10 playing cards are placed face-down on a table, forming a circle.
Your are allowed to turn over simultaneously 4 cards that either:
a.)all 4 are consecutive,
b.) 2 are on the left and 2 on the right of any specific card chosen by you (i.e. out of 5 consecutive cards only the middle one remains unturned) .
By executing a final number of the above (a. or b.) procedures can you reach an "all face-up" state?
If yes, explain how.
Otherwise, prove why not.
This was not very obvious, and deserves a higher difficulty than 2, IMHO. I vote for 3.
It is not possible, and the proof involves a parity argument. Color the cards black and red, alternating. In other words, each black card is surrounded by 2 red cards, and vice versa. It is clear that each procedure flips two red and two black cards. Therefore, the number of face-up red cards must always be even, and the number of face-up black cards must always be even, using a parity argument. Since there are 5 of each, we can never achieve all black cards or all red cards face up, much less having all cards face-up. This completes the proof.
Out of 32 possible black card states, half of them (16) are not ruled out by the parity argument. Same for the red cards. Altogether, one quarter of all positions (256 out of 1024) are not ruled out by the parity argument. And Charlie has shown that all 256 are achievable using the procedures.