Brooklyn Nine-Nine

The episode is literally called "White Whale." :P

Rosa sure did a lot of stuff before becoming a cop! I'm also just really amused that she did three years of med school and then just decided to stop. Guess she figured there wasn't much point if she wasn't interested anymore, so good on her for avoiding racking up more student loans than necessary! (Maybe she started fixing up old cars to sell to celebrities to pay off her student loans.) Also, I'm low key convinced that Rosa was a Child Prodigy because I don't see how else she could have done med school, business school, undergrad, ballet academy, and competitive gymnastics before going to police academy at the same time as Jake while being younger than him. Especially since I'm sure Jake would have gone to police academy ASAP since he's wanted to do that since he was a kid. Damn, Rosa!

I really loved the A-plot because Amy-Rosa subplots are always the best. I just love their dynamic! Also I'm glad they have a duo name: the Sleuth Sisters. And their handshake is very cool.

edited 17th May '18 5:00:47 AM by DeathsApprentice

Anyone else excited for the wedding episode on Sunday? Any predictions? I’m trying to avoid most of the promo material, but I know that Teddy shows up for some reason… God help us. These toit nups can’t be boring!

Oh, and Gina Rodriguez is guest starring!!!

edited 18th May '18 7:09:29 PM by wisewillow

I am very excited! (And I have seen the promos because I love spoilers, and they've just made me even more excited. )

this was great. while I could see this having worked as a series finale, i'm glad it is getting another season after how they just ended it.

"I'm sure you can tell from my facial expression what the result is." "JUST TELL US, YOU MONSTER"

I can't believe they left us with another cliffhanger, those monsters.

That was so freaking precious. I love Jake/Amy so goddamn much. Amy’s vows were beautiful (and I love that it was her nemesis! Though I kinda wish it had been Minsk, the counterfeiter they caught during Boyle Linetti wedding. The parallels and symbolism woulda been cool. Plus she hated Minsk! He spelled pluribus wrong!).

The look on Rosa’s face when Alicia got out of the car was hilarious, and a perfect example of how useless we bi women are when we see a pretty girl

Not gonna lie, I was completely expecting Amy to say her nemesis was Minsk. It definitely would have been a really cool call back if it had been him, but what the episode did instead works fine, too. (Speaking of callbacks, I was also expecting Jake to say, "I'd marry you in a dumpster" when listing all the worst places he'd still be willing to marry her in.)

All of Rosa's and Alicia's interactions were so cute. I especially loved Rosa's giggle-snort.

edited 21st May '18 6:21:33 AM by DeathsApprentice

Theory: Gina had no intention of wearing a wedding dress to Jake and Amy’s wedding. Gina bought Amy a backup wedding dress just in case.

The dress she gave Amy is 0% Gina’s style, but it’s definitely Amy’s style. Gina had a (gorgeous) dress for Amy at the Boyle Linetti Wedding “in case I hated your outfit” which Amy put on since her dress was filthy. Gina Linetti just didn’t want to admit that she did something so thoughtful and nice for Amy, hence the “I was gonna wear it” lie.

I do like that theory. But also note that it's been previously established that their clothes fit each other (when Gina stole one of Amy's pantsuits for a Halloween heist), so she wouldn't have had to fit it for Amy specifically.

Yeah, size isn’t a problem, but that dress design? Does not match Gina’s wardrobe/style on the show. Gina doesn’t go for lace; she goes for glam, not classy/delicate. It was totally a back up dress because Gina knows Jake and Amy are unlucky/disasters

That's what I'm choosing to believe too. I mean, Gina playing off a nice act by pretending it's actually selfish is her MO. (Also, if she was actually planning on wearing that dress to the wedding, that'd make her look unbelievably dickish, even for her. So I'd just prefer to assume that she was lying about why she had that wedding dress.)

Sadly, I can see Gina actually doing that. The only reason they knew she had it was because of how it was brought up, not so much as her stepping forward to help or anything like that. And I can see her capable of pulling something like that due to her self-centeredness.

Gina would never wear that dress in her life. It’s lace. What is she, a Boyle grandma???

And the woman who won Halloween IV can certainly mention having a white dress while in earshot of Charles, knowing he blabs EVERYTHING.

edited 22nd May '18 8:50:04 PM by wisewillow

So, I have just decided to give the show a shot after all the fuss made around its possible cancellation, and I am currently binge watching it. I have just seen the episode in which the Captain has this riddle for his people to solve, only to admit in the end that he doesn't know the solution either.

Now, what me confuses me is that in an earlier scene, he shots down a suggestion which actually could be one of the two possible solution.

A reminder: The riddle goes this way, you have 12 people on an island and a scale, 11 weight exactly the same, one weights something different, and you have to figure out which one by using the scale only three times.

After having thought about it for half a minute (frankly, it was way too easy for supposedly being unsolvable), I came up with no less than two possible solution.

In the first step you can either weight six against six, which shows you in which group the heavier (was it heavier or lighter? let's go with lighter) guy is, or you can weight three against three. In this case, if the scale is in balance you know that the one with the different weight is in the group you haven't weighted yet, and if it isn't you can go directly with the group of three which is lighter.

If you got lucky and your can narrow it down to three, you actually only need another step, just weight two of them against each other, and if there scale isn't in balance, you know who the right guy is based on that, and if it is in balance, it can only be the third one left.

But if you didn't got lucky and you are left with six possible candidates, it is still solvable. Just weight two against two. Again, you can then either narrow it down to either group or, if the scale is in balance, you know it is one the two left of the group. In any case you end up with two possible candidates which you can then weight against each other.

Now, why is Holt shutting down the six against six idea? If you start with it, you will need three tries for sure, but starting this way you can still solve it.

I remember googling this problem; that solution doesn’t work. I can’t remember why though.

Looking it up...oh...apparently you don't know if the person whose weight is off is heavier or lighter. Yeah, this would make the riddle sufficiently complicated.

Thanks, that explains it. Now I am not quite sure if the riddle is even solvable unless you kind of cheat and let the people shift positions while being on the scale.

Okay...I think I have half of the solution….

You weight four against four. If the scale is in balance, you know that all eight of them can't be "X". So you take two of the ones you know belong to the "even" group and weight them against two of the ones left. If the scale is in balance, you are down to two and you can weight one of the against one of the "even" ones. Then you know if the one on the scale (if the scale is out of whack) or the one you haven't weighted yet (if the scale is in balance) is it. Is the scale out of balance in the two again two weighting, you just remove one of both sides and you get an result.

Now I only have to figure out what to do with the scale is out of whack during the first round. because then you end up with eight possible candidates instead of just four….

Soo...I think I have it now….

In the worst case of having eight possible candidates, you take three candidates from one side and one from the other side and weight them against the fourth from the one side and three of the ones you know weight even.

If the scale is now heavier at the same side it was beforehand, you know for sure that the off one out is heavier than the others. And you can exclude everyone you didn't weight, the one you weighted with the ones you know weight the same, and the one you switched from one side to the other. Leaving three possible candidates, weight two of them against each other, and then pick whoever is heavier or, if the scales are then even, the one you didn't weight.

If the scale is now heavier at the other side it was beforehand, you know that it is one of the two persons you switched from left to right. Weight one of them against one of the persons you know weight even and you know which one weights off.

If the scale is in balance, you know that it has to be one of the three persons you took off the scale. And you know if the person you are searching for weights more or less (if they came from the side which initially weighted more, said person is heavier, if not, said person is lighter). So, weight two of the three left against each other and you know which one is the right person.

Yeah, this should work. But I guess it is a little bit more believable that they wouldn't figure it out.

I feel like I've seen quite a few posts about that same brain teaser recently (and all the posts referenced it as being from B99), which is weird considering how long ago that episode aired. But yeah, I think what you said is the solution, Swanpride. It's definitely rather tough, but I still find it weird that no one figured out the answer anyway.

How are you liking the show so far?

Edited by DeathsApprentice on Sep 3rd 2018 at 2:10:47 PM

Well, the show is on Netflix over here, so maybe other people are catching up, too.

I quite like it, which is surprising, because I am actually not into comedy shows at all. But the characters are so great drawn. I especially like Captain Hold (especially him interacting with Kev cracks me up all the time), but they are all great, even the ones which should be more one-note.

It is amusing because the brainteaser is a common one except for the fact you don't know whether the person is supposed to be heavier or lighter. That makes finding the solution functionally impossible, as you can only operate as the scale balances or imbalances. The closest I've come is thus:

• Separate the group into three sets of four. This may seem counterintuitive, but it provides a stronger chance at finding the person if the group is divided into smaller parts.
  • Scale Use #1- Weigh two groups against each other. If perfectly even, you know it is someone in the third group. If uneven, you can discard the third group altogether.
  • Scale Use #2- If the first two groups were even, take the last four and weigh one against one. If uneven it is one of them. If even it is one of the two remaining. If the first two groups were even, weigh three against three, leaving two people out. BUT, keep track of who is on which side and switch the places of two others. If even, it is one of the two left out. If uneven but balances differently, it is one of the two you switched. If uneven but balances the same, you can remove the extra two and the two you switched, leaving four.
  • Scale Use #3-
    • With the third group of four, switch out just one person who was previously on the scale. If it was previously balanced and still balanced, it is the fourth person, you just don't know if it's over/under. If it was previously unbalanced and it balances, it is the person removed from the scale, but at this point if they were high or low you know the over/under. If it was previously unbalanced and remains unbalanced, it is the person left on the scale and similarly know the over/under.
    • With the first two groups, if you narrowed it down to two (either the ones removed or the ones switched) simply weigh one person against someone who balanced out. If it balances the other person is the one and you don't know the over/under. If unbalanced you know the over/under of the person who remained on the scale.
    • With the first two groups, if you narrowed it down to four you can't solve it in this turn. But you can do something similar to the second step, switch two and if it is imbalanced the same way it is one of the two you didn't switch. If it balances differently it is one of the two you switched.

So after all that, best case scenario is still narrowed down to two people.

I do really enjoy the show overall, it has a number of gut-busting hilarious moments. The thing was that I watched all of Community on Hulu (I had seen a scattering of episodes before, but never saw the show beginning to end) and it wasn't until I started watching B99 that I realized what my main problem with Community was. Community had this "anything goes" attitude when it came to the story and the characters. It could be quite funny with its spot on genre-parody episodes, but it never really established a clear tone for its world and its characters were always being altered to fit the genre-characters they were parodying.

With B99, it's a show that is heavily grounded in the environment and the characters they are portraying. Any given episode doesn't quite work if the characters aren't acting consistent. It's rather fascinating to see Jake balance his childlike behavior with his hard work ethic as a detective, most other shows would just have him be Brilliant, but Lazy. Terry has to reconcile a lot of things, from being a husband and father in a dangerous profession as well as a black man in an organization that struggles the most with racial stereotyping. It is kind of sad that despite how goofy the characters are, they also represent what most people probably wish their local law enforcement was like.

Edited by KJMackley on Sep 3rd 2018 at 2:36:07 AM

Naturally the riddle is solvable...I just did it above. The trick is to weight the ones which might weight different against the people you know weight as much as everyone else. Once you have figured that out (and the trick with switching the position on the scale if you end up with 8 potentials in the first round), the rest is easy.

Or at least it is easy for me...I have a thing for stuff like this.

It took me a minute to parse out what you said. It's definitely a difficult puzzle, requiring you to keep track of four different groups at the same time. I did figure out early on that part of the solution would be having people switch sides to reduce options, but the 3-1 switch was where I got hung up.