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        riddles >> microsoft >> Billiard balls
        
(Message started by: Nick on Feb 13^th, 2003, 12:24am)

Title: Billiard balls
Post by Nick on Feb 13^th, 2003, 12:24am

Dear all,
I just found this site. Super debate going on..:)
A few other MS questions.
Feel free to debate over the answers.

Suppose you had 8 billiard balls, and one of them was slightly heavier, but the only way to tell was by putting it on a scale against another. What's the fewest number of times you'd have to use the scale to find the heavier ball?

Nick

Title: Re: Billiard balls
Post by redPEPPER on Feb 13^th, 2003, 4:48am

Two times.

This resembles very much another riddle that's somewhere on the site: 12 balls look identical. They weight the same, except 1. Find it and tell whether it's lighter or heavier with only three uses of the scale.

Title: Re: Billiard balls
Post by Alex Wright on Mar 3^rd, 2003, 9:22pm

Actually the way to be absolutely sure with out taking a random guess at the beginning as to which ball, then it would take three weighs.

First make two groups of four, wheigh the groups against each other, discard the light group.
Repeat the last step with two groups of two.
You are left with two balls, one of which must be the heavier ball.

Title: Re: Billiard balls
Post by BNC on Mar 4^th, 2003, 12:58am

on 03/03/03 at 21:22:36, Alex Wright wrote:

Actually the way to be absolutely sure with out taking a random guess at the beginning as to which ball, then it would take three weighs.

It can be done with only two weights without guessing.

Title: Re: Billiard balls
Post by Dard on Mar 17^th, 2003, 9:33pm

Here's how I would do the original 8 ball question assuming that you know for a fact one of the balls is heavier.
[hide]1. Split the balls up into 3 groups of sizes 3, 3, 2 [/hide]
[hide]2. Weigh the two groups of 3[/hide]
[hide]3. If they are equal, weigh the 2 remaining balls to find the heavier one.[/hide]
[hide]4. If one of the two groups of 3 is heavier, pick 2 balls from the heavier group to weigh.[/hide]
[hide]5. If they are balanced, the ball you omitted is the heaviest, otherwise you will know from the scale.[/hide]

Title: Re: Billiard balls
Post by AJShred on Mar 24^th, 2003, 7:55am

Actually, two is the correct answer. First split them into 2 groups of three with two balls leftover. Weigh the groups of three. If they are even, it's one of the leftover two balls. Weigh them against each other and you're done. That's the easier of two possible scenarios. The second senario is if the weighing of the groups of three balls are not even. If this is the case, discard the lighter group. You have narrowed it down to the three in the heavier group. Weigh two balls from the heavier group against each other. If they weigh even, it's the ball you didn't weigh from the group. If they are not even, it's the heavier of the two. There it is. Two weighs, no waiting. 8)

Title: Re: Billiard balls
Post by Rujith de Silva on Apr 10^th, 2003, 9:07am

Actually, it's possible to test NINE balls with only two weighings under these conditions. The problem is trivial, anyway.

Title: Re: Billiard balls
Post by THUDandBLUNDER on Apr 21^st, 2003, 2:10pm

I think the formula for the maximum number of balls we can handle (without knowing whether one is lighter or heavier) is, where w = the number of weighings, (3^w - 3)/2

Title: Re: Billiard balls
Post by Icarus on Apr 21^st, 2003, 4:19pm

That would mean that for two weighings, 3 balls is the max! I think I can better that result. :P

Title: Re: Billiard balls
Post by cho on Apr 21^st, 2003, 5:08pm

Thud is answering the other riddle, where you don't know if the oddball is lighter or heavier, and you have to plan all the weighings ahead of time. And you have to figure out whether the oddball is light or heavy (otherwise the max would be one more, a ball that's never weighed).

Title: Re: Billiard balls
Post by Icarus on Apr 24^th, 2003, 6:22pm

Ack! When will I ever learn to read the post again carefully when it doesn't make sense?

Thanks.

By the way, cho, why don't you register? I think you are by far the most prolific "visitor" on this site.

Title: register
Post by Boody on Apr 25^th, 2003, 2:19am

on 04/24/03 at 18:22:41, Icarus wrote:

By the way, cho, why don't you register? I think you are by far the most prolific "visitor" on this site.

I agree with Icarus, You should register.
Lets go, your great contributions will not be affected by the registration. :)

Title: Re: Billiard balls
Post by abalagoo on Sep 22^nd, 2005, 12:25pm

3 times.

Actually, what I did before is 12 balls without knowing the weight, the solution is pretty same. I think this forum must have that.

Hint is exchange.

Title: Re: Billiard balls
Post by VishalManocha on Aug 27^th, 2006, 4:12am

This can be solved in two weighings.

First name the balls as 1,2,3,4,5,6,7,8

Weigh 123 and 456 first.

1st Case : If both are identical we know the heavier ball is in 7 or 8.
Weigh them and will come to know.

2nd Case : If 456 weighs more , then Weigh 4 and 5 , will come to know which is heavier , if both are identical then 6 is heavier.

3rd Case 123 weighs more. Appy same as 456.

So with 2 weighings we can come to know which is heavier