Interview Puzzles

Chasing Dogs Puzzle

2011-07-24T11:40:00.000-07:00

Chasing Dogs Puzzle :

Puzzle :

There are four dogs each at the corner of a unit square. Each of the dogs starts chasing the dog in the clockwise direction. They all run at the same speed and continuously change their direction accordingly so that they are always heading straight towards the other dog. How long does it take for the dogs to catch each other and where?

Have a better solution? Let us know through the comments section!

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Heaven Puzzle

2011-07-24T07:05:00.000-07:00

Heaven Puzzle.

Puzzle :

A person dies, and he arrives at the gate to heaven. There are three doors in the heaven. one of them leads to heaven. another one leads to a 1-day stay at hell, and then back to the gate, and the other leads to a 2-day stay at hell, and then back to the gate. every time the person is back at the gate, the three doors are reshuffled. How long will it take the person to reach heaven?

this is a probability question - i.e. it is solvable and has nothing to do with religion, being sneaky, or how au dente the pasta might be ;-)

Answer:

1/3 of the time, the door to heaven will be chosen, so 1/3 of the time it will take zero days. 1/3 of the time, the 1-day door is chosen; of those, the right door will be chosen the next day, so 1/9 trips take 1 day. Similarly, 1/9 will take two days (choosing the 2-day door, then the right door).

After that, the cases split again, and again, and again. I can’t seem to make a nice infinite sum this way, so let’s try again.

Suppose the average days spent is X. 1/3 of the cases are done in zero days as before. 1/3 of the cases are 1 day plus X. 1/3 are 2 + X. So:

X = 1/3 * 0 + 1/3 * (1 + X) + 1/3 * (2 + X)

= 0 + 1/3 + X/3 + 2/3 + X/3

= 1 + 2X/3

Therefore,

X/3 = 1

X = 3

On average, it takes three days to get to heaven. Two if the noodles are limp.

Took me one blind alley, and about five minutes.

Chess Board Puzzle

2011-01-25T10:55:00.000-08:00

Chess

Chess Board Puzzle :

Imagine there are infinite number of Queens (Chess Game Piece) with u. Find the minimum number of queens required so that every square grid on the chess board is under the attack of a queen. Arrange this minimum no. of Queens on a chess board.

Answer :

Indicates Queen.

Tumblers on a Rotating Table

2010-09-22T09:43:00.000-07:00

Tumblers on a Rotating Table Puzzle :

A blind gnome and an evil goblin take turns to play a game. Four tumblers are placed at the corners of a square table. The initial configuration of the tumblers (facing up or facing down) is chosen by the evil goblin. When the blind gnome gets his turn, he is allowed to specify a subset of the four tumblers and flip them simultaneously. To be precise, he may choose “one tumbler”, “two diagonally opposites”, “two adjacent”, “three tumblers” or “four tumblers” lying in front of him, and flip them simultaneously. After flipping, if all four tumblers are upright, he wins the game! Otherwise, the game continues and the evil goblin is allowed to rotate the table by an amount of his choice. Can the blind gnome win the game with a deterministic strategy?

Solution :

If there were only two cards, then “flip both” — “flip one” — “flip both” would ensure a victory for the blind gnome. For the case of four cards, let F denote “all four cards”, A denote “two adjacent cards”, D denote “two diagonally opposite cards”, O denote “one card”. Then the following 15-step sequence ensures victory for the blind gnome: F, D, F, A, F, D, F, O, F, D, F, A, F, D, F. The procedure can be generalized to 2^n cards. This problem is studied in detail by Ehrenborg and Skinner (see bibliography below); they call it “The Blind Bartender’s Problem”.

A variant of the above problem allows the blind gnome to first “touch” any t out of n tumblers on a polygonal table with n sides. Touching allows the gnome to deduce which of the t tumblers he touched are upright, and then to decide which of these t tumblers to flip. It was this variant that was originally published in Martin Gardner’s article in Feb 1979. For four tumblers, the blind gnome can win in five moves, as Gardner showed in his subsequent article in Mar 1979. The problem generated quite some interest. Two pairs of mathematicians independently proved that the blind gnome can win with a deterministic strategy if and only if t ≥ n(p-1)/p where p is the largest prime factor of n (Lewis and Willard in 1980; Laaser and Ramshaw 1981). These proofs are quite involved

Mars Rovers thoughtworks puzzles

2010-09-08T10:23:00.000-07:00

Mars Rovers is one of the famous puzzle asked by many companies while recruiting people for them. Here the problem is defined and answers are given with Java program.

Mars Rovers thoughtworks puzzles

A squad of robotic rovers are to be landed by NASA on a plateau on Mars. This plateau, which is curiously rectangular, must be navigated by the rovers so that their on-board cameras can get a complete view of the surrounding terrain to send back to Earth.

A rover's position and location is represented by a combination of x and y co-ordinates and a letter representing one of the four cardinal compass points. The plateau is divided up into a grid to simplify navigation. An example position might be 0, 0, N, which means the rover is in the bottom left corner and facing North.

In order to control a rover, NASA sends a simple string of letters. The possible letters are 'L', 'R' and 'M'. 'L' and 'R' makes the rover spin 90 degrees left or right respectively, without moving from its current spot. 'M' means move forward one grid point, and maintain the same heading.

Assume that the square directly North from (x, y) is (x, y+1).

INPUT:

The first line of input is the upper-right coordinates of the plateau, the lower-left coordinates are assumed to be 0,0.

The rest of the input is information pertaining to the rovers that have been deployed. Each rover has two lines of input. The first line gives the rover's position, and the second line is a series of instructions telling the rover how to explore the plateau.

The position is made up of two integers and a letter separated by spaces, corresponding to the x and y co-ordinates and the rover's orientation.

Each rover will be finished sequentially, which means that the second rover won't start to move until the first one has finished moving.

OUTPUT:

The output for each rover should be its final co-ordinates and heading.

INPUT AND OUTPUT:

Test Input:

5 5

1 2 N

LMLMLMLMM

3 3 E

MMRMMRMRRM

Expected Output:

1 3 N

5 1 E

==========

SOLUTION :

Administrator removed the solution. Try your own.

Covent Garden Puzzle

2010-09-03T03:18:00.000-07:00

Covent Garden Puzzle

Mrs. Ram and Mrs. Jones had equal number of apples but Mrs. Jones had larger fruits and was selling hers at the rate of two for a penny, while Mrs. Ram sold three of hers for a penny.

Mrs. Ram was for some reason called away and asked Mrs. Jones to dispose of her stock. Upon accepting the responsibility of disposing her friend's stock, Mrs. Jones mixed them together and sold them of at the rate of five apples for two pence.

When Mrs. Ram returned the next day the apples had all been disposed of, but when they came to divide the proceeds they found that they were just seven pence short, and it is this shortage in the apple or financial market which has disturbed the mathematical equilibrium for such a long period.

Supposing that they divided the money equally, each taking one-half, the problem is to tell just how much money Mrs. Jones lost by the unfortunate partnership?

Puzzle Solution

The mixed apples were sold of at the rate of five apples for two pence. So they must have had a multiple of five i.e. 5, 10, 15, 20, 25, 30,…, 60, 65,… etc apples.

But the minimum number of apples they could have together is 60; so that 30 would have been of Mrs. Ram's that would fetch her 10 (an integer) pence and the other 30 of Mrs. Jones's that would fetch her 15 (also an integer) pence.

When sold separately it would fetch them 10+15=25 pence altogether. But when sold together it would fetch them 60X2/5=24 pence i.e. a loss of one (25-24=1) pence.

Since they lost 7 pence altogether; they had altogether 60X7=420 apples that fetched them only 420X2/5=168 pence and they shared 84 pence each of them. But Mrs. Jones could sell her 420/2=210 apples for 210/2=105 pence so she lost "21 pence".

Note: to solve it algebraically:
They lost 7 pence altogether
Suppose each lady has x apples
x/2 + x/3 - 2(2x/5) = 7

15x + 10x - 24x = 210

x = 210
Note: Mrs. Johns lost 21 pence.

But without working Mrs. Ram earned 14 extra pence!

(84 pence – 210/3 pence = 14 pence).

Not very fair!

(Perhaps Mrs. Johns was not very good at math)

Einstein’s Puzzle

2010-09-01T07:06:00.000-07:00

Puzzle :

Five friends have their gardens next to one another, where they grow three kinds of crops: fruits (apple, pear, nut, cherry), vegetables (carrot, parsley, gourd, onion) and flowers (aster, rose, tulip, lily).

1. They grow 12 different varieties.
2. Everybody grows exactly 4 different varieties
3. Each variety is at least in one garden.
4. Only one variety is in 4 gardens.
5. Only in one garden are all 3 kinds of crops.
6. Only in one garden are all 4 varieties of one kind of crops.
7. Pear is only in the two border gardens.
8. Paul's garden is in the middle with no lily.
9. Aster grower doesn't grow vegetables.
10. Rose grower doesn't grow parsley.
11. Nuts grower has also gourd and parsley.
12. In the first garden are apples and cherries.
13. Only in two gardens are cherries.
14. Sam has onions and cherries.
15. Luke grows exactly two kinds of fruit.
16. Tulip is only in two gardens.
17. Apple is in a single garden.
18. Only in one garden next to the Zick's is parsley.
19. Sam's garden is not on the border.
20. Hank grows neither vegetables nor asters.
21. Paul has exactly three kinds of vegetable.

Who has which garden and what is grown where?

Answer :

How to interview a person - An Article

2010-08-31T21:52:00.001-07:00

How to Conduct an Interview

Interviews have four stages that precede the writing of a story:

ARRANGEMENTS

Once an interviewer has decided to interview someone, He should call the candidate in advance to make an appointment. Identify himself by his name and the name of his publication. If the interviewer feels that he need to do so or asked to describe what the story is about, be brief and general. The shape of the story might change as him continue his reporting. If the interviewer interviewing several persons in connection with the story, interview the principal person last, because he will be better prepared based on what the learning from the earlier interviews.

PREPARATION

Try to prepare as much research as possible on the person in advance on topic working on. Sources might be the library, public records, the internet and the expert people in that technology for background information. Prepare the questions in advance if necessary writing and bring them into the interview. Refer to them but don't show them to the candidate, because it creates too formal an atmosphere. Ask other questions as they might arise, based on what the candidate says or something new that might come at the moment. Bring pencil (or pen) and paper. A stenographer's notebook is usually easier to handle than a large pad but use whatever is comfortable. Bring a tape recorder but be sure to get the permission to use it from the person you are interviewing.

THE INTERVIEW

            It is inadvisable to launch right into the interview unless it’s being given a few minutes. Some casual conversation to start with will relax both of you. Interviewers should ask Questions should be as short as possible. Give the respondent time to answer. Interviewers must be a good listener. If candidate prattles on, it is appropriate to move on as politely. Interviewer might say something such as: "Fine, but let me ask you this ". Try to draw out specifics: How long, how many, when, etc.?

          Try to absorb the atmospherics of the locale where the interview takes place, with particular attention to what might be a reflection of the candidate’s personality and interests, such as photos of children or bowling trophies or a paper-littered desk or a clean one, etc. Note characteristics of the candidate that might be worth mentioning in your story, such as pacing, looking out the window to think, hand gestures and the like. Invite the person to call you if she/he thinks of anything pertinent after the interview. It often happens, Interviewer should provide his name, email addresses and phone number on a card or piece of paper before to leave. If that person has a secretary, be sure to get that person's name and telephone number, too, in case there is some detail that needs follow up and, again, leave information as to how you may be contacted. If a photo is needed and is not taken during the interview, be sure to make arrangements then to have one taken at a later time.

RECONSTRUCTION

As soon as it's practical after the interview, find a quiet place to review the handwritten notes. While taking notes, that may have written abbreviations for words that won't mean anything in a day or two later. Or some scribbling may need deciphering, and, again, it is more likely to be better able to understand the scribbles soon after the interview. Underline or put stars alongside quotes that seemed most compelling. One star for a good quote, two stars for a very good one, etc. It will speed the process when you get to the writing stage. One other thing to look for in the notes: the quote wrote down might not make a lot of sense, unless to remember what specific question it was responding to. In short, fill in whatever gaps exist in the notes that will help better understand them when writing.

How to Interview a Programmer

Finding good programmers is hard because good programming is dependent on much more than just knowledge of programming language syntax.

         1. The Interviewer needs someone who, despite wearing striped pants with a polka dot shirt, has a good sense of taste in OO design.
        2. He need someone who the candidate is creative enough to find innovative solutions to problems, yet anal retentive enough to always line up their curly braces.
        3. The Interviewer needs someone who (the Programmer) is humble enough to be open to suggestions for improvement, but arrogant enough to stand firm and provide leadership when they are the best person to provide it.

How to tell all this about a stranger by spending 30 minutes with them in a conference room?

Summery:

Explore an Area of Expertise
Have Them Critique Something
Ask Them to Solve a Problem
Look at Their Code
Ask About a People Problem
Get to Know Them
Conclusion

Explore an Area of Expertise

Rather than simply look for expertise and experience in the exact area in which the candidate will work, it is necessary look for general programming talent and ability. One way to explore and judge a candidate's talents is to explore an area of their expertise:

Hire for talent. One of the biggest mistakes companies make is to recruit from a shopping list: I need a programmer with six years Java, three years Oracle, and two years EJBs. The world changes, so to hire folks who change with it. Look for people who know computing, not necessarily particular narrow niches. Not only will they adapt better in the future, they're also more likely to be innovative in the present.
To identify how good the candidates are technically, I let them choose an area in which they feel they have expertise. I need them to know something well, and I ask them about that. I ask them why. I want them to know why something in their area of expertise works the way it does. I'm not necessarily after an expert in the area I need. If they learned why in the past, I have confidence they'll learn why in the future.

Have Them Critique Something

Another technique involves the importance of creating a dialog with the candidate. To get to know the candidate's talents and personality, Do n’t merely ask questions that have short factual answers. Find a way to engage a conversation. To stimulate dialog, ask the candidate to critique some technology:

I ask candidates to critique a system or platform that we both have in common, preferably something they will use on the job. For example, I might ask, "What parts of Java don't you like and why?"
I give candidates samples of our current code and ask them to explain and critique it. This gives me a sense of their skills, but also lets them know what they can expect.

Ask Them to Solve a Problem

Another way to foster an open-ended dialog is to ask the candidate to perform a task: to solve a problem or create a design. Although everyone at the meeting seemed to agree that this was important and useful technique, it also generated a lot of concern. People felt that asking the candidate to solve puzzles and problems needed to be done with care:

Several expertise interview taken and their review is given below

I like to ask a candidate to solve a small-scale design problem, finger exercises, to see how they think and what their process is: "How would you write a function that tells me if its argument is a power of 2?" I'm not looking for the optimal bit-twiddling solution ((n & -n) == n). I'm looking to see if they get the method signature right, if they think about boundary cases, if their algorithm is reasonable and they can explain its workings, and if they can improve on their first attempt.
I ask candidates to create an object model of a chicken. This eliminates any problems with uncertainties about the problem domain, because everyone knows what a chicken is. I think it also jars people away from the technical details of a computer. It tests to see if they are capable of thinking about the big picture.
I hate anything that asks me to design on the spot. That's asking to demonstrate a skill rarely required on the job in a high-stress environment, where it is difficult for a candidate to accurately prove their abilities. I think it's fundamentally an unfair thing to request of a candidate.
I don't like when I'm asked to write a program that does X on a piece of paper. Don't ask the candidate to write a program on paper. That is a waste of time and sweat. People don't write software on paper, they do it with computers using auto-completion, macros, indexed API documentation, and context-sensitive help. They think about it, refractor it, and even rewrite it. If you want to see a person's work, ask them to write some small module or implement some interface before the interview and bring the code on a notebook PC or on hard copy. Then you can review it and discuss the design, coding style, and decisions that went into it. This will give you a much more realistic and useful assessment of a person's work and style.
I like design dialog questions that don't have a single fixed answer. That way they have to ask me questions, and this sparks a discussion. It's good to have a whiteboard available in the room. A dialog lets the interviewer see how the interviewee works, whereas a question of fact is just that: it is great for TV quiz shows, but doesn't tell you how someone will work and approach things over time. A puzzle question requires knowing a trick, which is in essence something that is either known or unknown. I dislike puzzle questions, because they don't require dialog.
What constitutes a reasonable question depends a lot on the candidate's experience and maturity.
I look for people with curiosity. Present problems, not puzzles.

Look at Their Code

          One technique we all seemed to like: Have the candidate bring a code portfolio to the interview. Look at the candidate's code and talk to them about it. Although we were concerned that some candidates may not have code they could legally bring to the interview, we figured most candidates could probably come up with something. It can't hurt at least to ask a candidate to bring to the interview a sample of code they had written in the past.

         Find Out What Books They Read Several people indicated that they ask candidates about the programming books they read to see if a programmer is self-motivated or concerned about improving their own programming skills:

Ask About a People Problem

            As important as technical ability, or perhaps more important, is personality. How well would the candidate fit the team? How well would they fit the work environment? People used various techniques to judge personality. Good citizenship is probably more important than technical prowess, because if you have people with the right kind of attitude and demeanor, you can help them gain the technical knowledge and software development habits. But if you have people who lack humility and maturity, it can be extremely difficult to get them to cooperate in reaching a goal, no matter how bright they are or what they've accomplished in the past.

Get to Know Them

          Perhaps the prominent theme of our discussion was that you need to try to get to know the candidate as best you can. Talk to the candidate in the interview. Try to get a feel for them. If possible, bring them in on a trial basis or for a probationary period. That would give you more time to get to know the candidate, and give the candidate more time to get to know you:

Chuck Allison: I talk to them. I get a feel for them. I always ask about what they've done. I have found that by discovering what a person is excited about technically, you can learn a lot of important things about them. In the past I've asked people to describe a project that was especially interesting to them, or that was challenging and successful. On occasion I've asked what they've done that they're the most proud of. This usually reveals the depth of one's understanding and mastery. It also gets them to turn on the fire hose verbally, and you can sit back and get most of the answers you need.

Randy Stafford: I look at past projects listed on their resume, and ask them to talk about those projects—how was the team organized, what the technologies and architectures were used, was the software successful in production, etc. In their answers I'm listening for what lessons they learned from those experiences, and whether those lessons match with lessons I've learned from my experiences and from professional literature. I get a glimpse into how they perceive themselves in relation to the world around them. Some come off as arrogant, some ignorant, some helpless. Others sound humble, intelligent, and motivated. I often ask them what software development literature they read. Continuous education is very important to me.

Angelika Langer: In Germany, hiring is like marrying someone. It's "until death do us part"—a marriage without the backdoor of a divorce, because you can't fire employees. The only chance for firing someone is during a three- to six-month probationary period, or when the company goes out of business. The major filtering is done before the interview, based on the curriculum vitae (CV) and submitted papers, such as evaluations from former employers. (In Germany, employers must provide every employee with a written evaluation when they leave the company.) The interview itself is usually brief. The main tool in filtering is scrutinizing the CV and papers; 98 percent of all applicants are disqualified in this phase. The interview should confirm the impression you gain from the applicant's papers and allows you to sense their personality. The lucky winner then goes on a probationary period. Probation definitely does not replace the filtering; it just keeps a last exit open until you really must commit.

Andy Hunt: We've hired people who interviewed well, but they were terrible at the job. If possible, hire them in for a trial period.

Dawn McGee: You could also bring candidates in for half a day, and have them do what they would be doing on the job.

Conclusion

             To sum up the overlying themes from our hour-long discussion in Portland: You should look for talent and fit more than specific skill sets. Ask open-ended questions to initiate revealing dialog. Ask candidates to critique something. Ask them to design something. Investigate their past experience. Review their code. And through conversation and, if possible, a trial period, you should try to become familiar with the candidate's technical abilities, talents, and personality.

Poisened Drink Puzzle

2010-08-31T21:50:00.000-07:00

Poisened Drink Puzzle

A rich old man has died. After his death, his children are surprised to learn that he has left all of his money to his oldest son Jeremiah, who loved him dearly, and ignored his other children, who hated him.

So, the funeral is a day or two later, and the other sons and daughters have decided to kill Jeremiah and take his inheritance. Since his father's death, Jeremiah has taken to drinking, and they know that, at the wake, he's going to be gulping down the liquor like it was nectar of the gods. So they decide to poison the drinks. One of the other sons, Wallace, tends bar, and gets the poison all ready. So Jeremiah comes up, crying and depressed, and orders a scotch on the rocks. Wallace serves him one, and he chugs it down in two seconds. "Give me another." Wallace gives him a second glass of scotch, which he also drinks in a matter of moments. The other siblings are puzzled...the poison is fast-acting; Jeremiah should be convulsing on the floor and retching his guts out. Finally, fifteen minutes later, a rather inebriated and very much alive Jeremiah orders one last glass of scotch, but as Wallace hands it to him, he changes his mind and leaves, sobbing. The other siblings come over to Wallace, and wonder what's going on. They talk about what could have gone wrong for a few minutes, and figure the poison's harmless. So Wallace sips the drink he poured for Jeremiah, and is pronounced DOA thirty minutes later.

Why did Jeremiah live? (He had no immunity to the poison, he didn't know it was coming, and the poison was obviously deadly.)

Answer :

Jeremiah orders a scotch on the rocks (ice cubes). Dropping the poison into his drink, it lands on top of the ice cubes. Jeremiah downs his two drinks very quickly, not giving enough time for the poison to mix with his drink. Conversely, the brother waits a "few minutes" before drinking, allowing the ice cubes to melt and giving ample time for the poison to mix with the scotch. He is soon dead as a result!

Four ships Puzzle

2010-08-28T21:51:00.000-07:00

Puzzle :

Four ships are sailing on a 2D planet in four different directions. Each ships traverses a straight line at constant speed. No two ships are traveling parallel to each other. Their journeys started at some time in the distant past. Sometimes, a pair of ships collides. A ship continues its journey even after a collision. However, it is strong enough only to survive two collisions; it dies when it collides a third time. The situation is grim. Five of six possible collisions have already taken place (no collision involved more than 2 ships) and two ships are out of commission. What fate awaits the remaining two?

Solution :

Let z-axis denote time. let x- and y- axes denote the 2D planet. Then the four trajectories are straight lines. Since no collision involved more than two ships, these four lines must all lie in a plane. So, one might be tempted to believe that the two other ships will also collide. But, they might have collided in negative time. So, it cannot be decided from the given information that the two ships will collide or not?

This can be made clearer through the following two diagrams. In the first one, the four ships move such that there is no collision between the two ships. In the other diagram, the four ships move such that there is a collision between them.

Three Boxes and a Ruby Puzzle with Solution

2010-08-28T21:41:00.000-07:00

Puzzle :

Alice places three identical boxes on a table. She has concealed a precious ruby in one of them. The other two boxes are empty. Bob is allowed to pick one of the boxes. Among the two boxes remaining on the table, at least one is empty. Alice then removes one empty box from the table. Bob is now allowed to open either the box he picked, or the box lying on the table. If he opens the box with the ruby, he gets a kiss from Alice (which he values more than the ruby, of course). What should Bob do?

Solution:

This problem is known as the Monty Hall Problem.

If Bob switches his choice, he wins with probability of 2/3. The YouTube video explaining this solution in detail. To get better intuition, it helps to consider a slightly different problem with 100 boxes with one box containing a ruby. Bob then picks one of the boxes at random. At least 98 of the remaining boxes are empty — these are removed by Alice. So now, we are left with two boxes: should Bob switch? Indeed, for he wins with probability 99/100 if he switches!

Eight Queens Puzzle

2010-08-15T11:54:00.000-07:00

Eight Queens Puzzle

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard such that none of them are able to capture any other using the standard chess queen's moves.

Available movement of the Queen is shown in above picture. The queens must be placed in such a way that no two queens attack each other. In chess, a queen can move as far as horizontally, vertically, or diagonally. The standard 8 by 8 Queen's problem asks how to place 8 queens on an ordinary chess board so that none of them can hit any other in one move.

This puzzle has twelve unique solution. The following image show one solution as using backtracking algorithm method. See wikipedia for more details about 8 Queen puzzle.

Algorithms :

Check the following links

1. http://www.mactech.com/articles/mactech/Vol.13/13.12/TheEightQueensProblem/
2. http://yuval.bar-or.org/index.php?item=9
3. http://firefang.net/english/n-queens
4. http://bridges.canterbury.ac.nz/features/eight.html
5. http://mathworld.wolfram.com/QueensProblem.html

for algorithm to solve this problem.

Source Code :

The solved java program is available here.

1. http://tszone.co.cc/NQueens.java

2. http://rooparam.blogspot.com/2009/12/eight-queens-solution-with-dfs-using.html

Hats and IIT students

2010-08-14T10:47:00.000-07:00

Puzzle:

This is another hats puzzle same like as prisoners and hats puzzle. The puzzle is Nine IIT students were sitting in a classroom. Their professor wanted them to test. Next day the professor told all of his 9 students that he has 9 hats, The hats either red or black color. He also added that he has at least one hat with red color and the no. of black hats is greater than the no. of red hats. The professor keeps those hats on their heads and ask them tell me how many red and black hats the professor have? Obviously students can not talk to each other or no written communication, or looking into each other eyes; no such stupid options and no tricks.

Professor goes out and comes back after 20 minutes but nobody was able to answer the question. So he gave them 10 more minuets but the result was the same. So he decides to give them final 5 minutes. When he comes everybody was able to answer him correctly.

So what is the answer? and why?

Answer :

After 20 minutes :
Lets assume that their is 1 hat of red color and 8 hats of black color. The student with red hat on his head can see all 8 black hats, so he knows that he must be wearing a red hat.
Now we know that after first interval nobody was able to answer the prof that means our assumption is wrong. So there can not be 1 red and 8 black hats.

After second interval of 10 minutes :
Assume that their are 2 hats of red color and 7 hats of black color. The students with red hat on their head can see all 7 black hats and 1 red hat, so they know that they must be wearing a red hat.
Now we know that after second interval nobody was able to answer the prof that means our assumption is again wrong. So there can not be 2 red and 7 black hats.

After third interval of final 5 minutes :
Now assume that their is 3 hats of red color and 6 hats of black color. The students with red hat on their head can see all 6 black hats and 2 red hats, so they know that they must be wearing a red hat.
Now we know that this time everybody was able to answer the prof that means our assumption is right.

So there are 3 red hats and 6 black hats.

Now as everybody gave the answer so there can be a doubt that only those 3 students know about it how everybody came to know ?
Then here is what i think, the professor gave them FINAL 5 minutes to answer, so other guys will think that the professor expects the answer after 3rd interval (according to prof it must be solved after 3 intervals), so this is the clue for others.

Engineers and Managers Puzzle

2010-08-12T10:07:00.000-07:00

Engineers and Managers Puzzle

Puzzle :

         In a city, The police has surrounded the Bank. There are 50 people in the building. Each person is either an engineer or a manager of the bank. All computer files have been deleted, and all documents have been shredded by the managers. The problem confronting the police is to separate the people into these two classes, so that all the managers are locked in a room and all the engineers are freed. every people knows the status of all others. The interrogation consists entirely of asking person i if person j is an engineer or a manager. The engineers always tell the truth. What makes it hard is that the managers may not tell the truth. In fact, the managers are evil geniuses who are conspiring to confuse the interrogators.

   1. Under the assumption that more than half of the people are engineers, can you find a strategy for the Police to find one engineer with at most 49 questions?
   2. Is this possible in any number of questions if half the people are managers?
   3. Once an engineer is found, he/she can classify everybody else. Is there a way to classify everybody in fewer questions?

Answer :

Part 1:

Here's an n-1 query solution to part 1. Maintain three sets of people: UNSEEN, STACK, and DISCARD. Initialize the process by picking one arbitrarily to be the STACK, everything else is UNSEEN. Repeat the following step until UNSEEN is empty:

Pick an UNSEEN element x, remove it from UNSEEN. Ask the top of the STACK y about x. If y says "manager" pop y off the stack and DISCARD both x and y. If it says "engineer" add x to the top of the STACK.

After all elements have been processed in this way (n-1 comparisons), the top of the stack must be an engineer.

Why does this work? First observe that whenever we discard a pair, at least one of them is a manager. So among the rest of them (STACK and UNSEEN) a majority must still be engineers. So at the end, when UNSEEN is empty, there must be an engineer in the stack, therefore the top of the stack must be an engineer.

This can be improved to 48 simply by stopping one earlier. When there's one UNSEEN left, if the stack is empty, that UNSEEN one is an engineer. Otherwise, the top of the stack must be an engineer.

first step we can just throw out one person, and appy this algorithm to the remaining 49 obtaining 47 comparisons. This gives the optimal algorithm.

This is optimal. The proof appears in the solution of homework assignment 7 of Steven Rudich's course 15-251 taught at CMU in the spring semester of 2002. See Solution 7.

Part 2: If half or more of the people are managers, then the problem cannot be solved. The managers can ensure this simply by always lying. Now there's way to separate the two sets of people. Each one simply claims the others are Managers.

Part 3: I don't know any better solution than to simply using the solution to Part 1 to identify everybody.

Lateral Thinking Puzzles

2010-08-08T04:36:00.000-07:00

Puzzle 1:

A man is trapped in a room. The room has only two possible exits: two doors. Through the first door there is a room constructed from magnifying glass. The blazing hot sun instantly fries anything or anyone that enters. Through the second door there is a fire-breathing dragon. How does the man escape?

Answer :

He waits until night time and goes out the first door.

Puzzle 2:

In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison.
The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted.
What exactly happened there?

Answer :

The treasurer's plan was to drink a weak poison prior to the meeting with the king, and then he would drink the pharmacist's strong poison, which would neutralize the weak poison. As his own poison he would bring water, which will have no effect on him, but the pharmacist who would drink the water, and then his poison would surely die. When the pharmacist figured out this plan, he decided to bring water as well. So the treasurer who drank poison earlier, drank the pharmacist's water, then his own water, and died of the poison he drank before. The pharmacist would drink only water, so nothing will happen to him. And because both of them brought the king water, he didn't get a strong poison like he wanted.

The Gold Chain -Interview Puzzle

2010-08-08T04:26:00.000-07:00

The Gold Chain - Puzzle :

A woman wants to buy a painting at an auction where you bid grams of gold instead of money. She owns a gold chain made of 23 interlocking loops, each weighing 1 gram. She wants to go to a jeweler before the auction to cut the minimum number of loops that would allow her to pay any sum from 1 to 23. For example, she could pay a 13 gram price with a 12 link chain and a single link. After much thought, she figures out a way to do it by cutting just 2 of the loops in the chain. How many loops are in the pieces of chains that she has after the 2 cuts?

Answer :

Make a cut in 4 th and loops of the Gold chain so you can get pieces of 1,1,3,6 and 12. from this you can weights from 1 to 23 grams.

Dragon and Knight - Trilogy interview puzzle

2010-08-08T04:08:00.000-07:00

Dragon and Knight - Trilogy interview puzzle

This is another one famous puzzle asked in many interview puzzle. This was asked in Trilogy interview.

Lets consider a dragon and knight live on an island. That island has seven poisoned wells, which is numbered 1 to 7. If you drink from a well, you can only save yourself by drinking from a higher numbered well. The Well whose is number 7 is located at the top of a high that mountain, so only the dragon can reach it.

One day they decide that the island isn't big enough for the two of them, and they have a duel. Each of them brings a glass of water to the duel, they exchange glasses, and drink. After the duel, the knight lives and the dragon dies.

Why did the knight live? Why did the dragon die?

Answer :

Dragon knows that knight cant reach well 7. So he thinks that if i give knight water from well 6 or 7, knight would die for sure. So he would get water from well 6 or 7 for knight.

Knight knew that no matter from which well he gives the water, dragon will go and drink from well 7 and live. So he got normal water for dragon and before duel drank water from well 1.

When they exchanged glasses and drank water, dragon rushed to well 7 and drank poison from it, thinking that it would cure the poison he just drank (but he drank normal water), so dragon died. Knight on the other hand already had poison from well 1 so what dragon gave him effectively cured him. So he lived.

comments if you have any other answers..

Blue-eyed Samurais Puzzle

2010-08-08T03:51:00.000-07:00

Puzzle :

A remote island contains a tribe whose members have either blue or brown eyes. Tribe members do not know the color of their eyes and if they learn it, they must kill themselves the same night. Every member of the tribe always behaves logically. So one day, a sailor visits the island and makes an observation that at least one of the members of the tribe has blue eyes. On the 10th night after that, all people with blue eyes kill themselves. How many blue-eyed people were there in the tribe?

Solution :

Ten. The logic goes as follows. If there were only one person with blue eyes on the island, then he would see that no one else has blue eyes and will kill himself on the first night. If there were two people with blue eyes, then one of the people with blue eyes would see that the second blue-eyed person did not kill himself on the first night, leading him to conclude that since he sees no one else with blue eyes, he must be one of them and will kill himself on the second night. The other blue-eyed person will follow the same logic and also kill himself on the second night. The same idea applies to the rest of the problem.

Comments if you have any other answers

Image Puzzles with Answers

2010-08-06T11:28:00.000-07:00

Puzzle Images

1.

Infosys Interview puzzles with answers

2010-08-06T11:07:00.000-07:00

Logical Puzzles

1. A clock with the hours round the face in Roman block numbers, as illustrated in the sketch fell down and the dial broke into four parts. The numerals in each part in every case summed to a total of 20. Can you show how the four parts of the clock face was broken?

2. Shown in the sketch are 6 matches. Can you rearrange them to make nothing?

3. Here is a sketch: Can you rearrange the position of the numbers 1 to 10 so that the sum of any two adjacent numbers is equal to the sum of the pair of numbers at the opposite ends of the diameter?

4. Can you place 10 coins in such a way that they lie in 5 straight lines and on each line there are 4 coins. There are at least two solutions.

5. Nine dots are arranged by 3 rows of 3 in the form of a square as shown in the sketch below:

Can you draw 4 straight lines, the second beginning where the first ends, the third beginning where the second ends and the fourth beginning where the third ends so that each dot is or at least one line?

Answers

Old Lady and her Eggs - Puzzle

2010-07-22T23:39:00.000-07:00

Old Lady and her Eggs - Puzzle

This puzzle can be solved based on simple math tactics..

Puzzle :

Once upon a time, There was an old lady has more number of eggs. She went to sell her vast quantity of eggs at the local market.
When a small boy asked how many she had, she replied:
Son, I can't count past 100 but I know that.
If you divide the number of eggs by 2 there will be one egg left.
If you divide the number of eggs by 3 there will be one egg left.
If you divide the number of eggs by 4 there will be one egg left.
If you divide the number of eggs by 5 there will be one egg left.
If you divide the number of eggs by 6 there will be one egg left.
If you divide the number of eggs by 7 there will be one egg left
If you divide the number of eggs by 8 there will be one egg left.
If you divide the number of eggs by 9 there will be one egg left.
If you divide the number of eggs by 10 there will be one egg left.
Finally. If you divide the Number of eggs by 11 there will be NO EGGS left.
How many eggs did the old lady have?

Einstein’s Riddle

2010-07-16T23:01:00.000-07:00

Einstein's Riddle - Interview puzzles

Einstein said that only 2% of the world could solve this problem. There are 5 houses in 5 different colors. In each house lives a person with a different nationality. The 5 owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet. No owners have the same pet, smoke the same brand of cigar, or drink the same beverage.

Hints:

· The Brit lives in the red house.
· The Swede keeps dogs as pets.
· The Dane drinks tea.
· The green house is on the left of the white house.
· The green homeowner drinks coffee.
· The person who smokes Pall Mall rears birds.
· The owner of the yellow house smokes Dunhill.
· The man living in the center house drinks milk.
· The Norwegian lives in the first house.
· The man who smokes Blend lives next to the one who keeps cats.
· The man who keeps the horse lives next to the man who smokes Dunhill.
· The owner who smokes Bluemaster drinks beer.
· The German smokes prince.
· The Norwegian lives next to the blue house.
· The man who smokes Blend has a neighbor who drinks water.
Who owns the fish?

Image puzzles with Answers - II

2010-07-08T04:24:00.000-07:00

Image puzzles with Answers

This puzzles are based on the images we need to find out the phrase. It shows the our identification based on the images. This is same as like the puzzles available in Image puzzles with Answers. The images are given below.

1.

Answers :

1. A) An outside chance b) Go for it c) Highway Overpass

2. A) Foreign Language b) Who’s in charge c) Neon Lights

3. A) I’m bigger than you b) Platinum c) Break in Custom

4. A) Ill in bed b) Safety in Numbers c) Read between the lines

5. A) Tip-toe through the tulips b) Alice in Wonderland c) Merry go round

Infosys interview Puzzles with Answers

2010-07-06T05:28:00.000-07:00

Infosys interview puzzles with Answers

Puzzle 1 : 9 cards are there. u have to arrange them in a 3*3 matrix. cards are of 4 colors.they are red,yellow,blue,green. conditions for arrangement: one red card must be in first row or second row.2 green cards should be in 3rd column. Yellow cards must be in the 3 corners only. Two blue cards must be in the 2nd row. Atleast one green card in each row.

Answer :

                        Yello     Red        Green
                         Blue     Blue        Green
                         Yellow Green     Yellow

Puzzle 2 : In a soap company a soap is manufactured with 11 parts. For making one soap you will get 1 part as scrap. At the end of the day u have 251 such scraps. From that how many soaps can be manufactured?

Answer : 22 + 2+ 1 = 25.

Puzzle 3 : There are five thieves, each loot a bakery one after the other such that the first one takes 1/2 of the total no. of the breads plus 1/2 of a bread. Similarly 2nd, 3rd,4th and 5fth also did the same. After the fifth one no. of breads remained are 3. Initially how many breads were there?

Answer : 31

Puzzle 4 : Two candles of equal lengths and of different thickness are there. The thicker one will last of six hours. The thinner 2 hours less than the thicker one. Ramesh light the two candles at the same time. When he went to bed he saw the thicker one is twice the length of the thinner one. For how long did Ramesh lit two candles .

Answer : 3 hours

Puzzle 5 : There are 5 persons who have won top 5 places in an event Olympics one of them asks all the 5 regarding their positions they reply as a: I am not the last b: c is in third place c: e is behind a d: b is in 1st place e: d is not the first find positions in order.

Answer : bdcae

Puzzle 6 : If one tyre of a car suddenly gets stolen…. and after sometime u find the tyre without the screws how will u make ur journey complete?

Answer : Open 3 screws, 1 from each tyre and fix the tyre.

Puzzle 7 : How will you recognize the magnet & magnetic material & non-magnetic material?

Answer : Drag one piece of material over another. There is no attractive force in the middle portion of the magnet. OR Get a piece of thread and tie up with the one bar and check for poles. If it iron bar then it moves freely and if it is magnetic bar then it fix in one direction according to poles.

Puzzle 8 : You are given a cake; one of its corner is broken. How will u cut the rest into Two equal parts?

Answer : Slice the cake

Puzzle 9 : What is the height of room if after entering the room with a watch ur head strikes a hanging bulb?
Answer: Oscillate the hanging bulb. Calculate the time period for one complete oscillation by Simple Harmonic Motion (SHM) of the handing bulb. Put it in the
formula T=2 * 3.14 * (L/G)^1/2

L will be the length of the hanging thread.
Add the L with ur height to get the height of the room.

OR
Drop it from the room and find the time at which it strikes the floor. Using physics formula s = (at^2)/2 (IM NOT SURE ABOUT THIS ONE)
Puzzle 10 : Can u make 120 with 5 zeros?
Answer : Factorial (factorial (0)+factorial (0)+factorial (0)+factorial (0)+factorial (0)) = 120

Puzzle 11 : There are three people A, B, C. Liars are of same type and Truth speaking people are of same type. Find out who is speaking truth and who is speaking false from the following statements:

a) A says: B is a liar.
b) B says: A and C are of same type.

Answer : lets assume A is speaking truth. It means B is a liar then it means A and C are not of same type.

Puzzle 12 : Ann, Boobie, Cathy and Dave are at their monthly business meeting. Their occupations are author, biologist, chemist and doctor, but not necessarily in that order. Dave just told the biologist that Cathy was on her way with doughnuts. Ann is sitting across from the doctor and next to the chemist. The doctor was thinking that Boobie was a goofy name for parent's to choose,but didn't say anything. What is each person's occupation?

Answer: Since Dave spoke to the biologist and Ann sat next to the chemist and across the doctor, Cathy must be the author and Ann the biologist. The doctor didn't speak, but David did, so Bobbie is the doctor and Dave the chemist.

Paul Sloane's list of Classic Lateral Thinking Puzzles with Answers

2010-06-28T06:29:00.000-07:00

The Puzzles

Puzzle 1. The man in the Elevator

A man lives on the tenth floor of a building. Every day he takes the elevator to go down to the ground floor to go to work or to go shopping. When he returns he takes the elevator to the seventh floor and walks up the stairs to reach his apartment on the tenth floor. He hates walking so why does he do it?

This is probably the best known and most celebrated of all lateral thinking puzzles. It is a true classic. Although there are many possible solutions which fit the initial conditions, only the canonical answer is truly satisfying.

Answer : The man is (of course) a dwarf. Variants of this puzzle include the clue that on rainy days he goes up in the elevator to the tenth floor (he uses his umbrella!)

Puzzle 2. 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.

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.

Answer : 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 simple puzzle to state 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.

Puzzle 3. The Man who Hanged Himself

Not far from Madrid, there is a large wooden barn. The barn is completely empty except for a dead man hanging from the middle of the central rafter. The rope around his neck is ten feet long and his feet are three feet off the ground. The nearest wall is 20 feet away from the man. It is not possible to climb up the walls or along the rafters. The man hanged himself. How did he do it?

Answer : He climbed on a block of ice which has since melted.
This one is often stated with the clue of a puddle of water, but surely this is too much assistance. It is one of several problems which depend on the change of state of water (snow or ice to water or steam).

Puzzle 4. Death in a Field

A man is lying dead in a field. Next to him there is an unopened package. There is no other creature in the field. How did he die?

Answer : The man had jumped from a plane but his parachute had failed to open. It is the unopened package. This is sometimes given with the following rather elegant clue - as he approached the centre of the field he knew he was going to die. This is another of the top classics which is right up there with 'The Man in the Bar'. If the solver is thinking along the wrong lines (i.e. in the two dimensions of the ground) then the lateral jump to the third dimension can be tough to make.

Puzzle 5. Anthony and Cleopatra

Anthony and Cleopatra are lying dead on the floor of a villa in Egypt. Nearby is a broken bowl. There is no mark on either of their bodies and they were not poisoned. How did they die?

Answer : Anthony and Cleopatra were goldfish whose bowl was knocked over by a clumsy dog. This is one of a set of puzzles which deceive by using human names for animals. This is not a very satisfactory basis for a good puzzle but despite that, the puzzle has enduring popularity.

Puzzle 6. The Coal, Carrot and Scarf

Five pieces of coal, a carrot and a scarf are lying on the lawn. Nobody put them on the lawn but there is a perfectly logical reason why they should be there. What is it?

Answer : They were used by children who made a snowman. The snow has now melted. Another change of state puzzle. After this you should be on the look-out for them!

Puzzle 7. 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?

Answer : 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?

Puzzle 8. Push that Car

A man pushed his car. He stopped when he reached a hotel at which point he knew he was bankrupt. Why?

Answer : He was playing Monopoly.

Puzzle 9. The Arm of the Postal Service

One day a man received a parcel in the post. Carefully packed inside was a human arm. He examined it, repacked it and then sent it on to another man. The second man also carefully examined the arm before taking it to the woods and burying it. Why did they do this?

This one probably has more variations than any other. A great one to puzzle out. It requires plenty of good questions.

Answer : The three men had been stranded on a desert island. Desperate for food, they had agreed to amputate their left arms in order to eat them. They swore an oath that each would have his left arm cut off. One of them was a doctor and he cut the arms off his two companions. They were then rescued. But his oath was still binding so he later had to have his arm amputated and sent to his colleagues.

This is often told with a further twist whereby a doctor pays a tramp a large sum in order to amputate the tramp's arm which the doctor then sends to another man who inspects it etc. This variation can make for a long night of questioning!

Puzzle 10. Heaven

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

Answer : 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!

Mice Puzzle

2010-06-25T05:08:00.000-07:00

Mice Puzzle :

Do you Know Mice can multiply very fast… This puzzle based on the Mice multiplication. Consider an Adult mice can give birth once every month. The baby mice grow into adult mice in just two months after they are also born. This is the condition for this puzzle.

Puzzle Question: If you would buy a baby mouse just after it was born, how many mice would you have in 10 months?

Answer :
Only One. You cannot have more mice if you start with one!

2 Egg Puzzle Secrets

2010-06-21T22:22:00.000-07:00

2 Egg Puzzle

This is another 2 egg puzzle same as like 2 egg puzzle asked by google. Here the number of storey is different instead of 100 storey it is 36. The puzzle is given as

Suppose that we wish to know which windows in a 36-story building are safe to drop eggs from, and which will cause the eggs to break on landing. We make a few assumptions:

An egg that survives a fall can be used again.
A broken egg must be discarded.
The effect of a fall is the same for all eggs.
If an egg breaks when dropped, then it would break if dropped from a higher window.
If an egg survives a fall then it would survive a shorter fall.

It is not ruled out that the first-floor windows break eggs, nor is it ruled out that the 36th-floor windows do not cause an egg to break.

If only one egg is available and we wish to be sure of obtaining the right result, the experiment can be carried out in only one way. Drop the egg from the first-floor window; if it survives, drop it from the second floor window. Continue upward until it breaks. In the worst case, this method may require 36 droppings. Suppose 2 eggs are available. What is the least number of egg-droppings that is guaranteed to work in all cases?

Solution :

The number of drop required is 8.

First i have to start dropping the egg on 8th floor, if not breaks, then 15th floor, if not breaks 21st floor...

the series goes like this...

8,15,21,26,30,33,35,36...

i.e (8), (8+7), (8+7+6), (8+7+6+5), ....(8+7+6+5+4+3+2+1)

so in general,

the minimum value of N such that,

n(n+1)/2 >=36

so n= 8.

let us assume it breaks in 20th floor...

put down from 8 th floor - will NOT be broken

put down from 15th floor - will NOT be broken

put down from 21st floor - it will be broken.

then u have one egg remaining.

start putting from 16th floor.. ie.. 16, 17,18,19, 20

19th floor it will not be broken and

at 20th floor it will be broken.

even for the worst case scenario the maximum number of droppings required is 8.

so the answer is 8!

Moshe Sniedovich from Department of Mathematics and Statistics, The University of Melbourne Parkville, Australia placed a paper in journal regarding this 2 egg problem. This is famous among math teachers.

According to that paper they proposed the problem , problem introduction , Problem analysis Best case secnario , worst case secnario etc..

Famous 2-egg puzzle from an OR/MS perspective and explore the structure of the optimal policies for this puzzle. It is shown that the puzzle provides a vivid environment for teaching/learning basic concepts related to decision making under uncertainty, including the differences and similarities between the Worst Case and Expected Value scenarios. The more general problem where N test eggs -rather than 2- are provided is also examined.

In that paper the proposed solution for 2 egg with N storey buildings they also solved the problem for N egg for N story building in appendix. All the solution are math oriented. Click here to download the paper as pdf.

Family and Dogs Puzzle

2010-06-21T02:41:00.000-07:00

Family and Dogs Puzzle :

Consider there are five families and their five dogs (each family member owned one of the dogs) were hiking when they encountered a river to cross. They rented a boat which could hold three living things people or dogs. Unfortunately, the dogs were tempermental: each was comfortable only with its owner and could not be near another (not even momentarily) unless its owner was present. Dogs could be with other dogs however. The crossing would have been impossible except that Lisa’s dog had attended a first-rate obedience school and knew how ro operate the boat. No other dogs were that well-educated. How was the crossing arranged and how many trips did it take?

Answer :
Solution is somewhat similar to River puzzle which is already present in this blog. Try to find and Comment your answers...

Interview Riddles

2010-06-13T22:51:00.000-07:00

These questions are riddles. What do you do if you get this type of question? First, assess what quality the interviewer is looking for. Secondly, decide whether or not you want to work for a company who’s famous at throwing you tricks. Some companies will ask you these kinds of questions to see if you’re prepared to handle off-the-wall kinds of questions and answers you may get when on the job. They also might want to assess how quickly you are on your feet. Or, they may want to see how you go through a line of thinking.

Frequently Asked famous Interview Riddles :

1. You wake up one morning and there’s been a power outage. You know you have 12 black socks and 8 blue ones. How many socks do you need to pull out before you’ve got a match?

2. Why are manhole covers round?

3. You’re trying to get to Truthtown. You come to a fork in the road. One road leads to Truthtown (where everyone tells the truth), the other to Liartown (where everyone lies). At the fork in the road is a man from one of those towns -- but which one? You get to ask him one question to discover the way. What’s the question?

4. How many cubes are at the center of a Rubik’s Cube?

5. You have two containers, one holds five gallons, the other holds three. You can have as much water as you want. Measure exactly four gallons of water into the five gallon container.

6. The government is building a highway through your neighborhood and you’re forced to sell your home. How do you arrive at your asking price?

7. There are eight balls, one of which is slightly heavier than the others. You have a two-armed scale, which you are allowed to use only twice. Find the ball that’s heavier.

8. Calculate the number of degrees between the hour hand and the minute hand of a clock (nondigital) that reads 3:15.

9. How many barbers are there in Chicago?

10. You are in solitary confinement. It is Friday afternoon and you absolutely must have a cigarette. The only person who can give you one is the guard outside your cell. What do you do?

Interview Riddles - Answers

1. Pick three -- there are only two colors.

2. Manhole covers are round so that they don’t fall into the manholes.

3. To find the way to Truthtown, ask the man, “Which way is your hometown?” Then, go whichever way he points: If he’s from Liartown, he’ll point to Truthtown, and if he’s from Truthtown, he’ll point to Truthtown.

4. There is only one cube at the center of a Rubik’s Cube.

5. Fill up the three-gallon container and pour it into the five-gallon container. Do it again -- there will be one gallon left in the three-gallon container. Empty the five, pour in the one, fill the three again and pour it into the firve-gallon container -- you have four!

6. Calculate the price of your home using conventional valuation methods - but remember to throw in the value you attach to your memories for however long you’ve lived there.

7. Put three balls on each side of the scale. If the arms are equal, you know the heavy ball is one of the two remaining. If the arms are unequal, take the three balls on the heavier side, pick two and weigh them against each other.

8. The hour hand will have moved one-fourth of an hour; therefore, there will be 7.5 degrees between the two hands.

9. The variables you’ll want to consider are the population of Chicago and the percentage that’s male; the number of haircuts the average male has per year divided by the number of days in the year, taking into account the number of days per year baber shops are open; and the number of haircuts an average barber can give per day. At the time of writing this up, there are 550 barber shops in Chicago; 6273 active barbers in Illinois; with 66% of the state’s population, Chicago has roughly 4140 barbers.

10. To get the guard to give you a cigarette (and this really is the preferred answer to this question), threaten to kill yourself by smashing your head against the wall of your cell. That gives you leverage with the guard - he’d be tied up by doing paperwork about your suicide, so he’d miss weekend time with his family (it’s Friday afternoon, remember?) -- so he’ll give you a cigarette.

MACHINE WEIGHING - COIN Puzzle

2010-06-08T21:21:00.000-07:00

Coin Puzzle :

This is another coin puzzle. As per the puzzle you have 20 coin machines, each of which produce the same kind of coin. you know how much a coin is supposed to weigh. one of the machines is defective, in that every coin it produces weighs 1 ounce less than it is supposed to. you also have an electronic weighing machine. how can you determine which of the 20 machines is defective with only one weighing? (by one use, we mean you put a bunch of stuff on the machine and read a number, and that's it -- you not allowed to accumulate weight onto the machine and watch the numbers ascend, because that's just like multiple weighings). you are allowed to crank out as many coins from each machine as you like.

This puzzle is like another coins puzzle (Box of Defective balls...) But It's solution is somewhat different. In other coins puzzle you can weigh number of times. But here it is only once. Try to think differently and post your answers..

Answer :

In case of normal coin puzzle they ask number if times required to use the balance to find the answer. Here it is different. Here you can use only one time the balance.

The solution is to take the coins in the following order. Take one coin from the first machine, two coins from the second machine, three coin from the third mechine etc.. So that totally you can take
(20 * (20+1) )/2 = 210 coins. Lets consider each coins have 10 ounce weight except one machine coins. So totally you have 2100 ounce weight for 210 coins. If you weigh it now you can get less weight because one machine produces less weight. From the weight you can find the machine which produces less weight coins. Suppose the resultant weight is 2099 then the first machine is fault. If it is 2098 then second machine is fault. That is the fault machine number = 2100 - Obtained weight from balance.

The Warden and 23 prisoners - Google interview puzzle

2010-06-05T01:06:00.000-07:00

The Warden and 23 prisoners

Puzzle :

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?

Solution:

The team nominates a leader. The group agrees upon the following rules:

The leader is the only person who will announce that everyone has visited the switch room. All the prisoners (except for the leader) will flip the first switch up at their very first opportunity, and again on the second opportunity. If the first switch is already up, or they have already flipped the first switch up two times, they will then flip the second switch. Only the leader may flip the first switch down, if the first switch is already down, then the leader will flip the second switch. The leader remembers how many times he has flipped the first switch down. Once the leader has flipped the first switch down 44 times, he announces that all have visited the room.

It does not matter how many times a prisoner has visited the room, in which order the prisoners were sent or even if the first switch was initially up. Once the leader has flipped the switch down 44 times then the leader knows everyone has visited the room. If the switch was initially down, then all 22 prisoners will flip the switch up twice. If the switch was initially up, then there will be one prisoner who only flips the switch up once and the rest will flip it up twice.

The prisoners can not be certain that all have visited the room after the leader flips the switch down 23 times, as the first 12 prisoners plus the leader might be taken to the room 24 times before anyone else is allowed into the room. Because the initial state of the switch might be up, the prisoners must flip the first switch up twice. If they decide to flip it up only once, the leader will not know if he should count to 22 or 23.

In the example of three prisoners, the leader must flip the first switch down three times to be sure all prisoners have visited the room, twice for the two other prisoners and once more in case the switch was initially up.

Amazon interview puzzle - Find Age Of Children

2010-05-24T21:16:00.000-07:00

This is another famous puzzle asked in many IT company writtern tests and interviews. The puzzle is asked by amazon is given below. This puzzle belongs to the category of logic puzzles.

Amazon interview puzzle - Find Age Of Children

A man walks into a bar, orders a drink, and starts chatting with the bartender. After a while, he learns that the bartender has three children. "How old are your children?" he asks. "Well," replies the bartender, "the product of their ages is 72." The man thinks for a moment and then says, "that's not enough information." "All right," continues the bartender, "if you go outside and look at the building number posted over the door to the bar, you'll see the sum of the ages." The man steps outside, and after a few moments he reenters and declares, "Still not enough!" The bartender smiles and says, "My youngest just loves strawberry ice cream." How old are the children? A variant of the problem is for the sum of the ages to be 13 and the product of the ages to be the number posted over the door. In this case, it is the oldest that loves ice cream. Then how old are they?

Answer :

First, determine all the ways that three ages can multiply together to get 72:

72 1 1 (quite a feat for the bartender)
36 2 1
24 3 1
18 4 1
18 2 2
12 6 1
12 3 2
9 4 2
9 8 1
8 3 3
6 6 2
6 4 3

As the man says, that's not enough information; there are many possibilities. So the bartender tells him where to find the sum of the ages--the man now knows the sum even though we don't. Yet he still insists that there isn't enough info. This must mean that there are two permutations with the same sum; otherwise the man could have easily deduced the ages.

The only pair of permutations with the same sum are 8 3 3 and 6 6 2, which both add up to 14 (the bar's address). Now the bartender mentions his "youngest"--telling us that there is one child who is younger than the other two. This is impossible with 8 3 3--there are two 3 year olds. Therefore the ages of the children are 6, 6, and 2.

Pedants have objected that the problem is insoluble because there could be a youngest between two three year olds (even twins are not born exactly at the same time). However, the word "age" is frequently used to denote the number of years since birth. For example, I am the same age as my wife, even though technically she is a few months older than I am. And using the word "youngest" to mean "of lesser age" is also in keeping with common parlance. So I think the solution is fine as stated.

In the sum-13 variant, the possibilities are:
11 1 1
10 2 1
9 3 1
9 2 2
8 4 1
8 3 2
7 5 1
7 4 2
7 3 3
6 6 1
6 5 2
6 4 3

The two that remain are 9 2 2 and 6 6 1 (both products equal 36). The final bit of info (oldest child) indicates that there is only one child with the highest age. This cancels out the 6 6 1 combination, leaving the childern with ages of 9, 2, and 2.

Puzzle Types

2010-05-19T21:06:00.000-07:00

Induction Puzzles or Logical Puzzles:

Induction Puzzles are logic puzzles. In most cases, the puzzle's scenario will involve several participants with reasoning capability (typically people) and the solution to the puzzle will be based on identifying what would happen in an obvious case, and then repeating the reasoning that: "as soon as one of the participants realizes that the obvious case has not happened, they can eliminate it from their reasoning, so creating a new obvious case".

Typical tell-tale features of these puzzles include any puzzle in which each participant has a given piece of information about all other participants but not themselves. Also, usually some kind of hint is given to suggest that the participants can trust each others intelligence.

               Examples :
                            1. Prisoners and hats puzzle.
                            2. 100 Prisoners in Solitary cells
               Reference :
                            1. http://en.wikipedia.org/wiki/Induction_puzzles

Dissection Puzzles

A dissection puzzle, also called a transformation puzzle, is a tiling puzzle where a solver is given a set of pieces that can be assembled in different ways to produce two or more distinct geometric shapes. The creation of new dissection puzzles is also considered to be a type of dissection puzzle. Puzzles may include various restraints, such as hinged pieces, pieces that can fold, or pieces that can twist. Creators of new dissection puzzles emphasize using a minimum number of pieces, or creating novel situations, such as ensuring that every piece connects to another with a hinge.

                 Examples :
                               1. Stick Matches Image Puzzle

                 Reference :
                               1. http://en.wikipedia.org/wiki/Dissection_puzzle

Lateral Thinking Puzzles

Lateral thinking puzzles are often strange situations which require an explanation. They are solved through a dialogue between the quiz master who sets the puzzle and the solver or solvers who try to figure out the answer. The puzzles as stated generally do not contain sufficient information for the solver to uncover the solution. So a key part of the process is the asking of questions. The questions can receive one of only three possible answers - yes, no or irrelevant.

When one line of enquiry reaches an end then another approach is needed, often from a completely new direction. This is where the lateral thinking comes in.

Some people find it frustrating that for any puzzle it is possible to construct various answers which fit the initial statement of the puzzle. However, for a good lateral thinking puzzle, the proper answer will be the best in the sense of the most apt and satisfying. When you hear the right answer to a good puzzle of this type you should want to kick yourself for not working it out!

This kind of puzzle teaches you to check your assumptions about any situation. You need to be open-minded, flexible and creative in your questioning and able to put lots of different clues and pieces of information together. Once you reach a viable solution you keep going in order to refine it or replace it with a better solution. This is lateral thinking!

Prisoners and hats puzzle

2010-05-19T20:56:00.000-07:00

The prisoners and hats puzzle is an induction puzzle (a kind of logic puzzle) that involves reasoning about the actions of other people, drawing in aspects of Game theory. Generaly interview questions asked from math oriented. Companies expect that IT students have the backround knowledge of problem solving. This puzzle comee under game theory.

Puzzle : Prisoners and three hats

Four prisoners are arrested for a crime, but the jail is full and the jailer has nowhere to put them. He eventually comes up with the solution of giving them a puzzle so if they succeed they can go free but if they fail they are executed.

The jailer puts three of the men sitting in a line. The fourth man is put behind a screen (or in a separate room). He gives all four men party hats (as in diagram). The jailer explains that there are two red and two blue hats. The prisoners can see the hats in front of them but not on themselves or behind. The fourth man behind the screen can't see or be seen by any other prisoner. No communication between the men is allowed.

If any prisoner can figure out and say (out loud) to the jailer what colour hat he has on his head all four prisoners go free. The puzzle is to find how the prisoners can escape.

The solution

For the sake of explanation let's label the prisoners in line order A B and C. Thus B can see A (and his hat colour) and C can see A and B.

The prisoners know that there are only two hats of each colour. So if C observes that A and B have hats of the same colour, C would deduce that his own hat is the opposite colour. However, If A and B have hats of different colours, then C can say nothing. The key is that prisoner B, after allowing an appropriate interval, and knowing what C would do, can deduce that if C says nothing the hats on A and B must be different. Being able to see A's hat he can deduce his own hat colour. (The fourth prisoner is irrelevant to the puzzle: his only purpose is to wear the fourth hat).

Puzzle : Prisoners and Four - hats.
Now the same puzzle with different condition. Instead of two red and two blue hats there are 3 hats of one colour and only 1 hat of another, and the 3 prisoners can see each other i.e. A sees B & C, B sees A & C and C sees A & B. As like shown in diagram (D again not to be seen and only there to wear the last hat)

The solution

There are two cases:

In first case, one of the three prisoners (in A, B, C) wears the single off-colour hat, thus the other two can easily deduce the colour of theirs. For example if A is single color then B can see that C and A have different color hats. From this he concludes that his color is not single so he shouts his color.

In second case, the three prisoners wear hats of the same colour, while D wears the off-colour hat. After a while, all four prisoners should be able to deduce that, since none of the others was able to state the colour of his own hat, D must wear the off-colour hat.

Puzzle : Prisoners and Five - hats.
Another different condition. only three prisoners and five hats (supposedly two black and three white) are involved. The three prisoners are ordered to stand in a straight line facing the front, with A in front and C at the back. They are told that there will be two black hats and three white hats. One hat is then put on each prisoner's head; each prisoner can only see the hats of the people in front of him and not on his own's. The first prisoner that is able to announce the colour of his hat correctly will be released. No communication between the prisoners is allowed. After some time, only A is able to announce (correctly) that his hat is white. Why is that so?

The solution

Assuming that A wears a black hat.

• If B wears a black hat as well, C can immediately tell that he is wearing a white hat after looking at the two black hats in front of him.

• If B does not wear a black hat, C will be unable to tell the colour of his hat (since there is a black and a white). Hence, B can deduce from A's black hat and C's response that he (B) is not wearing a black hat (otherwise the above situation will happen) and is therefore wearing a white hat.

This therefore proves that A must not be wearing a black hat.

Hundred numbers in a sheet puzzle

2010-05-16T21:03:00.000-07:00

This is another one frequently asked puzzle in writtern test taken by IT companies. Anyone can easily understand and try to answer this puzzle. A simple logic is behind this puzzle try to find out.

Puzzle :

A sheet of paper has statements numbered from 1 to 100. Statement n says "exactly n of the statements on this sheet are false." Which statements are true and which are false? What if we replace "exactly" by "at least"?

Answer :

In these 100 statements one and only one statement can be correct. That means 99 statements have to be incorrect and thus statement 99th is true

If replaced by "at least", and the "real" number of false statements is x, then statements x+1 to 100 will be false (since they falsely claim that there are more false statements than there actually are). So, 100-x are false, ie. x=100-x, so x=50. The first 50 statements are true, and statements 51 to 100 are false.

However, there is a hidden and incorrect assumption in this argument. To see this, suppose that there is one statement on the sheet and it says "One statement is false" or "At least one statement is false," either way it implies "this statement is false," which is a familiar paradoxical statement. We have learned that this paradox arises because of the false assumption that all statements are either true or false. This is the hidden assumption in the above reasoning.

If it is acknowledged that some of the statements on the page may be neither true nor false (i.e., meaningless), then nothing whatsoever can be concluded about which statements are true or false.

This problem has been carefully contrived to appear to be solvable (like the vacuous statement "this statement is true"). By changing the numbers in some statements and changing "true" to "false," various circular forms of the liar's paradox can be constructed. A much more complicated version of the same problem is:

1) At least one of the last two statements in this list is true.

2) This is either the first true or the first false statement in the list.

3) There exist three consecutive false statements.

4) The difference beween the numbers of the last true statement and the first true statement is a factor of the unknown number.

5) The sum of the numbers of the true statements is the unknown number.

6) This is not the last true statement.

7) Each true statement's number is a factor of the unknown number.

8) The unknown number equals the percentage of these statements which are true.

9) The number of different factors which the unknown number has (excluding 1 and itself) is more than the sum of the numbers of the true statements.

10) There are no three cosecutive true statements.

What is the number?

The incorrect but plausible solution is:

By 2, either way 1 must be false, and then so must both 9 and 10.

6, if false, says "This is the last true statement", which gives a paradox, thus 6 must be true, and so must 7 and/or 8.

7 and 8 cannot both be true, as the number had to be a multiple of 6,7,8 , that is a multiple of 168 (by 7), and less than 100 (by 8)

If 8 is false, then 3 is true (8,9,10 is false), if 8 is true, then 3 is false (3 cannot be true when both 6 and 8 are true, then there are no three consecutive statements left).

So we have either

A) F X F X X T F T F F or B) F X T X X T T F F F

In A), 4 and 5 must be true (by false 10), and 2 may be true or false. So by 5 the number shall be either 27 (2+3+4+5+6+7) or 25 (3+4+5+6+7). None of these can fullfill 8, though, so A) is out leaving us with B)

Now, (by 7) 3,6 and 7 are factors, the number must be a multiple of 42, as 2+3+4+5+6+7=27, 5 must be false.

By false 10, 2 and 4 must be true, that is 5 shall be a multiple of the number. Now the number must be a multiple of 2,3,4,5,6 and 7, that is a multiple of 3*4*5*7=420. 420 has 22 different factors (2,3,4,5,6,7,10,12,14,15,20,21,28,30,35,42,60,70,84,105,140,210) and the sum 2+3+4+6+7 = 22, so the only multiple of 420 that fulfills false 9 is 420.

If you know any other answers please comment here..

Stick Matches Image puzzles

2010-05-11T21:01:00.000-07:00

Stick Matches Image puzzles.

Move 1 stick from the image below to make a valid equation. (inequalities are not allowed).

Solution:

Google Interview puzzles with answers:

2010-05-09T03:09:00.000-07:00

Google Interview puzzles with answers:

Puzzle 1:

   The puzzle is if the shopkeeper can only place the weights in one side of the common balance. For example if shopkeeper has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names the weights you will need to measure all weights from 1 to 1000. This is a fairly simple problem and very easy to prove also. Answer for this puzzle is given below.

Solution :

   This is simply the numbers 2^0,2^1,2^2 ... that is 1,2,4,8,16... So for making 1000 kg we need up to 1, 2, 4, 8, 16, 32, 64, 128, and 512. Comments your suggestions or other answers.

Puzzle 2:

   This is same as the above puzzle with the condition of placing weights on only side of the common balance being removed. You can place weights on both side and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg. Answer for this puzzle is given below.

Solution:

   For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

Image puzzles with answers:

2010-05-04T08:23:00.000-07:00

Image puzzles with answers

In these types of puzzles, a word is behind on the image. We need to find out the word from the image given. The images are not difficult but the images symbolically represent the words behind it. Generally these types of puzzles are asked in general puzzle competitions. But now few IT companies also asked these questions in their written test.

Example:

From the above image we can easily read that “hi way pass” surrounded by red line. But the answer is “highway overpass” because the word “hiway” is above the word “pass” so make these two words to form the phrase “highway overpass”. Similarly try to answer the following images.

Image puzzles:

Answers :

     1 . a) painless operation        b) Excuse me     c) play in the yard
     2.   a) Life after death           b) crossbox          c) partly cloud
     3.   a) An after though           b) I'll be there in a minite          c) Tom apart
     4.   a) One way or the other   b) Too big too ignore             c) Easy on the eyes
     5.   a) hitting below the belt     b) Lie in wait                       c) search high and low

Adobe Interview puzzles with answers

2010-05-04T08:14:00.000-07:00

Adobe Interview puzzles with answers:

The given puzzles are asked in many written exams conducted by varies IT companies like Google, Adobe, Amazon, yahoo, Microsoft etc…

Adobe Interview Puzzles : Weight Puzzle

Puzzle 1 : (check this is already posted)

   There are 8 Balls available in a box, From that only one odd ball being heavier or Lighter. question is to find out the minimum number of weighings required to spot the odd heavier ball among 8 identically looking balls using a common balance.

Solution :

   This is a little trickier to solve.Let us name the 8 balls as A1,A2,A3,B1,B2,B3,C1 and C2.Now weigh with A1,A2,A3 on one side and B1,B2,B3 on the other side. If both weigh equal then the odd ball is among C1 and C2.Now we know that A's and B's are all normal ones we can weigh C1 with A1 and check whether C1 weighs the same as the normal ball. In this way, we can figure out in one weigh which of C1 and C2 is odd. Now if A's weighed more than B's then we know for sure C's are normal ones. Now let’s assume A's heavier than B's and we still don't know whether the odd is among A's or B's.
We know A1 A2 A3 < B1 B2 B3
Now we compare A1 B2 B3 to B1 C1 C2 . Now if the A1 B2 B3 is heavier than B1 C1 C2 Then it means the odd ball is among A1 and B1.If A1 B2 B3 is lighter than B1 C1 C2 then the odd ball is among B2 and B3.If A1 B2 B3 equal to B1 C1 C2 then odd ball is among A2 and A3.So we have zeroed down to 2 balls.Now its very easy as we can compare any of the normal ball with one of the 2 balls.So answer is 3,ie in 3 weighings we can find out the odd ball.

Puzzle 2:

   This is same like the above weight puzzle But instead of 8 balls here You are given 13 balls. The odd ball may be either heavier or lighter. Find out the odd ball in 3 weightings.

Puzzle 3:
   You are given a cake; one of its corner is broken. How will u cut the rest into Two equal parts?

Solution :
Slice the cake(Horigentally).

Puzzle 4 :
   How it is possible to place four points that are equidistance from each other?

Solution :
   place points in the shape of a pyramid.

Find House Address puzzle

2010-04-30T22:02:00.000-07:00

Find House Address puzzle

Puzzle :

This is another one rarely asked puzzle in interviews. The puzzle is Sally and Sue have a strong desire to date Sam. They all live on the same street yet neither Sally nor Sue knows where Sam lives. The houses on this street are numbered 1 to 99. Sally asks Sam ”Is your house number a perfect square?” He answers. Then Sally asks”Is is greater than 50?” He answers again. Sally thinks she now knows the address of Sam’s house and decides to visit. When she gets there, she finds out she is wrong. This is not surprising, considering Sam answered only the second question truthfully. Sue, unaware of Sally’s conversation, asks Sam two questions. Sue asks ”Is your house number a perfect cube?”. He answers. She then asks”Is it greater than 25?”. He answers again. Sue thinks she knows where Sam lives and decides to pay him a visit. She too is mistaken as Sam once again answered only the second question truthfully. If I tell you that Sam’s number is less than Sue’s or Sally’s, and that the sum of their numbers is a perfect square multiplied by two, you should be able to figure out where all three of them live. solution for this puzzle is given below. Write your comments.. and any different answers.

Answer :

As per the above puzzle Sally asks Sam "Is your house number a perfect square?". He answers. Then Sally asks "Is is greater than 50?". He answers again. Sally thinks she now knows the address of Sam's house and decides to visit. Since Sally thinks that she has enough information, I deduce that Sam answered that his house number was a perfect square greater than 50. There are two of these {64,81} and Sally must live in one of them in order to have decided she knew where Sam lives. When she gets there, she finds out she is wrong. This is not surprising, considering Sam answered only the second question truthfully. So Sam's house number is greater than 50, but not a perfect square. Sue, unaware of Sally's conversation, asks Sam two questions. Sue asks "Is your house number a perfect cube?". He answers. She then asks "Is it greater than 25?". He answers again. Observation: perfect cubes greater than 25 are {27, 64}, less than 25 are {1,8}. Sue thinks she knows where Sam lives and decides to pay him a visit. She too is mistaken as Sam once again answered only the second question truthfully. Since Sam's house number is greater than 50, he told Sue that it was greater than 25 as well. Since Sue thought she knew which house was his, she must live in either of {27,64}. If I tell you that Sam's number is less than Sue's or Sally's, Since Sam's number is greater than 50, and Sue's is even bigger, she must live in 64. Assuming Sue and Sally are not roommates (although awkward social situations of this kind are not without precedent), Sally lives in 81. and that the sum of their numbers is a perfect square multiplied by two, you should be able to figure out where all three of them live. Sue + Sally + Sam = 2 p^2 for p an integer 64 + 81 + Sam = 2 p^2 Applying the constraint 50 < Sam < 64, looks like Sam = 55 (p = 10). In summary, Sam = 55 Sue = 64 Sally = 81

I hope all of you understand puzzle with answers.. If your are not understand comments your queries..

Microsoft interview puzzles : Aeroplane

2010-04-24T07:01:00.000-07:00

Aeroplane.

Puzzle :

The puzzle question is : On Bagshot Island, there is an airport. The airport is the homebase of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel.

What is the fewest number of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer?

Notes:

(a) Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground.

(b) Each airplane must have enough fuel to return to airport.

(c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.)

(d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world)

solution for this puzzle is given below. Write your comments.. and any different answers.

Answer :

As per the puzzle given ablove The fewest number of aircraft is 3! Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refuelled and starts immediately again to follow A and C.

C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first "auxiliary" aircraft reaches it in time in order to refuel it, and both "auxiliary" aircraft are the able to return safely to the home base.

Now in the same manner as before both B and C fully refuelled fly towards A. Again B refuels C and returns home to be refuelled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refuelling process is applied analogously as for the first phase of the flight.

I hope all of you understand puzzle with answers.. If your are not understand comments your queries..

Four People on a Rickety Bridge

2010-04-19T08:43:00.000-07:00

Four People on a Rickety Bridge

About this Puzzle :

This is one of the frequently asked puzzle in writtern test and interview conntected by top IT companies like Google, Yahoo, Microsoft , Adobe , Syemantec , Thought works , CA, Macafee etc.. The puzzle question and its answer is given below.

Puzzle Question :

The puzzle is Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?

solution for this puzzle is given below. Write your comments.. and any different answers.

Answer :

Solution for the above puzzle is 17 minites. As per the puzzle to get everyone across in 17 minutes, we need get the two slowest people across together; otherwise we are wasting too much time. Once we get them across, how do we not make one of them walk back with the flashlight? Just have one of the faster people already there waiting to sprint the flashlight back across.

person A: 1 minute
person B: 2 minutes
person C: 6 minutes
person D: 10 minutes

1. A & B cross. total time: 2 minutes.

C || A
D || B
   || flashlight

2. B comes back. total time: 4 minutes.

C || A
D ||
B ||
flashlight

3. C & D cross. total time: 14 minutes.

B || A
   || C
   || D
   flashlight

4. A comes back. total time: 15 minutes.

A || C
B || D
   ||
flashlight

5. A & B cross. total time: 17 minutes.

   || A
   || B
   || C D
   flashlight

Another valid solution is to have A bring the flashlight back in step 2.

Q*** 10 10 2 1*** Sol- 17
Q*** 10 9 2 1 *** Sol- 17
Q*** 10 7 2 1 *** Sol- 17
Q*** 10 7 2 1 *** Sol- 17
Q*** 10 6 2 1 *** Sol- 17
Q*** 10 5 2 1 *** Sol- 17
Q*** 10 4 2 1 *** Sol- 17
Q*** 10 3 2 1 *** Sol- 17
Q*** 10 2 2 1 *** Sol- 17
**Quick Sol**- Suppose a,b,c,d are the take time one can cross the bridge such that a>b>c>d or a>b=c>d or a=b>c>d or a=b=c>d or a=b=c=d
Genreric Sol- a+3*c+d (independent of b. :) )

I hope all of you understand puzzle with answers.. If your are not understand comments your queries..

Find Defective ball

2010-04-18T06:02:00.000-07:00

This puzzle is similar to the Defective balls asked by microsoft. In this case instead of 10 balls here 12 balls. The puzzle question and answers are given below.

Puzzle Question :

You have 12 identical balls. One of the balls is heavier OR lighter than the rest (you don't know which). Using just a balance that can only show you which side of the tray is heavier, how can you determine which ball is the defective, how many measurements required??

Answer :

comment your answers...

Horse Race

2010-04-17T04:14:00.000-07:00

Horse Race

Puzzle :

The puzzle is there are 25 horses among which you need to find out the fastest 3 horses. You can conduct race among at most 5 horses to find out their relative speed. At no point you can find out the actual speed of the horse in a race. Find out how many races are required to get the top 3 horses. The answer for this puzzle is given below.

Answer :

Seven.
The first 5 to find the 3 fastest in each group and the sixth with the winners of each group.Now we know the fastest horse.
The second fastest horse must either be the second fastest in the sixth race or the second fastest horse in the final winners first race.
In the first case, the third fastest horse can either be the third fastest horse in the sixth race or the second fastest horse in either the fastest or second fastest horses first race.
Otherwise the third fastest horse could either be the second fastest horse in the sixth race or the third fastest horse in the final winners first race.
This makes for a total of five horses for the seventh, and final, race.

Microsoft puzzle with answer : Globe Walker

2010-04-16T07:59:00.000-07:00

Globe Walker

Puzzle :
This is another one puzzle asked from Micrsoft. How many points are there on the globe where, by walking one mile south, then one mile east and then one mile north, you would reach the place where you started? Answer for this puzzle is given below.

Answer:
The trivial answer to this question is one point, namely, the North Pole. But if you think that answer should suffice, you might want to think again!

Let’s think this through methodically. If we consider the southern hemisphere, there is a ring near the South Pole that has a circumference of one mile. So what if we were standing at any point one mile north of this ring? If we walked one mile south, we would be on the ring. Then one mile east would bring us back to same point on the ring (since it’s circumference is one mile). One mile north from that point would bring us back to the point were we started from. If we count, there would be an infinite number of points north of this one mile ring.

So what’s our running total of possible points? We have 1 + infinite points. But we’re not done yet!

Consider a ring that is half a mile in circumference near the South Pole. Walking a mile along this ring would cause us to circle twice, but still bring us to back to the point we started from. As a result, starting from a point that is one mile north of a half mile ring would also be valid. Similarly, for any positive integer n, there is a circle with radius

r = 1 / (2 * pi * n)

centered at the South Pole. Walking one mile along these rings would cause us to circle n times and return to the same point as we started. There are infinite possible values for n. Furthermore, there are infinite ways of determining a starting point that is one mile north of these n rings, thus giving us (infinity * infinity) possible points that satisfy the required condition.

So the real answer to this question is 1 + infinity * infinity = infinite possible points!

Visitors can write your comments and different answers as you think..

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An Honest Man and The Card

2010-04-12T07:56:00.000-07:00

Puzzle:

An Honest Man

An honest man holds a card with one of the three possible numbers on it “1”, “2” or “3”. You can ask one question and the man allowed to answer only “Yes”,”No” and “I don’t know”. The honest man obviously never lies. Which question would you ask to say with 100% certainty which number is on the card the honest man holds?

Answer :

Question should be: "If I substract 2 from your number, and then take squery root, would the result be greater than zero?"

Man's number 3 -> Sqrt(3-2)=Sqrt(1)=1, (1>0)==TRUE, so the answer is "YES"
Man's number 2 -> Sqrt(2-2)=Sqrt(0)=0, (0>0)==FALSE, so the answer is "NO"
Man's number 1 -> Sqrt(1-2)=Sqrt(-1)=i, (i>0)==???, so the answer is "I DON't KNOW"

Microsoft Puzzles : River , Solders and the boat

2010-04-10T02:57:00.000-07:00

This puzzle is asked by microsoft long ago in their interview. This puzzle is associated with how solders cross the river in a small boat. To solve this puzzle you do not need to be math genious simple thinking is enough. It is very simple puzzle. Try to solve this puzzle.

Puzzle :

Consider there are 10 soldiers on the one side of the river. They need to go to the over side of the rever. There is no bridge in the rever and no one can swin in the rever. One of the soldiers spots the boat with two boys inside. The boat is very small and the boys in the boats also very small. The boat can either hold two boys or one soldier. Now tell me how can all soldiers go to the other side of the river using this boat ?

Answer :

First you have the two boys take the boat to one side of the river and leave a boy on that side of the river. One boy takes the boat back to the other side and stands on the shore. Then a soldier gets in the boat and rides it to the other side. When he arrives on the other side, then the boy gets in the boat and takes it back to the other side and picks up the other boy. They ride back to the other shore and drop off one of the boys and continue this process until all the soldiers are on the other side of the river.

4 Quarts of Water

2010-04-05T08:55:00.000-07:00

Puzzle Question : 4 Quarts of Water

If you had an infinite supply of water and a 5 quart and 3 quart pails, how would you measure exactly 4 quarts? and What is the least number of steps you need?

Answer:

This question is very simple actually. Since we can’t hold 4 quarts in the 3 quart pail, we have to look to filling up the 5 quart pail with exactly 4 quarts. Lets count the steps as we move along

1. Fill 3 quart pail ( 5p – 0, 3p – 3)
2. Transfer to 5 quart pail (5p – 3, 3p – 0)
3. Fill 3 quart pail ( 5p – 3, 3p – 3)
4. Transfer to 5 quart pail (5p – 5, 3p – 1)
5. Empty 5 quart pail (5p – 0, 3p – 1)
6. Transfer to 5 quart pail (5p – 1, 3p – 0)
7. Fill 3 quart pail ( 5p – 1, 3p – 3)
8. Transfer to 5 quart pail (5p – 4, 3p – 0) We are done!!!

That was easy right. Now for those who are mathematical and need everything solved in terms of a formula, here comes a little more mathematical solution.

Now the general steps are to fill up the 3 quart pail and keep transferring to the 5 quart pail (empty if full) until we hit 4 quarts. Therefore, the total amount of water we filled in the 3 quart pail must be equal to 4 more than a multiple of 5 (since we discard 5 quarts of water at a time). From this, we can derive this formula.
5n + 4 = 3m, where n and m are arbitrary positive integers
n represents the number of times we had to empty the 5 quart pail and m represents the number of times we had to fill up the 3 quart pail.

Now all we have to do is solve for the lowest set of positive integer solutions for {n,m}. {1, 3} is the lowest set. Some of the other solutions are {4, 8}, {7, 13}, {10, 18} and so on.

Microsoft puzzle - A Box of Defective Balls

2010-03-29T20:47:00.000-07:00

This puzzle is asked by many companies while taking interviews. It is another one of the famous puzzle. The puzzle is given below..

Puzzle Question :

You have 10 boxes of balls (each ball weighing exactly10 gm) with one box with defective balls (each one of the defective balls weigh 9 gm). You are given an electronic weighing machine and only one chance at it. How will find out which box has the defective balls?

Answer :

For convenience sake, let’s name the boxes from 1 to 10. In order to solve this problem, you have to leverage the fact that you know exactly what each good ball is supposed to weigh and what each defective ball is supposed to weigh. Many of us instinctively will take one ball out of each box and try to find a way to make it work but the trick to take different number of balls from each box.
The number of balls you pick from each bag is equal to the box number. For example, pick 1 ball from box 1, 2 balls from box 2 and so on. In total you will have 55 balls. If all of the boxes have good balls, then the total weight of these balls would be 550gm.
If box 1 has defective balls, then the total weight should be 1gm less than expected (only one ball weighing 9 gm). If box 2 has defective balls, then the total weight should be 2gm less than expected (two balls weighing 9 gm). So once you weigh the set of chosen balls, find out the difference between the total weight and the expected weight. That number represents the box number which contains the defective balls.

Gold for 7 Days of Work: Asked by Microsoft

2010-03-25T08:55:00.000-07:00

This puzzle is also asked by Microsoft. But this is famous puzzle asked by almost all companies nowadays interviews while recruiting people. This puzzle is not math oriented. By using simple analytic logic we can answer this puzzle.

Puzzle :

Lets us consider someone working for you seven days. You have a Gold bar to pay him. You must pay the worker for their work at the end of every day. If you are only allowed to make two breaks in the gold bar, how do you pay your worker? (Assuming equal amount of work is done during each day thus requiring equal amount of pay for each day).

Answer :

The trick is not to try and how to cut in such a way to make 7 equal pieces but rather to make transactions with the worker. Make two cuts on the gold bar such that you have the following sizes of bars.

1/7, 2/7 and 4/7. For convenience sake, I would just refer to the bars as 1, 2 and 4.

At the end of Day 1: Give Bar 1 (You- 2 and 4, Worker- 1)

At the end of Day 2: Give Bar 2, Take back Bar 1 (You- 1 and 4, Worker- 2)

At the end of Day 3: Give Bar 1 (You- 4, Worker- 1 and 2)

At the end of Day 4: Give Bar 4, Take back Bar 1 and Bar 2 (You- 1 and 2, Worker- 4)

At the end of Day 5: Give Bar 1 (You- 2, Worker- 1 and 4)

At the end of Day 6: Give Bar 2, Take back Bar 1 (You- 1, Worker- 2 and 4)

At the end of Day 7: Give Bar 1 (You- Empty, Worker- 1, 2 and 4)

That should take care of everything.

Microsoft interview Puzzle: Three Ants on Triangle

2010-03-25T08:50:00.000-07:00

This puzzle is asked during interview by Microsoft while electing the people for their company. Now many companies ask this puzzle while recruiting the students. This puzzle somewhat mathematical oriented. This interview puzzle is famously known as “Three Ants on Triangle” puzzle. People who know the concepts of probability can easily solve this puzzle.

Puzzle Question:

Three ants are sitting at the three corners of an equilateral triangle. Each ant starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the ants collide?

Answer:

The ants can only avoid a collision if they all decide to move in the same direction (either clockwise or anti-clockwise). If the ants do not pick the same direction, there will definitely be a collision. Each ant has the option to either move clockwise or anti-clockwise. There is a one in two chance that an ant decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision.

P(No collision) = P(All ants go in a clockwise direction) + P( All ants go in an anti-clockwise direction) = 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5 = 0.25

Microsoft Puzzle : 100 Prisoners in Solitary Cells

2010-03-23T09:16:00.000-07:00

100 Prisoners in Solitary Cells

Puzzle :

100 prisoners are stuck in the prison in solitary cells. The warden of the prison got bored one day and offered them a challenge. He will put one prisoner per day, selected at random (a prisoner can be selected more than once), into a special room with a light bulb and a switch which controls the bulb. No other prisoners can see or control the light bulb. The prisoner in the special room can either turn on the bulb, turn off the bulb or do nothing. On any day, the prisoners can stop this process and say “Every prisoner has been in the special room at least once”. If that happens to be true, all the prisoners will be set free. If it is false, then all the prisoners will be executed. The prisoners are given some time to discuss and figure out a solution. How do they ensure they all go free?

Answer :

Since this is the only way they will EVER get out of that prison, they decide to work together and make a plan. They select one prisoner (Bob, easier to refer) as the counter.

Every time any prisoner is selected other than Bob, they follow these steps. If they have never turned on the light bulb before and the light bulb is off, they turn it on. If not, they don’t do anything (simple as that). Now if Bob is selected and the light bulb is already on, he adds one to his count and turns off the bulb. If the bulb is off, he just sits there meditates or whatever he wants to. The day his count reaches 99, he calls the warden and tells him “Every prisoner has been in the special room at least once”.

So how does this solution work? Every time a prisoner enters the room first, he turns on the bulb if it is off. This way every prisoner turns on the bulb only once. When Bob enters and sees the bulb on, he knows that one new prisoner has entered the room so he adds one to his count. So when his counter reaches 99, he knows the rest of them have all been in the special room and obviously, he has been in the special room.

Is Your Husband a Cheat? : Puzzle Asked by Microsoft

2010-03-23T09:14:00.000-07:00

Microsoft Riddle :

This puzzle is asked by Microsoft while selecting the people for their company. This puzzle or Riddle is famous for interviewers while they taking interviews. This puzzle is famously known as "Is Your Husband a Cheat". Visitors of this page can try to answer this puzzle.

Is Your Husband a Cheat?

Puzzle : A certain town comprises of 100 married couples. Everyone in the town lives by the following rule: If a husband cheats on his wife, the husband is executed as soon as his wife finds out about him. All the women in the town only gossip about the husbands of other women. No woman ever tells another woman if her husband is cheating on her. So every woman in the town knows about all the cheating husbands in the town except her own. It can also be assumed that a husband remains silent about his infidelity. One day, the mayor of the town announces to the whole town that there is at least 1 cheating husband in the town. What do you think happens?

Answer : Stumped? Let’s solve this methodically. Say there was only 1 cheating husband in the town. There will be 99 women who know exactly who the cheater is. The 1 remaining woman, who is being cheated on, would have assumed there are no cheaters. But now that the mayor has confirmed that there is at least one cheater, she realizes that her own husband must be cheating on her. So her husband gets executed on the day of the announcement.

Now let’s assume there are 2 cheaters in the town. There will be 98 women in the town who know who the 2 cheaters are. The 2 wives, who are being cheated on, would think that there is only 1 cheater in the town. Since neither of these 2 women know that their husbands are cheaters, they both do not report their husbands in on the day of the announcement. The next day, when the 2 women see that no husband was executed, they realize that there could only be one explanation – both their husbands are cheaters. Thus, on the second day, 2 husbands are executed.

Through induction, it can be proved that when this logic is applied to n cheating husbands, they all die on the n th day after the mayor’s announcement.

Google Interview Puzzle : 2 Egg Problem...

2010-03-16T05:52:00.000-07:00

This Puzzle is asked by Google while they had an interview to a select student

for their company. Then many companies asked this Puzzle during their

interview or written test in campus. This puzzle is famously known as 2 Egg

Problem. The puzzle is given below.

Puzzle : Two Egg Problem:

* You are given 2 eggs.

* You have access to a 100-storey building.

* Eggs can be very hard or very fragile means it may break if dropped from the first floor or may not even break if dropped from 100 th floor.Both eggs are identical.

* You need to figure out the highest floor of a 100-storey building an egg can be dropped without breaking.

* Now the question is how many drops you need to make. You are allowed to break 2 eggs in the process

If you are one of the people who likes to solve a puzzle before seeing the answer you must quit the blog now and come back later for checking the answer.

Now that this is a Google interview question I am taking the normal "Interview-Style" of solving a problem. Simply saying thinking aloud through the solution from worst to the best correcting the flows optimizing the solution or taking the 5-minute hard thinking acting pause to a problem, which you know already and just want to make your interviewer think that you are a challenge lover.

Answer :

Let x be the answer we want, the number of drops required.

So if the first egg breaks maximum we can have x-1 drops and so we must always put the first egg from height x. So we have determined that for a given x we must drop the first ball from x height. And now if the first drop of the first egg doesn’t breaks we can have x-2 drops for the second egg if the first egg breaks in the second drop.

Taking an example, lets say 16 is my answer. That I need 16 drops to find out the answer. Lets see whether we can find out the height in 16 drops. First we drop from height 16,and if it breaks we try all floors from 1 to 15.If the egg don’t break then we have left 15 drops, so we will drop it from 16+15+1 =32nd floor. The reason being if it breaks at 32nd floor we can try all the floors from 17 to 31 in 14 drops (total of 16 drops). Now if it did not break then we have left 13 drops. and we can figure out whether we can find out whether we can figure out the floor in 16 drops.

Lets take the case with 16 as the answer

1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45 .....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 We can easily do in the end as we have enough drops to accomplish the task

Now finding out the optimal one we can see that we could have done it in either 15 or 14 drops only but how can we find the optimal one. From the above table we can see that the optimal one will be needing 0 linear trials in the last step.

So we could write it as

(1+p) + (1+(p-1))+ (1+(p-2)) + .........+ (1+0) >= 100.

Let 1+p=q which is the answer we are looking for

q (q+1)/2 >=100

Solving for 100 you get q=14.
So the answer is: 14
Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (i.e. move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100).

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