scipy.optimize.quadratic_assignment(A, B, method='faq', options=None)[source]

Approximates solution to the quadratic assignment problem and the graph matching problem.

Quadratic assignment solves problems of the following form:

$\begin{split}\min_P & \ {\ \text{trace}(A^T P B P^T)}\\ \mbox{s.t. } & {P \ \epsilon \ \mathcal{P}}\\\end{split}$

where $$\mathcal{P}$$ is the set of all permutation matrices, and $$A$$ and $$B$$ are square matrices.

Graph matching tries to maximize the same objective function. This algorithm can be thought of as finding the alignment of the nodes of two graphs that minimizes the number of induced edge disagreements, or, in the case of weighted graphs, the sum of squared edge weight differences.

Note that the quadratic assignment problem is NP-hard. The results given here are approximations and are not guaranteed to be optimal.

Parameters
A2-D array, square

The square matrix $$A$$ in the objective function above.

B2-D array, square

The square matrix $$B$$ in the objective function above.

methodstr in {‘faq’, ‘2opt’} (default: ‘faq’)

The algorithm used to solve the problem. ‘faq’ (default) and ‘2opt’ are available.

optionsdict, optional

A dictionary of solver options. All solvers support the following:

maximizebool (default: False)

Maximizes the objective function if True.

partial_match2-D array of integers, optional (default: None)

Fixes part of the matching. Also known as a “seed” [2].

Each row of partial_match specifies a pair of matched nodes: node partial_match[i, 0] of A is matched to node partial_match[i, 1] of B. The array has shape (m, 2), where m is not greater than the number of nodes, $$n$$.

rng{None, int, numpy.random.Generator,

If seed is None (or np.random), the numpy.random.RandomState singleton is used. If seed is an int, a new RandomState instance is used, seeded with seed. If seed is already a Generator or RandomState instance then that instance is used.

For method-specific options, see show_options('quadratic_assignment').

Returns
resOptimizeResult

OptimizeResult containing the following fields.

col_ind1-D array

Column indices corresponding to the best permutation found of the nodes of B.

funfloat

The objective value of the solution.

nitint

The number of iterations performed during optimization.

Notes

The default method ‘faq’ uses the Fast Approximate QAP algorithm [1]; it typically offers the best combination of speed and accuracy. Method ‘2opt’ can be computationally expensive, but may be a useful alternative, or it can be used to refine the solution returned by another method.

References

1

J.T. Vogelstein, J.M. Conroy, V. Lyzinski, L.J. Podrazik, S.G. Kratzer, E.T. Harley, D.E. Fishkind, R.J. Vogelstein, and C.E. Priebe, “Fast approximate quadratic programming for graph matching,” PLOS one, vol. 10, no. 4, p. e0121002, 2015, DOI:10.1371/journal.pone.0121002

2

D. Fishkind, S. Adali, H. Patsolic, L. Meng, D. Singh, V. Lyzinski, C. Priebe, “Seeded graph matching”, Pattern Recognit. 87 (2019): 203-215, DOI:10.1016/j.patcog.2018.09.014

3

“2-opt,” Wikipedia. https://en.wikipedia.org/wiki/2-opt

Examples

>>> from scipy.optimize import quadratic_assignment
>>> A = np.array([[0, 80, 150, 170], [80, 0, 130, 100],
...               [150, 130, 0, 120], [170, 100, 120, 0]])
>>> B = np.array([[0, 5, 2, 7], [0, 0, 3, 8],
...               [0, 0, 0, 3], [0, 0, 0, 0]])
>>> res = quadratic_assignment(A, B)
>>> print(res)
col_ind: array([0, 3, 2, 1])
fun: 3260
nit: 9


The see the relationship between the returned col_ind and fun, use col_ind to form the best permutation matrix found, then evaluate the objective function $$f(P) = trace(A^T P B P^T )$$.

>>> perm = res['col_ind']
>>> P = np.eye(len(A), dtype=int)[perm]
>>> fun = np.trace(A.T @ P @ B @ P.T)
>>> print(fun)
3260


Alternatively, to avoid constructing the permutation matrix explicitly, directly permute the rows and columns of the distance matrix.

>>> fun = np.trace(A.T @ B[perm][:, perm])
>>> print(fun)
3260


Although not guaranteed in general, quadratic_assignment happens to have found the globally optimal solution.

>>> from itertools import permutations
>>> perm_opt, fun_opt = None, np.inf
>>> for perm in permutations([0, 1, 2, 3]):
...     perm = np.array(perm)
...     fun = np.trace(A.T @ B[perm][:, perm])
...     if fun < fun_opt:
...         fun_opt, perm_opt = fun, perm
>>> print(np.array_equal(perm_opt, res['col_ind']))
True


Here is an example for which the default method, ‘faq’, does not find the global optimum.

>>> A = np.array([[0, 5, 8, 6], [5, 0, 5, 1],
...               [8, 5, 0, 2], [6, 1, 2, 0]])
>>> B = np.array([[0, 1, 8, 4], [1, 0, 5, 2],
...               [8, 5, 0, 5], [4, 2, 5, 0]])
>>> res = quadratic_assignment(A, B)
>>> print(res)
col_ind: array([1, 0, 3, 2])
fun: 178
nit: 13


If accuracy is important, consider using ‘2opt’ to refine the solution.

>>> guess = np.array([np.arange(len(A)), res.col_ind]).T
>>> res = quadratic_assignment(A, B, method="2opt",
...                            options = {'partial_guess': guess})
>>> print(res)
col_ind: array([1, 2, 3, 0])
fun: 176
nit: 17