}$, and satisfy
$$
x_{i}=\left\{\begin{array}{l}
-1, \text { when } x_{i} \text { represents an odd number, } \\
+1, \text { when } x_{i} \text { represents an even number. }
\end{array}\right.
$$

Then $x_{1}+x_{2}+\cdots+x_{2 n}=0$,
and. $A_{i}=x_{i}+x_{i+1}+x_{i+2}$
$$
=\left\{\begin{array}{ll}
-3, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { are all odd, } \\
-1, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { exactly 2 are odd, } \\
+1, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { exactly 1 is odd, } \\
+3, \text { when } x_{i}, x_{i+1}, x_{i+2} & \text { are all even. }
\end{array}\right.
$$

where $x_{2 n+i}=x_{i}$.
$$
\text { By } \begin{aligned}
0 & =3\left(x_{1}+x_{2}+\cdots+x_{2 n}\right) \\
& =A_{1}+A_{2}+\cdots+A_{2 n} \\
& =(-3 a)-b+\cdots+3 d) \\
& =-3(a-c)-(b \cdots c)
\end{aligned}
$$

we get $\quad \frac{b-c}{a-d}=-3$.