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HI,I wanted to know the ans from You for this puzzle..here it is: have 10 marbles on the Jar. how many minimum no. of trials would you take to find a marble which is greater than others?regards

[228 byte] By [Pannara] at [2007-11-27 9:13:24]

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# 1

All marbles are equal. But some are more equal than others

Jamwaa at 2007-7-12 22:00:46 >

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# 2

sorry forgot to mention that all marbles exactly looks like same..

Pannara at 2007-7-12 22:00:46 >

# 3

> sorry forgot to mention that all marbles exactly> looks like same..In which case, you need to decide by what criteria you're deciding on relative greatness

georgemca at 2007-7-12 22:00:46 >

# 4

If there is one that weights more than the others, and want to discover it, I think that with 10 marbles you can discover it using the balance 6 times.

Xtremebcna at 2007-7-12 22:00:46 >

# 5

how will you achive this in real time? whats the solution? do we need to apply some Permutaion formula to find the higer weight marble?

Pannara at 2007-7-12 22:00:46 >

# 6

> how will you achive this in real time? whats the> solution? do we need to apply some Permutaion formula> to find the higer weight marble?What does "real time" have to do with it?

georgemca at 2007-7-12 22:00:46 >

# 7

From what am understanding, this is how the number of times occurs;

Suppose you have 5 identical looking marbles. But one of the five marbles is different from the others, say, weightwise. To find this marble in the quickest way, we may divide the marbles into three groups of 2, 2 and 1.

We then put the groups on the scales. If the first group counter-balances with the second then the odd marble is the lone one. Else one of the groups on the scales contains the rogue marble.

Is the general algorithm?

Jamwaa at 2007-7-12 22:00:46 >

# 8

sorry i meant to say "real life situation", assume am a buyer , wanted to choose the greater one among others.

Pannara at 2007-7-12 22:00:46 >

# 9

> sorry i meant to say "real life situation", assume> am a buyer , wanted to choose the greater one among> others.Grab 'em all!

prometheuzza at 2007-7-12 22:00:46 >

# 10

Jamwa, i think your hints will give the solution and right . There is no need to concern abt algorithm or program.. just puzzle..So how many minimum number of trials will you make to find the greater one?

Pannara at 2007-7-12 22:00:46 >

# 11

Try dividing into 3 groups of 4, 4, 2. then try the "balancing act". Try with different numbers making up the groups (for instance 3, 3, 3, 1) and find out which is the group number combination giving the least number of trials. Over to you now Pannar.

Jamwaa at 2007-7-12 22:00:46 >

# 12

CEIL{LOG[3](N)}

prometheuzza at 2007-7-12 22:00:46 >

# 13

Will tryMessage was edited by: jdolphin

jdolphina at 2007-7-12 22:00:46 >

# 14

> May I know what is the relevance of the puzzle with> this forum?Yes, you may: there is no relevance.

prometheuzza at 2007-7-12 22:00:46 >

# 15

I got it.. Thanks very much to all the Great people whoever spent time for me to give the solution !! :-)

Pannara at 2007-7-21 22:57:22 >

# 16

Just a non-Java brain exercise, dude! Cheer up.

Jamwaa at 2007-7-21 22:57:22 >

# 17

> Just a non-Java brain exercise, dude! Cheer up.He did cheer up: he edited his post.

prometheuzza at 2007-7-21 22:57:22 >

# 18

> I got it.. > Thanks very much to all the Great people whoever> spent time for me to give the solution !! :-)No problem. Feel free to ask another.

Jamwaa at 2007-7-21 22:57:22 >

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