### FactFinder's Super Genius IQ Test - Cannonballs Revisited

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Many of you did excellent work in solving Problem 3 of my MiniIQ Test thread below, in some cases, using methods I didn't think would work! Now, if you're really so smart, solve it in ....3 weighings. (Yes, it can be done)

Problem 3.

There are 12 cannonballs.  All look alike, but one is the "oddball." The oddball is either heavier or lighter than the other balls. You are supplied with a balance scale that can hold as many cannonballs as you would like on each side of the scale.  The problem is, in 3weighings (4 uses of the scale), find the oddball.

-- FactFinder (FactFinder@bzn.com), July 09, 2000

This was in the puzzles section of Scientific American back when I could understand the questions. Nice find.

Hint: Number the balls and don't forget what you learn about each.

Please too, don't publish the answer, just let us know when you solve it.

-- Carlos (riffraff1@cybertime.net), July 09, 2000.

As usual, "Carlos" never reads or understands a thing. Solutions were already posted. http://hv.greenspun.com/bboard/q-and-a-fetch-msg.tcl? msg_id=003SZk

-- cpr (buytexas@swbell.net), July 09, 2000.

My dear little CEEP. Think you missed the "oddball" part, again.

-- Carlos (riffraff1@cybertime.net), July 09, 2000.

The answer is still there if you think about it. Like the distance the bird travels solution is found outside the box.

Side bar, the speed of the bird is not equal to the Velocity which approximates Hawk's IQ.

-- cpr (buytexas@swbell.net), July 09, 2000.

OK. For cpr ALONE. The ball is EITHER light or heavy. Got it TWIT?

-- Carlos (riffraff1@cybertime.net), July 09, 2000.

Carlos,

Just remind cpr that IT'S A SUBSET OF THE SET!!!!!!!!!!

-- J (Y2J@home.comm), July 09, 2000.

Doesn't matter. Everything is relative. You need to "think" about it and that is always your big problem.

-- cpr (buytexas@swbell.net), July 09, 2000.

cpr,

Do you mean like "thinking" about Heller's math in the cell phone thread before declaring that it was wrong?

-- J (Y2J@home.comm), July 09, 2000.

Carlos, I didn't see the puzzle in the magazine, as I mentioned in the earlier thread (Mini IQ Test) I got the problems from a book "The Ideal Problem Solver" by Scientific American.

Here's a hint to solving the cannonball question in three steps. Eliminate as many balls as possible at each weigh. Weighing all 12 during the first weigh eliminates nothing! Try other numbers...3, 4, 5 each side. As carlos points out, the "oddball" is either heavier OR lighter.

I worked the problem with sketches and listing the possible results for various methods. I found one solution. I doubt that there are more than one, but I am sure if there are, you guys will find it...

-- FactFinder (FactFinder@bzn.com), July 09, 2000.

Hey Factfinder I posted this about a month ago. I think chimming my two cents in got it right. Good puzzler.

-- Maria (anon@ymous.com), July 10, 2000.

Ahh, I didn't see the post, Maria, thanks for letting me know. Did chiming in do it in three weighs? (The original from the book I have gives 4 weighings)

-- FactFinder (FactFinder@bzn.com), July 10, 2000.

Must be serious hint time. As with the previous problem No3 you can afford no more than 3 unknowns at the final weighing. Forced to weigh 4 vs 4 initially assume an imbalance. Still leaves 8 unknowns but also some useful info about them. How to pare down to 3 for the 2nd weighing is the crux. You can weight 3, leave 3 unweighed but what of the pesky other 2? Try swapping them.

-- Carlos (riffraff1@cybertime.net), July 10, 2000.

I must be slipping, of the 24 possible permutations, in my best solution I can only determine 20 of them with complete certainty after 3 weighings. The remaining 4 permutations are centered on 2 of the cannonballs.

After 3 weighings, if "A" is heavy the result is indistinguishable from "B" being light. Again, if "B" is heavy, it gives the same result after 3 as if "A" were light. In 4 weighings I can do it fine.

I may give up before I solve it. Give me another day.

-- Brian McLaughlin (brianm@ims.com), July 11, 2000.

I solved it in 3 weighings. I understand we are not to post it?

-- Debra (...@....), July 11, 2000.

LOL ... hold on ... maybe I didn't :)

-- Debra (...@....), July 11, 2000.

Yes it was in three weighings. But I gave a hint:

You have nine balls, one of them weighs heavy. Find the odd ball in two weighings. The way you solve this puzzler is a hint for this one. You will have to know certain things about the balls on the last try. Flint got this one but I don't believe he solved the one with twelve.

-- Maria (anon@ymous.com), July 11, 2000.

If you know in advance whether the oddball is heavy or light, then it is easy to do in three weighings.

In the "hint" problem you know in advance that the oddball is heavy, so when you get to the second weighing (piece of cake to be left with three suspects in one weighing), you can find the oddball by which way the scale moves. Not so in the 12-ball problem.

There's no dificulty to reduce twelve balls to three with two weighings (thus making it equivalent to the nine-ball problem with one extra move), but nothing in those two weighings is guaranteed to tell you whether the oddball is heavy or light. All you can go by is whether the scale moved or not, and you might well get through the first two weighs and never had the scale move at all!

Unless the oddball is known in advance to be heavy or light, it cannot be done in three. That's the difference between the two problems - three balls can be tested in one weigh (one on each side of the scale, and one not weighed at all), only if the direction that the scale moves will give you information. As stated, it will not do so.

-- Paul (anonymous@coward.com), July 11, 2000.

Paul, you'll need to keep thinking about it. Actually there's more than one way to "skin this cat". But all solutions have a "theme" on the technique. You know "direction" (heavy or light) after the first weighings and you must associate it with each object.

After my brother gave me the hints, I took about 30 minutes to solve the puzzler.

-- Maria (anon@ymous.com), July 11, 2000.

Doh!

It wasn't a matter of a cat-skinning problem (As I said, it was obvious to me how to get down to three balls after the second weighing). What I didn't realize at the time is that the second weighing also tells you whether the oddball is heavy or light! In other words, you do it the same way as the four-weighing solution, just pay a bit more attention to the results.

This is one of the best logic problems I've come across in a long time. It almost makes all the annoyed lurking I've done in these forums worthwhile. Thanks factfinder!

-- paul (anonymous@coward.com), July 11, 2000.

Sorry for not getting back to the thread sooner, Carlos has asked if he can post the answer and since his hints were right on target, I'm sure he will post the right answer...he better, lol.

For anyone who still wants to work for themselves: *Weighing 4 on each side for the first weigh might be a good idea ;) * As Carlos suggested, numbering the balls will help you keep up with what you learn about each ball i.e., ball 1 is either "heavier or good" or "lighter or good", etc.- remember that the oddball can be light or heavy). * For the second and third (final) weigh, you also *might* want to swap some balls around...

-- FactFinder (FactFinder@bzn.com), July 12, 2000.