Study Notes on Pigeon nest principles

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• Simple Form of Pigeon nest Theorem
• Strengthening the concept of Pigeon nest

Simple Form of Pigeon nest Theorem

Pigeon nest principle (Drawer principle): If n + 1 pigeons need to enter n pigeon nests, at least one Pigeon nest contains two or more pigeons.

It seems like a "nonsense", but this "nonsense" is quite attractive in proving the existence of an arrangement or a phenomenon!

A more abstract description of the concept of Pigeon nest:

• If X has more elements than Y, then f is not one-to-one.
• If X and Y have the same number of elements and F is onto, then f is one-to-one.
• If X and Y have the same number of elements and F is one-to-one, then f is.

 

• Proof of the Pigeon nest Theorem in China's Residue Theorem

Let M, N be a positive integer of the reciprocal, and make a and B two integers, and 0 <= A <= n-1, 0 <= B <= n-1. Therefore, there is a positive integer x, so that X mod m = A and X mod n = B; that is, x = PM + A, x = qn + B. Here, p and q are two integers.

Proof:

Consider n integers first

A, M + A, 2 * m + A, 3 * m + A,..., (n-1) * m +;

Each number in these integers is divided by M and all are. Let two of them divide by N and have the same remainder R. Make these two numbers I * m + A, J * m +. (0 <= I <j <= n-1 ). Therefore, there are two integers, Qi, Qj

I * m + A = qI * n + R ---------- (1)

J * m + A = Qj * n + R ----------- (2)

Sorted

(J-I) * m = (Qj-qi) * n ------------ (3)

From the formula (3), we can see that N is a factor of (J-I) * m, because N and m are mutually Prime. Therefore, n is a factor of (J-I. However, 0 <= I <j <= n-1 is 0 <j-I <= n-1, that is, N cannot be a factor of J-I. This is the same as the previous assumption: N integers a, m + A, 2 * m + A, 3 * m + ,..., (n-1) * m + A; there are two divided by N, there will be the same remainder. Therefore, we can determine that each of the N numbers divided by N has a different remainder. According to the Pigeon nest principle: N numbers, 0, 1,..., each number in n-1 is used as the remainder. A special number B also appears. So that P is an integer that satisfies 0 <= P <= n-1, and the remainder of number x = PM + A divided by N is B. Q:

X = Q * n + B

This proves the Chinese residue theorem.

 

Strengthening the concept of Pigeon nest

Theorem: Q1, q2, Q3,..., qN is a positive integer. If you place Q1, q2, Q3,..., qN-n + 1 objects in N boxes, there is an I so that the I box contains at leastQI items;

Proof: Assume that Q1, q2, Q3,..., qN-n + 1 items are placed in N boxes respectively. If each I (I = {1, 2, .. n}) is put in lessQThe total number of items in all boxes cannot exceed

(Q1-1) + (Q2-1) +... + (qN-1) = Q1 + q2 +... + qn-n

Obviously, it is one less than the total number to be put. Therefore, it can be determined that for an I (I = {1, 2,. n}), the I-th box contains at least Qi items.