\ (Catalan \) the number of columns
Given \ (n-\) a \ (0 \) and \ (n-\) a \ (1 \) , are arranged in a certain order of length \ (2N \) sequence satisfies any prefix \ ( 0 \) number of less than all \ (1 \) number is the sequence number \ (Catalan \) columns.
prove:
Order \ (n-\) a \ (0 \) and \ (n-\) a \ (0 \) freely arranged in a length \ (2N \) sequence \ (S \) , if the \ (S \) and does not satisfy any of the prefix \ (0 \) number is not less than \ (1 \) number, there is a minimum position \ (2P + 1 \) , so \ (S \) [ \ (1 , 2P 1 + \) ] have \ (P \) a \ (0 \) , \ (P + 1 \) a \ (1 \) . And the \ (S \) [ \ (2P + 2 \) , \ (\ 2N) ] numbers are negated, comprising \ (np-1 \) a \ (0 \) and \ (NP \) th \ (1 \) . So, we get \ (n-1 \)A \ (0 \) and \ (n + 1 \) a \ (1 \) arranged in sequence.
Similarly, so \ (n-1 \) a \ (0 \) and \ (n + 1 \) a \ (1 \) freely arranged in a length \ (2N \) sequence \ (B \) . Also there must be a minimum position \ (2P +. 1 \) , so \ (s [1,2p + 1] \) have \ (P \) a \ (0 \) , \ (P +. 1 \) one \ (1 \) , the sequence take back counter, to give a \ (n-\) a \ (0 \) , \ (n-\) a \ (1 \) is present consisting of a prefix \ (0 \) ratio \ (1 \) and more sequences.
So, in fact, the number of two sequences is the same:
- There \ (n-\) a \ (0 \) , \ (n-\) a \ (1 \) is present consisting of a prefix \ (0 \) than \ (1 \) multiple sequence.
- By the \ (n-1 \) a 0, \ (n-1 + \) a \ (1 \) sequences.
So, with all cases minus the illegal situation to get a few Cartland.
inference
The following problems are related to (Catalan \) \ For the number
1. 括号匹配的合法括号序列数量
2. 合法出栈序列数量
3. $n$ 个节点构成的不同二叉树的数量
4. 平面直角坐标系中，每一步只能向上或向右走，从$(0,0)$到$(n,n)$并且除了两个端点外不接触直线$y=x$的路线数量为$2Cat_{n-1}$。