Flipping Cups on a Rotating Table
Four cups are placed at the corners of a square table, each either face-up or face-down in some unknown configuration. You are blindfolded. On each move, you select two cups (by their geometric relationship -- e.g., two diagonal or two adjacent corners) and flip both of them. After each move, the table is silently rotated by an unknown amount, so you lose track of which physical cup is where.
Your goal is to reach a state where all four cups face the same way (all up or all down), at which point the game ends and you win.
- Is there a strategy that guarantees a win? If so, describe it and prove it works.
- What is the maximum number of moves your strategy requires?
- Are there any starting configurations for which no strategy can ever win? If so, identify them and prove why.
Hints
- Assign $+1$ and $-1$ to the two cup orientations and look at what quantity is preserved when you flip exactly two cups.
- The product of all four values is invariant under two-cup flips. What does this tell you about which starting configurations are reachable from the goal state?
- For solvable (2-2) configurations, you only have two geometric moves available: flip a diagonal pair or flip an adjacent pair. Try the sequence Diagonal, Adjacent, Diagonal and track what happens in each case.
Worked Solution
How to Think About It: Since you are blindfolded and the table rotates after each move, you have no way to track individual cups. The only moves available to you are defined by geometry: flip two diagonal cups or flip two adjacent cups. The key question is whether there is some invariant of the system that prevents you from reaching the goal. Before designing a strategy, check what flipping exactly two cups preserves.
Key Insight: Assign $+1$ to face-up and $-1$ to face-down. The product of all four cup values is invariant under any move that flips exactly two cups, because each flip multiplies the product by $(-1)^2 = 1$. The target state (all same) always has product $+1$ (either
The Method:
There are three types of starting configuration (up to symmetry and rotation):
- All same (product $+1$): Already won.
- 2-2 diagonal (product $+1$): Opposite corners match, e.g., cups 1,3 up and cups 2,4 down.
- 2-2 adjacent (product $+1$): Neighboring corners match, e.g., cups 1,2 up and cups 3,4 down.
- 3-1 split (product $-1$): Three one way and one the other.
For the 2-2 configurations, the following three-move strategy guarantees a win:
Move 1 -- Flip diagonal. Pick any pair of diagonally opposite cups and flip both. - If the configuration is 2-2 diagonal, the two cups you flip are in opposite states. Flipping both makes them match, so all four cups are now the same. Win. - If the configuration is 2-2 adjacent, flipping a diagonal pair just shuffles the adjacent pattern (it remains 2-2 adjacent). No win yet.
Move 2 -- Flip adjacent. Pick any pair of adjacent cups and flip both. - You are still in a 2-2 adjacent configuration (if Move 1 did not win). There are two sub-cases depending on the unknown rotation: - If you happen to flip two cups that are both in the same state, you win (all four match). - If you flip two cups in opposite states, the configuration converts from 2-2 adjacent to 2-2 diagonal.
Move 3 -- Flip diagonal. Same as Move 1. - If Move 2 converted the configuration to 2-2 diagonal, this move wins (same reasoning as Move 1). - If Move 2 already won, the game ended before reaching this step.
So the strategy is: Diagonal, Adjacent, Diagonal. At most 3 moves.
Why this covers all cases. After Move 1, either you have won (diagonal case) or you are in 2-2 adjacent. After Move 2, either you have won (lucky adjacent flip) or you are in 2-2 diagonal. After Move 3, you win (diagonal case). Every path terminates in a win within 3 moves.
Part 3 -- Unsolvable configurations. A 3-1 starting configuration (three cups one way, one the other) has product $-1$. Since every two-cup flip preserves this product, and the goal states all have product $+1$, no sequence of moves can ever reach the goal from a 3-1 start. This is the only family of unsolvable configurations. If you are allowed to flip 1 or 3 cups per move instead, the parity changes and a different strategy can handle 3-1 starts -- but under the stated rules (exactly two flips per move), 3-1 is provably impossible.
Answer: Yes, a winning strategy exists for any 2-2 starting configuration: play Diagonal, Adjacent, Diagonal. This guarantees a win in at most 3 moves. However, if the starting configuration is a 3-1 split, no strategy can ever win -- the parity invariant (product of all cup values) prevents it.
Intuition
This problem is really about recognizing a parity invariant hiding beneath a physical puzzle. The product of the four cup values is unchanged by any legal move, which immediately splits all configurations into two classes: even-parity (0, 2, or 4 cups face-up) and odd-parity (1 or 3 cups face-up). The goal state lives in the even class, so odd-parity starts are provably unreachable. Once you see this, the puzzle reduces to designing a short strategy that covers the two remaining non-trivial cases (2-2 diagonal and 2-2 adjacent) despite the adversarial rotations.
This kind of invariant argument shows up constantly in quantitative interviews and in practice. Whenever a system has a conserved quantity, it constrains what states are reachable -- whether it is parity in a combinatorial puzzle, a martingale property in pricing, or a conservation law in a physical system. The instinct to check "what does this operation preserve?" before trying to solve anything is one of the most powerful tools in a quant's toolkit. The secondary lesson is about adversarial uncertainty: you cannot track individual cups, so your strategy must work for every possible rotation. This is analogous to designing strategies that are robust to unknown market states -- you plan for the worst case, not a specific scenario.