Riddle Me This 5: All That Glitters Is Gold (Coins)

These are less straightforward than the card-flipping riddles, but it all comes down to flipping in the end! Answers (along with a hint for the harder one) can be found below each question, but feel free to ask for hints or let me know if you have the solution! :D

Easier:
You are blindfolded with 72 coins on a table in front of you. Exactly 1/4 of them are heads up, though you don't know which they are. Separate the 72 coins into 2 piles, each containing the same number of heads up. You can flip as many coins as you like, though you won't be able to tell if they are heads or tails up.

This is a fun math trick that always blows my mind. There are 18 heads up. Say we take 18 coins from the 72 coin pile, and there are x heads up in the remaining 54. Then there are 18-x heads up in the 18-coin pile, and x tails. But then if we flip all of those 18 coins, there will be x heads up in that pile too, leaving x heads in each pile. :D

 

Harder (first read on the riddle page of Dr. Miller here):
You sit blindfolded in front of a square with a coin in each corner. You want to get all coins heads up or all tails up. You have no idea what the starting formation is, of course; they could even be all heads up to begin with. You may flip however many you want, then ask if you are done (this constitutes a turn). If you are not done, the square is then spun an undisclosed amount clockwise or counterclockwise. You then get another turn and so the game continues. Is there a strategy that is guaranteed to work in a finite number of moves, and if so, what is that smallest number of moves you need?

Think about what possible states the four coins could be in. For example, there could be 1 head and 3 tails up. What other states could they be in? Do any of these states have a "guaranteed win" move?

 

This one is really tricky. The key is to enumerate all the possible states and work through them to see which you can solve fastest. First, all the states:
All 4 of one kind (4:0)
3 of one, and 1 of the other (3:1)
2 each, with same-sided coins next to each other (2:2 adjacent)
2 each, with same-sided coins diagonally opposite each other (2:2 diagonal)
Next, let's define some flips:
No flip (N)
Just one coin (O)
2 adjacent coins (A)
2 diagonally opposite coins (D)
The first thing to notice is that if we do a diagonal flip (D) from the 2:2 diagonal state, we win! This is good news. Even better, this has no effect on the 3:1 state or the other 2:2 state!
Now, what happens if we try to solve the 2:2 adjacent state by using A? Either we win (yay!) or we get to the 2:2 diagonal state! Then we can do D again and win!
Fortunately, neither of these 2 coin flips affect the 3:1 state. If we try to solve this state by using O, either we win or we get to one of the 2:2 states. Then we can do the same sequence of D, A, D to be sure of success!
Since we also want to check the case where we're done to begin with, we'll check that first by flipping no coins. In sum, we have N, D, A, D, O, D, A, D. Just 8 flips and we're sure to win! :)