Dove hole principle or drawer principle

Principle Overview:

There are n pigeons and M pigeon caves, all pigeons live in the pigeon Cave, and if n>m, then at least two pigeons must live in the same pigeon cave.

Function View:

Think of the pigeon as a definition of the element ai in domain A, the pigeon cave as the element in the range B BJ, pigeon Live pigeon Cave as a function relationship.

Pigeon Cave principle:

Set F is a function from finite set A to finite set B, if |a|>|b|, then there must be a1,a2∈a,a1≠a2, so that f (A1) =f (A2) =B∈BF is included in B (Bf is the elephant domain).

Contrary: if to any a1,a2∈a,a1≠a2,f (A1) ≠f (A2), |a|=|bf|≤|b| and |a|>|b| contradiction

Note:

The principle of pigeon cave is to judge the adequacy of a non-one function.

This principle looks easy to understand and has a wide range of applications.

Example 1: In any permutation of n2+1 different integers, it is proved that there must be an ascending or descending subsequence with a length of n+1.

Proof :

Set this sequence as: A1,a2,...,ak, ..., starting from AK the length of the subsequence is XK, the descending subsequence length is YK, and each AK (K=1,2,...,N2+1) corresponds to (Xk,yk).

If there is no ascending or descending subsequence with a length of n+1, then xk≤n,yk≤n, the shape (Xk,yk) of the different points to a maximum of N2, and AK has n2+1, there must be Ai,aj (1≤i

Because Ai≠aj, if ai aj, then Yi is at least 1 larger than YJ, with (xi,yj) = (Xj,yj) contradiction.

Example 2: 132 balls placed in 77 boxes, each box at least one ball, verify: there must be 21 balls in a few adjacent boxes.

Proof :

Set the K box to put the ball for BK, get b1,b2,...,b77, press test instructions, looking for I and j,i>j≥0, make bj+1+bj+2+ ... +bi=21, which is the completion certificate.

Set ak=, Get A1,a2,...,a77, obviously, {AK} is strictly monotonically Rising, Ai≠aj (i≠

Set ck=, Get C1,c2,...,c77,{ck} is also strictly monotonous rise, CI≠CJ (I≠j), and c77=153,ci=ai+21.

{AK} and {CK} together a total of 154, but only in the 1~153 value, by the pigeon cave principle, there will be two the same, and not with the AK, also will not be CK in, may wish ai=cj=aj+21,ai-aj=21, that, get bj+1 +bj+2+ ... +bi=21.

Example 3:

The X1,X2,...,XN is any integer of arbitrary arrangement, proving that there are several successive numbers, and their sum is a multiple of n.

proof : Set ai=

If there is a multiple of n in the AI, the proposition is established

If there is not a multiple of n in the AI, the remainder of the AI is only taken in 1~n-1 by n

With AI (i=1,2,...,n) as pigeons, 1~n-1 as pigeon holes, Ai,aj, so Ai≡aj (MODN) (i>j), then n| (Ai-aj), xj+1+xj+2+ ... +xi is a multiple of n.

Excerpt from Baidu Library: dove Hole principle

Dove hole principle or drawer principle