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If you could solve this puzzle in less than a minute,…

You are a genius!

If you could solve it within 2 minutes…,

You are gifted!!

If you are able to solve it within 3 minutes…,

You are sharp!!

If you could crack it with 4 minutes,...

You are clever!!

And if you could solve it between 4 and 5 minutes,…

You have average intelligence!!

But if you needed more than 5 minutes…,

Well…., you’re OK, but remember almost everyone should be able to solve it given the time.

If you've seen it before and already know the answer, please do not participate.

Here it is:

You have 8 similar golf balls, but one of the balls is slightly heavier than the rest of the balls. You can’t tell that by holding the balls with your hands.

Given a balance scale, your mission is to find out which ball is the heaviest (out of the 8 balls) by using the balance scale only TWICE. Remember: only two times!!

When you solve it, please don’t post the solution (in order not to spoil it for the others). Post only how long it took you to solve it, or what your rank is as above.

Hey... Be truthful!!

I'm as stupid as hell

Got two balls left, want to hit them to see which one comes furthest, but am only allowed to use a damn balance scale...

RobbyG wrote:I'm as stupid as hell

Got two balls left, want to hit them to see which one comes furthest, but am only allowed to use a damn balance scale...

are you sure?

Figured it out in about 5 minutes (Give or take 2 mins ).

I guess I have average intelligence!!!

Yahooooooo!!!

I've heard about this a long time ago. Though the balls were 12, and u don't know whether the different one is heavier or lighter. A bit harder but at the end it's a simple tree algorithm.

bedro wrote:I've heard about this a long time ago. Though the balls were 12, and u don't know whether the different one is heavier or lighter. A bit harder but at the end it's a simple tree algorithm.

No.., this one is 8 balls not 12, and no algorithm is involved. It’s pretty simple!

Tip:

Most people tend to try to solve it by dividing all the balls equally and weighing them at the first run, to see which 4 have the heaviest ball. That’s not correct.

You don’t need to use all the balls at the first run. Think of leaving some out.

Many people, especially people rooted in computer science, are quick to settle for 3 weighings. This is most probably due to the O( log2 n ) trail of thought. Or mistakingly thinking the answer is to first weigh balls {1,2,3,4} in one bowl, and {5,6,7,8} in the other, pick the heavier bowl. Then as weighing number 2, weigh the 4 heavy balls, e.g. {1,2} against {3,4}. Finally, the third weighing weighing the two remaining balls one in each bowl, thus identifying the heavier ball, e.g. {3} against {4}.

Anyways I'm sharp... I took only 3 minutes.

Vera.

vera.chandler wrote:Many people, especially people rooted in computer science, are quick to settle for 3 weighings. This is most probably due to the O( log2 n ) trail of thought. Or mistakingly thinking the answer is to first weigh balls {1,2,3,4} in one bowl, and {5,6,7,8} in the other, pick the heavier bowl. Then as weighing number 2, weigh the 4 heavy balls, e.g. {1,2} against {3,4}. Finally, the third weighing weighing the two remaining balls one in each bowl, thus identifying the heavier ball, e.g. {3} against {4}.

Anyways I'm sharp... I took only 3 minutes.

Vera.

It took you 3 minutes solving it using 3 weighings or 2?

I assume you’re sharp because you solved it with only 2 weighings, as required. Right??

Or...

Cmon guys, i'm feeling stupid here...when does the right answer come by?

;(

ofcourse 2

vera.chandler wrote:Many people, especially people rooted in computer science, are quick to settle for 3 weighings. This is most probably due to the O( log2 n ) trail of thought. Or mistakingly thinking the answer is to first weigh balls {1,2,3,4} in one bowl, and {5,6,7,8} in the other, pick the heavier bowl. Then as weighing number 2, weigh the 4 heavy balls, e.g. {1,2} against {3,4}. Finally, the third weighing weighing the two remaining balls one in each bowl, thus identifying the heavier ball, e.g. {3} against {4}.

Anyways I'm sharp... I took only 3 minutes.

Vera.

no need for the third weighing after {1,2} against {3,4}... you could do it without {3} against {4}... and also using his TIP.....

RobbyG wrote:Cmon guys, i'm feeling stupid here...when does the right answer come by?

;(

See my PM to you with the full solution!!

Cheers!

vera.chandler wrote:Many people, especially people rooted in computer science, are quick to settle for 3 weighings. This is most probably due to the O( log2 n ) trail of thought. Or mistakingly thinking the answer is to first weigh balls {1,2,3,4} in one bowl, and {5,6,7,8} in the other, pick the heavier bowl. Then as weighing number 2, weigh the 4 heavy balls, e.g. {1,2} against {3,4}. Finally, the third weighing weighing the two remaining balls one in each bowl, thus identifying the heavier ball, e.g. {3} against {4}.

Anyways I'm sharp... I took only 3 minutes.

Vera.

Vera: A CS princess in the city

The word "algorithm" was misunderstood. So how about O(log2 n)?

Have you seen the lisp proggy that solves it?

For Tom:

I mean by "algorithm" that it's not difficult to generalize a method to solve the problem for any number of balls. E.g.: 8,12,16,... etc

I've solved the 12 balls version so the 8 balls didn't take a min. It doesn't work on 4 ball rounds since u'd have a ball left with no clue in some cases (possibilities, nodes in the tree)...

bedro wrote:

vera.chandler wrote:Many people, especially people rooted in computer science, are quick to settle for 3 weighings. This is most probably due to the O( log2 n ) trail of thought. Or mistakingly thinking the answer is to first weigh balls {1,2,3,4} in one bowl, and {5,6,7,8} in the other, pick the heavier bowl. Then as weighing number 2, weigh the 4 heavy balls, e.g. {1,2} against {3,4}. Finally, the third weighing weighing the two remaining balls one in each bowl, thus identifying the heavier ball, e.g. {3} against {4}.

Anyways I'm sharp... I took only 3 minutes.

Vera.

Vera: A CS princess in the city

The word "algorithm" was misunderstood. So how about O(log2 n)?

Have you seen the lisp proggy that solves it?

For Tom:

I mean by "algorithm" that it's not difficult to generalize a method to solve the problem for any number of balls. E.g.: 8,12,16,... etc

I've solved the 12 balls version so the 8 balls didn't take a min. It doesn't work on 4 ball rounds since u'd have a ball left with no clue in some cases (possibilities, nodes in the tree)...

Thanks Bedro for the explanation.

Understood,.. even though I hated math in school!!!

Ur welcome buddy,

I hate maths now since it's only accounting figures, liabilities, payables and sh1t