Google - Microsoft - amzon, yahoo interview puzzles

1.
Title: Blind Man and Cards

ID: blind
Puzzle:
A blind man was handed a deck of 52 cards with exactly 10
cards facing up. How could he divide it into two piles, each
of which having the same number of cards facing up?
Hint:
Yet to write.
Solution:
He divides the cards into two piles with 10 and 42 cards each. He
then flips all cards in the smaller pile.

2.
Title: Rope
ID: rope
Puzzle:
Rajeev is trapped atop a building 200m high. He has with him
a rope 150m long. There is a hook at the top where he
stands. Looking down, he notices that midway between him and
the ground, at a height of 100m, there is a ledge with
another hook. In his pocket lies a Swiss knife. Hmm... how
might he be able to come down using the rope, the two hooks
and the Swiss knife?
Hint:
Yet to write.
Solution:
Cut rope into 50m and 100m pieces. Tie one end of the 50m piece to
the top hook and make a noose at the other end. Pass the 100m piece
through the noose and tie its two ends.

3.
Title: Three boxes and a Ruby
ID: ruby
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?
Hint:
Yet to write.
Solution:
If Bob switches his choice, he wins with probability 2/3.

Title: Choice of Three
ID: three
Puzzle:
In the previous problem, Bob had to pick one of the three
boxes lying on the table. If he wished to select them with
equal probability, how could he do it by using a penny in
his pocket? What if the penny was biased?
Hint:
Yet to write.
Solution:
Toss the coin twice. Let TH, HT and TT correspond to the three
boxes. If HH, repeat.

Title: Treasure Island
ID: island
Puzzle:
An old parchment contains directions to a treasure chest
buried in an island:

"There is an old unmarked grave
and two tall oak trees. Walk from the grave to the left tree,
counting the number of steps. Upon reaching the left tree, turn
left and walk the same number of steps. Mark the point with a
counting the number of steps. Upon reaching the right tree, turn
right and walk the same number of steps. Mark this point with
another flag. The treasure lies at the midpoint of the two
flags"

A party of sailors reached the island. They find a pair of
tall oak trees merrily swaying in the wind. However, the
unmarked grave is nowhere to be found. They are planning to
dig up the entire island. It'll take a month. Can they do any
better?
Hint:
Yet to write.
Solution:
Yet to write.

Title: 30 Coins
ID: coins
Puzzle:
30 coins of arbitrary denominations are laid out in a row.
Ram and Maya alternately pick one of the two coins at the
ends of the row. Could Maya ever collect more money than Ram?
Hint:
Yet to write.
Solution:
The first player could pick all coins in odd-numbered positions or
all coins in even-numbered positions, whichever set is larger in
value.

Title: Cake Cutting
ID: cake
Puzzle:
Mary baked a rectangular cake. Merlin secretly carved out a
small rectangular piece, ate it and vanished! The remaining
cake has to be split evenly between Mary's two kids. How could
this be done with only one cut through the cake?
Hint:
Yet to write.
Solution:
Cut along the line joining the centers of the two rectangles.

Title: Cube Problems
ID: cube
Puzzle:
Imagine a cube on a flat table, tantalizingly balanced on one of
its vertices such that the vertex most distant from it is
vertically above it. (a) What is the length of the shortest path an
ant could take to go from the topmost vertex to the bottommost
vertex? (b) What will be the projection on the table
if there is a light source right above the cube? (c) What would be
the cross-section obtained if we slice the cube along a plane
parallel to the table, passing through the midpoint of the
topmost and the bottommost points of the cube?
Hint:
Yet to write.
Solution:
(a) Square-root of five times the side of the cube, (b) Regular
hexagon, (c) Regular hexagon.

Title: Tiling a Chessboard
ID: chessboard
Puzzle:
An 8x8 chessboard has had two of its diagonally opposite squares
removed, leaving it with sixty-two squares. Is it possible to
tile it with non-overlapping 2x1 rectangles such that all
squares are covered?
Hint:
Yet to write.
Solution:
Both squares have the same color.

Title: Cubes in Cube
ID: cubeslice
Puzzle:
Imagine a 3x3x3 solid cube. How many cuts do we need to break it
into twenty-seven 1x1x1 cubes? How about an n x n x n cube? What
if we are not allowed to stack pieces and cut them together?

Is there a way to repeatedly hop from a small cube to an
adjacent small cube (sharing a surface) within the large 3x3x3
cube such that there is a closed path covering all small cubes,
without ever passing through a small cube twice?
Hint:
Yet to write.
Solution:

The central 1x1x1 cube in a 3x3x3 cube has six faces, no two of
which can be revealed in a single cut. For the general a x b x c
cuboid, we need f(a) + f(b) + f(c) cuts, where f(y) = ⌈ log y
/ log 2 ⌉. This can be proved by induction. Note that at any point,
the "largest" cuboid requires the most steps; others can be sliced
in parallel with this cuboid. When the "largest" cuboid has
dimensions a x b x c, there are exactly 3 choices: to slice along
one of the three sides. For a lying in the range (2^i, 2^(i+1)], any
cut that results in the new "largest cuboid" being z x b x c, where
z lies in the range (2^(i-1), 2^i], is optimal. This is because f(z)
for all these z's is identical. For example, for 10 x 34 x 98
cuboid, any of the following could be the maximal cuboid for the
remaining steps:

 5 x 34 x 1006 x 34 x 1007 x 34 x 1008 x 34 x 100 10 x 17 x 10010 x 18 x 10010 x 19 x 100... and so on till10 x 32 x 100 10 x 34 x 4910 x 34 x 5010 x 34 x 5110 x 34 x 52.. and so on till10 x 34 x 64.

If a slice can pass through exactly one existing cuboid, then the
number of slices is n3 - 1 because a slice increases the
total number of cuboids by exactly 1.

Title: Forty-five Minutes
ID: fortyfive
Puzzle:
How do we measure forty-five minutes using two identical
wires, each of which takes an hour to burn. We have
matchsticks with us. And yeah, the wires burn non-uniformly.
Hint:
Yet to write.
Solution:
Light three out of four ends. When two ends meet, light the fourth.

Section: Easy

Title: Bigger or Smaller
ID: biggersmaller
Puzzle:

Alice writes two distinct real numbers between 0 and 1 on
two chits of paper. Bob selects one of the chits randomly to
inspect it. He then has to declare whether the number he
sees is the bigger or smaller of the two. Is there any way
he can expect to be correct more than half the times Alice
plays this game with him?

Hint:
Yet to write.
Solution:
Let the number revealed to Bob be x. Then Bob should say "bigger"
with probability x, "smaller" otherwise.

Title: Thousand Jars
ID: thousandjars
Puzzle:
Imagine an array of one thousand jars, labeled one thru
thousand. Jars can be colored red or black. Initially, all jars
are red. The color of a jar changes on successive days as
follows: On the i-th day, jars whose labels are divisible
exactly by i, switch their color. How many jars are red at
the end of the thousandth day?
Hint:
Yet to write.
Solution:
A number has an odd number of divisors iff it is square.

Title: Working Computer
ID: workingcomputer
Puzzle:
A room has n computers, less than half of which are
damaged. It is possible to query a computer about the status of
any computer. A damaged computer could give wrong answers. The
goal is to discover an undamaged computer in as few queries as
possible.
Hint:
Yet to write.
Solution:
(a) Pick a computer at random. Ask other computer whether it is
damaged or not.

(b) Yet to write a second solution.

Title: Average Salary
ID: averagesalary
Puzzle:
Four honest and hard-working computer engineers are sipping
coffee at Starbucks. They wish to compute their average
salary. However, nobody is willing to reveal an iota of
information about his/her own salary to anybody else. How do
they do it?
Hint:
Yet to write.
Solution:
The first engineer picks a random k-digit integer for some large k,
adds his salary to it and writes the sum on a chit. The chit is
passed around. When it returns to the first engineer, he subtracts
the k-digit integer.

Title: Number Guessing Game I
ID: numberguessingi
Puzzle:
Shankar chooses a number between 1 and 1000. Geeta has to guess the
chosen number as quickly as possible. Shankar will let Geeta know
whether her guess is smaller than, larger than or equal to the
number. (a) What should Geeta's strategy be? (B) In a modified
version of the game, Geeta loses if her guess is "larger than the
number" two or more times. (c) What if Shankar is allowed to choose
an arbitrarily large number?
Hint:
Yet to write.
Solution:
(a) Binary search. (b) and (c) Guess 1, 4, 9, 16, 25, and so on, to
discover k such that Shankar's number lies between k2 and
(k+1)2. Then guess k+1, k+2, k+3 and so on. On the
whole, this requires O(√n) steps.

Title: Number Guessing Game II
ID: numberguessingii
Puzzle:
Shankar chooses a number uniformly at random between 1 and
1000. Geeta has to guess the chosen number as quickly as
possible. Shankar will let Geeta know whether her guess is smaller
than, larger than or equal to the number. If Geeta's guess is larger
than the number, Shankar replaces the number with another number
chosen uniformly at random [1, 1000]. (a) What should Geeta's strategy
be? (b) In a modified version of the game, Shankar gets to choose a
number arbitrarily / adversarially. (c) And finally, what if Shankar
just says "equal to" or "not equal to" when Geeta guesses (the rest
of the setup remains the same)?
Hint:
Yet to write.
Solution:
(a) Repeatedly guess g = n - n/k until Shankar says "greater
than". Then guess g+1, g+2, g+3 and so on. The expected number of
steps is ≈ k + n/2k, which is minimized for k ≈
√(n/2). (b) and (c) Yet to write.

Title: Four Ships
ID: fourships
Puzzle:
Four ships are sailing on a 2D planet. Each ships traverses
a straight line at constant speed. 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 and two
ships are out of commission. What fate awaits the remaining two?
Hint:
Yet to write.
Solution:
Let z-axis denote time. Then the four trajectories are straight
lines. Three of the lines are known to be coplanar. The fourth line
intersects two of these three. Still, the sixth collision may or may
not happen!

Title: f(f(x))
ID: ffx
Puzzle:
Is it possible to write a function int f(int x) in
C that satisfies f(f(x)) == -x? Without globals and static
variables? Is it possible to construct a function f mapping
rationals to rationals such that f(f(x)) = 1/x?
Hint:
Yet to write.
Solution:
Function f can be defined iff we can divide non-zero integers into
quadruples of the form (a, b, -a, -b). This is because f(0) must be
0. For any other integer "a", let f(a) = b. Then f(b) = -a, f(-a) =
-b and f(-b) = a. Now, C integers are either 32-bit or 64-bit. So
it is impossible to create such quadruples because the number of
positive and the number of negative integers are not the same!

Title: Measuring Weights
ID: measuringweights
Puzzle:
(a) Customers at a grocer's shop always want an integral
number pounds of wheat, between 1 pound and 40 pounds. The
grocer prefers to measure wheat in
exactly one weighing with a beam balance. What is the least
number of weights he
needs? (b) Customers come to a pawn shop with antiques. An
antique always weighs an integral number of pounds,
somewhere between 1 pound and 80 pounds. The owner of the
pawn shop is free to do as many weighings as necessary to
ascertain the unknown integral weight by using a beam
balance. What is the least number of weights he needs?
Hint:
Yet to write.
Solution:
(a) Four weights: 1, 3, 9 and 27. (b) Four weights: 2, 6, 18 and 54.

Title: Counting with a Magical Bird
ID: magicalbird
Puzzle:

An evil goblin assembles 100 gnomes together. He tells them
that they will be locked up into individual cells. Each
cell will have a window and a large supply of
grains. Thereafter, a magical bird will hop from window to
window, ad infinitum. Initially, the bird is white in
color, and it will switch its color from white to black and
vice versa if a grain is fed to it. The bird will be fair
in the sense that it will visit every cell infinitely often.
As soon as some gnome is sure that the bird has visited
every cell, he should say "abracadabra" aloud. If the bird
has indeed visited every cell, all gnomes will be
released. Otherwise, all of them will be killed. The gnomes
have ten minutes to arrive at a strategy. How can they save
themselves?

Hint:
Yet to write.
Solution:
Yet to write.

Title: Troll n Gnomes
ID: trollgnomes
Puzzle:
An evil troll once captured a bunch of gnomes and told them,

"Tomorrow, I will make you stand
in a file, ordered by height such that the tallest gnome can
see everybody in front of him. I will place either a white
cap or a black cap on each head. Then, starting from the
tallest, each gnome has to declare aloud what he thinks the
color of his own cap is. In the end, those who were correct
will be spared; others will be eaten, silently."

The gnomes set thinking and came up with a strategy. How many
of them survived? What if hats come in 10 different colors?
Hint:
Yet to write.
Solution:
The tallest gnome should declare the "parity". For example, he could
say "white" iff the number of white caps he sees is odd.

Section: Moderate

Title: Red-black Squares
ID: redblacksquares
Puzzle:
Given k arbitrary points in a grid of size m by n, is it
always possible to color them either red or black such that
each row and each column is balanced? A row or column is
said to be balanced if the difference in the number of red
and black points in it is at most one.
Hint:
Yet to write.
Solution:
Thanks to Iman
Hajirasouliha
for this elegant solution!
Model the grid with a bipartite graph G[X, Y], a vertex in
part X for each row and a vertex in part Y for each
column. Add edges between xi and yj
iff there is a point in the corresponding place. If G has a
cycle (it must be an even cycle), we color the edges
alternatively and remove those edges, continuing this
procedure we can assume that G is a tree. When G is a tree,
we can solve the problem by induction on the number of edges
as follows: remove e a leaf edge, color the remaining edges
balanced, now look at the parent of that leaf and choose the
right color for e to maintain the induction hypothesis.

Title: Four Cards
ID: fourcards
Puzzle:

Four cards are placed on a square table, one card at each
corner. A blind gnome and an evil goblin take turns to play
the following game. The blind gnome gets to choose a subset
of the four cards and flip them simultaneously. If all four
cards are face up, he wins the game. Otherwise, the evil
goblin get to rotate the table by an amount of his choice.
Show that for any initial configuration of cards chosen by
the evil goblin, the blind gnome can win the game in a
finite number of steps with a deterministic
strategy.

Hint:
Solution: Yet to write.

Title: And-Or Gates
ID: andorgates
Puzzle:
Alice and Bob were working in the hardware design lab one
morning giving final touches to their 4-bit microprocessor,
when suddenly they discovered that three of their NOT gates
were malfunctioning! They opened the inventory cupboard and
received another shock: only two NOT gates were available.

"Look. All we have is hundreds of AND
and OR gates but just two NOT's! Damn! Just six hours before
the deadline. We're out of luck. Any ideas?
Alice.

"Well, we obviously cannot hope to
get three NOTs out of two NOTs
", replied Bob.

"I guess you're right ...", said Alice
with a frown on her face and her shoulders dropping. Then
suddenly she jumped and said, "No, wait!
It can be done! Here it it!
" And pulling out a pen
from her pocket, she sketched a circuit of a piece of paper.

"Wow! You're a genius, Alice!",
cried Bob.
Are you as smart as Alice? In general, how
many signals can we invert using n NOT gates and any number
of AND and OR gates? No other gates may be used.
Hint:
Yet to write.
Solution:
Yet to write.

Title: Dijkstra's Problem
ID: dijkstra
Puzzle:
There are n+1 processors named 0, 1, ..., n. Processor i has
a counter C(i) that takes values in the range [0, n]. Its
initial value is arbitrarily chosen from [0, n]. Processor 0
is said to be privileged if C(0) = C(n). Processor
i, where i > 0, is said to be privileged if C(i) is not equal
to C(i-1).
At successive clock ticks, exactly one out of
possibly several privileged processors is arbitrarily chosen
and its counter is updated as follows:

If processor 0 is chosen, we set C(0) =
(C(0) + 1) mod (n+1).
Otherwise, we set
C(i) = C(i+1).
Prove that after a bounded number of clock
ticks, exactly one processor will be privileged. And that
this will continue to hold forever.
Hint:
Yet to write.
Solution:
Yet to write.

Title: Tiling Problem
ID: tiling
Puzzle:
A big rectangle is composed of smaller rectangles, each having an
integral width or integral height or both. Does the big
rectangle enjoy the same property?
Hint:
Yet to write.
Solution:
Yet to write.

Title: Marked Squares
ID: markedsquares
Puzzle:
Consider an n by n grid of squares. A square is said to be a
neighbour of another one if it lies directly above/below or
to its right/left. Thus, each square has at most four
neighbours. Initially, some squares are marked. At
successive clock ticks, an unmarked square marks itself if
at least two of its neighbours are marked. What is the
minimum number of squares we need to mark initially so that
all squares eventually get marked?
Hint:
Yet to write.
Solution:
Yet to write.

Title: Horses on Auction
ID: horsesauction
Puzzle:
You are the chief guest at an auction, where an unknown
number of horses will be revealed and auctioned, one after
the other, in no particular order. You are a connoiseur of
horses, and can judge whether one horse is 'better' than
another. Being the chief guest, you have a one-time
privilege of selecting a horse, after it is revealed, but
before it gets auctioned off. You get to keep this horse
for yourself. Your objective is to maximize the probability
of selecting the best horse. What do you do?
Hint:
Yet to write.
Solution:
Strategy: Pick the first horse that is superior to all horses that
have been revealed so far. Analysis: If the very first horse is the
k-th best horse, then the probability of picking the best horse with
this strategy is 1 / (k - 1). The overall probability of picking
the best horse is approximately [c + log (n - 1)] / n where c is
Euler's constant. One could extend this idea to wait even more,
i.e., pick the s-th horse that beats all horses that went
by. Identifying the optimal value of s is an interesting
exercise. Is s = 1 optimal?

Title: Red Card
ID: redcard
Puzzle:
Alice repeatedly draws a card randomly, without replacement,
from a pack of fifty-two cards. Bob has a one-time privilege to
raise his hand just before a card is about to be drawn. If the
card drawn is Red just after Bob raises his hand, Bob wins;
otherwise he loses. Is there any way for Bob to be correct more
than half the times he plays this game with Alice?

Hint:
Yet to write.
Solution:
No, Bob cannot be right more than half the times. Yet to write proof.

Title: Geometry without a Ruler
ID: noruler
Puzzle:
Using only a compass (and without a straight edge or a ruler),
is it possible to identify (a) the midpoint of two points? (b)
the center of a circle? (c) all four corners of a square, given
two of them?
Hint:
Yet to write.
Solution:
Using only a compass, all those points can be constructed which can
be constructed using a compass and a ruler. Surprising, isn't it!
The constructions are far more complicated though.

Title: Chessboard with Signs
ID: chessboardsigns
Puzzle:
An 8x8 chessboard has either a plus or a minus written in each
of its sixty-four squares. You are allowed to repeatedly choose
a 3x3 or a 4x4 block of squares and invert all signs within
it. How would you go about getting rid of all the minuses?
Hint:
Yet to write.
Solution:
There are 36 3x3 blocks and 25 4x4 blocks. An arbitrarily long
sequence of inversions is equivalent to inverting some subset of
these 36 + 25 = 61 blocks in any order. Thus at most 2^61 distinct

Title: Five Card Trick
ID: fivecardtrick
Puzzle:
A mathemagician asks a volunteer to give him five cards drawn
from a pack of fifty-two. He hands one card back to the
volunteer and arranges the remaining four in some sequence he
chooses. He then hands the sequence to a second volunteer and
leaves the room. His assistant enters. The assistant asks the
second volunteer to read out aloud the sequence handed to
him. The assistant ponders a little and correctly announces the
identity of the card held by the first volunteer. How could
this be done? In general, how large a deck of cards can be
handled if n cards are drawn initially?
Hint:
Yet to write.
Solution:
Yet to write.

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