Name: Necklace Splitting (a lesson for jewel thieves) - Numberphile
Uploaded: 2020-03-27T18:22:17.000Z
Duration: 15 min 14 s
Description: Thousands of YouTube videos with English-Chinese subtitles! Now you can learn to understand native speakers, expand your vocabulary, and improve your pronunciation...

Okay, so I'm going to talk about face pleating, necklaces and about known constructive proofs.

Adverbs of degree

They still together a necklace, but they are not just any thief.

So imagine that we have two thieves and they still together A necklace, and this necklace has a beads off two types say, Let's say, diamonds and Rubies.

They want toe split the necklace evenly among them, and he wants it.

Each off them will get the same number off diamonds and also the same number off Rubies.

It's a class, and then they have the Interfax speeds, and the necklace could be complicated.

Let's assume for simplistically that the number of diamonds and also the number of Rubies, is even so that it is possible to split it evenly among the two thieves.

Now they want to minimize the number off cuts that they do in the Nicolas.

What they want to do is to cut the necklace in several places.

This will split it in tow, intervals off speeds and then they will take the intervals and will distribute it among them.

And they want in these examples that each off them we get four diamonds and in three off the roof or together seven.

Minimal cuts, not necessarily the minimum possible number off completely, but they want to have a small number.

Each of them should get seven beads so we can try toe cut the necklace in the middle.

Then we will have two intervals off with the left part and they write part.

We have too few rupees, only one and each of them should get three.

What we can do is we can take this Cinderella seven beats and shift it one to the right in vain.

So this corresponds to having the cuts here and here.

And then in this interval we have two Rubies, which is still too few.

But we're making progress and then we can try to shift it great lately and hopes at some point it will be okay if we shifted upto this place here.

So 31234 So the two cuts will be here and here.

Then in the middle interval, we have exactly four diamonds in three Rubies.

Therefore, also we have four diamonds and three Ruiz in what's left.

Now he claims that actually what we proved here, they pay cuts are always enough.

And wise is the case because we started by splitting the necklace.

So imagine now that you say, and even, maybe longer and very complicated necklace with diamonds and Rubies we split it in tow two halfs by cutting the middle.

Now, if on the left cut there are too few rupees then on the right card, they will be too many, right?

Because all together we have the right amount.

And then we use what is called the discreet version off the intermediate value serum.

Remain with my kicks with these are just words.

What this means so we observe here is that whenever we shift, they inter filed here by want with right Then the number off Rubies in it can change only by one because we lose one bead and we gain one beat so we can gain at most one ruby and therefore we started maybe was to fuel its the end.

And every time we change the number off them, just one.

We must heat exactly the right number, for this means IQ to cuts always so five.

So the really numbers only ever changing by one.

But the diamonds are taking care of themselves as well.

Say the diamond will be OK as well, because the total number off beads each one gets is a correct number.

So once we have the right number of Rubies, then the number of time ones will be will be correct is what we saw in this example.

That's kind of obvious if it if all the diamonds appear continuously on the necklace somewhere and all the Rubies also appear continuously, it is clear that the number of cuts we need is it least too, because we have to have some cut in the middle off interval off diamonds and we have to have some cart in the middle of faint of ALF Rubies.

So it means that sometimes we need to cut.

And what we just proved is that two cuts are always sufficient.

That's only when it's symmetric or yeah, so you don't have to be symmetric.

What it means that exactly in the left side we have the correct number, Ruby.

Sometimes one cart is enough, but what we proved about to cut is a two cuts are always so.

Now we can try to make it a little bit more complicated.

Let's say that's still the number off types off me.

You know, sometimes it's better to have a bigger group.

So So we have four small gang of thieves and they still are mathematically oriented.

Still, they want to split it evenly and still they are lucky.

And the number of diamonds and also the number off Rubies is divisible by four.

We have this necklace which is an interval off by what we sold before we can have two carts, maybe here and here, so they tend this interval and those two together, we have the correct number off Rubies in the correct number off that we split it into two fair ships.

Now we can take this interval that contains 1/2 that went toe to toe with thieves.

They split themselves into two groups off toothy.

And then once these two thief's got the fair shares that they deserve, then the use another two cuts in their part in order to split it evenly among the poor fame in thes two intervals off other two, we can shift them to somewhere.

Let's say, here in this interval we put the pier and which forget that they are two intervals for review it again as an interval feeds and again, this other group off.

Two additional thieves can cut this part in two places in split it evenly among them.

So if we go baked with originally place, this corresponds to six cuts.

The two original ones this toe end this okay, so it means with four thief's in two types.

Six cuts are always enough, and it is good to observe.

So again the simply case, Where's the diamonds?

Appear continuously and afterwards they Ruby's also appear continuously.

If now we have four thieves, then off course.

We need at least three carts inside the interval off the diamonds because we have to cut it into four parts and we need at least three cuts inside.

Six cuts are always enough was very convenient because you are able to use what we learned from two.

Let's leave it as an exercise to tow the viewers, but it's not much more complicated to seize it For two types, off beads with three, then four cuts will always be enough for a similar reason toward, we said before, I mean with an extra simple tweet.

Sometimes we need for so so it will be possible.

It's getting significantly more complicated when the number off types off beads increases.

So maybe we can have diamonds and sapphires and Rubies.

It's a group off K thieves we denote by T is a number off types off beat in the game they told together as a necklace.

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Necklace Splitting (a lesson for jewel thieves) - Numberphile

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