When Is Cheryl's Birthday?

Seesaw...

Do 6 and 6, that narrows it down to 6, then do 2 and 2. If the 2 and 2 is balanced, then you can do the remaining 1 v 1 that you didn't check to get the solution. If the 2v2 is not balanced, you 1v1 the 2 side that was not balanced.

I usually suck at these, but that wasn't too bad. The bday thing confused me ;p

edit: Fuck, I'm wrong. I forgot the weight isnt fixed and can be heavier or lighter.

Answer: Cheryl is an attention whore who needs to shut the fuck up.

Seesaw: when loading a seesaw, it's a little silly to have 6 get on one side and then 6 get on the other. The people would gradually get on each side at the same time, and the moment it tips you separate the people added last. Grab one additional person from the non-tipping group and have them weigh against each person, so if you get a level seesaw on the first go you still have another to see if the person is heavier or lighter. 3 uses, guaranteed results regardless of weight. Unless the stipulation is that you have to load one side at a time.

The birthday one just sounds like a bunch of assholes trying to be clever.

His is easy to explain. You split the group in half and weigh them, but it assumes that one is either definitely lighter or definitely heavier, not ambiguously either. If the dude is lighter, then you split the group that is on the raised end into 3s and then grab the side that flips into the air. Any grouping of the remaining 3 people would result in whoever is lighter. But again, it assumes that the guy is either lighter or heavier, not ambiguously either.

The weight of our malformed guy being ambiguous is what makes his solution kind of not work.

Hoss_sl said:

I don't understand this, can you explain? 3 items with one measure? 552, 221, 11?

Yeah, it's easy peasey if you missed that the odd person could be heavier or lighter. Frankly, I suspect they intentionally made an unsolvable puzzle. I did come up with a solution, but I think it would be considered cheating.

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1) 5 and 5 narrows it down to either one group of 5, another group of 5 or a group of 2. If it's a group of 2, skip to step 3.

2) Group of 5 breaks down to 2, 2 and 1. If 2 are equal it is the 1 left over.

3) If it's a group of 2, pick one and measure him against someone disqualified. If equal it's the remaining guy, if not equal he's your guy.

same reason my retarded solution didn't work, you don't know if the offender is lighter or heavier than the rest

Drinsic_sl said:

same reason my retarded solution didn't work, you don't know if the offender is lighter or heavier than the rest

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edit: shit. step 2 doesn't work.

Okay let's try this.
Using the following terms H = Measured Heavy. L = Measured Light. D = disqualified.

1) Measure 4v4 4 remaining.

If 4 v 4 is equal:
DDDDDDDD ????

2a) Measure remaining 2v2. You get HH v LL
3a) Measure (HLD) vs (HDD). If they are even, remaining unmeasured L is your guy. If HLD is heavier, H from that group is your guy. If HLD is lighter L from that group is your guy. If HDD is heavier, H from that group is your guy.

If 4v4 yields difference:
HHHH LLLL DDDD

2b) HL v HL. If groups are even, go to step 3a you have narrowed down to between remaining HH and LL. (Edit: I added D's on both sides for no reason)

3b) If one group is heavier measure the H from that group vs a D. If they are even, L from other group is your guy, if H is heavier H is your guy.

Hoss_sl said:

c'mon don't be editing out your wrong answers.

So why all the extra Ds? In step 2b, why HLDD v HLDD instead of just HL v HL, and in Step 3a, why HLD v HDD instead of just HL v HD?

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You are correct, the extra D's are unnecessary. They were part of solutions I backed away from I guess. I don't think it matters as far as the validity though. My only edits were for clarity.

If the one is just randomly unequal instead of unequal in some defined way, I think it adds 1 measure. If it is randomly unequal then halves isn't a useful first measure. 5/12 is a luck based first measure (if 5 = 5 then you can do it in 3), Thirds is also a luck based first measure (but if you're lucky you can do it in 3). And no matter how lucky you are fourths takes more than 3 measures.

Unless you get lucky on your initial grouping or on your subsequent guess as to which two are the equal groups, assuming that the inequality is random, it's just going to take an extra measure. That measure is to define what the inequality is.

Unless it's just subtle and I'm dumb.

Someone pm khalid.

I'm gonna go out on a limb and guess that it isn't actually solvable in 3 attempts with the odd man out being able to weigh more OR less than the rest.

Cheryl is a woman and lied about her birthday. None of those are the correct answer.

Palum_sl said:

Cheryl is a woman and lied about her birthday. None of those are the correct answer.

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This

radditsu_sl said:

I would ruin Cheryl. TBH I would ruin Judy Greer too. As long as she gets her rocks off on the rough play.

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100% to both. And I don't doubt for a second that Judy Greer gets off to rough play.

that episode was possibly the only time I've been even slightly turned on by a cartoon.

Gavinmad_sl said:

I'm gonna go out on a limb and guess that it isn't actually solvable in 3 attempts with the odd man out being able to weigh more OR less than the rest.

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Literally 3 posts above yours is the solution. No one has discredited as far as I can tell.

But it won't work for every permutation. The only reason it won't is because there is only 1 of the 12 which is unequal, but you don't know if it's heavier or lighter. You only know that it's unequal.

So if you split thirds, the only way you're going to identify it within 3 measures is if your initial 2 groups of 4 on the measure are equal. Then you know that the unequal member is in the 3rd grouping. If your initial 2 groups of 4 are unequal, you know that your unequal member is on the measure and NOT off of it, but you don't know which grouping it's in. You can find out, and you can also find out if it's heavier or lighter -- but doing it takes a measure and you're limited to 3.

At best either can be solved in 3, but at worst you have to get lucky twice with 4/12. You only have to get lucky once with 5/12. I don't see a way that you don't have to get lucky in your groupings and still identify the unequal member in 3 measures. But with 4 measures, you can do it 2 different ways.

If you know the way that it's unequal beforehand, then it's trivial.