Riddles!

[TheGreatErenan]

Me too.

[Eijolend]

It's probably just extrapolating the answer to the 9 objects 2 weighings puzzle. But I didn't calculate it through.

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Hmm actually not knowing if it weighs more or less is a real problem. I'll have to think about that tomorrow.^^

[TheGoldenGriffin]

I didn't know that we don't know if the odd one was lighter or heavier so i had to make up a new one. Gf!sh stared with 4 vs 4 so i tried starting with 2 vs 2. Everything worked perfectly until i got to this.
Is this possible to solve?

You have 5 heads vs 5 heads
4 of the 5 heads on the left are safe but one of the other 6 HAS to be the odd one.
The 2 left out are safe.
You have one weigh left.

[TheGreatErenan]

EDIT: You ninjaed me, Griffin. I am responding to Eijolend. I will think about what you're saying, but I'm not sure I really understand what you're asking.

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Well, I didn't closely examine Gf!sh's answer to verify correctness, but I think he basically got it already. At a glance, it looks like he didn't technically fully explore every possibility, but the other possibilities besides the ones he describes are essentially the same situations but with different objects (and really, it would be slightly annoying to go through and list out literally every possibility). For example, if in the second weighing we see that the CGIJ group is lighter than the ABEF group, then either E or F is heavier or C is lighter, which is logically equivalent to the situation Gf!sh gave, and a similar third measurement will yield the answer.

So I'd say Gf!sh provided a good representative description of how to solve the problem.

[TheGoldenGriffin]

Im asking if my situation is solvable because i tried a bunch of ways to solve it but failed. If you guys CAN find a way to find the odd head from that, then my solution is valid (hopefully), its the missing step that will have everything fit together perfectly(possibly), like the Higgs boson(in a way).

If it helps, i have these steps: (its easier to read if you quote, remove the white color code, and preview)

Weigh 2 vs 2. If tilt, go to A. If even, go to B
A: If tilt, swap a pair of heads. If…
… it tilts the same way, you didn't move the odd one. Remove one side, replace with 2 unweighed ones (you know those are safe) and weigh. If even, you took the odd one out, which is the one you didn't swap. If tilt, the one you swapped that is still on the balance is the odd one.
… it tilts the other way, you moved the odd one to the other side, the head you move is the odd one. Remove one side, replace with 2 unweighed ones (you know those are safe) and weigh. If even, you took the odd one out, which is the one you swap. If tilt, the one you swapped that is still on the balance is the odd one.

B: If even, group the 4 heads together, you know all 4 is safe. There are 8 left, take 6 of them and put 1 with the 4 and other 5 on the other side, so you have 5 vs 5 and weigh. If…
… even, the 1 of the 2 you left out is the of one and you know all 10 weighted is safe. Weigh 1 safe vs 1 left. If even, the one left unweighted is the odd one. If tilt, the one left weighted is the odd one
… tilt, this is where the problem is

[awpertunity]

The 12 balls weighing problem is definitely a tough one. I actually solved it 2 weeks ago but it took my almost a week (not that I was working on it continuously).

As for Eijolend saying it is probably similar to 9 balls with 2 weighings, that is actually not true at all.

Amazingly, even with 4 balls, you need 3 weighings to determine which one is the the different one and if it is heavier/lighter.

@GreatErenan, I'm not sure if Gfish did fully solve the puzzle, it looks like he only was able to find out which one is different? Maybe I'm wrong since it definitely is a pain to go through haha.
The difficulty is not that you only need to find out which ball is different, but whether it is heavier or lighter than the rest!

[Gf!sh]

Actually, I think I have fully solved the problem. After you've done the first two weightings, you know that 2 could be too light or 1 too heavy, or vice verca. In any case, there are no heads left that could be either way, so you can determine in what way which head is odd in a single measurement.

[TheGreatErenan]

Yes.

Here's Gf!sh's solution in greater detail:


Label the twelve tiki heads with letters A - L.

1. Weigh ABCD vs. EFGH (1st weighing). If...
   
   1. ABCD < EFGH, then IJKL are clean, and either ABCD contains a light head or EFGH contains a heavy head. Go to 2.
   2. ABCD > EFGH, then IJKL are clean, and either ABCD contains a heavy head or EFGH contains a light head. Go to 3.
   3. ABCD = EFGH, then ABCDEFGH are clean, and EFGH contains either a heavy head or a light head. Go to 4.
      
2. Weigh ABEF vs. CGIJ (2nd weighing). If...
   
   1. ABEF < CGIJ, then CEF are clean, and either AB contains a light head or G is a heavy head. Go to 5.
   2. ABEF > CGIJ, then ABG are clean, and either EF contains a heavy head or C is a light head. Go to 6.
   3. ABEF = CGIJ, then ABCEFG are clean, and either D is a light head or H is a heavy head. Go to 7.
      
3. Weigh ABEF vs. CGIJ (2nd weighing). If...
   
   1. ABEF > CGIJ, then CEF are clean, and either AB contains a heavy head or G is a light head. Go to 8.
   2. ABEF < CGIJ, then ABG are clean, and either EF contains a light head or C is a heavy head. Go to 9.
   3. ABEF = CGIJ, then ABCEFG are clean, and either D is a heavy head or H is a light head. Go to 10.
      
4. Weigh IJ vs. AK (2nd weighing). If...
   
   1. IJ < AK, then either IJ contains a light head, or K is a heavy head. Go to 11.
   2. IJ > AK, then either IJ contains a heavy head, or K is a light head. Go to 12.
   3. IJ = AK, then IJK are clean, and L is either a light head or a heavy head. Go to 13.
      
5. Weigh A vs. B (3rd weighing). If...
   
   1. A < B, then A is a light head.
   2. A > B, then B is a light head.
   3. A = B, then G is a heavy head.
      
6. Weigh E vs. F (3rd weighing). If...
   
   1. E < F, then F is a heavy head.
   2. E > F, then E is a heavy head.
   3. E = F, then C is a light head.
      
7. Weigh A vs. D (3rd weighing). If...
   
   1. A > D, then D is a light head.
   2. A = D, then H is a heavy head.
   3. A < D, then either the universe broke or you have been spontaneously transported into an alternate universe in which the only difference is which one of your tiki heads is heavy or light.
      
8. Weigh A vs. B (3rd weighing). If...
   
   1. A < B, then B is a heavy head.
   2. A > B, then A is a heavy head.
   3. A = B, then G is a light head.
      
9. Weigh E vs. F (3rd weighing). If...
   
   1. E < F, then E is a light head.
   2. E > F, then F is a light head.
   3. E = F, then C is a heavy head.
      
10. Weigh A vs. D (3rd weighing). If...
    
    1. A < D, then D is a heavy head.
    2. A = D, then H is a light head.
    3. A > D, then either the universe broke or you have been spontaneously transported into an alternate universe in which the only difference is which one of your tiki heads is heavy or light.
       
11. Weigh I vs. J (3rd weighing). If...
    
    1. I < J, then I is a light head.
    2. I > J, then J is a light head.
    3. I = J, then K is a heavy head.
       
12. Weigh I vs. J (3rd weighing). If...
    
    1. I < J, then J is a heavy head.
    2. I > J, then I is a heavy head.
    3. I = J, then K is a light head.
       
13. Weigh A vs. L (3rd weighing). If...
    
    1. A < L, then L is a heavy head.
    2. A > L, then L is a light head.
    3. A = L, then whoever told you that exactly one of your heads weighed differently from the others was lying.
       

[TheGoldenGriffin]

You have 2 red balls, 2 white balls, and 2 blue balls. Each pair has a lighter one and heavier one. Figure out which ball is the light one and which is the heavier one for each pair in 2 weighings.

Oh and 2 weighings to determine all the balls, not 2 weighings for each pair

[TheGreatErenan]

Are there only two weights here? I mean, for example, is the light blue ball equal in weight to the light red ball?

Furthermore, do we know the difference in weight between light and heavy balls?