4. For any two points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right)$ in a Cartesian coordinate system, we call $d\left(P_{1}, P_{2}\right)=$ $\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|$ the Manhattan distance between $P_{1}$ and $P_{2}$. If $P_{0}\left(x_{0}, y_{0}\right)$ is a fixed point and $Q(x, y)$ is a moving point on the curve $F(x, y)=0$, we call the minimum value of $d\left(P_{0}, Q\right)$ the Manhattan distance from point $P_{0}$ to the curve $F(x, y)=0$. Then the Manhattan distance from $M(-1,0)$ to the parabola $x^{2}=y$ is $\qquad$ .
4. $\frac{3}{4}$.

Let $Q\left(t, t^{2}\right)$ be a moving point on the parabola $x^{2}=y$, then the perpendicular distance $d$ from $M(-1,0)$ to the parabola $x^{2}=y$ is $S=|t-(-1)|+\left|t^{2}-0\right|=t^{2}+|t+1|$. Therefore,
$$
S=\left\{\begin{array}{ll}
t^{2}+t+1, & t>-1 ; \\
t^{2}-t-1, & t \leqslant-1 .
\end{array}\right.
$$

When $t>-1$,
$$
S=t^{2}+t+1=\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4},
$$

with equality when $t=-\frac{1}{2}$.
When $t \leqslant-1$,
$$
S=t^{2}-t-1=\left(t-\frac{1}{2}\right)^{2}-\frac{5}{4} \geqslant 1,
$$

with equality when $t=-1$.
Therefore, when $t=-\frac{1}{2}$, $S$ has its minimum value $\frac{3}{4}$. That is, the perpendicular distance from $M(-1,0)$ to the parabola $x^{2}=y$ is $\frac{3}{4}$.