The city has just opened its

one-of-a-kind Fabergé Egg Museum with a single egg displayed on each floor

of a 100-story building. And the world’s most notorious jewel thief

already has her eyes on the prize. Because security is tight

and the eggs are so large, she’ll only get the chance to steal one by dropping it out the window

into her waiting truck and repelling down

before the police can arrive. All eggs are identical in weight

and construction, but each floor’s egg is more rare

and valuable than the one below it. While the thief would naturally like

to take the priceless egg at the top, she suspects it won’t survive

a 100-story drop. Being pragmatic, she decides to settle

for the most expensive egg she can get. In the museum’s gift shop,

she finds two souvenir eggs, perfect replicas

that are perfectly worthless. The plan is to test drop them to find the highest floor

at which an egg will survive the fall without breaking. Of course, the experiment

can only be repeated until both replica eggs are smashed. And throwing souvenirs out the window

too many times is probably going to draw

the guards’ attention. What’s the least number of tries

it would take to guarantee that

she find the right floor? Pause here if you want

to figure it out for yourself! Answer in: 3 Answer in: 2 Answer in: 1 If you’re having trouble getting started

on the solution, it might help to start

with a simpler scenario. Imagine our thief only

had one replica egg. She’d have a single option: To start by dropping it

from the first floor and go up one by one until it breaks. Then she’d know that the floor below that is the one she needs

to target for the real heist. But this could require

as many as 100 tries. Having an additional replica egg

gives the thief a better option. She can drop the first egg from different

floors at larger intervals in order to narrow down the range

where the critical floor can be found. And once the first breaks, she can use the second egg to explore

that interval floor by floor. Large floor intervals don’t work great. In the worst case scenario, they require

many tests with the second egg. Smaller intervals work much better. For example, if she starts by dropping

the first egg from every 10th floor, once it breaks, she’ll only have to test

the nine floors below. That means it’ll take at most

19 tries to find the right floor. But can she do even better? After all, there’s no reason

every interval has to be the same size. Let’s say there were only ten floors. The thief could test this whole building

with just four total throws by dropping the first egg

at floors four, seven, and nine. If it broke at floor four, it would take

up to three throws of the second egg to find the exact floor. If it broke at seven, it would take up to two throws

with the second egg. And if it broke at floor nine, it would take just one more

throw of the second egg. Intuitively, what we’re trying to do here

is divide the building into sections where no matter which floor is correct, it takes up to the same number

of throws to find it. We want each interval to be one floor

smaller than the last. This equation can help us solve

for the first floor we need to start with in the 100 floor building. There are several ways

to solve this equation, including trial and error. If we plug in two for n,

that equation would look like this. If we plug in three, we get this. So we can find the first n to pass 100 by adding more terms

until we get to our answer, which is 14. And so our thief starts on the 14th floor, moving up to the 27th, the 39th, and so on, for a maximum of 14 drops. Like the old saying goes, you can’t

pull a heist without breaking a few eggs.

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I have another solution.

What if we start from midway through 50th if 1st egg broke then above all r waste if not below r waste and we can use them for more trail, and we do same thing again and again.

I think that will take less than 10 trail.

50th 25th 12th 6th 3rd and finally the ans. 6 trails

She believes the priceless egg won't survive the 100 – story drop

I don't think any of the eggs will survive the drop. THEY'RE EGGS

Also why are eggs valuable?

City: let's build a 100 – floor building. And display just ONE EGG on every floor.

Builder: sounds good to me

What if we throw an egg at f50, then f75 if the egg from the previous attempt didn't break or f25 if it did? If you repeat the process of throwing the egg from the middle height of the remaining floors, we can find the answer with 6 attempts.(I think)

Good you use a girl in this example, to show they exist too.. but do NOT do it very often then they will complain "they portrays theives etc etc as women always why why why" so for now you are good.

why are we helping a criminal get EGGS?

The thief is HowToBasic confirmed

If the replicas are perfectly worthless than why do they cost 15$

The solution:

Get floor 💯 and put the 🥚 in the thief’s 💼 and slide down the rope

🧠

Y to yeeet the egg ,when she can just carry it along with her and slide down onto the rope. Now don't say she need both hands to slide down…😂.

Sure I'll help the thief

I think i've might have a better solution using dichotomic sort but it need much ressources but l'ESS try ( start a 50th floor if it break try 25th floor if not try 75th then you affine the interval until you round the one. Since you divide the interval by two each try , and since 100 IS beteween 2^6(64) and 2^7(128) you only need seven try (but you need seven egg)

Or repel down with egg in hand🤷🏽♂️

What if the eggs were rubber?

I didn't get it

I love how criminals can learn from these riddles without doing any work

Instead

Why did you just take the 100th egg then slide down while holding the egg

Y'all are overthinking it. Just put a purple matress in your truck. Boom. Or just climb down with the egg

Me:

tries to buy 14 eggsGift Shop Owner: Soooo… Whatcha doin with 14 eggs. Ya know, just the other day some lunatic threw 19 eggs of the 20th floor

i got it to 6

Just find some other building and drop these replicas from there.

TED-Ed: Can yo-

Me:No but I’ll watch it

Who else just binge watches these riddle videos and can’t figure out the answers, so they just watch the answers??

Lmao just throw out every egg from the lowest floor to the highest until you ee3 that the eggs can't survive anymore

4/10 just slide down the rope with it

If you can walk between the floors, why not just take the top egg and throw it from 1 floor?

But how do you drop it,

from the first floorMy idea was to use interpolation to get the answer. It would take atmost 8 tries

Why would that thief need to steal expensive eggs when she has an IQ of like 230

Did no one else think about starting on floor 50, if it breaks go to floor 25, if that breaks go to floor 12, if that breaks go to floor 6, if that breaks go to floor 3 if that breaks you only have 2 more tries. This works for either side. Only takes about 7 tries I believe.?😕🙃

Amazing riddle loved it

Cool! First S3 riddle I wasn't able to solve!

the power of recursion!!!!

I'm with everyone here says that this ted-ed's problem could have several solutions that doesn't require dropping eggs after reading comments, so here's my answer: Freaking jump out through the 100th story building window holding the egg, while having the pick up truck really open (and I mean the whole top of the truck open) and have that truck have foams or pillars filling up the whole truck so that way, nothing is harmed.

But the real question is: why are we helping the thief figure out a way to perfect an egg heist?

Binary search is a much better option in my opinion

But did she steal it tho?

Btw, couldn't solve it. You guys did great

Wait why are we helping a criminal?

Can you sol-Me=Nope, imma watch anywaysIs not more easy just tie the egg in the rope you gonna descend after?

Is there a proof why the 14th method is the best? I get it intuitively, but a rigorous proof would be nice

Why can't she get down the rope with an egg in a bag in one hand🤔🤔🤔

This is a good riddle, but one question: Why are we helping a thief?

Why don't you just get the 100th egg then go down in the elevated with it then drop it on a low floor

Why not hold it in a back pack a repel down

Faberge made fewer than 100 eggs in his life

And EGGzactly who would want to steal an egg may I ask

Search in her phone : How high can an egg survive before breaking?. Then divide the ans she get by 2.8 (Average height of a floor) then go to that floor. Easy

But if you did 19 tries that would be too suspicious??

How does thief know math? She is not going to waste money on studies.

So your helping people steal stuff?

(I’m kidding)

Couldn’t she get the egg on the 100 floor go down a few floors then drop it? If she’s a thief she could go through vents :/

In conclusions this video was pointless any true thief could take the 100 floor egg even a robber this vid was nice but it was easier to solve with logic

CS student: Shouldn't he just go with binary search strategy?

Cant u just drop the egg at 100th and if it breaks, drop at 50th, of it didnt break, drop at 75 and if it breaks, drop at 63 and so on

Y u no kill security camera

If its a perfect replica then sell it for the price of the priceless 100 story egg

the first one is easy to do get a soulvenear and then replace the egg that looks the same with the soulvenear and no onw will suspect a thing they will just think its a soulvenear that you bought

You can reduce the number of throws. Once you find the interval, you can throw the egg every other or every third floor, starting from the top of that interval. If the egg breaks, you are too high. If it survives, work your way up one floor at a time, till the most recent break.

I believe that the answer is wrong….i looked at it this way…..

1) Throw the first egg from 50th floor….we will have 2 outcomes…a)it breaks b) it doesn't break

if a) it breaks- then we have to move to lower floor(half of it)

else b) it doesn't break – we have to move to higher floor(half of the number)

2) So the second egg will be thrown at the 75th floor or 25th floor …..again we will have two outcomes….a)break b) no break

again if a) breaks- move to a lower floor( but above 50th( in case of 75) )

else b) no break- move to higher floor( but till 50th ( in case of 25))

………in this way select half each time and move up or down accordingly……..

and we can do this in 6-7 throws

just an idea…give your views

wait wouldn't throwing eggs repeatedly out of windows weaken them?

Other way to solve the riddle ,is put a pillow were the egg will fall and tada ,no damage

Title: Can you solve…

Me: Well no I can’t but I’ll watch the 5 min. Video anyway

If you split the available options in half then it will only take 6 eggs to find the highest floor. Start at 50, if it breaks try 25. If it doesn't break go to 75, and so on….

simple: use Dynamic programming

1 rent a helicopter

2 make it stay out the window

3 steal the 100th egg

4 escape

Just put it in a bag and slide down the rope. Easy.

So did we just help someone steal a valuable egg?

I better be paid for this.

i can do it with 0 throws. just bring a parachute for the egg.

“ You can’t pull a heist without breaking a few egg’s “

binary search? will take 7 tries

just wrap the 100th egg with bubble wrap then throw it out the window..

Thank you for helping me on my journey for doing crimes tedED I rly appreciate it

That thief is a bad egg

Okay but wouldn’t the egg be weaker after every throw so the experiment would be influenced…

Also, just stuff the egg in your clothes so that you don’t have to throw it???

FAH BURR JAYY

"Throwing souvenirs out the window too many times is probably going to draw the guards' (or maybe it's 'guard's' if they just have the one) attention."

Guard 1:

Hears noiseWhat was that?Guard 2: Oh, that must be that girl throwing another replica egg from the gift shop out of a window again.

Guard 1: Isn't that kind of suspicious? I mean, we have all these valuable eggs in here, and somebody could probably steal one by dropping it out of a window into a truck or something.

Guard 2: Eh, maybe. But I don't read too much into that kind of stuff unless it happens at least 15 times. So far, she's only done it 12 times in the last hour, so I don't see a problem.

What if the egg weakens because of all your dropping?

Also the guards apparently don't notice 19 drops.

Why are we helping a thief.

Can be done with 7

If u watch this video u should know that she could have taken the priceless egg

https://m.youtube.com/watch?v=3oIakJekCZU

Wont the structural integrity of the eggs lessen with every drop?

Wait, why is a diamond thief stealing eggs !!?

Thief:

tries to drop egg multiple times by climbing different floorsMe:

can't even climb from my gradesDidn't you just say you have a 100 story rope with you? Grab the egg and just climb down that.

ans should be 9

There's one problem here: the replica eggs may survive a fall from a certain height, but they are still fragile. Therefore, if it survives a fall, it still loses some of its structural integrity and becomes less likely to survive being dropped again from the same height.

Have a matress inside the goddamn car.

You only have 2 eggs. Throw one at 50, and if it breaks throw at 25. If not at 50, throw at 75.

Correct me if im wrong what if she throws the egg at 50% then either she goes up or down 50% again then she lower how many intervals and lessen the drops

Can’t she just go down the stairs

Step one: go to college kids

No step two

Start at 1 and go all the way to 100 whenever the egg breaks use the second egg to throw under that level

For example you drop at 1 2 3 and it breaks at 4 so then u should throw the egg at 3

Technically how did she enter the building and go through floor by floor ?? I mean security?

If the thief is going to jump the same way the egg will be thrown, than why can't she just put the 100th floor egg in her pocket and jump with it…

If the security is non-tight enough that she could throw 14 fake eggs, she could've thrown 14 real eggs.

Hay a thief will not do that much mathematics calculation

Why not pocket the egg and slide down with it

Answer is log100 base 2 = 7

Used in binary Searching