**The Problem**Edit

- You have an n episode TV series. You want to watch the episodes in every order possible. What is the least number of episodes that you would have to watch?
- Overlapping is allowed. For example, in the case of n=2, watching episode 1, then 2, then 1 again, would fit the criteria.
- The orders must be continuous. For example, (1,2,1,3) does NOT contain the sequence (1,2,3)

**Current Collaboration**Edit

### Algorithm and BoundsEdit

Credit: Anonymous (/sci/)

4 permutations are contained within each n-cycle. So the goal of this algorithm is to systematically generate the sequence, and show that any other method would give a less efficient method. In the long run, I'd like to create some axioms/theorems for the proof, such as more overlap=more efficiency. I think this will call for some modular arithmetic to generalize it for all n but I'm not sure how to do so.

*by follow, I mean use it as a "rule" to tell you the next number in the sequence*

Start with 1 and follow (1 2 3 4) 6 times. We do this by convention. 1,[2,3,4,1,2,3]

Follow (1 4 2 3) 5 times. This group must be chosen, as (23) is in it. 1,2,3,4,1,2,3,[1,4,2,3,1]

Follow (1 2 4 3) 5 times. This is the last group with (3 1) in it. 1,2,3,4,1,2,3,1,4,2,3,1,[2,4,3,1,2]

Since no remaining groups have (1 2) in them, we have to choose one

A): Following (1 4 3 2) 6 times would end the sequence in wrong: 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2[1,4,3,2,1,4] No group left has (1 4) in it, so this option looses efficiency

B): Following (1 3 2 4) 6 times would end the sequence in wrong: 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2[1,3,2,4,3,2] No group left has (3 2) in it, so this option looses efficiency

C): Following (1 3 4 2) 6 times works: 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2[1,3,4,2,1,3]

Follow (1 3 2 4) 5 times. This is the last group with (1 3) in it. 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2,1,3,4,2,1,3,[2,4,1,3,2] Follow the last group (1 4 3 2) 5 times, to complete the sequence. 1,2,3,4,1,2,3,1,4,2,3,1,2,4,3,1,2,1,3,4,2,1,3,2,4,1,3,2,1,4,3,2,1

### The Lower BoundEdit

I think I have a proof of the lower bound n! + (n-1)! + (n-2)! + n-3 (for

$ n \geq 2 $). I'll need to do this in multiple posts. Please look it over for any loopholes I might have missed.As in other posts, let n (lowercase) = the number of symbols; there are n! permutations to iterate through. The obvious lower bound is n! + n-1. We can obtain this as follows:Let L = the running length of the string

$ N_0 $ = the number of permutations visited $ X_0 = L - N_0 $ When you write down the first permutation, $ X_0 $ is already n-1. For each new permutation you visit, the length of the string must increase by at least 1. So $ X_0 $ can never decrease. At the end, $ N_0 = n! $, giving us $ L \geq n! + n-1 $.I'll use similar methods to go further, but first I'll need to explain my terminology...

Edges:

I'm picturing the ways to get from one permutation to the next as a directed graph where the nodes correspond to permutations and the edges to ways to get from one to the next. A k-edge is an edge in which you move k symbols from the beginning of the permutation to the end; for example,1234567 -> 4567321

would be a 3-edge. Note that I don't include edges like

12345 -> 34512

in which you pass through through a permutation in the middle (in this case 23451). This example would be considered two edges:

12345 -> 23451 23451 -> 34512

From every node there is exactly one 1-edge, e.g.:

12345 -> 23451

These take you around in a cycle of length n. A 2-edge moves the first two symbols to the end. A priori, it could either reverse or maintain the order of those two symbols:

12345 -> 34521 12345 -> 34512

But the second, as already stated, is not counted as a 2-edge because it is a composition of two 1-edges. So there is exactly one 2-edge from every node.

1-loops:

I call the set of n permutations connected by a cyclic path of 1-edges a 1-loop. There are (n-1)! 1-loops.The concept of 1-loops is enough to get the next easiest lower bound of n! + (n-1)! + n-2. That's because to pass from one 1-loop to another, it is necessary to take a 2-edge or higher. Let us define:

$ N_1 $ = the number of 1-cycles completed or that we are currently in $ X_1 = L - N_1 - N_2 $ The definition of $ N_1 $ is a bit more complicated than we need for this proof, but we'll need it later. You might ask, isn't $ N_1 $ just one more than the number of completed 1-cycles? No! When we have just completed a 1-cycle, it is equal to the number of completed 1-cycles. In order to increment $ N_1 $, we have to take a 2-edge, which increases L by 2 instead of 1. Therefore $ X_1 $ can never decrease. Since $ X_1 $ starts out with the value n-2, and we have to complete all (n-1)! 1-loops, we get the lower bound n! + (n-1)! + n-2.

2-loops:

Suppose we enter a 1-loop, iterate through all n nodes (as is done in the greedy palindrome algorithm), and then take a 2-edge out. The edge we exit by is determined by the entry point. The permutation that the 2-edge takes us to is determined by taking the entry point and rotating the first n-1 characters, e.g.:12345 is taken by n-1 1-edges to 51234 which is taken by a 2-edge to 23415

If we repeat this process, it takes us around in a larger loop passing through n(n-1) permutations. I call this greater loop a 2-loop.

The greedy palindrome algorithm uses ever-larger loops; it connects (n-k+1) k-loops via (k+1)-edges to make (k+1)-loops. But I haven't been able to prove anything about these larger loops yet.

The tricky thing about 2-loops is that which 2-loop you're in depends on the point at which you entered the current 1-loop. Each of the n possible entry points to a 1-loop gives you a different 2-loop, so there are n*(n-2)! 2-loops, which overlap.

And now for the proof of the n! + (n-1)! + (n-2)! lower bound...-Written by Anonymous

To review: n = alphabet length L = running string length$ N_0 $ = number of permutations visited $ X_0 = L - N_0 $ $ N_1 $ = number of 1-cycles completed or that we are currently in $ X_1 = L - N_1 - N_2 $ In order to increase $ N_1 $, you must jump to a new 1-cycle -- having completed the one you are leaving. That means the next permutation P' in the 1-cycle (following your exit point P) is one you have already visited. Either you have at some point entered the 1-cycle at P', or this is the second or greater time you've visited P. If you have ever entered the 1-cycle at P', leaving at P by a 2-edge will not take you to a new 2-cycle; you will be in the same 2-cycle you were in when you entered at P'.So these are the available ways to enter a 2-cycle you've never been in before:* take a 3-edge or higher

- take a 2-edge but don't increase$ N_1 $
- take a 2-edge from a permutation P that you were visiting for the second or greater time
In the first two cases, $ X_1 $ increases by 1 in the step under consideration. In the third case, $ X_1 $ must have increased by 1 in the previous step. Because of this third case, it is convenient to regard any series of edge traversals that takes you through permutations you've already visited as a single step. Then if we define $ N_2 $ = number of 2-cycles visited $ X_2 = L - N_0 - N_1 - N_2 $ the quantity $ X_2 $ does not decrease in any step. Since 2-cycles are n(n-1) long, you must visit at least (n-2)! 2-cycles. $ X_2 $ is initially n-3, giving us the lower bound $ L \geq n! + (n-1)! + (n-2)! + n-3 $.

## Resources and InformationEdit

- http://pastebin.com/aNwANugC -Python algorithm by Anonymous
- http://www.notatt.com/permutations.pdf -proof for upper bound
- https://warosu.org/sci/thread/3751105 - source thread
- Latex'ed write up

### OtherEdit

- http://www.reddit.com/r/math/related/foj1l/the_shortest_string_containing_all_permutations/ http://oeis.org/A180632
- http://stackoverflow.com/questions/2253232/generate-sequence-with-all-permutations/2274978 http://forums.xkcd.com/viewtopic.php?f=17&t=68643

## HardmodeEdit

Define H(n) as the number of sequences that are most efficient. for n=2, h(n)=2 {(1,2,1), (2,1,2)} what is the H(n)? Note that it simply isn’t n!, for n=5, there are at least 2 efficient sequences that start with 12345