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Riley Thomas wrote:1: Weigh 6 balls on each side. The heavier side (obviously) contains the irregular ball.
Riley Thomas wrote:Assuming there is no limit to the amount of balls that can be placed on the scale, the answer is 3 weighs.
1: Weigh 6 balls on each side. The heavier side (obviously) contains the irregular ball.
2: Of those 6, weight 3 on each side. Again, the heavier side has the irregular ball.
3: Of those 3, weigh any 2. If they're equal, the third ball is the irregular.
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where 'n' is the no. of balls.log n to the base 2
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Orton K Randy wrote:If the left is heavier than the right, the conclusion is that the culprit is in the set we placed in the left tray(assuming we're looking for 1 ball that's heavier than the rest).
Mike Simmons wrote:
Orton K Randy wrote:If the left is heavier than the right, the conclusion is that the culprit is in the set we placed in the left tray(assuming we're looking for 1 ball that's heavier than the rest).
That particular conclusion only makes sense if we know, in advance, that the ball we're looking for is heavier than the rest. You're making an assumption not present in the problem statement  one I've pointed out previously in this thread.
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Deepak Lal wrote:which is the correct answer?
Tariik el berrak wrote:My answer is at the end of the explanation:
Algorithm :
Step 1:
=======
...
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Randall Twede wrote:i fail to see what it matters how much different the ball weighs..
Roger Rammer wrote:Superb! great minds think alike ;)
Riley Thomas wrote:Assuming there is no limit to the amount of balls that can be placed on the scale, the answer is 3 weighs.
1: Weigh 6 balls on each side. The heavier side (obviously) contains the irregular ball.
2: Of those 6, weight 3 on each side. Again, the heavier side has the irregular ball.
3: Of those 3, weigh any 2. If they're equal, the third ball is the irregular.
Orton K Randy wrote:I was asked the same question in my interview months ago. I am surprised no one in here mentioned the log/exponentiation? After trying out for several no. of balls, you'd have noticed a pattern. Ya don't even think twice. Min no. of weighings(in a physical balance of course) required is
where 'n' is the no. of balls.log n to the base 2
So if the no. of balls is 4, you need 2 weighings, for 8 you need 3, for 16 you need 4 and so on. Note that for 12 you need 3 and NOT 4. Likewise for 5,6,7 you need only 2 and NOT 3. So in programming terms, it's floor(log n to the base 2).
You might well be aware of this, still this one's for those trying to work out everytime.
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