A black dog stands in the middle of an intersection in a town painted black. None of the street lights are working due to a power failure caused by a storm. A car with two broken headlights drives towards the dog but turns in time to avoid hitting him. How could the driver have seen the dog in time?
Jake and his friend Paco had very famous challenge sessions at their school. One would suggest something they could do, and the other would prove it wrong somehow.
One day, Jake surprised Paco by stating: "I can answer any question in the world."
Sure that he would win the challenge, Paco accepted the task of proving it wrong. He wrote up a test full of impossible questions. After a while, Jake returned the test. Paco unbelievably lost the challenge and told Jake he could indeed answer any question. How did Jake win?
For all the impossible questions, Jake simply wrote "I don't know".
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
First Player picks 14, now row of coins is
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:
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.
There is a low railroad bridge in your town. One day you see a large truck stopped just before the underpass. When you ask what has happened, the driver tells you that his truck is half of inch higher than the indicated height of the opening. This is the only road to his destination. What can he do to get through the underpass the easiest way?
Let enough air out of the tires to lower the truck.