# Forming Combinations with conditions

Suppose B is a finite collection of distinct square matrices of nth order.

And , A a subcollection of B.

I want unique combinations of four distinct elements -three from A and one from B.

I tried this

X=Combinations(A,3)
Y=Combinations(B,1)
for i in range(len(X)):
for j in range(3):
for k in range(len(Y)):
if ((X[i])[j])!=(Y[k]):
show(X[i]+Y[k])


I know problem is with my "if" command.

My question is how to change the “if” command so that none of the matrices in X[i] equals Y[k], so that we get distinct matrices in a combination.

Further, how to get unique combinations.

Thanks.

edit retag close merge delete

Let $C$ be the set difference of $B$ and $A$. Then construct combinations of 3 elements from $A$ and one from $C$, or all 4 elements from $A$.

( 2022-09-20 17:01:29 +0200 )edit

I think that it is acceptable if all four elements end up coming from A, so either choose the combination c from A first and then take the set difference of B and c and choose one element from that, or choose one element y from B and take combinations from B \setminus {y}.

( 2022-09-20 18:51:34 +0200 )edit

Yes, it is acceptable if all elements come from A. Thanks for your further insights into the problem.

( 2022-09-20 19:38:16 +0200 )edit

Sort by » oldest newest most voted

Let's use sets. Matrices must be made "immutable" to be members of sets, so:

B_list = [random_matrix(ZZ, 3) for n in range(8)]
for b in B_list:
b.set_immutable()
B = set(B_list)

A_list = B_list[:5]
A = set(A_list)


A combination of size 1 is just an element of the set, or if you like, Y will range over all subsets of B of size 1. Then we can choose X to range over all size 3 subsets of A but with Y removed:

for y in B:
X = Combinations(A.difference(set([y])), 3)
for x in X:
# now combine the combination x with the matrix y however you would like
print(x, y)


You can also form a set of these combinations and update it with a new set formed with x and y — a set since presumably you don't care about the order among y and the elements of x. Since sets ignore duplicates, you will only get unique combinations.

more

Thanks for your invaluable insights into the problem and for your crystal clear explanation .

( 2022-09-20 19:42:55 +0200 )edit