[portlander12345]

The prisoner starting with their own number box ensures that they are in the chain with their number ticket in it.

[itp]

This point was obvious once pointed out, and is definitely the reason this link clicked for me and the Wikipedia explanation did nothing.

[thaumasiotes]

As I pointed out in a sibling comment, this is also stated in the Wikipedia article.

[thaumasiotes]

> The cycle notation is not unique since a cycle of length l can be written in l different ways depending on the starting number of the cycle. During the opening the drawers in the above strategy, each prisoner follows a single cycle which always ends with his own number.

[MrBuddyCasino]

And thats the part I still don't understand.

[alkonaut]

Think of it this way: you are always in a chain. It can be between 1 and 100 boxes long. But it can’t be a dead end or not include the 1-note. Worse case it visits all boxes and gets back to 1.

Try forming a scenario that doesn’t. E.g a “loop”: Prisoner #1 opens box 1. Finds the number 2. Opens box 2 finds the number 3. Opens box 3... Could he now find the “2” that would lead to him being stuck in the 2-3 loop? No. He already found the 2. It can’t appear again. In the third box he’ll find a number he hasn’t seen. The loop cannot form in any other way than to come to where he started - which it will do when he finds the note with “1” on it.

So by starting with box 1, he is on a loop of length 1..100 including the pointer TO box 1 (which is what he is looking for).

[thaumasiotes]

So, following your cycle looks like this:

    Open box 7, find number 2
    Open box 2, find number 2700
    Open box 2700, find number -16
    Open box -16, find number 0.9
    Open box 0.9, find number ℵ
    Open box ℵ, find number 赢
    Open box 赢, find number two
    ...

As you can see, it's not necessary to use numeric digits to identify the boxes; any identifier at all will do.

But this is a cycle; if we had started by opening box 0.9, it would look like this:

    Open box 0.9, find number ℵ
    Open box ℵ, find number 赢
    Open box 赢, find number two
    ...
    Open box 7, find number 2
    Open box 2, find number 2700
    Open box 2700, find number -16
    Open box -16, find number 0.9

Exercise: given that this cycle starts with "open box 7", how does it end? (Or in other words, what is the second half of the line before "Open box 7"?) If this cycle instead started with "Open box ᚠ", how would it end in that case?

[nebulous1]

If you consider that each box only has one box that "points" to it, and that last box must necessarily be part of the box's cycle (by definition, there's no way to avoid it as if you hypothetically swapped the last box to point somewhere else then all you're doing is linking the cycles and you'll end up in the same place but later)

[jlarocco]

For me it helped to think of it as a collection of linked lists. At most, there can be 100 lists of length 1, where each number points to itself. On the other extreme, there are many ways to create a list of length 100.

The strategy takes advantage of the fact that randomly permuting the numbers is less likely to create one or two long lists, and more likely to create a bunch of shorter lists.

Since each prisoner knows which list he belongs to, he can use that information to reduce his search space from all of the boxes to just the boxes in his list.