The Wire Identification Problem

There is a bunch of 120 electrical cables laid under the ground. The cables are 10 KM in length and both ends of each wire are out in open.




We need to identify and label the cables 1-120. All you have is a circuit completion detector bulb.(i.e. if you join same side ends of any two cables and then on the other side apply the bulb to corresponding 2 ends then bulb will light up, but to detect this way you have to make a 10 KM trip)

How would you identify each cable in minimum number of trips.

Solution

We need to identify and label the cables 1-120. As you might have guessed, the only way to gain information in this setup is to connect some set of wires at one end, walk up to the other end, and test for conductivity. The process may have to be repeated many times before a complete labeling can be constructed. And since each such step involves walking across 10 KM, we wish to solve the problem with the least number of trips.

How would you identify each cable in minimum number of trips?

Answer:
Boundary Condition: If we have just 2 wires, there’s really nothing we can use to break the symmetry of the situation.

Solution: At the first end, connect pairs of wires together, leaving two wires unconnected. Go to the other end. Find a pair of connected wires, andnumber them #2 and #3. Find another pair and label them #4 and #5. Repeat for all of the pairs, with the last pair labeled #118 and #119. There remain two wires that are not connected to each other. Label one of these #1 and the other #120.

Now at second end, connect #1 to #2, #3 to #4, etc, leaving #119 and 120 unconnected. Go back to the first end. One of the originally unconnected wires still is unconnected. Label it #120 and label the other originally connected wire #1. Now find the wire connected to #1 and label it #2. The wire that originally was connected with new wire #2 can be labeled #3. The wire that is now connected to the newly labeled #3 is #4. In this way, all of the wires can be identified on both ends in two trips (one round trip). If it is necessary to disconnect the connections at the other end, a third trip is necessary.

Below is an execution diagram for 6 wires. We have to label wires from 1-6.





Source

http://ashutoshmehra.net/blog/2009/09/wire-identification-problem/

6 Riddles

QUESTIONS:

1) The Elder Twin
One day Kerry celebrated her birthday. Two days later her older twin brother, Terry, celebrated his birthday.
How could this happen?

2) Manhole Covers
Why is it better to have round manhole covers than square ones?

This is logical rather than lateral, but it is a good puzzle which can be solved by lateral thinking techniques. It is supposedly used by a very well-known software company as an interview question for prospective employees.

3) The Deadly Party
A man went to a party and drank some of the punch. He then left early. Everyone else at the party who drank the punch subsequently died of poisoning.
Why wasn't the first guy poisoned?

4) Heaven
A man died and went to Heaven. There were thousands of other people there. They were all naked and all looked as they died at the age of 21. He looked around to see if there was anyone he recognized. He saw a couple and he knew immediately that they were Adam and Eve.
How did he know?

5) Trouble with Sons
A woman had two sons who were born on the same hour of the same day of the same year. But they were not twins.
How could this be so?

6) The Man in the Bar
A man walks into a bar and asks the barman for a glass of water. The barman pulls out a gun and points it at the man. The man says `Thank you' and walks out.
Why?

This puzzle has claims to be the best of the genre. It is simple in its statement, absolutely baffling and yet with a completely satisfying solution. Most people struggle very hard to solve this one yet they like the answer when they hear it or have the satisfaction of figuring it out.





ANSWERS:

1) At the time she went into labor, the mother of the twins was traveling by boat. The older twin, Terry, was born first early on March 1st. The boat then crossed the International Date line (or anytime zone line) and Kerry, the younger twin, was born on February the 28th. In a leap year the younger twin celebrates her birthday two days before her older brother.

2) A square manhole cover can be turned and dropped down the diagonal of the manhole. A round manhole cannot be dropped down the manhole. So for safety and practicality, all manhole covers should be round.

3) The poison in the punch came from the ice cubes. When the man drank the punch the ice was fully frozen. Gradually it melted, poisoning the punch.

4) He recognized Adam and Eve as the only people without navels. Because they were not born of women, they had never had umbilical cords and therefore they never had navels.

This one seems perfectly logical but it can sometimes spark fierce theological arguments!

5) They were two of a set of triplets (or quadruplets etc.)

This simple little puzzle stumps many people. They try outlandish solutions involving test-tube babies or surrogate mothers. Why does the brain search for complex solutions when there is a much simpler one available?

6) The man had hiccups. The barman recognized this from his speech and drew the gun in order to give him a shock. It worked and cured the hiccups - so the man no longer needed the water.

The is a but a difficult one to solve. It is a perfect example of a seemingly irrational and incongruous situation having a simple and complete explanation. Amazingly this classic puzzle seems to work in different cultures and languages.

King Bear & Poison ..Very Common & Tough Puzzle

Problem : A king has a wine cellar with 100 drums of wine. A traitor sneaked into his wine cellar to poison the drums. He could mix poison in only one of the drums when guards caught him. However, guards could not see which one drum was poisoned. Our king being a drunkard wishes to be able to drink wine as soon as possible but safely. The poison guarantees that the person drinking the wine would die in 10 days.

Being a king, he can order his servants to drink wine from the drums to find out poisonous drum. However, he wishes does not want to risk lives of too many servants.

1) King wishes to be able to drink some wine on 11th day. How many servant would be needed to take the risk of drinking wine before figuring out a harmless wine drum?

2) King wishes to throw a grand party on 11th day. How many servant would be needed for figuring out the poisonous wine drum?

Solution :

The solution to the first problem is extremely simple and obvious. Only one servant need to drink any one of the wine drum. If the servant dies, all the other drums are harmless. If he does not die, we can guarantee that at least that one wine-drum is harmless.

Well, second problem has actually two sides to it. Say for example, king wishes that number of servant to be put on stake can be arbitrarily large, but the number of servant died during the process should be less. Then, we need 99 servant for each wine drum. At most one servant would die in this case.

However, what if we want to minimize the number of servants to be put on stake? Initially I went on a very wrong direction. That was to use prime factorization of every number from 1 to 100. For example, there will be a servant drinking from wine-drums which are numbered in multiple of 2, and similarly another servant for multiple of 5. So, we would be able to detect number 10 as the poisonous drum if both of them dies. Using this multiple scheme, number of servant needed would be

2,4,8,16,32,64

3,9,27,81

5,25

7,49

all primes number between 1 to 100 which has not appeared in the above list.
Basically, the number comes out to be huge. Certainly a very very bad solution.

A good solution would be to arrange wine-drums in 10×10 maze. We would need only 20 servant. 10 servant for the row and 10 for the column. So when in 10 days i-th servant for row and j-th servant for column dies, we would know that the drum located at (i,j) is the culprit. But still we are far from the best solution.

We can visualize the maze in three dimension. 5x5x5 would need only 15 servants. Of course, there would be some spots empty in the maze, but that does not matter.6x6x3 = 15. Going ahead we can have 5x5x4=14.

It seems we can still push it further. We can visualize the maze in four dimension. 4x3x3x3 = 13. This is the least possible number I could find. I don’t know whether we can push it still further down. If you can figure out a still smaller number, please comment on this post.


Soln By –Saurabh Joshi from his blog me.maths & puzzle
i found its interesting so i pasted here

Who Can Name the Bigger Number?

http://www.scottaaronson.com/writings/bignumbers.html

Another Hat Color Problem

The Island - One puzzle a day - Puzzle Buddies
100 men were captured and prisoned in an island.the island is guarded by a genie.
Everyday genie takes prisoners out.genie places them in a circle so that everyone can see each other,then all the prisoners are blindfolded. Genie then places hats on prisoners head,some of them are white and some of them are red.their folds will be opened after the genie places hats.

On genie's command the prisoners with white hat should step forward.They will be set free if only the men with white hat steps forward.
If any one without white hat steps forward,genie executes all of them
If no one steps forward,the same game will be repeated the next day,if the game is repeated,each prisoner will get the same hat that they got on the first day

Prisoners are not allowed to communicate with each prisoner.if communicated all of them will be executed.prisoners are not allowed to see the hat that they are wearing
At least one white hat is given, and all prisoners are aware of this.

how many days did it take for the prisoners to get out if genie gave "n" white hats where 1<=n<=100, and how did they do it?



Solution:
Let us say there is one white hat,the answer is simple.the prisoner wearing white hat sees no other white hat and he steps forward.
Now let us take two white hats.No body comes forward on the first day,because the prisoners wearing white hat sees one white hat.On the second day since no body has come forward they both know there must be two white hats and one he can see and other must be his own.so they come forward on second day.
Similarly if there are three white hats they will not come out on second day.it will take three days.
and so on..for "n" it will take n days

Monty Hall Problem

You are a contestant on a game show. You have three closed doors in front of you. One of the doors has a car behind it and the other two doors have nothing. You have to choose a door to open.

You have with you a Magic Watch.
- Whenever you ask the watch a question with a yes/no answer, it will blink either red or blue.
- One of the colors represents 'yes' and the other 'no'.
- You don't know which color means what. (gotcha!)
- You are allowed to ask the watch only 2 questions before you make your decision.

Which questions would you ask and how would you choose the door to open?

The puzzle might have a number of solutions and this is one of them.

Solution:

The contestant should ask the following two questions:
1) If I asked whether the car was behind door 1, would you blink red?
2) If I asked whether the car was behind door 2, would you blink red?

There are two cases here:
a) red = no, blue = yes:

The watch can answer in 4 different ways:
1) red, blue:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2st question, car is not behind door 2.
2) blue, red:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is behind door 2.
3) blue, blue:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is not behind door 2.
4) red, red:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2nd question, car is behind door 2.
b) red = yes, blue = no:

The watch can answer in 4 different ways:
1) red, blue:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2st question, car is not behind door 2.
2) blue, red:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is behind door 2.
3) blue, blue:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is not behind door 2.
4) red, red:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2nd question, car is behind door 2.

Observe that the conclusions are exactly the same for the same answer set irrespective of the color code.
Hence, when the watch answers:
1) red, blue: Pick door 1.
2) blue, red: Pick door 2.
3) blue, blue: Pick door 3.
4) red, red: This outcome is impossible since it implies that the car is behind door 1 as well as door 2.

How can you get a fair coin toss

Interview question: How can you get a fair coin toss if someone hands you a coin that is weighted to come up heads more often than tails?

The story of these types of questions starts at the software giant Microsoft who pioneered the use of puzzle type question, where the aim is to test the interviewers problem solving skills and creativity. It's ironic that these type of questions are no longer asked in Microsoft job interviews, but the rest of the software industry hasn't caught on.



The solution is to toss the coin twice, which gives four possible outcomes:
H (heads) followed by H, probability of this is pH * pH
H followed by T (tails), probability of this is pH * pT
T followed by H, probability of this is pT * pH
T followed by H, probability of this is pT * pT
Looking at the above outcomes we can see that HT (H followed by T) is just as likely as TH,
So to get a fain coin toss, toss the coins and use HT or TH as fair outcomes of the toss, and discard the results of TT and HH.