I need to generate the following adjacency matrices:
No of Nodes = 3
A B C AB AC BC
A 0 1 1 0 0 1
B 1 0 1 0 1 0
C 1 1 0 1 0 0
AB 0 0 1 0 0 0
AC 0 1 0 0 0 0
BC 1 0 0 0 0 0
To generate an adjacency matrix for 3 nodes, I can use the code available here, which is
out = np.block([
[1 - np.eye(3), np.eye(3) ],
[ np.eye(3), np.zeros((3, 3))]
]).astype(int)
But it cannot use for different number of nodes, for example if we have 5 nodes then:
No of Nodes = 5
A B C D E AB AC AD AE BC BD BE CD CE DE
A 0 1 1 1 1 0 0 0 0 1 1 1 1 1 1
B 1 0 1 1 1 0 1 1 1 0 0 0 1 1 1
C 1 1 0 1 1 1 0 1 1 0 1 1 0 0 1
D 1 1 1 0 1 1 1 0 1 1 0 1 0 1 0
E 1 1 1 1 0 1 1 1 0 1 1 0 1 0 0
AB 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0
AC 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0
AD 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0
AE 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0
BC 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0
BD 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0
BE 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0
CD 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0
CE 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0
DE 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0
Is there any simple and easiest way to implement these adjacency matrices?
CodePudding user response：
Inferring the logic of the output from your two examples, I think something like this might do what you want:
import numpy as np
def make_matrix(N, dtype=int):
n_comb = N * (N - 1) // 2
upper_left = 1 - np.eye(N, dtype=dtype)
lower_right = np.zeros((n_comb, n_comb), dtype=dtype)
cross = np.ones((N, N, N), dtype=dtype)
i = np.arange(N)
cross[i, :, i] = 0
cross[:, i, i] = 0
cross = cross[(np.triu_indices(N, k=1))]
return np.block([[upper_left, cross.T],
[cross, lower_right]])
print(make_matrix(3))
[[0 1 1 0 0 1]
[1 0 1 0 1 0]
[1 1 0 1 0 0]
[0 0 1 0 0 0]
[0 1 0 0 0 0]
[1 0 0 0 0 0]]
print(make_matrix(5))
[[0 1 1 1 1 0 0 0 0 1 1 1 1 1 1]
[1 0 1 1 1 0 1 1 1 0 0 0 1 1 1]
[1 1 0 1 1 1 0 1 1 0 1 1 0 0 1]
[1 1 1 0 1 1 1 0 1 1 0 1 0 1 0]
[1 1 1 1 0 1 1 1 0 1 1 0 1 0 0]
[0 0 1 1 1 0 0 0 0 0 0 0 0 0 0]
[0 1 0 1 1 0 0 0 0 0 0 0 0 0 0]
[0 1 1 0 1 0 0 0 0 0 0 0 0 0 0]
[0 1 1 1 0 0 0 0 0 0 0 0 0 0 0]
[1 0 0 1 1 0 0 0 0 0 0 0 0 0 0]
[1 0 1 0 1 0 0 0 0 0 0 0 0 0 0]
[1 0 1 1 0 0 0 0 0 0 0 0 0 0 0]
[1 1 0 0 1 0 0 0 0 0 0 0 0 0 0]
[1 1 0 1 0 0 0 0 0 0 0 0 0 0 0]
[1 1 1 0 0 0 0 0 0 0 0 0 0 0 0]]
The logic here is that each row of the upper-right matrix is like the flattened upper-triangle of a straightforward NxN matrix. For example:
# construct second row of upper-right matrix for 5x5 case:
x = np.ones((5, 5), dtype=int)
x[1] = 0
x[:, 1] = 0
print(x)
# [[1 0 1 1 1]
# [0 0 0 0 0]
# [1 0 1 1 1]
# [1 0 1 1 1]
# [1 0 1 1 1]]
print(x[np.triu_indices(5, k=1)])
# [0 1 1 1 0 0 0 1 1 1]