I interviewed 68 applicants, none of them could answer this question

I interviewed 68 applicants, none of them could answer this question.

Can Sup Forums do better?

Other urls found in this thread:

youtube.com/watch?v=tE2dZLDJSjA
curiouser.co.uk/puzzles/12balls.htm
youtube.com/watch?v=cB8GNQuyMPc
twitter.com/NSFWRedditImage

Not doing your homework for you.

/thread

You scale the balls against each other and see what ball is abnormal.

They give you the answer right there cshitter Pajeet

If they're identical, they all have to weigh the same, too.

That implies looks only.
All of those questions are just gotcha bullshit.

The weights are irrelevant, they could all be heavier than the balls. If you're going to trick people, do it well, this way you're just being a dick to your applicants.

Weigh the balls against each other, B1 vs {B2, ..., Bn}.

3 weighs, not 3 weights. it's fucking simple if you can't extrapolate from there.

Split in 4 groups, take 2, weigh, take other 2, weigh, put the groups that were even aside, compare one of the uneven groups with one of the even groups and draw your conclusion.

compare 6 to 6, which one has the different weight ball because it doesn't specify is it heavier or lighter?

First I weigh 6 vs 6. The lighter side contains the lighter ball.
Then I weigh 3 vs 3 with the same outcome.

3 balls left. Then I weigh 1 vs 1. Either I get the result from that or I don't, in which case I know that the left over ball has to be the one I am looking for.

What the fuck, how could anyone not solve this? It's even a "riddle" where the most obvious solution turns out to be the right one when you think it through, too.

You need to find the ball that has the different weight and determine if it is heavier or lighter than the other balls.

You haven't determined if the ball is heavier or lighter, you've just assumed that the target ball is lighter.

who said it was lighter :^)
also you only have 3 attempts on the scales

Right, my bad.

>tell my students to carefully read the question first
>don't do it myself

I'm trash.

Divide and conquer. You can even find this out with 2 weighs. 3rd is really just to confirm the result. Weight 6v6 then then 1v1, done.

fucking uneducated Sup Forums neets

>education
>kid riddle
I don't even think this riddle has to do anything with logic.

>Weight 6v6 then then 1v1, done.

This doesn't sound right at all

Is that even possible? With 3 weighs you can only determine either different ball of whenever it lighter or heavier, but not both.

...

Its not a fucking riddle. Its a task with hard constraint.

shit question, interview dropped

It's impossible, it's a riddle from youtube. But they don't play by their own rules. Using a scale implies putting something on both sides and weighing it. But on youtube they use it 4 times not 3.
The become very vague about what "using" a scale implies.

Ex for yt:
4BALLS, 4BALLS, 4BALLS

Pick out odd group, scale uses: 2 times

2BALLS (ABNORMAL), 2BALLS (NORMAL) Pick out odd group, if present else you know one of the other 2 balls has an issue. scale uses: 3 times

1BALL, 1BALL If the balls weighed are both equal the remaining ball is the culprit else the current will be. Scale uses: 4

As you can see the riddle is just plain wrong. The author of the riddle only counted comparing the 4 groups 2 times as a 1 time use of the scale.

lol

...

>Weight 6v6 then then 1v1, done.
You have about a 1/12 chance of getting the correct answer from that.

You don't know if the odd ball is heavier or lighter

You don't know if different ball is lighter or heavier.

But what if the ball is lighter? Then you'd be weighing two balls that weigh the same in the last step.

What if the abnormal ball is lighter, and it's in the first groups you weigh?

That would only work if the target ball is heavier, not lighter.

Next.

Thanks. That's the conclusion I came to as well. If the weight difference of the unknown ball is given it can be done in 3 uses of the scale, otherwise you need the 'test' group of 4 balls.

This one is wrong there is no way to figure out which group is the odd one without comparing the groups of 4 balls twice. Or at least not as illustrated in the picture.

12 / 2 = 6
6 / 2 = 3
compare any of the three balls. If you pick 2 which are the same weight, the third is the different one

>You have 12 identical balls.
>One of them is slightly different

>all identical
>one is different
???

Are you fucking retarded? How the fuck do you know if it's heavier or not you dumb piece of shit?

>12 identical balls
>one is heavier
hmmmm

Deconstruct the scale to make a ramp. Roll all the balls from the same point. The one with the different weight will be obvious.

Done in 0 fucking weighs.

>"There is no need to weight, they're identical."
>"You're hired!"

It won't matter if they are equally solid.

Drop the balls from the window, the one with the abnormal terminal velocity is the ball we're looking for.

If the weight was different enough you could accurately measure the difference in terminal velocity then you would have no trouble feeling the weight difference in your hands.

are you retarded

That only measures aerodynamics, which is equal.

Pick each ball up one by one
Drop them one by one
One that falls the fastest or the slowest is the heaviest or lightest ball

Its impossible to know without it being specified whether the odd ball is either heavier or lighter.

You can make 3 groups of four balls, and if you know if it's supposed to be heavier or lighter you can eliminate two groups with one weigh.

So now you have four balls left. Simply weigh them against each other individually with your two remaining weighs and pick out the odd ball.

This, but do it twice. First pick the group that goes down, if that evens up pick the group that goes up.

>The author of the riddle only counted comparing the 4 groups 2 times as a 1 time use of the scale.

You don't have to. Compare two of the three sets: if it's the same the third is the odd one out, if they're not it's one of the 8 you weighed.

I don't know about the youtube video but I had this in a real interview. The key is to use all the information you get from a weighing:
>Are they different?
>Which direction does it go in?

Why not just compare the balls to each other?
There is no need to use the weights

Drop each ball from equal height. A weight disparity will be reflected in the recorded bounce height.

The balls will all reach the ground at the same time idiot

This is LITERALLY grade school physics

Read the question again mong

>Hurr it's impossible

You failed the interview before even getting to the whiteboard IDE

You still can't though. Even if you get to 1v1 at the third weigh, if all you get is a scale tipping, you'd still don't know which one is the odd one, you'd need yet another weigh.

No, because you also know which direction it went in and use that information.

Just drop them all from the same height. The heavier ball will fall faster.

woah dude

That only measures which one is heavier, which is an assumption, the odd one could be lighter.

But you don't know whether the off ball is supposed to be lighter or heavier... So you have no way of knowing whether your supposed to weight H1 v H2 or H3 v H4 on your third weighing.

youtube.com/watch?v=tE2dZLDJSjA

Your post made me remember about this video I watched a few months ago.

It tells you exactly how to solve the problem.

doesn't matter because the H's are from the heavier group.

Are you actually going to put any thought into the diagram or just assume you're right and this -famous puzzle- is impossible?

You don't need to know because that's solved using the information from earlier about which direction it went in.

curiouser.co.uk/puzzles/12balls.htm

A ball that isn't solid will roll differently, but if they are equally solid, it doesn't matter their mass, they will roll the same speed.

youtube.com/watch?v=cB8GNQuyMPc