QUBO Problem

A QUBO problem is often defined by the following expression $f$:

\[f(X) = \sum_{i=0}^{n-1}\sum_{j=0}^{n-1}w_{i,j}\, x_i x_j\]

Here, $X = (x_0, x_1, \ldots, x_{n-1})$ denotes $n$ binary variables, and $W = (w_{i,j})$ ($0 \leq i, j \leq n-1$) is an $n \times n$ matrix of coefficients. In other words, the QUBO expression is defined by the matrix $W$. When a QUBO expression is given in this form, PyQBPP can construct and solve it as follows:

import pyqbpp as qbpp

w = qbpp.array([[1, -2, 1], [-4, 3, 2], [4, 2, -1]])
x = qbpp.var("x", shape=3)
f = qbpp.expr()
for i in range(3):
    for j in range(3):
        f += w[i][j] * x[i] * x[j]
f.simplify_as_binary()
print("f =", f)

solver = qbpp.EasySolver(f)
sol = solver.search(time_limit=1)
print("sol =", sol)

This program demonstrates an example with $n = 3$. A $3 \times 3$ integer array w is defined with qbpp.array(...), and the expression f is constructed from it. After applying simplify_as_binary() to simplify the expression using the binary variable rule ($x_i^2 = x_i$), the EasySolver searches for the optimal solution. Running this program produces the following output:

f = x[0] +3*x[1] -x[2] -6*x[0]*x[1] +5*x[0]*x[2] +4*x[1]*x[2]
sol = Sol(energy=-2, {x[0]: 1, x[1]: 1, x[2]: 0})

A more concise formulation with einsum

The double for-loop above is a direct translation of the mathematical definition, but the same expression can be written as a single call to qbpp.einsum:

import pyqbpp as qbpp

W = qbpp.array([[1, -2, 1], [-4, 3, 2], [4, 2, -1]])
x = qbpp.var("x", shape=3)
f = qbpp.einsum("ij,i,j->", W, x, x)
f.simplify_as_binary()
print("f =", f)

solver = qbpp.EasySolver(f)
sol = solver.search(time_limit=1)
print("sol =", sol)

Here, qbpp.array(...) builds a $3 \times 3$ integer array W (corresponding to the matrix $W = (w_{i,j})$), and qbpp.var("x", shape=3) creates the binary variable vector $X = (x_0, x_1, x_2)$.

The subscript "ij,i,j->" reads almost like the mathematical formula $\sum_{i,j} W_{ij}\, x_i\, x_j$:

• The first input W is labeled ij (the rows and columns of the matrix).
• The second input x is labeled i (matched to the rows of W).
• The third input x is labeled j (matched to the columns of W).
• The right-hand side is empty, so both i and j are summed (contracted) and the result is a scalar Expr.

The resulting expression f, the simplified form, and the solution are all identical to the for-loop version. For larger $n$ the einsum formulation is also much faster, since it runs entirely inside the C++ backend with multithreading — avoiding the per-iteration Python ctypes overhead of the for-loop version.