My Solution to the 12 Balls Problem

This is a really cool problem. If you like math and don’t want it spoiled, take a look at the other post (The Math Problem Post.)

I suppose I better talk about something for a while to fill up a bit of space before revealing the answer… AH YES! Beer last night was GRAND! I, like I promised myself I would do, took a taxi to go out as to reduce likelyness of injury. The grandness of last night was because of a nice group of people, plus the eventual addition of a table of Taiwanese that was sitting behind our group.

One of the guys at that table was a regular that we see often, so that sort of sparked things. Then we got offered some Taiwan Chewing Gum (aka Betel Nut) all of at the table took some and things went… down hill? (the good kind) from there.

Anyhow, I played some drinking games with’m, chatted it up and had a really good time. Then, on the walk home (from the taxi) I got to see a guy knocked the fuck out from drinking too much. I took a movie with you guys in mind. šŸ˜€ (the movie is being processed by Revver, I’ll put it in a separate post.)

@ The Factory

My Solution

OK, this is how I did it. First the photo. Click the photo to see a larger size:

My Solution Notes

Here’s the run down. The picture might help my explanation… Actually, I can use a bit of what I gathered from the solutions I found online to help.

You have 12 balls. We’ll label them a,b,c,d,e,f,g,h,i,j,k and l.

Your first attemp puts 4 vs. 4, so: abcd vs. efgh (and ijkl sits to the side.)

If it is uneven

Start like this: abcd is the UP/light side, efgh is the DOWN/heavy side. Take off abc, move efg to abc’s spot and put ijk where efg was. Why? You’ll see in a sec, but if the scale was unbalanced, you’ll know that ijk and l are OK, the other 8 would be unknown.

So now we have:

  • abc siting on the side (and unknown)
  • efg d vs ijkh (with ijk all being OK)
  • and l sitting on the other side (and known to be OK.)

I bolded the unknowns, I think that’ll make it a bit easier.

Now 3 things can happen from here

  1. The weight shifts. If the weight shifts, we know the unknown ball is in the “efg” set (and heavey.)
  2. Nothing happens. If nothing happens, the odd ball is either d or h. We don’t know if it’s heavy or light, but we do know that “h” is currently down and “d” is up.
  3. Things go equal. If things get equal, we know the unknown ball is in the “abc” set and light.

Now I know it may look like I’m assuming one side is heavy, but I’m not. If you did this with real balls and a scale, you would see one side down and one up. You can always start with the down (assumed “heavy”) side.

This gives us 3 possible sets

  1. an “efg” heavy set
  2. an “abc” light set
  3. a “d” (up/light) set “h” (down/heavy)

With the 3 balls (efg (heavey), or abc (light)) just match 2 against each other. If it balances, the odd ball is sitting out (and you know whether it’s heavy or not.) If it doesn’t balance, you’ll be able to see which it is (since you already know whether it’s heavy or light.)

With the 2 balls, just match one against an OK ball. d (up) vs. h (down.) Replace d. If no change, h is the ball (and heavy.) If things go to equal, d is the ball (and light.)

WOAH! It’s not done yet, but that was the longer part of the explanation.

What happens if on that first weigh attempt things are equal?

You have abcd vs efgh and ijkl sitting to the side. Now you know the odd ball is in the ijkl set.

Weigh ij vs ka (visualizee ka being heavy/down.) You know a is OK. That’s #2.

Now move i off, k to i and bring another OK ball in. So i is on the side and it’s kj vs. ab.

Like this i | kj vs. ab | cdefgh (bolded unknown.) (this is #3)

Three things will happen (is this part familiar?)

  1. If things change k is the odd ball (and heavy.)
  2. If things go equal i is the odd ball (and light.)
  3. If nothing happens j is the odd ball (and light.)

BOOM! It’s done. If that was hard to follow, I can draw a really good visual description now.

I think it took me longer to write this post than it took me to figure it out šŸ˜›

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4 Responses to My Solution to the 12 Balls Problem

  1. James Lick says:

    It turns out that there’s a lot of ways to do it. Mine was needlessly more complex than yours, however there is a solution that’s a bit simpler than yours for the “if the first test doesn’t balance” case. See the second post in this URL:

    http://mathforum.org/library/drmath/view/55618.html

    His “if the first test balances” case isn’t as good as yours though because it doesn’t determine the weight of the different item. The third post shows how to do that test. Finally there is a fourth post by some dude named Lars which gives a big complex explanation of the problem but I don’t understand a word of it.

  2. Twocs says:

    I have a better solution. Save up some money and buy a scale that tells you the actual weights. Weigh each of the balls individually.

    This method is superior to the one above, because it will help you to find a solution even if there are more than one counterfeit balls.

  3. Frost says:

    Twocs is right, buy a machine. I can’t tell you how many times I have run into counterfeit balls! Umm… forget that last part. (runs away…)

  4. Mark says:

    Oh dear. Did you post this before I posted my nearly identical solution last night? I sure feel silly.

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