# Archive for category Puzzle

### Pirates !

Five pirates have 100 gold coins.They have to divide up the

loot.In order of seniority,each pirate proposes a distribution

of the loot.All the pirates vote,and if at least half accept the

proposal,the loot is divided as proposed.If not,the most senior

pirate is eliminated,and they start over again with the next

senior pirate.What solution does the most senior pirate

propose? Assume they are very intelligent,extremely greedy,

and interested in surviving.

Solution:

The most senior pirate knows that he needs to get two other pirates

to vote for his solution in order for him not to be eliminated. So why

would any pirate vote for him? The next most senior pirate would

surely see it in his self-interest for the senior pirate to be eliminated.

Letís start simplifying. If there were only one pirate, there would

be no puzzle. The pirate would take all the loot, and no one

would complain.

Now consider the situation with two pirates. Same outcome. The

senior pirate takes all the loot, and the other pirate can’t do a thing

about it as the senior pirate’s vote represents half of the voters.

It gets more complicated for the senior pirate when there are

three pirates. Let’s number the pirates from least to most senior:

1, 2, and 3. With three pirates, pirate 3 has to convince at least one

other pirate to join his collation. Pirate 3 realizes that if his plan

is not adopted, he will be eliminated, and they will be left with two

pirates. All of them know what happens when there are two pirates:

pirate 2 takes all the loot himself and pirate 1 gets nothing. So pirate

3 proposes that he will take 99 gold coins and give 1 coin to pirate

1. If pirate 1 has any self-interest at all, he really has no choice. If

pirate 1 rejects the offer, he gets nothing. So pirate 3ís plan will

pass two to one over pirate 2ís objection.

With four pirates, an even number, the senior pirate needs just one

vote other than his own to impose his will. His question now is which

one of the other pirate’s votes can be exchanged for the fewest number

of coins. Pirate 2 recognizes that he is most vulnerable. Therefore, pirate

4 knows that if he gives pirate 2 anything at all, he will vote for it.

A rule is emerging here. In each case, the senor pirate should buy

only the votes he needs, and buy them as cheaply as possible. Apply

the rule to the five-pirate case. Pirate 5 needs two votes plus his own.

The goal is to toss a coin or two to pirate 1 and pirate 3, the two

pirates in the most vulnerable positions. Both will be empty-handed .

if the senior pirate is eliminated and four pirates remain. So pirate

5 offers one coin to pirate 3 and one coin to pirate 1. Pirates 2 and

4 get nothing.

Click here for detailed solution.

### Why nine minutes?

Many consumer alarm clocks have a snooze button that is

calibrated for 9 minutes.Does that sound like an odd

decision? Many people would think a 10-minute snooze

interval would make more sense.So were the designers asleep?

Why 9 minutes?

By setting the snooze time to 9 minutes, the alarm clock only needs to

watch the last digit of the time. Whether implemented by physical circuitry

or software, it is easier to manipulate just the terminal digit.

### Programmers and HR !

One hundred programmers are lined up one in front of the

other by an evil human resources person such that

programmers can see only the programmers in front of

them.The HR person puts a red hat or a blue hat on every

programmer.The programmers cannot see their own hats,

nor can they see the hats of those behind them,but they can

see the hats of the people in front of them.The HR person

starts with the last programmer in the back and says,

“What color is your hat?”

Programmers can answer with only one of two words,red

or blue.If the answer does not match the color of the hat on

the person’s head,the candidate is dismissed.Then the evil

HR person moves on to the next programmer,going from the

rear to the front of the row.Programmers in front get to hear

the answers of the programmers behind them,but not whether

they are dismissed or not.They can consult and agree on a

strategy before being lined up,but after being lined up and

having the hats put on,they can’t communicate in any way

other than by what has been specified.

What strategy should the programmers select to guarantee

the maximum number of surviving programmers?

**Solution: **

Is the best strategy for every programmer to say the same color, let’s

say “red” ? Assuming the hats were randomly distributed, half of them

would survive. Another strategy calls for every programmer to identify

the color of the hat on the head of the programmer immediately in front

of him or her. This guarantees the survival of that programmer, but does

nothing to promote the programmer’s own chances for survival. This

strategy also guarantees that at least half survive, plus a little more since

chance predicts that some of the programmers would have hats of the

same color as the person in front of them, thus sparing them both. In

fact, assuming random distribution of hats, this strategy yields a survival

rate of 75 percent. Getting better.

Some candidates might try to game the puzzle by suggesting that the people

in the puzzle could agree to give clues to each other by using a certain tone

of voice or drawing out the words. One candidate suggested that they could

agree that if they said their hat color in a soft voice, it means the hat in front

of them is the same color, and if they say it in a loud voice, it means the

hat in front is a different color. Recruiters think this is definitely good and

on the correct track. Another option is they could say “reeeeeeeeeeed” for

x number of seconds, where x represented the distribution of hats where a

hat was a bit in a binary number (red = 1, blue = 0).

But the ideal solution, that the programmers can say only red or blue and cannot alter their

voice in such a convincing way as to signal any information other than

the word they said. A good way to get this point across is simply to

change the problem slightly by saying ìthe evil HR person gets to hear

their plan beforehand and will thwart it if it is not to the rules.î

But even with the HR person hearing their plan in advance, there is

still a way to save almost everyone. We know that the first programmer

is never going to have any information about the color of his or her hat,

so this person cannot be guaranteed to survive. But every other person can

be saved with certainty. The programmers simply agree that if the number

of red hats that the rear-most person can see is even, then the programmer

will say “red” If the number of red hats that the rear-most person can see

is odd, the programmer will say “blue” This way, programmer number

99 can look ahead and count the red hats. If the sum of red hats is an

even number and number 100 said “red” then 99 must be wearing a blue

hat. If they add up to an even number and number 100 said “blue” signal-

ing an odd number of red hats, number 99 must also be wearing a red hat.

Number 98 knows that 99 said the correct hat, and so uses that information along with the 97 hats in front to figure out what color hat is on

the head of programmer 98.

Even if the evil HR person knows the plan, the person can’t thwart

it. The plan doesn’t rely on any specific ordering of the hats. The worst

outcome is that the evil HR person will ensure that programmer number

100 is dismissed.

### Bucket and water!

You have a 3-quart bucket,a 5-quart bucket,and an infinite

supply of water.How can you measure out exactly 4 quarts?

Solution:

Step 1- Fill 3-quart bucket and pour it into 5-quart bucket.

Step 2- Fill the 3-quart bucket again and pour it into 5-quart bucket , this leaves you with 1-quart water in the 3-quart bucket.

Step 3- Empty the 5-quart bucket and pour the 1 quart water from the 3-quart bucket.

Step 4- Fill the 3-quart bucket once again and pour it into the 5-quart bucket and you have 4 quart water in the bucket.

### Contaminated pill!

You have ten jars of pills.One jar of pills is contaminated.The

only way to tell which pills are contaminated is by weight.A

regular pill weighs 10 grams; a contaminated pill weighs 1.1 grams.

You are given a scale and allowed to make just one measurement

with it.How do you tell which jar is contaminated?

**Solution: **

1- Take 1 pill from 1st jar,2 pills from 2nd jar and so on. .. …

n pills from nth jar.

2-Weight them

3- If the total weight is :

10.1 –> 1st jar had contaminated pill.

10.2 –>2nd jar had contaminated pill.

10.n –> nth jar had contaminated pill.