Notifications
Q&A

# organizing a library

+3
−0

Suppose you have $n>1$ books lined up on a shelf, numbered $1$ to $n$, not in the correct order, and you wish to put them in order. Here's your method: Choose a misplaced book[1] at random, and put it in its correct spot. For example, if $n=5$ and you pick book number $2$ out of spot number $4$, there are now four books left, and you put the book back between the first two, since it's book number $2$.

What's the maximum number of times you might have to do the pick-and-replace before the books are in order?

[1] meaning, a book numbered $k$ which is not in position $k$

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

#### 2 comment threads

what I know so far (2 comments)

## 1 answer

+2
−0

$2^{n-1}-1$ is a lower bound on the maximum, at least. For $n$ books, if you start with the ordering $n, 1, 2, \ldots, n - 1$ (I'm following your convention of 1-indexing the books), then the books could be reordered via the sequence $S_n$ defined as follows: \begin{align} S_1 &= () \\ S_{n + 1} &= ([S_{n} + 1], 1, [S_{n} + 1]) \end{align}

(Here $[S_n + 1]$ means to include the sequence $S_n$, adding 1 to each element.)

I'll walk through $n = 4$, where $S_4 = (3, 2, 3, 1, 3, 2, 3)$:

4 1 2 3 (choose 3)
4 1 3 2 (choose 2)
4 2 1 3 (choose 3)
4 2 3 1 (choose 1)
1 4 2 3 (choose 3)
1 4 3 2 (choose 2)
1 2 4 3 (choose 3)
1 2 3 4

It should be clear that the length of $S_n$ is $2^{n-1} - 1$. Slightly less clear is why this strategy is always valid, though an inductive argument should be within reach.

I don't know if this is the worst case, though.

Why does this post require moderator attention?
You might want to add some details to your flag.

#### 0 comment threads

This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!

Like what we're doing? Support us!