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Unformatted text preview: MATH 681 Notes Combinatorics and Graph Theory I 1 Catalan numbers Previously, we used generating functions to discover the closed form C n = 2 ( 1 / 2 n +1 ) ( 4) n . This will actually turn out to be marvelously simplifiable: C n = 2 1 / 2 n + 1 ( 4) n = 2( 4) n ( 1 2 )( 1 2 )( 3 2 ) ( 1 2 n 2 ) ( n + 1)! = ( 4) n ( 1)( 3)( 5) (1 2 n ) 2 n ( n + 1)! = 2 n 1 3 5 (2 n 1) ( n + 1)! = 2 n 1 2 3 4 5 (2 n 1)(2 n ) 2 4 6 (2 n 2)(2 n )( n + 1)! = 2 n (2 n )! 2 n n !( n + 1)! = (2 n )! n ! n !( n + 1) = 1 n + 1 2 n n 1.1 A purely combinatorial approach If wed rather not have this much algebra, take heart: theres a simpler, elegant, combinatorial argument. Let us interpret C n as the number of paths with n upsteps and n downsteps which dont go below their starting point (such a path is called a Dyck path . Let us consider such a path with n = 4, drawn with the yaxis representing height and the xaxis representing time: If we rotate this 45 degrees counterclockwise, and impose a grid structure, we see that this structure corresponds to a combinatorial object much like some weve counted before: So we might note that C 4 is exactly the number of 8step walks on the following grid: Page 1 of 7 October 15, 2009 MATH 681 Notes Combinatorics and Graph Theory I We could do this using inclusionexclusion enumeration of walks through (1 , 0), (2 , 1), (3 , 2) and (4 , 3), but unfortunately, there are walks going through each of those in every combination, each of them different in number, so such an inclusionexclusion would be a hideous 16term monstrosity! Fortunately, there is a simple way to count these after all, by building a bijection between walks to ( n,n ) which touch the subdiagonal and walks to ( n + 1 ,n 1). Consider a walk from (0 , 0) to ( n,n ) which is not always above the subdiagonal. Then, there is at least one point ( x,y ) on it where x > y ; consider the first such point, which can easily be seen to have the form ( x,x 1). We may consider the walk from (0 , 0) to ( n,n ) as being a composition of two walks: one from (0 , 0) to ( x,x 1), and one from ( x,x 1) to ( n,n ). This latter walk consists of n x rightsteps and n x + 1 upsteps. Now let us flip this walk, replacing upsteps with rightsteps and viceversa, so now we have a walk from ( x,x 1) consisting of n x +1 rightsteps and n x upsteps; this will be a walk to ( n + 1 ,n 1). Thus, we may map each walk from (0 , 0) to ( n,n ) which goes below the diagonal to a walk from (0 , 0) to ( h + 1 ,n 1). To demonstrate that this is a bijection, let us observe how this process could be reversed. A walk from (0 , 0) to ( n +1 ,n 1) must necessarily pass through a point ( x,y ) where x > y : the destination is such a point, even if no other point is. Consider the first such point in the walk; as above, its coordinates must be ( x,x 1), and now we consider this walk as a composition of a walk from (0...
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 Fall '09
 WILDSTROM
 Combinatorics, Graph Theory

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