Interview Questions And Answers 
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There are four dogs, each at the counter of a large square. Each of the dogs begins chasing the dog clockwise from it. All of the dogs run at the same speed. All continously adjust their direction so that they are always heading straight towards their clockwise neighbor. How long does it take for the dogs to catch each other? Where does this happen? (Hint: Dog’s are moving in a symmetrical fashion, not along the edges of the square).
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when they all reach the center of the square.
Since the dogs are moving symmetrically, the distance seperating them will decrease evenly as they case each other, so the dogs should catch eachother in the center of the room. Their path will be curved, but for sake of approximation, lets se it’s a straigt line to the center.
Assuming that the dogs run at 1m/sec, and room is y x y in diamater.
h^2 = (y/2)^2 + (y/2)^2
h^2 = 2(y/2)^2
h = sqrt(2) * y/2
h = y/sqrt(2) meters
thus it would take the dogs approximtly y/sqrt(2) seconds to catch each other. This is a lower bounds, as the dog will not be traveling in a straight line, but along a curve.
Each dog starts at the corner and moving symmetrically. So each dogs start moving perpendicular to the adjacent dogs. Lets assume v.
So each one start moving with v towards the next dog.
If we see realtive speed of the dog1 (v1), w.r.t dog 2, it changes perpendicularly. So it’ll not affect the time taken along the direction of the dog1 to dog2 and the speed will be v only always.
So it they have started at corners with the distance of the length of the square(s).
Time = s/v. They’ll mee at the center.
Each dog starts at the corner and moving symmetrically. So each dogs start moving perpendicular to the adjacent dogs. Lets assume v.
So each one start moving with v towards the next dog.
If we see realtive speed of the dog1 (v1), w.r.t dog 2, it changes perpendicularly. So it’ll not affect the time taken along the direction of the dog1 to dog2 and the speed will be v only always.
So it they have started at corners with the distance of the length of the square(s).
Time = s/v. They’ll mee at the center.
They are dogs - unless the master says “move” they will not move.
Assuming the master said move at time “t” and they all run at the same speed.
time when the reach the center = t+ (PI x r / 2)/s
PI = 22/7
r = radius of circle = half one side of the square
s = speed
t = sqrt(d/2v); meet at centre;
The path of the dogs will be the same of an ever shrinking square which rotates at a fixed rate inside the bigger square. In other words they will trace a spiral towards the centre.
The time taken would be the time taken for the square’s side to shrink from its original size (say ‘d’) to zero. This is equal to:
t = d/(shrinking velocity)
Now the shrinking velocity can be calculated to bear a proportion of sqrt(2) to the speed of the dogs. In other words, if the speed of the dogs is ‘v’, the shrinking speed is sqrt(2)*v.
So the time taken will be
t = d/(sqrt(2)*v) where
d => length of square side
v => Speed of the dogs.
After time t, 4 dogs arived at 4 points in the square, sinece all dogs runs at same
speed, these 4 points consists a smaller square with same center as original one. At the end 4 dogs meets at center.
If they have an instant reaction time they will never meet. They will deadlock. As soon as a dog faces the dog in it’s clockwise direction it must immediately turn around to face the dog behind it.
If all dogs are moving at the same speed, and kept their movement perpendicular to the dog clockwise from it, they would never meet. But, moving symmetrical, should one dog stray from the perpendicular then the others would as well, causing an inward spiral effect, thus, they would all meet at a point in the center.
It’s obvious that they make a spiral trajectory and all will meet in the center.
Now what time did they spend?
Obviously they all spend the same amount of time because they run the same distance with same speed.
Let’s designate the speeds:
v1, v2, v3, v4
and the distances:
d1, d2, d3, d4
Now:
t = d/v;
t=d1/v1=d2/v2=d3/v3=d4/v4;
Let’s have a closer look for example at:
t=d1/v1=d2/v2;
We can make a transofrmation:
t=d1/d2=v1/v2;
since d1=d2 and v1=v2 …
The conclusion: time is always equal to 1 (units).
Proof for Anand’s distance traveled by a dog = d/sqrt(2):
let D be the side of the rectangle
let s(x) be the side of the shrinking rectangle after the dog advanced a distance x from its initial corner
let du be an incremental advance in the distance x travelled by the dog
by applying geometry,
s(x+du)^2 = (s(x) - du)^2 + du^2
= s(x)^2 + du^2 -2*s(x)*du +du^2
= s(x)^2 + 2*du^2 - 2*s(x)*du
therefore,
s(x+du) - s(x) = sqrt((s(x+du) - s(x))^2)
= sqrt(s(x+du)^2 + s(x)^2 - 2*s(x)*s(x+du))
= sqrt(s(x)^2 + 2*du^2 - 2*s(x)*du + s(x)^2 - 2*s(x)*sqrt(s(x)^2 + 2*du^2 - 2*s(x)*du))
= sqrt(2*s(x)^2 + 2*du^2 - 2*s(x)*du - 2*s(x)*sqrt(s(x)^2 + 2*du^2 - 2*s(x)*du))
deducing the derivative:
ds/du = lim(du->0) (s(x+du) - s(x))/du
[substituting s(x+du)]
= lim(du->0) sqrt(2*s(x)^2 + 2*du^2 - 2*s(x)*du - 2*s(x)*sqrt(s(x)^2 + 2*du^2 - 2*s(x)*du))/du
[reducing du to 0 where appropriate]
= sqrt(2*s(x)^2 + 2*du^2 - 2*s(x)*0 - 2*s(x)*sqrt(s(x)^2 + 2*0 - 2*s(x)*0))/du
= sqrt(2*s(x)^2 + 2*du^2 - 2*s(x)*sqrt(s(x)^2))/du
= sqrt(2*s(x)^2 + 2*du^2 - 2*s(x)*s(x))/du
= sqrt(2*du^2)/du
= sqrt(2)
Integrating to find the distance U traveled by a given dog,
du = ds/sqrt(2)
U = integrate(from s(x)=D to s(x)=0)(ds/sqrt(2)) = D/sqrt(2)
Time taken by the dog T = U/v(dog) = D/(v * sqrt(2))
I think the meeting time should be:
{(2*pi)*(d/2)*(1/4)}/v = (d*pi)/4
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