Check common element in tuples

[deanhystad]

Ts = [(45,50,30,21), (30,9,56,3), (56,10,17,4), (41,30,50,56)]

The third tuple starts with 56 which also appears in the second tuple. If that shouldn't result in setting result[2] = True then I don't understand your tuple matching rule.

You may not understand "result" in the first example because there is an error. I messed up what True and False meant:

def find_matches(tuples):
    result = [False]  # <- Was incorrectly set to True
    for i in range(len(tuples)-1):
        pair = list(set(tuples[i])) + list(set(tuples[i+1]))  // Protect against duplicates in a list
        result.append(len(set(pair)) != len(pair))  # And corresponding error here.
    return result
                    
print(find_matches([(45,50,30,21), (30,9,56,3), (57,10,21,4), (41,30,50,56)]))

What this program does is grab two consecutive tuples and add them together. First time through it creates a list from the first two tuples: (45,50,30,21) and (30,9,56,3) and calls this pair.

pair = [45,50,30,21,30,9,56,3]

Then it compares the length of pair to the lenght of set(pair). Creating a set from a list removes any duplicate values, so the only way len(pair) == len(set(pair)) is if all values in pair are unique.

pair = [45,50,30,21,30,9,56,3], set(pair) = {3,9,45,50,21,56,30}. len(pair) != len(set(pair))

I ran the code on a few different test cases and noticed it failed for tuples like this:

Ts = [(1,1,1,1), (2,2,2,2), (3,3,3,3), 4,4,4,4)]

The result for this should be [False, False, False, False] but it came up [False, True, True, True]. The reason is that set(tuple) removes all duplicates, so (1,1,1,1,2,2,2,2) becomes (1,2). To guard against this I need to remove all duplicates from each tuple before combining them in pair.
pair = list(set((1,1,1,1)) + list(set((2,2,2,2)) == [1] + [2] == [1, 2]