Solve 2016:19 p1-2 "An Elephant Named Joseph"

Learned a lot about the Josephus' Problem today!

Solved part 1 by using a dict, but eventually
ended up just adding the mathematical shortcut
and rewriting both parts to use deque() for
performance.

Part 2 was tricky, since k (the elf to remove
after all presents were stolen from them) is a
index that changes over time. No tries for a
solution that was performant enough using lists
and dicts were succesfull, so by inspiration from
the subreddit the final solution code is based on
2 deque() that pops and appends between them.

There are 2 part 1 solutions.

- A correct implementation of the Josephus' Problem,
  using deque(). Recursion would have worked as well,
  but Python do not like recursions.
- The mathematical superior version, with a link
  to the Youtube video were it is introduced.

2016-python2/output/day_19.py

@@ -0,0 +1,52 @@
+from collections import deque
+
+
+def solve(data):
+    elfes = int(data)
+    p1 = left_adjacent_rule(elfes)
+    p1b = mathematical_superiority(elfes)
+    p2 = opposite_side_rule(elfes)
+    assert p1 == p1b
+    return p1, p2
+
+
+def mathematical_superiority(num_elfes):
+    # Josephus' problem, quick method:
+    # https://www.youtube.com/watch?v=uCsD3ZGzMgE
+    b = format(num_elfes, "b")
+    return int(b[1:] + b[0], 2)
+
+
+def left_adjacent_rule(num_elfes):
+    # https://en.wikipedia.org/wiki/Josephus_problem
+    q = deque(list(range(1, num_elfes + 1)))
+    while len(q) > 1:
+        q.rotate(-1)
+        q.popleft()
+    return q.pop()
+
+
+def opposite_side_rule(num_elfes):
+    elfes = list(range(1, num_elfes + 1))
+    separator = num_elfes // 2
+    L, R = deque(elfes[:separator]), deque(elfes[separator:])
+
+    while L and R:
+        R.popleft()
+        l2r = L.popleft()
+        R.append(l2r)
+        if len(R) - len(L) != 1:
+            r2l = R.popleft()
+            L.append(r2l)
+
+    return R.pop()
+
+
+if __name__ == "__main__":
+    with open("./input/19.txt", "r") as f:
+        inp = f.read().strip()
+
+    p1, p2 = solve(inp)
+
+    print(p1)
+    print(p2)