1. How could Java classes direct program messages to the system console, but error messages, say to a file?
The class System has a variable out that represents the standard output, and the variable err that represents the standard error device. By default, they both point at the system console. This how the standard output could be re-directed:
Stream st =
new Stream (new
FileOutputStream ("techinterviews_com.txt"));
System.setErr(st);
System.setOut(st);
2. What’s the difference between an interface and an abstract class?
An abstract class may contain code in method bodies, which is not allowed in an interface. With abstract classes, you have to inherit your class from it and Java does not allow multiple inheritance. On the other hand, you can implement multiple interfaces in your class.
3. Why would you use a synchronized block vs. synchronized method?
Synchronized blocks place locks for shorter periods than synchronized methods.
4. Explain the usage of the keyword transient?
This keyword indicates that the value of this member variable does not have to be serialized with the object. When the class will be de-serialized, this variable will be initialized with a default value of its data type (i.e. zero for integers).
5. How can you force garbage collection?
You can’t force GC, but could request it by calling System.gc(). JVM does not guarantee that GC will be started immediately.
6. How do you know if an explicit object casting is needed?
If you assign a superclass object to a variable of a subclass’s data type, you need to do explicit casting. For example:
Object a;Customer b; b = (Customer) a;
When you assign a subclass to a variable having a supeclass type, the casting is performed automatically.
7. What’s the difference between the methods sleep() and wait()
The code sleep(1000); puts thread aside for exactly one second. The code wait(1000), causes a wait of up to one second. A thread could stop waiting earlier if it receives the notify() or notifyAll() call. The method wait() is defined in the class Object and the method sleep() is defined in the class Thread.
8. Can you write a Java class that could be used both as an applet as well as an application?
Yes. Add a main() method to the applet.
9. What’s the difference between constructors and other methods?
Constructors must have the same name as the class and can not return a value. They are only called once while regular methods could be called many times.
10. Can you call one constructor from another if a class has multiple constructors
Yes. Use this() syntax.
11. Explain the usage of Java packages.
This is a way to organize files when a project consists of multiple modules. It also helps resolve naming conflicts when different packages have classes with the same names. Packages access level also allows you to protect data from being used by the non-authorized classes.
12. If a class is located in a package, what do you need to change in the OS environment to be able to use it?
You need to add a directory or a jar file that contains the package directories to the CLASSPATH environment variable. Let’s say a class Employee belongs to a package com.xyz.hr; and is located in the file c:/dev/com.xyz.hr.Employee.java. In this case, you’d need to add c:/dev to the variable CLASSPATH. If this class contains the method main(), you could test it from a command prompt window as follows:
c:\>java com.xyz.hr.Employee
13. What’s the difference between J2SDK 1.5 and J2SDK 5.0?
There’s no difference, Sun Microsystems just re-branded this version.
14. What would you use to compare two String variables - the operator == or the method equals()?
I’d use the method equals() to compare the values of the Strings and the = = to check if two variables point at the same instance of a String object.
15. Does it matter in what order catch statements for FileNotFoundException and IOExceptipon are written?
A. Yes, it does. The FileNoFoundException is inherited from the IOException. Exception’s subclasses have to be caught first.
16. Can an inner class declared inside of a method access local variables of this method?
It’s possible if these variables are final.
17. What can go wrong if you replace && with & in the following code:
String a=null;
if (a!=null && a.length()>10)
{...}
A single ampersand here would lead to a NullPointerException.
18. What’s the main difference between a Vector and an ArrayList
Java Vector class is internally synchronized and ArrayList is not.
19. When should the method invokeLater()be used?
This method is used to ensure that Swing components are updated through the event-dispatching thread.
20. How can a subclass call a method or a constructor defined in a superclass?
Use the following syntax: super.myMethod(); To call a constructor of the superclass, just write super(); in the first line of the subclass’s constructor.
21. What’s the difference between a queue and a stack?
Stacks works by last-in-first-out rule (LIFO), while queues use the FIFO rule.
22. You can create an abstract class that contains only abstract methods. On the other hand, you can create an interface that declares the same methods. So can you use abstract classes instead of interfaces?
Sometimes. But your class may be a descendent of another class and in this case the interface is your only option.
23. What comes to mind when you hear about a young generation in Java?
Garbage collection.
24. What comes to mind when someone mentions a shallow copy in Java?
Object cloning.
25. If you’re overriding the method equals() of an object, which other method you might also consider?
hashCode()
26. You are planning to do an indexed search in a list of objects. Which of the two Java collections should you use: ArrayList or LinkedList?
ArrayList
27. How would you make a copy of an entire Java object with its state?
Have this class implement Cloneable interface and call its method clone().
28. How can you minimize the need of garbage collection and make the memory use more effective?
Use object pooling and weak object references.
29. There are two classes: A and B. The class B need to inform a class A when some important event has happened. What Java technique would you use to implement it?
If these classes are threads I’d consider notify() or notifyAll(). For regular classes you can use the Observer interface.
30. What access level do you need to specify in the class declaration to ensure that only classes from the same directory can access it?
You do not need to specify any access level, and Java will use a default package access level.
Friday, October 26, 2007
Friday, October 12, 2007
General Knowledge Questions for Interview
red marbles, blue marbles
problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
solution: red marbles, blue marbles
problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
solution: chance! chance is easy if you know how to do the formula. we know that we have two choices to make. first we'll pick a jar, and each jar will have a 1/2 chance of being picked. then we'll pick a marble, and depending how we stack the marbles, we'll have a (# of red marbles in jar)/(# of total marbles in jar) chance of getting a red one.
for example, say we put all the red marbles into jar A and all the blue ones into jar B. then our chances for picking a red one are:
1/2 chance we pick jar A * 50/50 chance we pick a red marble
1/2 chance we pick jar B * 0/50 chance we pick a red marble
do the math and you get 1/2 chance for a red marble from jar A and a 0/2 chance for a red marble from jar B. add 'em up and you get the result = 1/2 chance for picking a red marble.
think about it for awhile and see if you can figure out the right combination. we had a 50/50 (guaranteed) chance in picking a red marble from jar A, but we didn't have to have 50 red marbles in there to guarantee those fantastic odds, did we? we could've just left 1 red marble in there and the odds are still 1/1. then we can take all those other marbles and throw them in jar B to help the odds out there.
let's look at those chances:
1/2 we pick jar A * 1/1 we pick a red marble
1/2 we pick jar B * 49/99 we pick a red marble
do the math and add them up to get 1/2 + 49/198 = 148/198, which is almost 3/4.
we can prove these are the best odds in a somewhat non-formal way as follows. our goal is to maximize the odds of picking a red marble. therefore we can subdivide this goal into maximizing the odds of picking a red marble in jar A and maximizing the odds of picking a red marble in jar B. if we do that, then we will have achieved our goal. it is true that by placing more red marbles into a jar we will increase the chances of picking a red marble. it is also true that by reducing the number of blue marbles in a jar we will increase the odds also. we've maximized the odds in jar A since 1/1 is the maximum odds by reducing the number of blue marbles to 0 (the minimum). we've also maximized the number of red marbles in jar B. if we added any more red marbles to jar B we would have to take them out of jar A which reduce the odds there to 0 (very bad). if we took any more blue ones out of jar B we would have to put them in jar A which reduce the odds there by 50% (very bad).
problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
solution: red marbles, blue marbles
problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
solution: chance! chance is easy if you know how to do the formula. we know that we have two choices to make. first we'll pick a jar, and each jar will have a 1/2 chance of being picked. then we'll pick a marble, and depending how we stack the marbles, we'll have a (# of red marbles in jar)/(# of total marbles in jar) chance of getting a red one.
for example, say we put all the red marbles into jar A and all the blue ones into jar B. then our chances for picking a red one are:
1/2 chance we pick jar A * 50/50 chance we pick a red marble
1/2 chance we pick jar B * 0/50 chance we pick a red marble
do the math and you get 1/2 chance for a red marble from jar A and a 0/2 chance for a red marble from jar B. add 'em up and you get the result = 1/2 chance for picking a red marble.
think about it for awhile and see if you can figure out the right combination. we had a 50/50 (guaranteed) chance in picking a red marble from jar A, but we didn't have to have 50 red marbles in there to guarantee those fantastic odds, did we? we could've just left 1 red marble in there and the odds are still 1/1. then we can take all those other marbles and throw them in jar B to help the odds out there.
let's look at those chances:
1/2 we pick jar A * 1/1 we pick a red marble
1/2 we pick jar B * 49/99 we pick a red marble
do the math and add them up to get 1/2 + 49/198 = 148/198, which is almost 3/4.
we can prove these are the best odds in a somewhat non-formal way as follows. our goal is to maximize the odds of picking a red marble. therefore we can subdivide this goal into maximizing the odds of picking a red marble in jar A and maximizing the odds of picking a red marble in jar B. if we do that, then we will have achieved our goal. it is true that by placing more red marbles into a jar we will increase the chances of picking a red marble. it is also true that by reducing the number of blue marbles in a jar we will increase the odds also. we've maximized the odds in jar A since 1/1 is the maximum odds by reducing the number of blue marbles to 0 (the minimum). we've also maximized the number of red marbles in jar B. if we added any more red marbles to jar B we would have to take them out of jar A which reduce the odds there to 0 (very bad). if we took any more blue ones out of jar B we would have to put them in jar A which reduce the odds there by 50% (very bad).
General Knowledge Questions for Interview
100 doors in a row
100 doors in a row aha:!
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
question: what state are the doors in after the last pass? which are open which are closed?
solution: 100 doors in a row
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
for example, after the first pass every door is open. on the second pass you only visit the even doors (2,4,6,8...) so now the even doors are closed and the odd ones are opened. the third time through you will close door 3 (opened from the first pass), open door 6 (closed from the second pass), etc..
question: what state are the doors in after the last pass? which are open which are closed?
solution: you can figure out that for any given door, say door #42, you will visit it for every divisor it has. so 42 has 1 & 42, 2 & 21, 3 & 14, 6 & 7. so on pass 1 i will open the door, pass 2 i will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. for every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9... leaving it open at the end. only perfect square doors will be open at the end.
100 doors in a row aha:!
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
question: what state are the doors in after the last pass? which are open which are closed?
solution: 100 doors in a row
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
for example, after the first pass every door is open. on the second pass you only visit the even doors (2,4,6,8...) so now the even doors are closed and the odd ones are opened. the third time through you will close door 3 (opened from the first pass), open door 6 (closed from the second pass), etc..
question: what state are the doors in after the last pass? which are open which are closed?
solution: you can figure out that for any given door, say door #42, you will visit it for every divisor it has. so 42 has 1 & 42, 2 & 21, 3 & 14, 6 & 7. so on pass 1 i will open the door, pass 2 i will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. for every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9... leaving it open at the end. only perfect square doors will be open at the end.
General Knowledge Questions for Interview
reverse a string - word by word
a typical programming interview question is "reverse a string, in place". if you understand pointers, the solution is simple. even if you don't, it can be accomplished using array indices. i usually ask candidates this question first, so they get the algorithm in their head. then i play dirty by asking them to reverse the string word by word, in place. for example if our string is "the house is blue", the return value would be "blue is house the". the words are reversed, but the letters are still in order (within the word).
solution: reverse a string - word by word
problem: reverse "the house is blue", the answer should be "blue is house the". the words are reversed, but the letters are still in order (within the word).
solution: solving the initial problem of just reversing a string can either be a huge help or a frustrating hinderance. most likely the first attempt will be to solve it the same way, by swapping letters at the front of the string with letters at the back, and then adding some logic to keep the words in order. this attempt will lead to confusion pretty quickly.
for example, if we start by figuring out that "the" is 3 letters long and then try to put the "t" from "the" where the "l" from "blue" is, we encounter a problem. where do we put the "l" from "blue"? hmm... well we could have also figured out how long "blue" was and that would tell us where to put the "l" at... but the "e" from "blue" needs to go into the space after "the". argh. its getting quite confusing. in fact, i would be delighted to even see a solution to this problem using this attack method. i don't think its impossible, but i think it is so complex that it's not worth pursuing.
here's a hint. remember before when we just reversed "the house is blue"? what happened?
initial: the house is blue
reverse: eulb si esuoh eht
look at the result for a minute. notice anything? if you still don't see it, try this.
initial: the house is blue
reverse: eulb si esuoh eht
wanted : blue is house the
the solution can be attained by first reversing the string normally, and then just reversing each word.
a typical programming interview question is "reverse a string, in place". if you understand pointers, the solution is simple. even if you don't, it can be accomplished using array indices. i usually ask candidates this question first, so they get the algorithm in their head. then i play dirty by asking them to reverse the string word by word, in place. for example if our string is "the house is blue", the return value would be "blue is house the". the words are reversed, but the letters are still in order (within the word).
solution: reverse a string - word by word
problem: reverse "the house is blue", the answer should be "blue is house the". the words are reversed, but the letters are still in order (within the word).
solution: solving the initial problem of just reversing a string can either be a huge help or a frustrating hinderance. most likely the first attempt will be to solve it the same way, by swapping letters at the front of the string with letters at the back, and then adding some logic to keep the words in order. this attempt will lead to confusion pretty quickly.
for example, if we start by figuring out that "the" is 3 letters long and then try to put the "t" from "the" where the "l" from "blue" is, we encounter a problem. where do we put the "l" from "blue"? hmm... well we could have also figured out how long "blue" was and that would tell us where to put the "l" at... but the "e" from "blue" needs to go into the space after "the". argh. its getting quite confusing. in fact, i would be delighted to even see a solution to this problem using this attack method. i don't think its impossible, but i think it is so complex that it's not worth pursuing.
here's a hint. remember before when we just reversed "the house is blue"? what happened?
initial: the house is blue
reverse: eulb si esuoh eht
look at the result for a minute. notice anything? if you still don't see it, try this.
initial: the house is blue
reverse: eulb si esuoh eht
wanted : blue is house the
the solution can be attained by first reversing the string normally, and then just reversing each word.
General Knowledge Questions for Interview
salary
three coworkers would like to know their average salary. how can they do it, without disclosing their own salaries?
solution: salary
How about: Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B, without showing C. B then adds his salary plus a random amount (BS + BR) and passes to C (at each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes to A. Now A subtracts her random number (AR), passes to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.
As has been noted already, there's no way to liar-proof the scheme.
It's also worth noting that once they know the average, any of the three knows the sum of the other 2 salaries.
three coworkers would like to know their average salary. how can they do it, without disclosing their own salaries?
solution: salary
How about: Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B, without showing C. B then adds his salary plus a random amount (BS + BR) and passes to C (at each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes to A. Now A subtracts her random number (AR), passes to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.
As has been noted already, there's no way to liar-proof the scheme.
It's also worth noting that once they know the average, any of the three knows the sum of the other 2 salaries.
General Knowledge Questions for Interview
world series
you have $10,000 dollars to place a double-or-nothing bet on the Yankees in the World Series (max 7 games, series is over once a team wins 4 games).
unfortunately, you can only bet on each individual game, not the series as a whole. how much should you bet on each game, so that, if the yanks win the whole series, you expect to get 20k, and if they lose, you expect 0?
basically, you know that there may be between 4 and 7 games, and you need to decide on a strategy so that whenever the series is over, your final outcome is the same as an overall double-or-nothing bet on the series.
solution: world series
this probably isn't the cleanest solution, but...
a dynamic-programming type solution is:
(1) Create a 5x5 matrix P.
So, P[i,j] holds your pile of money when the yanks have won i games and the mets have won j games.
initialize P[4,j] := 20 for j from 0 to 3 initialize P[i,4] := 0 for i from 0 to 3
fill P in bottom-right to top left by averaging bottom and right adjacent cells:
P[i,j] := (P[i+1,j]+P[i,j+1]) / 2
(2) Make another 5x5 matrix, B, which represets your bet at any-time.
So, B[i,j] represents your bet when the yanks have won i games and the Mets j games.
fill this top-left to bottom right by:
B[i,j] = P[i+1,j] - P[i,j]
(3) Look in matrix B for your bet at any time.
The final matricies are:
Pile-Matrix
0.00
1.00
2.00
3.00
4.00
0
10.00
6.88
3.75
1.25
0.00
1
13.13
10.00
6.25
2.50
0.00
2
16.25
13.75
10.00
5.00
0.00
3
18.75
17.50
15.00
10.00
0.00
4
20.00
20.00
20.00
20.00
Bet-Matrix
0
1
2
3
4
0
3.13
3.13
2.50
1.25
1
3.13
3.75
3.75
2.50
2
2.50
3.75
5.00
5.00
3
1.25
2.50
5.00
10.00
4
you have $10,000 dollars to place a double-or-nothing bet on the Yankees in the World Series (max 7 games, series is over once a team wins 4 games).
unfortunately, you can only bet on each individual game, not the series as a whole. how much should you bet on each game, so that, if the yanks win the whole series, you expect to get 20k, and if they lose, you expect 0?
basically, you know that there may be between 4 and 7 games, and you need to decide on a strategy so that whenever the series is over, your final outcome is the same as an overall double-or-nothing bet on the series.
solution: world series
this probably isn't the cleanest solution, but...
a dynamic-programming type solution is:
(1) Create a 5x5 matrix P.
So, P[i,j] holds your pile of money when the yanks have won i games and the mets have won j games.
initialize P[4,j] := 20 for j from 0 to 3 initialize P[i,4] := 0 for i from 0 to 3
fill P in bottom-right to top left by averaging bottom and right adjacent cells:
P[i,j] := (P[i+1,j]+P[i,j+1]) / 2
(2) Make another 5x5 matrix, B, which represets your bet at any-time.
So, B[i,j] represents your bet when the yanks have won i games and the Mets j games.
fill this top-left to bottom right by:
B[i,j] = P[i+1,j] - P[i,j]
(3) Look in matrix B for your bet at any time.
The final matricies are:
Pile-Matrix
0.00
1.00
2.00
3.00
4.00
0
10.00
6.88
3.75
1.25
0.00
1
13.13
10.00
6.25
2.50
0.00
2
16.25
13.75
10.00
5.00
0.00
3
18.75
17.50
15.00
10.00
0.00
4
20.00
20.00
20.00
20.00
Bet-Matrix
0
1
2
3
4
0
3.13
3.13
2.50
1.25
1
3.13
3.75
3.75
2.50
2
2.50
3.75
5.00
5.00
3
1.25
2.50
5.00
10.00
4
General Knowledge Questions for Interview
100 factorial aha:!
how many trailing zeroes are there in 100! (100 factorial)?
solution: 100 factorial
One per factor of 10, and one per factor of 5 (there are more than enough 2's to pair with the 5's), plus one per factor of ten squared (one occurrence) and one per factor of 5 squared (three occurrences).
So if I'm counting correctly, that'd be 10 + 10 + 1 + 3== 24 zeroes.
Assuming the question meant *trailing* zeroes. It'd be much harder to also count the intermingled zero digits in the entire expansion.
how many trailing zeroes are there in 100! (100 factorial)?
solution: 100 factorial
One per factor of 10, and one per factor of 5 (there are more than enough 2's to pair with the 5's), plus one per factor of ten squared (one occurrence) and one per factor of 5 squared (three occurrences).
So if I'm counting correctly, that'd be 10 + 10 + 1 + 3== 24 zeroes.
Assuming the question meant *trailing* zeroes. It'd be much harder to also count the intermingled zero digits in the entire expansion.
Thursday, October 11, 2007
General Knowledge Questions for Interview
oil mogul
you are an oil mogul considering the purchase of drilling rights to an as yet unexplored tract of land.
the well's expected value to its current owners is uniformly distributed over [$1..$100]. (i.e., a 1% chance it's worth each value b/w $1..$100, inclusive).
bcause you have greater economies of scale than the current owners, the well will actually be worth 50% more to you than to them (but they don't know this).
the catch: although you must bid on the well before drilling starts (and hence, before the actual yield of the well is known), the current owner can wait until *after* the well's actual value is ascertained before accepting your bid or not.
what should you bid?
solution: oil mogul
This problem amounts to properly defining the expected value of the well to you.
The following equation does it:
(1%) * [(1.5 - 1)] +
(1%) * [(3 - 2) + (1.50 - 2)] +
(1%) * [(4.5 - 3) + (3 - 3) + (1.5 - 3)] +
. . .
(1%) * [(150-100) + ... + (3-100) + (1.5-100)]
Each line represents your expected value from a bid of 1$, 2$, ..., 100$, respectively.
eg, consider line 2 above. if you bid $2...
With 98% probability you won't win the contract, so your profit is 0. With 1% probability, you will win something worth (150%*1) = 1.5, for which you paid 2$ With 1% probability, you will something worth worth (150%*2) = 3, for which you paid 2$
So, your goal is to maximize the following function of x, where x is your bid.
f(x) = 1% * Sum_{i = 1 to floor(x)}{ x - 1.5*i }
There's no benefit to non-integer bets, so re-write the maximization function as :
ARGMAX(k) {1% * Sum_{i = 1 to k}{1.5*i - k}}
(=) ARGMAX(k) {Sum_{i=1 to k}{1.5*i - k}} /* 1% isn't a function of k or i, so toss it */
(=) ARGMAX(k) {Sum_{i=1 to k}{1.5*i} - Sum_{i=1 to k}{k}} /* Split the summation */
(=) ARGMAX(k) {(0.75)(K)(K+1) - K^2 }} /* Closed form for summations */
(=) ARGMAX(k) {(0.75)(k)-(0.25)(K^2)}} /* Algebra */
And that function is maximized at k = 1 and k = 2.
When choosing b/w $1 and $2, you should bid $1 because of time-value, reinvestment risk, etc, of the extra dollar.
(ie, if you don't have to spend the extra $$ now, don't)
That's my solution.
you are an oil mogul considering the purchase of drilling rights to an as yet unexplored tract of land.
the well's expected value to its current owners is uniformly distributed over [$1..$100]. (i.e., a 1% chance it's worth each value b/w $1..$100, inclusive).
bcause you have greater economies of scale than the current owners, the well will actually be worth 50% more to you than to them (but they don't know this).
the catch: although you must bid on the well before drilling starts (and hence, before the actual yield of the well is known), the current owner can wait until *after* the well's actual value is ascertained before accepting your bid or not.
what should you bid?
solution: oil mogul
This problem amounts to properly defining the expected value of the well to you.
The following equation does it:
(1%) * [(1.5 - 1)] +
(1%) * [(3 - 2) + (1.50 - 2)] +
(1%) * [(4.5 - 3) + (3 - 3) + (1.5 - 3)] +
. . .
(1%) * [(150-100) + ... + (3-100) + (1.5-100)]
Each line represents your expected value from a bid of 1$, 2$, ..., 100$, respectively.
eg, consider line 2 above. if you bid $2...
With 98% probability you won't win the contract, so your profit is 0. With 1% probability, you will win something worth (150%*1) = 1.5, for which you paid 2$ With 1% probability, you will something worth worth (150%*2) = 3, for which you paid 2$
So, your goal is to maximize the following function of x, where x is your bid.
f(x) = 1% * Sum_{i = 1 to floor(x)}{ x - 1.5*i }
There's no benefit to non-integer bets, so re-write the maximization function as :
ARGMAX(k) {1% * Sum_{i = 1 to k}{1.5*i - k}}
(=) ARGMAX(k) {Sum_{i=1 to k}{1.5*i - k}} /* 1% isn't a function of k or i, so toss it */
(=) ARGMAX(k) {Sum_{i=1 to k}{1.5*i} - Sum_{i=1 to k}{k}} /* Split the summation */
(=) ARGMAX(k) {(0.75)(K)(K+1) - K^2 }} /* Closed form for summations */
(=) ARGMAX(k) {(0.75)(k)-(0.25)(K^2)}} /* Algebra */
And that function is maximized at k = 1 and k = 2.
When choosing b/w $1 and $2, you should bid $1 because of time-value, reinvestment risk, etc, of the extra dollar.
(ie, if you don't have to spend the extra $$ now, don't)
That's my solution.
General Knowledge Questions for Interview
vienna aha:!!
it's the middle ages, you're travelling across europe and you want to find the way to vienna. you come to a crossroads, now there are two ways to go. at the crossroads stand a knight and a knave. the knight answers every question truthfully. the knave answers every question falsely. you don't know which guy is which. how can you figure out which road leads to Vienna by only asking one question?
solution: vienna
Many people try "what would the other man say if I asked him which way to Vienna?" But for that to work, the knight would have to know that the other man is a knave, and the knave would have to know that the other man is a knight; it's not clear that either of these is a given.
To cope with that, you can ask "assuming the other man has the opposite predilection regarding truth-telling from what you do, what would he say if I asked him which way to Vienna?"
More interestingly, you can ask "if you were to ask yourself which way to Vienna, what would you say?" But before risking that question, I'd want a few more details about how the knave is wired up. What does it mean, to lie to oneself? Is such a thing even possible, regardless of what psychoanalysts might claim?
Furthermore, travelling through Europe in the middle ages, it's a bit of a rash assumption that people you meet at a crossroad would understand English. If you only have 1 question, you have to frame a question that is meaningful in all locally spoken languages (like writing a program that is both legal Fortran and legal C). Your question doesn't have to mean exactly the same thing in each language, as long as each possible answer carries in itself an indication of which language it is in.
Or you could try the universal language of mime. Say "Vienna" and open your arms wide with a questioning look. But how would you mime "if you were to ask the other man ..." (or "if you were to ask yourself ...") ?
it's the middle ages, you're travelling across europe and you want to find the way to vienna. you come to a crossroads, now there are two ways to go. at the crossroads stand a knight and a knave. the knight answers every question truthfully. the knave answers every question falsely. you don't know which guy is which. how can you figure out which road leads to Vienna by only asking one question?
solution: vienna
Many people try "what would the other man say if I asked him which way to Vienna?" But for that to work, the knight would have to know that the other man is a knave, and the knave would have to know that the other man is a knight; it's not clear that either of these is a given.
To cope with that, you can ask "assuming the other man has the opposite predilection regarding truth-telling from what you do, what would he say if I asked him which way to Vienna?"
More interestingly, you can ask "if you were to ask yourself which way to Vienna, what would you say?" But before risking that question, I'd want a few more details about how the knave is wired up. What does it mean, to lie to oneself? Is such a thing even possible, regardless of what psychoanalysts might claim?
Furthermore, travelling through Europe in the middle ages, it's a bit of a rash assumption that people you meet at a crossroad would understand English. If you only have 1 question, you have to frame a question that is meaningful in all locally spoken languages (like writing a program that is both legal Fortran and legal C). Your question doesn't have to mean exactly the same thing in each language, as long as each possible answer carries in itself an indication of which language it is in.
Or you could try the universal language of mime. Say "Vienna" and open your arms wide with a questioning look. But how would you mime "if you were to ask the other man ..." (or "if you were to ask yourself ...") ?
General Knowledge Questions for Interview
duel aha:!!
you find yourself in a duel with two other gunmen. you shoot with 33% accuracy, and the other two shoot with 100% and 50% accuracy, respectively. the rules of the duel are one shot per-person per-round. the shooting order is from worst shooter to best shooter, so you go first, the 50% guy goes second, and the 100% guy goes third.
where or who should you shoot at in round 1?
solution: duel
You have 3 options to consider:
(1) Shoot at 50% guy. Death comes with: 72% likilhood (2) Shoot at the 100% guy. Death comes with: 63.64% likihood (3) Shoot into the air (purposely miss). Death comew with 58.33% likihood.
So, you shoot into the air and hope for the best.
Calculations:
The p in the probabilities below represents your probability of dying at some point if you MISS either guy in the first round. It'll be filled in later.
(1) if you shoot at the 50% guy and get him, you're guaranteed to get shot the next round.
prob of dying at some point by aiming at the 50% guy:
(=) (33.3%)(100%)+(66.67%)*(p)
(=) (33.3%)+(66.67%)(p)
(2) if you shoot at the 100% guy and get him, you're left with this geometric sum of probability of getting shot by the 50% guy at some point in the future...it converges.
(=) [(50%)] +
[(50%)(66.67%)(50%)] +
[(50%)(66.67%)(50%)(66.67%)(50%)] + ...
(=) 50% * { 1 + (1/3) + (1/3)^2 + ... }
(=) 50% * {3/2)
(=) 75%
so, your prob of getting shot if you shoot at the 100% guy is:
(=) (33.3%)(75%)+(66.67%)(p)
(=) (24.75%) + (66.67%)(p)
still crummy, but it dominates the alternative, for positive p.
(3) What happens if you miss? ie, what's (p)?
if you miss, the second guy has a choice to make...does he shoot at you, or the other guy? his options:
(1) Shoot you: if he gets you, he's guaranteed to die on the next shot.
if he misses, he has q chance of dying (which we'll get later).
(=) (50%)(100%)+(50%)(q)
(=) (50%)+(0.5)(q)
(2) Shoot at the 100% dude:
If he gets the 100% guy, there some infinite sum representing his probability of getting shot by you:
[(33.33%)] +
[(66.67%)(50%)(33.33%)] +
[(66.67%)(50%)(66.67%)(50%)(33.33%)] + ...
(=) 33.33% * [ 1 + (1/3) + (1/3)^2 + ... ]
(=) 33.33% * [(3/2)]
(=) (50.0%)
So, he chances of getting shot if he shoots at the 100% guy are:
(=) (50%)*(50%) + (50%)(q)
(=) (25%)+(.5)q
No matter what q is, he'd prefer shooting the 100% guy to shooting you.
Now, what happens if *he* misses (the 100%) guy? ie, what's q? if he misses, then the 100% guy has to make a decision:
(1) Shoot you:
he's guaranteed to get you, so his chances of dying are just 50%. (game ends after the next round...the 50% gets only one shot)
prob of dying: 50%
(2) Shoot the 50% guy:
by the same logic, he'd have a 33.33% chance of death (getting shot by you).
So, he prefers to shoot the 50% guy.
So, q is 100% (ie, if the 50% guy misses, the 100% guy shoots him immediately). From that, we know the 50% guy's optimal move, if everyone's around on his first shot. He'd prefer shooting at the 100% guy to shooting at you or purposely missing.
Now, we need to get p, which is our probability of dying if we purposely miss either guy (shoot into the air).
We know the 50% guy shoots at the 100% guy if we miss.
(1) With 50% probability, the 50% guy kills the 100% guy, resulting in a shootout b/w us and the 50% guy.
(=) (50%) * [ (66.67%)*(50%) + (66.67%)*(50%)*(66.67%)(50%) + ...
(=) (50%) * [ (1/3) + (1/3)^2 + ... ]
(=) (50%) * [0.5]
(=) (25%)
(2) He misses him. Then, we get one shot at the 100% guy (after the 100% guy shoots the 50% guy).
Our chances of death are:
(50%)*(66.67%) = (33.33%)
So, p is (25%)+(33.33%) = 58.3%
So, again, our three choices are
(1) Shoot at 50% guy. Death comes with:
(=) (33.3%)+(66.67%)(p)
(=) 72% likilhood
(2) Shoot at the 100% guy. Death comes with:
(=) (24.75%) + (66.67%)(p)
(=) 63.64% likihood
(3) Shoot into the air (purposely miss).
(=) p
(=) 58.33% likelihood.
you find yourself in a duel with two other gunmen. you shoot with 33% accuracy, and the other two shoot with 100% and 50% accuracy, respectively. the rules of the duel are one shot per-person per-round. the shooting order is from worst shooter to best shooter, so you go first, the 50% guy goes second, and the 100% guy goes third.
where or who should you shoot at in round 1?
solution: duel
You have 3 options to consider:
(1) Shoot at 50% guy. Death comes with: 72% likilhood (2) Shoot at the 100% guy. Death comes with: 63.64% likihood (3) Shoot into the air (purposely miss). Death comew with 58.33% likihood.
So, you shoot into the air and hope for the best.
Calculations:
The p in the probabilities below represents your probability of dying at some point if you MISS either guy in the first round. It'll be filled in later.
(1) if you shoot at the 50% guy and get him, you're guaranteed to get shot the next round.
prob of dying at some point by aiming at the 50% guy:
(=) (33.3%)(100%)+(66.67%)*(p)
(=) (33.3%)+(66.67%)(p)
(2) if you shoot at the 100% guy and get him, you're left with this geometric sum of probability of getting shot by the 50% guy at some point in the future...it converges.
(=) [(50%)] +
[(50%)(66.67%)(50%)] +
[(50%)(66.67%)(50%)(66.67%)(50%)] + ...
(=) 50% * { 1 + (1/3) + (1/3)^2 + ... }
(=) 50% * {3/2)
(=) 75%
so, your prob of getting shot if you shoot at the 100% guy is:
(=) (33.3%)(75%)+(66.67%)(p)
(=) (24.75%) + (66.67%)(p)
still crummy, but it dominates the alternative, for positive p.
(3) What happens if you miss? ie, what's (p)?
if you miss, the second guy has a choice to make...does he shoot at you, or the other guy? his options:
(1) Shoot you: if he gets you, he's guaranteed to die on the next shot.
if he misses, he has q chance of dying (which we'll get later).
(=) (50%)(100%)+(50%)(q)
(=) (50%)+(0.5)(q)
(2) Shoot at the 100% dude:
If he gets the 100% guy, there some infinite sum representing his probability of getting shot by you:
[(33.33%)] +
[(66.67%)(50%)(33.33%)] +
[(66.67%)(50%)(66.67%)(50%)(33.33%)] + ...
(=) 33.33% * [ 1 + (1/3) + (1/3)^2 + ... ]
(=) 33.33% * [(3/2)]
(=) (50.0%)
So, he chances of getting shot if he shoots at the 100% guy are:
(=) (50%)*(50%) + (50%)(q)
(=) (25%)+(.5)q
No matter what q is, he'd prefer shooting the 100% guy to shooting you.
Now, what happens if *he* misses (the 100%) guy? ie, what's q? if he misses, then the 100% guy has to make a decision:
(1) Shoot you:
he's guaranteed to get you, so his chances of dying are just 50%. (game ends after the next round...the 50% gets only one shot)
prob of dying: 50%
(2) Shoot the 50% guy:
by the same logic, he'd have a 33.33% chance of death (getting shot by you).
So, he prefers to shoot the 50% guy.
So, q is 100% (ie, if the 50% guy misses, the 100% guy shoots him immediately). From that, we know the 50% guy's optimal move, if everyone's around on his first shot. He'd prefer shooting at the 100% guy to shooting at you or purposely missing.
Now, we need to get p, which is our probability of dying if we purposely miss either guy (shoot into the air).
We know the 50% guy shoots at the 100% guy if we miss.
(1) With 50% probability, the 50% guy kills the 100% guy, resulting in a shootout b/w us and the 50% guy.
(=) (50%) * [ (66.67%)*(50%) + (66.67%)*(50%)*(66.67%)(50%) + ...
(=) (50%) * [ (1/3) + (1/3)^2 + ... ]
(=) (50%) * [0.5]
(=) (25%)
(2) He misses him. Then, we get one shot at the 100% guy (after the 100% guy shoots the 50% guy).
Our chances of death are:
(50%)*(66.67%) = (33.33%)
So, p is (25%)+(33.33%) = 58.3%
So, again, our three choices are
(1) Shoot at 50% guy. Death comes with:
(=) (33.3%)+(66.67%)(p)
(=) 72% likilhood
(2) Shoot at the 100% guy. Death comes with:
(=) (24.75%) + (66.67%)(p)
(=) 63.64% likihood
(3) Shoot into the air (purposely miss).
(=) p
(=) 58.33% likelihood.
General Knowledge Questions for Interview
hen aha:!!!
if a hen and a half lay an egg and a half in a day and a half, how many hens does it take to lay six eggs in six days?
solution: hen
if 1.5 hens lay 1.5 eggs in 1.5 days (or 36 hours) then: 1 hen lays 1 egg in 1,5 days or 4 eggs in six days thus 1.5 hens lay 6 eggs in 6 days
if a hen and a half lay an egg and a half in a day and a half, how many hens does it take to lay six eggs in six days?
solution: hen
if 1.5 hens lay 1.5 eggs in 1.5 days (or 36 hours) then: 1 hen lays 1 egg in 1,5 days or 4 eggs in six days thus 1.5 hens lay 6 eggs in 6 days
mensa
if "24 H in a D" means "24 hours in a day", what does "26 L of the A" mean? what about the other 32 puzzles here on my mensa forward reproduction page?
see the others at mensa
thanks matt
see the others at mensa
thanks matt
box o' numbers
arrange the numbers 1 to 8 in the grid below such that adjacent numbers are not in adjacent boxes (horizontally, vertically, or diagonally).
___
| 1 |
=============
| 6 | 4 | 3 |
=============
| 2 | 7 | 5 |
=============
| 8 |
=====
the arrangement above, for example, is wrong because 3 & 4, 4 & 5, 6 & 7, and 7 & 8 are adjacent.
thanks to VJ for this one
solution: box o' numbers
The key (as I see it) is putting the 1 & 8 in the centre spots - which is required, because those spots both border all but one of the other spots, and 1 & 8 are the only numbers that are only adjacent to one number.
From there, the 2 & 7 are forced, then you have your choice of the next number, which then forces the rest. Really, though there are only two valid solutions, and they are mirror images of each other.
___
| 1 |
=============
| 6 | 4 | 3 |
=============
| 2 | 7 | 5 |
=============
| 8 |
=====
the arrangement above, for example, is wrong because 3 & 4, 4 & 5, 6 & 7, and 7 & 8 are adjacent.
thanks to VJ for this one
solution: box o' numbers
The key (as I see it) is putting the 1 & 8 in the centre spots - which is required, because those spots both border all but one of the other spots, and 1 & 8 are the only numbers that are only adjacent to one number.
From there, the 2 & 7 are forced, then you have your choice of the next number, which then forces the rest. Really, though there are only two valid solutions, and they are mirror images of each other.
Even Harder Coin Problem
it is similar but much more difficult.
from my buddy tom:
> ok, here's a tough one (i thought). there are no "aha!" tricks - it
> requires straightforward deductive-reasoning.
>
> you have 12 coins. one of them is counterfeit. all the good coins weigh
> the same, while the counterfeit one weights either more or less than a
> good coin.
>
> your task is to find the counterfeit coin using a balance-scale in 3
> weighs. moreover, you want to say whether the coin weighs more or less
> than is should and, and this is the real kicker, your weighs must be
> non-adaptive.
>
> that is, your choice of what to put on the balance for your
> second weigh cannot depend on the outcome of the first weigh and your
> decision about what to weigh for round 3 cannot depend on what happened on
> either your first or second weigh.
>
> for example, you can't say something like "take coin #1 and coin #2 and
> weigh them. if they balance, then take coins 3,4,5 and weight them
> against 6,7,8...if 1 and 2 don't balance, then weigh #1 vs #12..." you
> have to say something like:
>
> round #1: do this
> round #2: do this
> round #3: do this
>
> if the results are left tilt, balanced, and left tilt, respectively, then
> coin #11 is heavier than it should be.
>
> this problem is solveable...it took me about 1-2 hours of working on it to
> get it. i think even finding the counterfeit using an adaptive solution
> is tough. then non-adaptive constraint makes it quite hard and having to
> find whether it's heavier and lighter is cruel and unusual riddling ;-)
>
> have fun...
from my buddy tom:
> ok, here's a tough one (i thought). there are no "aha!" tricks - it
> requires straightforward deductive-reasoning.
>
> you have 12 coins. one of them is counterfeit. all the good coins weigh
> the same, while the counterfeit one weights either more or less than a
> good coin.
>
> your task is to find the counterfeit coin using a balance-scale in 3
> weighs. moreover, you want to say whether the coin weighs more or less
> than is should and, and this is the real kicker, your weighs must be
> non-adaptive.
>
> that is, your choice of what to put on the balance for your
> second weigh cannot depend on the outcome of the first weigh and your
> decision about what to weigh for round 3 cannot depend on what happened on
> either your first or second weigh.
>
> for example, you can't say something like "take coin #1 and coin #2 and
> weigh them. if they balance, then take coins 3,4,5 and weight them
> against 6,7,8...if 1 and 2 don't balance, then weigh #1 vs #12..." you
> have to say something like:
>
> round #1: do this
> round #2: do this
> round #3: do this
>
> if the results are left tilt, balanced, and left tilt, respectively, then
> coin #11 is heavier than it should be.
>
> this problem is solveable...it took me about 1-2 hours of working on it to
> get it. i think even finding the counterfeit using an adaptive solution
> is tough. then non-adaptive constraint makes it quite hard and having to
> find whether it's heavier and lighter is cruel and unusual riddling ;-)
>
> have fun...
Monday, October 8, 2007
Linux Admin Questions for Interview
1. Advantages/disadvantages of script vs compiled program.
2. Name a replacement for PHP/Perl/MySQL/Linux/Apache and show main differences.
3. Why have you choosen such a combination of products?
4. Differences between two last MySQL versions. Which one would you choose and when/why?
5. Main differences between Apache 1.x and 2.x. Why is 2.x not so popular? Which one would you choose and when/why?
6. Which Linux distros do you have experience with?
7. Which distro you prefer? Why?
8. Which tool would you use to update Debian / Slackware / RedHat / Mandrake / SuSE ?
9. You’re asked to write an Apache module. What would you do?
10. Which tool do you prefer for Apache log reports?
11. Your portfolio. (even a PHP guest book may work well)
12. What does ‘route’ command do?
13. Differences between ipchains and iptables.
14. What’s eth0, ppp0, wlan0, ttyS0, etc.
15. What are different directories in / for?
16. Partitioning scheme for new webserver. Why?
2. Name a replacement for PHP/Perl/MySQL/Linux/Apache and show main differences.
3. Why have you choosen such a combination of products?
4. Differences between two last MySQL versions. Which one would you choose and when/why?
5. Main differences between Apache 1.x and 2.x. Why is 2.x not so popular? Which one would you choose and when/why?
6. Which Linux distros do you have experience with?
7. Which distro you prefer? Why?
8. Which tool would you use to update Debian / Slackware / RedHat / Mandrake / SuSE ?
9. You’re asked to write an Apache module. What would you do?
10. Which tool do you prefer for Apache log reports?
11. Your portfolio. (even a PHP guest book may work well)
12. What does ‘route’ command do?
13. Differences between ipchains and iptables.
14. What’s eth0, ppp0, wlan0, ttyS0, etc.
15. What are different directories in / for?
16. Partitioning scheme for new webserver. Why?
Linux Admin Questions for Interview
1. How do you take a single line of input from the user in a shell script?
2. Write a script to convert all DOS style backslashes to UNIX style slashes in a list of files.
3. Write a regular expression (or sed script) to replace all occurrences of the letter ‘f’, followed by any number of characters, followed by the letter ‘a’, followed by one or more numeric characters, followed by the letter ‘n’, and replace what’s found with the string “UNIX”.
4. Write a script to list all the differences between two directories.
5. Write a program in any language you choose, to reverse a file.
6. What are the fields of the password file?
7. What does a plus at the beginning of a line in the password file signify?
8. Using the man pages, find the correct ioctl to send console output to an arbitrary pty.
9. What is an MX record?
10. What is the prom command on a Sun that shows the SCSI devices?
11. What is the factory default SCSI target for /dev/sd0?
12. Where is that value controlled?
13. What happens to a child process that dies and has no parent process to wait for it and what’s bad about this?
14. What’s wrong with sendmail? What would you fix?
15. What command do you run to check file system consistency?
16. What’s wrong with running shutdown on a network?
17. What can be wrong with setuid scripts?
18. What value does spawn return?
19. Write a script to send mail from three other machines on the network to root at the machine you’re on. Use a ‘here doc’, but include in the mail message the name of the machine the mail is sent from and the disk utilization statistics on each machine?
20. Why can’t root just cd to someone’s home directory and run a program called a.out sitting there by typing “a.out”, and why is this good?
21. What is the difference between UDP and TCP?
22. What is DNS?
23. What does nslookup do?
24. How do you create a swapfile?
25. How would you check the route table on a workstation/server?
26. How do you find which ypmaster you are bound to?
27. How do you fix a problem where a printer will cutoff anything over 1MB?
28. What is the largest file system size in solaris? SunOS?
29. What are the different RAID levels?
2. Write a script to convert all DOS style backslashes to UNIX style slashes in a list of files.
3. Write a regular expression (or sed script) to replace all occurrences of the letter ‘f’, followed by any number of characters, followed by the letter ‘a’, followed by one or more numeric characters, followed by the letter ‘n’, and replace what’s found with the string “UNIX”.
4. Write a script to list all the differences between two directories.
5. Write a program in any language you choose, to reverse a file.
6. What are the fields of the password file?
7. What does a plus at the beginning of a line in the password file signify?
8. Using the man pages, find the correct ioctl to send console output to an arbitrary pty.
9. What is an MX record?
10. What is the prom command on a Sun that shows the SCSI devices?
11. What is the factory default SCSI target for /dev/sd0?
12. Where is that value controlled?
13. What happens to a child process that dies and has no parent process to wait for it and what’s bad about this?
14. What’s wrong with sendmail? What would you fix?
15. What command do you run to check file system consistency?
16. What’s wrong with running shutdown on a network?
17. What can be wrong with setuid scripts?
18. What value does spawn return?
19. Write a script to send mail from three other machines on the network to root at the machine you’re on. Use a ‘here doc’, but include in the mail message the name of the machine the mail is sent from and the disk utilization statistics on each machine?
20. Why can’t root just cd to someone’s home directory and run a program called a.out sitting there by typing “a.out”, and why is this good?
21. What is the difference between UDP and TCP?
22. What is DNS?
23. What does nslookup do?
24. How do you create a swapfile?
25. How would you check the route table on a workstation/server?
26. How do you find which ypmaster you are bound to?
27. How do you fix a problem where a printer will cutoff anything over 1MB?
28. What is the largest file system size in solaris? SunOS?
29. What are the different RAID levels?
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