I made this mostly as a fun challenge :)

You are right that there is some arbitrariness involved when picking a solution, however it's a bit more subtle than that.

Let's say our problem has N free variables.

Step 1 is finding the subset of R^N that is the solution to the root finding problem. If this subset is a point, we are done (return that point). Note that if there is no solution at all bidicalc should correctly report it.

Step 2 is if the solution subset is not a point. Then there is multiple (maybe even an infinity of) solutions, and picking one is indeed arbitrary.

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kolarski a day ago | parent [-]

does the algorithm tries to make minimal changes to the free variables ? If we have 1 + 1 = 2 and change 2 -> 4 then -100000 + 100004 = 4 is also a valid solution. When I tried it it changed it to 2 + 2 so perhaps there is optimization but also a valid optimization can be minimal free variable changes in which case it would be 1+3 = 4 and we update 1 free variable instead of 2. I have no idea which is better just curios how it works. I like the idea very much.

	▲	fouronnes3 a day ago \| parent [-]
		The actual heuristic used to pick a solution from an infinite solution subspace is a bit too complex to explain in a comment. I really need a blackboard :D The main goal was actually to find a solution, any solution at all, and fast. I wanted the backwards update to be very fast to feel as magic as possible. So the heuristic is pretty simple and could definitely be improved!