Why does the separable filter reduce the cost of computing the operator?
Assume you have a $N\times M$ sized image.

If you know take what is classically used, a square filter kernel, of let's say size $L\times L$, you'd need to convolve that with the picture – which gives you $N\times M$ pixels, each needing $L^2$ multiply-accumulates. So you end up with $A_{2D}=L^2MN$ operations.

Now, if you can decompose that filter into an $L$-sized horizontal and an L-sized vertical 1D-filter, you could first do all rows – that's $M$ values per row, each needing $L$ operations, so $LMN$ for all rows – and then you'd do the same with the vertical filter, so $LNM$ for all columns – and you end up with $A_{1D}=2LMN$, and you'd only need to show that

$$\begin{align}
A_{1D} &amp;&lt; A_{2D}\\
\iff\\
2LMN &amp;&lt; L^2MN &amp;&amp;||:LMN, \text{ legal since $L,M,N &gt;0$}\\
\iff\\
2 &amp;&lt; L
\end{align}$$

most filters are larger than 2.