Unveiling Liu Qian’s Final Magic Trick: The Hidden Josephus Puzzle

The trick starts with a deck of cards cut in half and reassembled to form a repeating ABCDABCD sequence. No matter how many top cards are moved to the bottom, the fourth card always matches the last card.

Periodic Function View

Viewing ABCD as a function with a minimal period of 4 gives the relation F(x) = F(x+4) . Moving N cards corresponds to F(x+N) = F(x+4+N) , preserving the periodic equality.

Group Theory Interpretation

The set of all permutations of the ABCDABCD string forms a cyclic group. The operation “move the first card to the end” (and its powers) serves as the group’s binary operation, which is isomorphic to the operation “move the last card forward one position”.

Reduction to the Josephus Problem

After a series of specific moves (inserting cards, discarding, etc.), the situation reduces to the classic Josephus problem: N people stand in a circle, every second person ( M = 2 ) is eliminated until one remains.

For the male side, N = 6, M = 2 , the survivor is position 5; for the female side, N = 5, M = 2 , the survivor is position 3.

Binary Solution for M = 2

When M = 2 , the survivor can be found without recursion. Write N as 2ⁿ + R , where R is the remainder after removing the highest power of two. The surviving position is 2R + 1 . In binary, this is equivalent to removing the leading ‘1’ of N and appending a ‘1’ at the end.

Example: N = 6 → binary 110. Removing the leading ‘1’ yields 10; left‑shifting (multiply by 2) gives 100, adding 1 results in 101, which is decimal 5, matching the observed survivor.

Conclusion

The magic trick’s apparent mystery is fully explained by periodic functions, cyclic group theory, and the Josephus problem, with a concise binary formula 2R + 1 that predicts the final card for any N when M = 2 .