4 way flip flop relay?
I think I get it. Assuming you want to do this with flip flops, it would look like:







simulate this circuit &ndash; Schematic created using CircuitLab

If your Hall Effect sensor provides one full clock pulse, then the twisted Johnson counter will advance one step. It goes through four states: 00, 10, 11, 01. These are then translated into the pairs you expressed.

I would have used the \$\overline{Q}\$ outputs had the schematic editor allowed it for their D-flops, so something like this does the same thing without those NOT gates:







simulate this circuit

It produces the following:

$$\begin{array}{cccc}
1A&amp;1B&amp;2A&amp;2B \\
1&amp;0&amp;0&amp;1 \\
1&amp;0&amp;1&amp;0 \\
0&amp;1&amp;1&amp;0 \\
0&amp;1&amp;0&amp;1
\end{array}$$

And then returns to the top.

Do something to ensure that both flip flops are reset correctly at power up, too. You'll also have to figure out how to drive your solenoids from these values. That's another subject, though. And since I have NO IDEA at all what your circumstances are, I can't even begin to help on that score.



Try out the following circuit:







simulate this circuit

I've included one example LED output driver, too. You'll need four of those if you plan to look at four LED-visible outputs at the same time. I assumed a standard red LED operating at \$20\:\textrm{mA}\$ and requiring only \$2\:\textrm{V}\$ to operate. I think the CD4013 will power up okay with both FF cleared. But that's for testing. Technically, you probably should include a small RC circuit (at least) to create a reset pulse for your final device.