I thought this was an Archer thing, stopped reading two lines in. Disappointed.
Don't cheat you fucks:
Albert and Bernard just met Cheryl. “When’s your birthday?” Albert asked Cheryl.
Cheryl thought a second and said, “I’m not going to tell you, but I’ll give you some clues.” She wrote down a list of 10 dates:
May 15, May 16, May 19
June 17, June 18
July 14, July 16
August 14, August 15, August 17
“My birthday is one of these,” she said.
Then Cheryl whispered in Albert’s ear the month — and only the month — of her birthday. To Bernard, she whispered the day, and only the day.
“Can you figure it out now?” she asked Albert.
Albert: I don’t know when your birthday is, but I know Bernard doesn’t know, either.
Bernard: I didn’t know originally, but now I do.
Albert: Well, now I know, too!
When is Cheryl’s birthday?
I thought this was an Archer thing, stopped reading two lines in. Disappointed.
I would ruin Cheryl. TBH I would ruin Judy Greer too. As long as she gets her rocks off on the rough play.
Going from 3 to 1 remaining is flimsy but technically valid. 10 to 5 and 5 to 3 is easy though.
Also, thank god Caps tied it up 1-1
Albert still pulled the final deduction out of his ass though.
I'm sure we can choke it out of her. #GG
Bleh, Caps losing now. Bad logic puzzle.
Last edited by Malakriss; 04-16-2015 at 12:40 AM.
Albert knows that Bernard used his answer to arrive at a conclusion. That tells him that the information he provided Bernard was enough for Bernard to solve it and therefore which day it was.
All I care about is the year, because I just want to know if she is legal to bang. Because she is clearly crazy and we all know crazy chicks are great in the sack!
Anyone watch Brooklyn 99? Was that a real puzzle that Holt asked? IIRC, it went like this.
You're on an island with 12 people. They all weigh exactly the same except for 1, who is slightly heavier or lighter than the rest. The only tool you can use is a teeter totter, and you can only use it 3 times. How do you find the one person who is different?
Your solution does not work, because you do not know if the one guy is lighter or heavier, so you don't know in what set of 3 he is.
You can measure 3 items with 1 measure. It is either 1 != 1 = 1 or it is 1 = 1 != 1. If you get lucky you can do it in 2 measures, but the most it should ever take is 3. 5 5 2, 2 2 1, 1 1.
The dates puzzle was pretty good, I thought. I actually had to jot down a fast table for it to become obvious.
This is a good one to throw at someone at lunch... it's not difficult so much as it's confusing if you're distracted. If one and a half eggs costs one and a half cents, how much do a dozen cost? You'll get a lot of 18 cents.
Last edited by Iannis; 04-16-2015 at 06:49 AM.
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.
Last edited by Cybsled; 04-16-2015 at 03:03 PM.
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.
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.
Last edited by Rezz; 04-16-2015 at 04:06 PM.
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.
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.
But that takes a 4th measurement in some cases. You are only allowed 3.
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:
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.
Last edited by The Ancient; 04-16-2015 at 08:07 PM.
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?
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.
Last edited by Iannis; 04-16-2015 at 11:49 PM.
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.
Draegan is a faggoty piece of shit who sold the forum to mmorpg.com just to spite us. Register at the new site.
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Cheryl is a woman and lied about her birthday. None of those are the correct answer.
Last edited by Azrayne; 04-17-2015 at 05:54 AM.
" I wonder if you are destined to be forgotten. Will your life fade in the shadow of greater beings?"
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.
Last edited by Iannis; 04-17-2015 at 01:17 PM.
Iannis: I think you need to reread The Ancient's solution carefully and realize that through clever use of post-measure grouping the outlier is identified even if we don't know at first whether he is lighter or heavier.
WAIT... wait... 3a) is messed up I think. Let me reread it carefully.
Yeah 3a) is fubared because a single weighting is given 4 different outcomes instead of 3.
Last edited by Szlia; 04-17-2015 at 01:38 PM.
There is an extra D on both sides in 3a. You can disregard it.
Yeah you are right, 3a does fail. Oh well.
Last edited by The Ancient; 04-17-2015 at 01:47 PM.
Scenario 1: Teeter 4 vs 4. If balanced, discard 5, keeping 3 on one side. Then teeter 3 from untouched group against the 3 normal people still on the teeter. If balanced, the last untouched man is the oddball but you won't know if he is heavier or lighter without comparing to one of the 11 normal people. If the 3v3 is unbalanced, you'll know the 3 freshly teetered men contain the oddball and you will know if he is heavier or lighter. Discard all but those 3 and teeter 2 of them. Now you know who the oddball is.
Scenario 2: In progress..
If you really want to know the answer just google twelve coins problem. Pretty old logic puzzle.
I checked the wikipedia article to be sure we have the right formulation for the problem and we do. My ego is hurt a little by the fact the following paragraph started with "There is more than one solution to this problem..."
If the dude can be heavier OR lighter. Then with two people its unanswerable. Best you can do is 50%. More people doesn't make it any easier to solve.
Admittedly, I became stumped with a 4v4 scenario that was unbalanced from the start. There is another solution I came up with that isn't in the wiki article:
Teeter 3v3. If unbalanced, slide 3 men to the opposite side and teeter those original 6 containing the oddball against the 6 normal men. You will then know which 3 contain the oddball and whether he is heavier or lighter after witnessing the rise or fall of the original 6. Then teeter 2 of those 3 against each other to reveal the oddball.
However, I do not think there is a 3-step solution for 3v3 that is balanced from the start. So, 4v4 may be only sure way to solve the problem every time.
I get 3 v3 needing 4 steps to solve if equal initially ...3 Equal, move to one side,6 v 6 the heavy or light, 3 knock in half, 1+1+1
5 v5 if equal, take one off each side add one if equal last guy is odd weight if heavy or light it is guy who just got on
5 v5 unequal, takes up to like 5 or 6 steps
4 v4 equal 3 steps -4 v equal, 4 v for h/l(using 5 from first group 3 remainder) if equal it is last guy, if not then 1+1+1 for step 3
4 v4 unequal, 4 steps - 4 v4, take off light group, 4 v other4 for h/l(if equal 4 took off is the l group or group still on the h), 2 v2, 1 v1
Wait a tic, this Cheryl's birthday puzzle doesn't make sense.
I get that Albert can cross off May 19 and June 18 because Bernard was told the day, and since 18 and 19 are exclusive, if he was told 18 or 19 he would know right away. Since he didn't know right away, you get rid of May 19 and June 18.
I get that it can't be June 17, because Albert can determine that Bernard wasn't told 18 or 19, and if Albert had been told June, June 17 would be the only day left.
But I don't understand how you guys are writing off May 15 and May 16th. I don't see how May 19 not being a possibility also stops May 15 and 16 from being possibilities.
Like, here's how I'm following this if I translate what they're saying into a real conversation (assuming they're not allowed to tell each other the month or day they heard):
Bernard: "I have no idea when her birthday is, but I know it's not May 19th or June 18th, because I wasn't told 19 or 18 and those numbers are exclusive to those days."
Albert: "Well I wasn't told June, so I know it can't be June 17th, because after you eliminated June 18th, June 17th would be the only day left."
Where does the part about May come in?
Last edited by Kriptini; 04-19-2015 at 01:02 AM.
Albert knows that Bernard doesn't know, initially. He knows this because he was told neither May or June, which are the only two months with unique days listed.
I think what confuses people (and confused me for the last step), is that you are not trying to be in Albert's or Bernard's shoes, trying to solve Cheryl's riddle, you are an outside observer trying to find the only date for which this specific conversation between Albert and Bernard is possible.
- Albert knows Bernard does not know -> he was not told May nor June.
- Bernard knows the birthday if the month is not May nor June -> he was not told 14.
- Albert knows the birthday knowing the month and that Bernard was not told 14 -> the birthday is July 16.
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