N=[1,2,3,4]
P={(1,2),(2,3),(3,4)} # Make a set of forbidden combinations.

cartesian_product = set()

# You could also use itertools.product to find all permutations.
for i in N:
    for j in N:
        cartesian_product.add((i,j))

print("DEBUG ", cartesian_product)

result = cartesian_product - P
print("RESULT", result)

Output:
DEBUG  {(4, 4), (2, 4), (1, 2), (2, 1), (3, 4), (4, 1), (3, 1), (4, 3), (1, 1), (1, 4), (4, 2), (2, 3), (3, 3), (2, 2), (3, 2), (1, 3)}
RESULT {(2, 4), (2, 1), (4, 3), (3, 1), (2, 2), (3, 2), (4, 1), (1, 3), (4, 4), (1, 1), (1, 4), (4, 2), (3, 3)}

(Apr-05-2023, 02:04 AM)Deonvek Wrote: algorithm that won't test all the possible combinations

n = (1, 2, 3, 4)
p = ((1, 2), (2, 3), (3, 4))

# Convert p to dictionary
bad = {a: {a} for a in n}
for a, b in p:
    bad[a].add(b)
    bad[b].add(a)
print("Bad", bad)

# Create a dictionary of numbers that can be paired.
good = {a: [b for b in n if b not in bad[a]] for a in n}
print("Good", good)

# Create all good pairs.  Assumes order is not important: (1, 2) == (2, 1).
# If order matters, leave out "if b > a".
good_pairs = [(a, b) for a, c in good.items() for b in c if b > a]
print("Good pairs", good_pairs)

Output:
Bad {1: {1, 2}, 2: {1, 2, 3}, 3: {2, 3, 4}, 4: {3, 4}}
Good {1: [3, 4], 2: [4], 3: [1], 4: [1, 2]}
Good pairs [(1, 3), (1, 4), (2, 4)]

from itertools import permutations

res = []
nums = [1,2,3,4]
not_allowed = [(1,2),(2,3),(3,4)]

for perm in permutations(nums,2):
    if perm not in not_allowed:
        res.append(perm)
print(res)

Output:
[(1, 3), (1, 4), (2, 1), (2, 4), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)]