Problem: Express a formula using . I.e., if an input variable $x Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ is set to $0 Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$, that is interpreted as false, while $x=1 Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ is interpreted as true. The output of the polynomial should be 0 or 1 according to whether the formula is true or false as a whole.

Solution: You can do this using a single polynomial.

Illustrating with an example: the formula is $neg[(a vee b) wedge (neg c vee d)] Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ also known as

```not((a or b) and (not c or d))
```

The trick is to use multiplication for “and” and $1-x Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ for “not.” So $a wedge b Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ would be $x_1 x_2 Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$, and $neg z Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ would be $1-z Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$. Indeed, if you have two binary variables $x Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ and $y Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ then $xy Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ is 1 precisely when both are 1, and zero when either variable is zero. Likewise, $1-x = 1 Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ if $x Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ is zero and zero if $x Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ is one.

Combine this with deMorgan’s rule to get any formula. $a vee b = neg(neg a wedge neg b) Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ translates to $1 - (1-a)(1-b) Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$. For our example above,

$displaystyle f(x_1, x_2, x_3, x_4) = 1 - (1 - (1-a)(1-b))(1 - c(1-d)) Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$

Which expands to

$displaystyle 1 - a - b + ab + (1-d)(ac + bc - abc) Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$

If you plug in $a = 1, b = 0, c = 1, d = 0 Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ you get True in the original formula (because “not c or d” is False), and likewise the polynomial is

$displaystyle 1 - 1 - 0 + 0 + (1-0)(1 + 0 - 0) = 1 Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$

You can verify the work yourself, using the following table as a guide:

```0, 0, 0, 0 -> 1
0, 0, 0, 1 -> 1
0, 0, 1, 0 -> 1
0, 0, 1, 1 -> 1
0, 1, 0, 0 -> 0
0, 1, 0, 1 -> 0
0, 1, 1, 0 -> 1
0, 1, 1, 1 -> 0
1, 0, 0, 0 -> 0
1, 0, 0, 1 -> 0
1, 0, 1, 0 -> 1
1, 0, 1, 1 -> 0
1, 1, 0, 0 -> 0
1, 1, 0, 1 -> 0
1, 1, 1, 0 -> 1
1, 1, 1, 1 -> 0
```

Discussion: This trick is used all over CS theory to embed boolean logic within polynomials, and it makes the name “boolean algebra” obvious, because it’s just a subset of normal algebra.

Moreover, since boolean satisfiability—the problem of algorithmically determining if a boolean formula has a satisfying assignment (a choice of variables evaluating to true)—is NP-hard, this can be used to show certain problems relating to multivariable polynomials is also hard. For example, finding roots of multivariable polynomials (even if you knew nothing about algebraic geometry) is hard because you’d run into NP-hardness by simply considering the subset of polynomials coming from boolean formulas.

Here’s a more interesting example, related to the kinds of optimization problems that show up in modern machine learning. Say you want to optimize a polynomial $f(x) Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ subject to a set of quadratic equality constraints. This is NP-hard. Here’s why.

Let $varphi Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ be a boolean formula, and $f_varphi Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ its corresponding polynomial. First, each variable $x_i Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$ used in the polynomial can be restricted to binary values via the constraint $x_i(x_i - 1) = 0 Boolean Logic in Polynomials | Math ∩ Programming - latex - Boolean Logic in Polynomials | Math ∩ Programming$.

You can even show NP-hardness if the target function to optimize is only quadratic. As an exercise, one can express the subset sum problem as a quadratic problem using similar choices for the constraints. According to this writeup you even express subset sum as a quadratic with linear constraints.

The moral of the story is simply that multivariable polynomials can encode arbitrary boolean logic.