I don't think there is a simple criterion, no, assuming your parallelograms are arbitrary. An approach less general than that suggested by HenrikRüping, but likely easier to code is this. Check if one of the four vertices of $A$ is inside $B$. If so, return yesYes. Next check each edge of $A$ for intersection with each edge of $B$. If an edge intersection is detected, return yesYes. If all these tests fail, return noNo. How to perform the primitive intersection tests is all over the web, and in particularincluding herehere.