Suppose I have a set of $n$ equations (each equation can be linear or quadratic).

I want to determine whether the system is underdetermined or overdetermined. For example, $a_1=2,a_1=3$ is overdetermined since I already know $a_1$ from the first equation, while $a_1+a_2+a_3=1,a_1^2+a_2=5$ is underdetermined.

If I know which variables are presented in each equation, is it possible to tell whether the system is underdetermined or overdetermined?

For now I'm thinking about the following algorithm:

  1. Determine the number of variables in each equation (let it be $m$).
  2. Add weight $\frac{1}{m}$ to each variable presented.
  3. After examining all equations the system is overdetermined when any weight is greater than two or sum of all weights is greater than $n$. It is underdetermined if any weight is less than one or sum of all weights is less than $n$.

For example, consider the above system $a_1+a_2+a_3=1,a_1^2+a_2=5$.

First equation: $w_{a_1}=\frac{1}{3}$, $w_{a_2}=\frac{1}{3}$, $w_{a_3}=\frac{1}{3}$.

Second equation: $w_{a_2}=\frac{1}{2}$, $w_{a_2}=\frac{1}{2}$.

Total: $w_{a_1}=\frac{5}{6}$, $w_{a_2}=\frac{5}{6}$, $w_{a_3}=\frac{1}{3}$.

Thus, the system is underdetermined.

There are cases, that due to the above definition, the system can be both overdetermined and underdetermined: $a_1=2,a_1=3,a_1+a_2+a_3=3$. We consider such equations overdetermined.

Are there any other ideas?


1 Answer 1


If your system is linear, Gaussian elimination will tell you everything you want.

In contrast, given a system of quadratic equations, it is NP-hard to determine whether it has any solution. I'm not sure what is your definition of overdetermined and underdetermined, but the corresponding problems are likely also NP-hard, by reduction to the above-mentioned problem.

To see that this problem is indeed NP-hard, we show that quadratic equations can encode Boolean circuits. Suppose that we have a Boolean circuit on inputs $x_1,\ldots,x_n$ with output $r$. We will have equations $x_i^2 = x_i$ for all $i$, and for each gate $g$, the equation $g^2 = g$ together with:

  • If $g = \lnot h$ then $g = 1-h$.
  • If $g = a \land b$, the equations $g + G = a+b$ (where $G$ is a new variable), $G^2 = G$, and the two equations $g(1-a) = g(1-b) = 0$.
  • If $g = a \lor b$, the equations $g + G = a+b$ (where $G$ is a new variable), $G^2 = G$, and the two equations $a(1-g) = b(1-g) = 0$.

If we add the equation $r = 1$, then every satisfying assignment for the circuit (that is, an assignment which causes the circuit to output $1$) lifts to a unique solution of the system, and these are the only solutions of the system. In particular, this gives a reduction from SAT to our problem.


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