Monotonic sequence proofs from the book chapter22

Chapter 22 of this book describes the concept of a pigeon cage. One of the questions is quite interesting.

In any sequence of Mn + 1, A1, A2,..., each real number is different. Either an incremental sequence with a size of m + 1, or an ascending sequence with a size of N + 1, or both.

The answer in the book is as follows:

First define: For an AI, there is a Ti representing the length of the Maximum ascending sequence starting from AI.

For example, the sequence m = 2, n = 2

1 2-1 0 3

The corresponding Ti sequence is

3 2 3 2 1

1) if Ti> = m + 1, there is an incremental sequence of m + 1, and the problem is proved.

2) assume that all Ti <= m. There are Mn + 1 t in total, and its value range is {1,..., m }. According to the pigeon cage principle, n + 1 has the same tvalue.

Assume that the value is S and 1 <= S <= m.

Evaluate the knowledge of the n + 1 count {A (J, 1), a (J, 2),..., A (J, n + 1 )},We can conclude that a (J, 1)> A (J, 2)>...> A (J, n + 1).

Why? If two adjacent numbers a (J, I) <A (J, I + 1 ). An incremental Sequence starting with a (J, I + 1) with a length of S. With a (J, I), we can find the sequence starting with a (I, j) the length of an incremental Sequence starting with "S + 1. Conflicts with previous assumptions!

According to a (J, 1)> A (J, 2)>...> A (J, n + 1), we can obtain a descending sequence with a length of N + 1. Proof of Problem

Let's talk about my solutions:

For each AI, the maximum increment ending with it is defined. The length of the descending sequence is (XI, Yi)

1) if xi> = m + 1, or Yi <= n + 1, the problem is proved!

2) Assume 1 <= xi <= m, and 1 <= Yi <= n.

First, we can obtain that any two pairs (XI, Yi) (XJ, YJ) are not exactly the same.If they are identical, there will be conflicts.

For example, AI is (3, 5), and AJ is (3, 5 ). If AI <AJ, a monotonically increasing sequence with a length of 4 ending with AJ can be found. The same is true for AI> AJ.

Because of the value range of Xi and Yi, there is only a completely different number of Mn pairs.

For example, m = 2, n = 3, only)

However, if we have Mn + 1 pairs (XI, Yi), there must be two pairs that are exactly the same.Any two pairs (XI, Yi) (XJ, YJ) are not exactly the same.. Because there are Mn cages and Mn + 1 pigeon.

So the proof of the proposition

Monotonic sequence proofs from the book chapter22