Try out the ODDBALL logic test?

Discussion in 'Intelligence & Machines' started by Alan McDougall, Jul 15, 2010.

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1. DywyddyrPenguinaciously duckalicious.Valued Senior Member

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You really, really really, do have comprehension problems don't you?
A total of 3 weighings. That's what ONLY means. No more than three. Three is the maximum allowed. You cannot do more three weighings. Four or more is not permitted.
Which part of this is giving you trouble?

3. John99BannedBanned

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It is a loophole. It is a technicality. You get a total of three times to weigh twelve balls. He did not say HOW to weigh them.

5. StryderKeeper of "good" ideas.Valued Senior Member

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Actually the problem was mathematical and some basic conclusions like how you weigh them were probably drawn. In fact you've actually hit upon one of the main problems with people getting the wrong answers in exams for not understanding the question properly. It's not your fault, it's just the way you process the question is different from the person that posed it.

7. DywyddyrPenguinaciously duckalicious.Valued Senior Member

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Bollocks.
It's a part of the problem. It's a logic problem, hence the restriction.

Correct. You can weigh them any way you like, providing
A) you use only the balance scale and
B) you do not use more than a total of three weighings.

8. John99BannedBanned

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Well i am quoting the original post so it does not get edited.

9. DywyddyrPenguinaciously duckalicious.Valued Senior Member

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So rather than try to defend your indefensible position you've resorted to obfuscation.
The quotes I gave to show where and how you were wrong were direct (unmodified) quotes.

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11. StryderKeeper of "good" ideas.Valued Senior Member

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My initial conclusions looked at a multipath solution, where by after step 1, paths would potentially split based upon outcome. This wasn't the correct way to analyse it because trying to create such a solution through this method was actually making it more difficult.

That's why the simplest outcome existed in a Matrix array as it covered the steps. Obviously if you were to use it for testing, you'd have to cross off the letters that appear when the scales are used for testing the balls

Don't mouseover if you want to work it out yourself.

12. DywyddyrPenguinaciously duckalicious.Valued Senior Member

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@ Cifo.
Huh?
That looked pretty good, except for.... (check your PMs).

13. Captain KremmenAll aboard, me Hearties!Valued Senior Member

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Which formerly banned member are you anyway?

Re Matrixes.
I never learned matrixes at school. Too Old.
If matrixes can solve the problem simply, I'll make the effort and depart from my formerly matrix free existence.
Will I have to have big tubes stuck in my back?

Last edited: Jul 15, 2010
14. DywyddyrPenguinaciously duckalicious.Valued Senior Member

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:spank:
Innocent until proven guilty, remember?
We do get new members now and again you know.
Not everyone in the world is a Sciforummer or former Sciforummer.

Nah, we'll do that for you.

Last edited: Jul 15, 2010
15. Captain KremmenAll aboard, me Hearties!Valued Senior Member

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I won't reply to that or I'll be banned myself.
Cifo. Next time try working your way in slowly. Start with the Star Trek thread or something. Use a crap avatar. Don't be so witty.

I'll be interested when the mathematical solution is given, because I don't think it is a succession of doing the same calculation.
I think some of it is trial and error.

Last edited: Jul 15, 2010
16. StryderKeeper of "good" ideas.Valued Senior Member

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Meh.. I put in the wrong answer, fixed now

The following is a test Matrix, the idea being that no two letters should occur together more than once.
The values represent 1 means Pan A, 2 means Pan B and 0 means Not in a Pan. Each column of numbers represents a Measure attempt.

Code:
[color=white]
a	1	1	0
b	1	0	2
c	1	0	0
d	1	2	1
e	2	2	2
f	2	1	0
g	2	1	1
h	2	1	2
I	0	2	0
j	0	0	1
k	0	2	1
l	0	0	2

I've added this test matrix so you can see how to test it, or not as the case may be.
[/color]

Last edited: Jul 15, 2010
17. John99BannedBanned

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I knocked the wind out of everyones sails.

There like 'oh...yeah, thats how it is solved'

18. DywyddyrPenguinaciously duckalicious.Valued Senior Member

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Bullshit.

Nope, more like "How can anyone be so stupid as to think weighing 12 times fulfils the parameters of the question?"

19. James RJust this guy, you know?Staff Member

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Moderator note:

John99 has been banned from sciforums for 2 week for trolling, following an explicit warning not to do so.

20. domesticated omCartoon characterValued Senior Member

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I remember someone posting something like this a while back, except it was 12 coins and one was counterfeit, and you could only weigh them a certain number of times.

This question has nothing to do with the riddle itself....but I wonder if its possible to create a computer program that can solve riddles like this without giving it a pre-defined set of solution routines?

By pre-defined, I mean its easy to create a checklist of things it could try, and the script simply runs down the list - but given no list, I wonder how it would come up with answers?

21. Alan McDougallAlan McDougallRegistered Senior Member

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Hi guys

You are all wrong, you are making assumptions like weigh six bails against six six

clue weighing six by six you will never be able to isolate the oddball and at the same tell us that the oddball is lighter or heavier

Don't base your attempt on luck like some of you have been doing. The answer must be thought out and water tight

This test of logic is quite difficult and not as easy some of you think

22. Captain KremmenAll aboard, me Hearties!Valued Senior Member

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Have you read my attempted solution. Post #4?
Where is it wrong?

23. tablariddimforexU2Valued Senior Member

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There is a way assuming you're lucky and find equal weighing sets on the first weigh

Weigh 5x5 (1), if equal, weigh remaining 2 balls 1x1 (2) they should be unequal, remove 1 ball from each set of original 10 already weighed and replace with the last 2 you just weighed in step 2 one on each side (3), You will see the imbalance immediately and therefore know which ball was different.

Or probably not, because you still don't know if one ball is heavier or lighter hmm. Hang on let's work through this. Ok, you found 10 balls that were equal, the last two didn't balance, say the one on the right was heavier. Now, you don't know that it's heavier than the first 10 or just heavier of the two, so it might in fact weigh equal to the ten and the ball on the left is actually lighter than the others.

Soooo...you replaced 1 ball from each set of 5 with the remaining 2, you put the 'heavier' one on the left pile. If that ball was actually heavier than the others it would tip the scales to the left but if the other ball was actually lighter than the others it would still tip the scales to the..um left. Bloody hell! So, if the 'heavy ball' wasn't actually heavy but equal to the original; because you put it on the left er er oh fuck it that's not working out.

What would Buddha do?

Ok, think I got it. You just replace 1 of the last 2 balls on one of the piles, if both piles are equal it means the remaining ball is different. If the scales tip one way or the other then it's the ball you just replaced which is different.

That seems to work out, but as I said, it depends on you getting the 1st weighing 5x5 equal.

I'd love to know the true answer to this.

Last edited: Jul 16, 2010