On Tue, Sep 8, 2009 at 9:52 AM, Kiran K Karthikeyan <
[email protected]> wrote:

> I do a lot of interviews, both for Product Managers and for Tech profiles.
> I've had two questions which I've continuously asked, but need to change.
> I've enjoyed the responses of various candidates for these questions for
> almost 3 years now and it has been very interesting to see how different
> people think through them.
>
> Question 1:
>
> There are 12 steel (or any other material, the material is immaterial :) )
> balls (yes, I make it cubes when interviewing women) which were
> manufactured
> to be identical in every way and hence indistinguishable. However, one of
> them has a manufacturing defect and has either less or more weight that the
> other 11. Given a weighing scale (with no standard weights), you have to
> find out which of the balls is defective as well as whether it weighs
> lesser
> or more than the others.
>
> Obviously, there is no solution in 3 weighings



Hope you did not reject candidates for saying this is possible in 3
weighings.  Haven't verified this completely -- it is possible I missed
something.  The approach should be obvious, though.  (As Thats mentioned,
this is something I first encountered and solved in school.)

(Sorry about the rich text mail.  It is an attempt to keep the formatting
for the ASCII art below to keep from getting messed up.)

Divide the balls into 3 groups of 4 each -- a1-a4, b1-b4, c1-c4.  Keep aX
balls on the left pan, bX balls on the right and cX ones outside.

  a1 a2 a3 a4       b1 b2 b3 b4
  ____________  W1  ____________
     ------------------------
                /\                      c1 c2 c3 c4

Possibilities:
    - Pans balance
      Implies: One of the cX balls is defective

     c1 c2            c3 X
   ___________  W2  ___________
     ------------------------
                /\                      c4

    (X is one of the other balls we know is not defective.)

    Possibilities:
        - Pans balance
          Implies: c4 is the defective ball
                   A weighing against any other will tell you
                   if it is lighter or heavier

        - Left pan is heavier
          Implies one of c1, c2 is heavier or c3 is lighter

       c1               c2
   ___________  W3  ___________
     ------------------------
                /\                      c3

         If the balance now shifts to the other pan, c2 is the
         defective ball and it is heavier.  If the left pan
         continues to be heavier, c1 is the defective ball and
         is heavier.  If the pans balance, c3 is the lighter
         ball.

Getting back to the other possibilities for the first
weighing:

Possibility:
    - Left pan is heavier  (Same analysis applies if the left
      is lighter.)
      Implies: Either one of the aX balls is
      heavier or one of the bX balls is lighter.

   a1 a2 b1         a3 b2 X
   ___________  W2  ___________
     ------------------------
                /\                  a4 b3 b4

    Possibilities:
        - Left pan stays heavier:
          Implies: a1 or a2 is heavier or b2 is lighter.
          (Those are the only balls that did not switch
          sides.)  Weigh a1 against a2.  If they are equal, b2
          is lighter.  Or else, the heavier ball is defective.

        - Right pan is now heavier
          Implies: One of the balls that switched sides is
          defective, i.e., b1 or a3.  Put them on the same pan
          and weigh them against two of the other
          non-defective balls.  If the pan with a3 and b1 is
          heavier, a3 is the defective and heavier ball.
          Otherwise, b1 is the defective and lighter ball.

        - Pans balance:
          Implies: a4 is heavier or one of b3 or b4 is
          lighter.  Weigh b3 against b4.  If the pans balance,
          a4 is defective.  Otherwise, the lighter one of b3
          and b4 is defective.


-- 
One hundred thousand lemmings can't be wrong.

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