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+# take-aways:
+#   - Use integers as lists of boolean values
+#   - Use 1 << n to compute 2^n where n = len(xs)
+
+def set_from_int(xs, n):
+    result = []
+    for i in range(len(xs)):
+        if n & (1 << i) != 0:
+            result.append(xs[i])
+    return result
+
+# subsets :: Set a -> List (Set a)
+def subsets(xs):
+    n = len(xs)
+    return [set_from_int(xs, i) for i in range(1 << n)]
+
+#   0 1 2
+# 0 N Y Y
+# 1 _ N Y
+# 2 _ _ N
+
+# For my interview, be able to compute *permutations* and *combinations*
+
+# This differs from permutations because this is about finding combinations...
+#
+# bottom-up
+# 0 =>        { }
+# 1 =>  {3}   {4}   {3}
+# 2 => {5,4} {5,3} {4,3}
+
+xs = [
+    ([], [[]]),
+    ([5], [[], [5]]),
+    ([5,4], [[],[5],[4],[5,4]]),
+]
+
+for x, expected in xs:
+    result = subsets(x)
+    print("subsets({}) => {} == {}".format(x, result, expected))
+    assert result == expected
+    print("Success!")