Four glasses on a rotating table
There is a square table whose top can rotate. A glass is placed at each corner.
Your eyes are covered. Someone spins the table at random. When it stops, you may reach out with both hands and choose two glasses at once, feel whether each is upright or upside down, and then flip any subset of the two you touched (none, one, or both). After you finish, the table is spun again; you may choose and operate again, and so on.
A bell rings when all four glasses match—either all upright or all upside down.
Design a strategy that succeeds in finitely many rounds, using as few rounds as possible in the worst case. How many rounds does your strategy guarantee?
Only three initial types matter (0 = upside down, 1 = upright):
Type ①
0 1
1 0
Type ②
0 0
1 1
Type ③
0 0
0 1
Round 1: Flip any diagonal pair. Type ① succeeds immediately; otherwise you get:
② → 1 0 / 1 0, or ③ → 1 0 / 0 0.
Round 2: Flip any adjacent pair. Type ② may succeed; otherwise → ② as 0 1 / 1 0 or ③ as 0 0 / 1 0.
Round 3: Flip any diagonal. Type ② succeeds; otherwise → ③ as 0 1 / 0 0.
If still not done after three rounds, the state must be that ③ pattern.
Round 4: Flip any single glass. Sometimes succeeds; otherwise → ① as 1 1 / 0 0 or ② as 0 1 / 1 0.
Round 5: Flip any diagonal. Type ② succeeds; otherwise only ① as 1 0 / 1 0 remains. Then repeat alternating adjacent and diagonal flips.
So five rounds suffice in the worst case.