There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.
Would you rather go first or second? Does it matter?
Assume that you go first, describe an algorithm to compute the maximum amount of money you can win.
Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.
Example 18 20 15 30 10 14
First Player picks 18, now row of coins is
20 15 30 10 14
Second player picks 20, now row of coins is
15 30 10 14
First Player picks 15, now row of coins is
30 10 14
Second player picks 30, now row of coins is
10 14
First Player picks 14, now row of coins is
10
Second player picks 10, game over.
The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.
Going first will guarantee that you will not lose. By following the strategy below, you will always win the game (or get a possible tie).
(1) Count the sum of all coins that are odd-numbered. (Call this X)
(2) Count the sum of all coins that are even-numbered. (Call this Y)
(3) If X > Y, take the left-most coin first. Choose all odd-numbered coins in subsequent moves.
(4) If X < Y, take the right-most coin first. Choose all even-numbered coins in subsequent moves.
(5) If X == Y, you will guarantee to get a tie if you stick with taking only even-numbered/odd-numbered coins.
You might be wondering how you can always choose odd-numbered/even-numbered coins. Let me illustrate this using an example where you have 6 coins:
Example
18 20 15 30 10 14
Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.
Since the sum of even coins is more, the first player decides to collect all even coins. He first picks 14, now the other player can only pick a coin (10 or 18). Whichever is picked the other player, the first player again gets an opportunity to pick an even coin and block all even coins.
One day a really rich old man with two sons died. In his will he said that he would give one of his sons all of his fortune. He gave each of his sons a horse and said they would compete in a horse race from Los Angeles to Sacramento, but the son whose horse came in second would get the money.
So one day they started the race. After one whole day they had only ridden one mile. At night they decided they should stop at a hotel. While they were booking in they told their problem to the wise old clerk, who made a suggestion. The next day the two brothers rode as fast as they could. What did the clerk suggest that they do?
The clerk told them to swap horses. The father said that whoever's horse crossed the finish line second would get the money. He didn't say that the owner of the horse had to be on it.
Two men ride their horses to the town blacksmith to ask for his daughter's hand in marriage. To help decide who will get to marry her, the blacksmith proposes a very strange race:
"You will race your horses down the mile-long road from here to the center of town, and the man whose horse passes through city hall's gates LAST will get to marry my daughter."
The men have no idea how to proceed, but after a few minutes of thinking, they come up with a great idea to abide by the blacksmith's rules. 30 minutes later, one of the men is gloating, having won the daughter's hand in marriage.
What was the idea the men had?
Each man rides the other man's horse. They race as they normally would. The blacksmith said the man whose horse crosses last would win, so the man who wins the race would have his horse finish last.
