It turns out that parallel parking is surprisingly simple.
I first encountered this as a homework problem in general relativity. Graduate level.
Since I believe it is still assigned as a homework problem, I will merely highlight the key points, and not provide the solution.
Consider the infinite real two dimensional plane (or a Mall parking lot, after closing).
You are in a vehicle, which can be idealised as a rectangle, width a, length b, with four wheels. For simplicity, assume only the front wheels turn, on a rigid axle, and the rear wheels provide traction displacing the vehicle.
Then the entire degree of freedom of your vehicle is well described by two differential operators:
R, the infinitesimal rotation of the front wheels, and
D, the infinitesimal displacement of the wheels, translating the vehicle.
Pick your own sign convention.
Clearly, R and D do not commute: “RD” does not lead to the same effect as “DR”
Equally clearly, this is adequate to cover the entire plane, through repeated operation of R and D in some succession.
There are a number of interesting properties, notably the vehicle is not oriented on the plane, and by assumption R and D can have either sign.
Now consider the “parallel parking problem”:
| | a
| | W
Through a succession of R, and D operations, the vehicle can in general be translated into the “space” limited by W >= a, and L >= b
Clearly there is some minimum L=Lmin for which this is possible.
Calculating Lmin, is left as an exercise for the reader.
For any L’ > Lmin, parallel parking the vehicle is trivial, requiring merely some algorithmically determined sequence of R and D operations, depending on the initial conditions only.