Once upon a time there was a dad and 3 kids. When the kids were adults, the dad was old and Death came to take the dad. The first son, who became a lawyer, begged Death to let the dad live a few more years. Death agreed. When Death came back, the second son, who became a doctor begged Death to let his father live a few more days. Death agreed. When Death came back the third son, who became a priest begged Death to let the dad live till that candle wick burned out and he pointed to a candle. Death agreed. The third son knew Death wouldn't come back, and he didn't. Why not?
The third son went over and blew out the candle after Death left because the son said "till the candle wick burns out", not "till the candle burns out".
There are 100 ants on a board that is 1 meter long, each facing either left or right and walking at a pace of 1 meter per minute.
The board is so narrow that the ants cannot pass each other; when two ants walk into each other, they each instantly turn around and continue walking in the opposite direction. When an ant reaches the end of the board, it falls off the edge.
From the moment the ants start walking, what is the longest amount of time that could pass before all the ants have fallen off the plank? You can assume that each ant has infinitely small length.
The longest amount of time that could pass would be 1 minute.
If you were looking at the board from the side and could only see the silhouettes of the board and the ants, then when two ants walked into each other and turned around, it would look to you as if the ants had walked right by each other.
In fact, the effect of two ants walking into each other and then turning around is essentially the same as two ants walking past one another: we just have two ants at that point walking in opposite directions.
So we can treat the board as if the ants are walking past each other. In this case, the longest any ant can be on the board is 1 minute (since the board is 1 meter long and the ants walk at 1 meter per minute). Thus, after 1 minute, all the ants will be off the board.
Mr. Jason was walking along the sea shore. Suddenly it started drizzling and turned into a heavy rain. He wasn’t carrying any umbrella, not even any cap. He was completely wet and all his clothes were soaked in rain. Yet not even a single strand of his hair was wet! How was that possible?
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.
Once there was a night watchman who had been caught several times sleeping on the job. The boss issued the final warning. On the next night he was caught with his head on his hand and his elbows on the desk. "Aha, I've caught you again," exclaimed the boss. The watchman's eyes popped open immediately and he knew what had happened. Being a quick thinking man, he said one word before looking up at the boss. The boss apologized profusely and went home. What was the one word?
The one word was "AMEN", thus making the Boss believe he was praying rather than sleeping.