Solved the pandigits, tried to do it way more complicated than necessary at first

32/main.py

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+# We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.
+# The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
+# Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
+# HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
+
+# So, we only need inputs where the summands are of size 5 and the output of size 4
+
+from itertools import combinations
+from itertools import permutations
+
+def splits(input, left, right):
+    assert len(input) == left + right
+
+    result = []
+
+    combis = combinations(input, left)
+    for c in combis:
+        r = list(set(input) - set(c))
+        result.append((list(c), r))
+
+    return result
+
+def check(series):
+    result = set()
+
+    for i in range(4):
+        # print("Testing", toInt(series[0:i+1]) , toInt(series[i+1:5]) , toInt(series[5:]))
+        if toInt(series[0:i+1]) * toInt(series[i+1:5]) == toInt(series[5:]):
+            result.add(toInt(series[5:]))
+
+    return result
+
+def toInt(l):
+    return int("".join(list(map(str,l))))
+
+def main():
+    print("Hello, this is Patrick")
+
+    digits = range(1, 10)
+
+    perms = permutations(digits)
+
+    products = set()
+    for perm in perms:
+        products.update(check(perm))
+    print(products)
+
+    print(sum(products))
+
+
+
+if __name__ == "__main__":
+    main()