Puzzles and More

Answer: 1000 Bottles of Wine

2009-09-01T21:36:00.000-07:00

My friend Qin told me this puzzle, but her version involves using 10 condemned prisoners to test the wines. I took the liberty of softening the story a little bit by replacing the 10 condemned prisoners with 10 samples of testing agent. The solution to this puzzles stays the same.

A result-dependent solution can be: first divide the 1000 bottles into two groups of 500 each and then test the sample from one group, the result of this testing tells us which group was contaminated and has to be tested again. Next divide this group of 500 bottles into two groups of 250 bottles each and test one group... and so on. Eventually the contaminated bottle will be singled out.

Here is a more elegant solution which uses the binary system.

1000 Bottles of Wine

2009-09-01T21:19:00.000-07:00

The King has 1000 bottles of wine one of which was contaminated with a deadly poison. A chemist offered a small vial of agent that can detect traces of the poison, but it can be used for only 10 samples.

How can the King find the contaminated bottle?

Answer: Bridge Crossing

2009-07-01T21:51:00.000-07:00

In the river crossing puzzle, the farmer is the only person who could operate the boat. Therefore the solution is to find minimum number of trips the farmer has to make to bring the wolf, the goat and the cabbage to the other side of the river without leaving the wolf and the goat or the goat and cabbage alone.

In this bridge crossing puzzle, however, we know they have to make 5 trips altogether, so the question is to find the trips of the shortest time. Read the answer here.

Bridge Crossing

2009-07-01T21:44:00.000-07:00

Remember the puzzle of a farmer crossing a river with a wolf, a goat and a cabbage? It is really a question of how the farmer ferries the wolf, goat and cabbage one at a time to the other side of the river without leaving the wolf and goat or the goat and the cabbage alone during the process.

Here is a puzzle of bridge crossing with some time constraints:

Four people A, B, C and D have to cross a bridge at night. The bridge is narrow so it can only allow two persons at a time. Since it is dark they have to use a flashlight when crossing the bridge. The speed of crossing the bridge for A, B, C and D are 1, 2, 5 and 10 minutes respectively. Their flash light can last only 17 minutes.
How do they cross the bridge?

Answer: Black Hats and White Hats

2009-03-01T23:11:00.000-08:00

I first read this puzzle in a book titled Fun with Mathematics (趣味数学) in Chinese. It was a translation of the book Moscow Puzzles, which became one of my favorite books.

The answer to this puzzle is not hard to get if we follow the logic thinking of C and B.
Read the answer here.

Black Hats and White Hats

2008-10-13T12:09:00.001-07:00

In a dark room there were 3 black hats and 2 white hats. Three people, A, B and C were asked to go inside the dark room and put a hat on and then leave the room to stand in a queue. No body could see the color of their own hats. A stood in the front so he could not see B or C. B could see A but not C and C could see both A and B.
When C was asked if he could tell the color of his hat, the answer was "no." And so was the answer from B. However when the same question was asked of A the answer was "yes."
What was the color of A's hat? How did A figure this out?

Solution: Twelve Coins

2008-10-11T22:57:00.000-07:00

Hint: First number the coins from 1 to 12, and divide them into 3 groups...

I first heard this puzzle from Anthony while Peter and I were visiting his antique shop in Grays Market in London. In the beginning I thought this should be easy. But I could not solve it there. I was still thinking about this puzzle when we were visiting the British Museum later on that day.

The next morning we were at the Heathow Airport waiting for our flight home. On a piece of paper I numbered the coins and started a flow chart of weighing results. The solution came and it was beautiful!

See the solution to the 12 coins puzzle.

Twelve Coins

2008-09-12T16:10:00.000-07:00

Given 12 coins that look identical, with an odd coin either lighter or heavier than the rest, use a balance to weigh the coins three times to determine which is the odd coin and whether it is lighter or heavier than the rest.