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imqwerty
 one year ago
REALLY fun question
imqwerty
 one year ago
REALLY fun question

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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4you have got 32 ball and all of them have different weights and you also have a scale to weigh them.What is the minimum number of weighings in which you can find the second heaviest ball?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1The questions gives away a few hints. The number of balls is a power of 2, and an arbitrary one at that. Can we generalise this for a given power of 2?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Or better yet, can it be done for any \(n\)?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.11 ball: does not apply. 2 balls: one time. 3 balls: two times. 4 balls: three times. (Right?)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1OK, so suppose that we have a collection of \(n\) balls. Is it true that with \(n+1\) balls, we would need only one more weight?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Or better yet, if we have a collection of \(n\) balls whose secondhighest weight is known, and we add another ball to it, would it be possible to determine if the newly added weight is the secondhighest weight in only one weighing?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4hey guys so heres the answer 1st  we divide the balls into pair of 2s and therefore we get 16 pairs. We weigh the balls of each pair and take the heavier ball from all the pairs. TOTAL WEIGHINGS = 16 2ndwe divide the 16 balls that we got frm above step into pairs of 2s and we get 8 pairs, we weigh each the balls of each pair and take the heavier balls.TOTAL WEIGHINGS = 8 3rd  we divide the 8 balls into pair of 2s and get 4 pairs and weigh them and get the 4 heavier balls.TOTAL WEIGHINGS = 4 4thwe divide the 4 balls into pair of 2s and weigh them and get 2 heaviest balls.TOTAL WEIGHINGS = 2 5TH we weigh the 2 balls so obtained and get the heaviest ball. TOTAL WEIGHINGS=1 now you might be wondering that we got the heaviest ball and not the second heaviest one. So now heres the way to find it Lets 1st think that how can that second heaviest ball can be eliminated the only way to eliminate the 2nd heaviest ball is to compete it with the heaviest ball which we got in the 5th step. But there can be a case that the second heaviest ball did not get the chance to compete with the heaviest ball and then it came to the final and got out, so this means that the ball who weighed less in the final was the second heaviest ball. So now to make sure that the ball which weighed less in the final was the second heaviest ball we weigh it with the ball who got eliminated by the heaviest ball in the 1st,2nd,3rd and 4th rounds. If the loser of 5th round weighs less than the balls who got eliminated by heaviest ball, then we'll take that winner ball and we'll compete it with the other balls who got eliminated by the heaviest ball so TOTAL WEIGHINGS= 4 OVERALL TOTAL WEIGHINGS = 16+8+4+2+1+4 = 35

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Oh, so it did have to do something with being a power of 2.

rational
 one year ago
Best ResponseYou've already chosen the best response.2This is actually a dynamic programming problem, the answer for general case that parth has started is \[n+\log_2 n2\]

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.0It seems more like a logic problem, like a mind bender type question.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Oh my, I don't think I'd ever have been able to do that. Haha.

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.0the scale must be a comparison scale

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4yes the scale was a comparison scale @Miracrown

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1So this problem can't be solved for an \(n\) that is not a power of 2 because it can have many cases. Hmm.

rational
 one year ago
Best ResponseYou've already chosen the best response.2Here is a slightly easier way to think about this problem in context to dynamic programming : It takes \(n1\) comparisons to find out the first heaviest ball. And the first heaviest ball is compared against exactly \(\log_2 n\) times : dw:1433838513485:dw The second heaviest ball must be one of the \(\log_2 n\) balls that got compared against the first heaviest ball. So it neds \(\log_2n1\) more comparisons.

rational
 one year ago
Best ResponseYou've already chosen the best response.2this problem can be solved for any \(n\), just replace \(\log_2n\) by \(\lceil \log_2 n\rceil \)

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.0I was reading this  something similar but the question is different ... http://classicpuzzles.blogspot.com.au/2006/12/13ballsproblemoneofhardest.html

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.0All your balls are different weights  so you need new logic. I would try putting all the balls into pairs, and doing a weighing of EACH one. and maybe placing them down in a way you remember their partner

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4yes @Miracrown the link has similar questions but i mixed the question with slight twist.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4@Miracrown yes this is the actual approach

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.0Another solution is here... to the same question... but a better answer https://weteachscience.org/mentoring/resources/lessonplans/eightballsweighingproblemlogic

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4@Miracrown how is the question same :)

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.0Well, the goal is to find "the odd weighted ball" but your question is different. If you combine them all on the scale, there is no guarantee you will have the heaviest of lightest ball in any given set...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1The question isn't the same, but the problem is. :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Actually not really.

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.0That is where the popping of a brain vessel comes in

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4i didn't combined all the ball nd weighed. what i did was that i weighed 2 balls at a time @Miracrown

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.0Whatever you did, it was SMART
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