Finding Catalan Numbers
Date: 12/15/1999 at 21:34:28 From: Bradley Subject: Catalan Numbers Dear Dr. Math, For a research paper, I must study Catalan numbers. Could you possibly explain them (how to find them, applications, etc.)? Thanks a million.
Date: 12/16/1999 at 10:36:52 From: Doctor Anthony Subject: Re: Catalan Numbers A physical interpretation is to consider an n x n square grid of streets and to find the number of routes from one corner of the square to the corner diagonally opposite that do not cross the diagonal. That is, while making the journey, the number of blocks north is always equal to or ahead of the number of blocks east - and of course vice versa. Catalan Numbers --------------- The number of sequences a(1), a(2), a(3), ..., a(2n) of 2n terms that can be formed by using n +1's and n -1's whose partial sums satisfy a(1) + a(2) + ... + a(k) >= 0 (where k = 1, 2, ..., 2n) .... equals the nth Catalan number C_n = 1/(n+1) C(2n,n) (n >= 0) We call a sequence of n +1's and n -1's acceptable if it satisfies  and unacceptable otherwise. [This is equivalent to the number of routes consisting of n northerly blocks (+1) and n easterly blocks (-1) that never cross the diagonal.] Let A(n) denote the number of acceptable sequences of n +1's and n -1's and let U(n) denote the number of unacceptable ones. The TOTAL number of sequences of n +1's and n -1's is (2n)!/(n!n!) = C(2n,n) So we must have A(n) + U(n) = C(2n,n) and we evaluate A(n) by first evaluating U(n) and subtracting from C(2n,n). Consider an unacceptable sequence of n +1's and n -1's. Because the sequence is unacceptable there is a smallest k such that the partial sum: a(1) + a(2) + ... + a(k) < 0 Because k is smallest there is an equal number of +1's and -1's preceding a(k) and we have: a(1) + a(2) + ... + a(k-1) = 0 and a(k) = -1 In particular, k is an ODD integer. We now reverse the signs of each of the first k terms; that is we replace a(i) by -a(i) for each i = 1, 2, 3, ..., k and leave unchanged the remaining terms. The resulting sequence a'(1), a'(2), ..., a'(2n) is a sequence of (n+1) +1's and (n-1) -1's. This process is reversible: Given a sequence of (n+1) +1's and (n-1) -1's there is a first instance when the number of +1's exceeds the number of -1's (since there are more +1's than -1's). Reversing the +1's and -1's up to that point results in an unacceptable sequence of n +1's and n -1's. Thus there are as many unacceptable sequences as there are sequences of (n+1) +1's and (n-1) -1's. The number of sequences of (n+1) +1's and (n-1) -1's is the number (2n)!/[(n+1)!(n-1)!] = C(2n,n-1). Therefore (2n)! U(n) = ------------ (n+1)!(n-1)! and thus (2n)! (2n)! A(n) = ----- - ------------ n! n! (n+1)!(n-1)! (2n)! 1 1 = -------- [--- - -----] n!(n-1)! n (n+1) (2n)! 1 = -------- [------] n!(n-1)! n(n+1) 1 (2n)! = ----- [-----] (n+1) n! n! = 1/(n+1) C(2n,n) There are many different interpretations of this theorem. The following are two examples. Example (1) ------------ There are 2n people in line to get into a theatre. Admission is 50 cents. Of the 2n people, n have 50-cent pieces and n have a $1 bill. The box office rather foolishly begins with an empty cash register. In how many ways can the people line up so whenever a person with a $1 bill buys a ticket, the box office has a 50-cent piece in order to give change? If we regard the people as indistinguishable and identify a 50-cent piece with a +1 and a dollar bill with -1, then the answer is the number C_n = 1/(n+1) C(2n,n) of acceptable sequences. If the people are regarded as distinguishable the answer is: (n!)(n!) 1/(n+1) C(2n,n) = (2n)!/(n+1) Example (2) ----------- A city lawyer works n blocks north and n blocks east of his place of residence. Every day he walks 2n blocks to work. How many routes are possible if he never crosses (but may touch) the diagonal line from home to office? Each acceptable route is either above the diagonal or below the diagonal. We find the number of acceptable routes above the diagonal and multiply by 2. Each route is a sequence of n northerly blocks and n easterly blocks. We identify north with +1 and east with -1. Thus each route corresponds to a sequence a(1), a(2) , a(3), ..., a(2n) of n +1's and n -1's and in order that the route not dip below the diagonal we must have a(1) + a(2) + a(3) + ... + a(k) >= 0 for k = 0, 1, 2, ..., 2n Hence by theorem proved above the total number of acceptable routes is 2.C_n = 2/(n+1) C(2n,n) Recurrence Relation for Catalan Numbers ---------------------------------------- We have C_n = 1/(n+1) C(2n,n) = 1/(n+1) (2n)!/(n!n!) and C_(n-1) = (1/n) (2n-2)!/[(n-1)!(n-1)!] dividing we obtain C_n (4n-2) ------- = ------ C_(n-1) (n+1) Therefore the Catalan sequence is determined by the following recurrence relation and initial condition: 4n-2 C_n = ------ C_(n-1) and C(1) = 1 n+1 Parenthesizing an Expression ---------------------------- If we have to find all the ways to parenthesize the letters abcd we could have: (((ab)c)d) (((abc 111000 ((a(bc))d) ((a(bc 110100 ((ab)(cd)) ((ab(c 110010 (a((bc)d)) (a((bc 101100 (a(b(cd))) (a(b(c 101010 The connection with Catalan numbers is that the information about parenthesizing does not require the letter d and we replace each left parenthesis '(' with '1' and each letter with '0'. Then an acceptable sequence must have the number of 1's equal to or exceeding the number of 0's. In general one can parenthesize the product of n+1 letters in C_n ways. So for the letters abcd we can parentesize in C_3 ways C(3) = 1/4 C(6,3) = (1/4) 20 = 5 ways You can find additional information about Catalan numbers here: Catalan Numbers http://mathforum.org/advanced/robertd/catalan.html - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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