Vector of variables and vector functions

PyQBPP supports vectors of variables and vector operations.

Defining vector of variables

A vector of binary variables can be created using the var() function.

  • var("name", size) returns a Vector of size variables with the given name.

The following program defines a vector of 5 variables with the name x. By printing x, we can confirm that it contains the 5 variables x[0], x[1], x[2], x[3], and x[4]. Next, using the expr() function, we create an Expr object f whose initial value is 0. In the for-loop from i = 0 to 4, each variable x[i] is added to f using the compound operator +=. Finally, f is simplified and printed.

from pyqbpp import var, expr

x = var("x", 5)
print(x)
f = expr()
for i in range(5):
    f += x[i]
print("f =", f.simplify_as_binary())

The output of this program is as follows:

[x[0], x[1], x[2], x[3], x[4]]
f = x[0] +x[1] +x[2] +x[3] +x[4]

NOTE var(name, size) returns a Vector object that contains size elements of type Var. The Vector class provides overloaded operators that support element-wise operations.

Sum function

Using the utility function sum(), you can obtain the sum of a vector of binary variables. The following program uses sum() to compute the sum of all variables in the vector x:

from pyqbpp import var, sum

x = var("x", 5)
print(x)
f = sum(x)
print("f =", f.simplify_as_binary())

The output of this program is exactly the same as that of the previous program.

QUBO for one-hot constraint

A vector of binary variables is one-hot if it has exactly one entry equal to 1, that is, the sum of its elements is equal to 1. Let $X = (x_0, x_1, \ldots, x_{n-1})$ denote a vector of $n$ binary variables. The following expression $f(X)$ takes the minimum value of 0 if and only if $X$ is one-hot:

\[\begin{align} f(X) &= \left(1 - \sum_{i=0}^{n-1}x_i\right)^2 \end{align}\]

The following program creates the expression $f$ and finds all optimal solutions:

from pyqbpp import var, sum, sqr, ExhaustiveSolver

x = var("x", 5)
f = sqr(sum(x) - 1)
print("f =", f.simplify_as_binary())

solver = ExhaustiveSolver(f)
sols = solver.search_optimal_solutions()
for i, sol in enumerate(sols):
    print(f"({i}) {sol}")

The function sum() computes the sum of all variables in the vector. The function sqr() computes the square of its argument. The Exhaustive Solver finds all optimal solutions with energy value 0 as follows:

f = 1 -x[0] -x[1] -x[2] -x[3] -x[4] +2*x[0]*x[1] +2*x[0]*x[2] +2*x[0]*x[3] +2*x[0]*x[4] +2*x[1]*x[2] +2*x[1]*x[3] +2*x[1]*x[4] +2*x[2]*x[3] +2*x[2]*x[4] +2*x[3]*x[4]
(0) Sol(energy=0, x[0]=0, x[1]=0, x[2]=0, x[3]=0, x[4]=1)
(1) Sol(energy=0, x[0]=0, x[1]=0, x[2]=0, x[3]=1, x[4]=0)
(2) Sol(energy=0, x[0]=0, x[1]=0, x[2]=1, x[3]=0, x[4]=0)
(3) Sol(energy=0, x[0]=0, x[1]=1, x[2]=0, x[3]=0, x[4]=0)
(4) Sol(energy=0, x[0]=1, x[1]=0, x[2]=0, x[3]=0, x[4]=0)

All 5 optimal solutions are displayed.

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