Catalan number: Encyclopedia II - Catalan number - Proof of the formula

Catalan number - Proof of the formula

There are several ways of explaining why the formula given for C_n is correct; that is, why it solves the combinatorial problems listed above. The first proof below uses a generating function, and is not particularly illuminating. The second and third proofs are examples of bijective proofs; they involve literally counting a collection of some kind of object to arrive at the correct formula.

Catalan number - First proof: using generating functions

The Catalan numbers satisfy the recurrence relation

This follows from the fact that every Dyck word w of length ≥ 2 can be written in a unique way in the form

with (possibly empty) Dyck words w₁ and w₂.

The generating function for the Catalan numbers is defined by

and using the above recurrence relation we see that

and hence

The square root term can be expanded as a power series using the identity

which can be proved, for example, by the binomial theorem, together with judicious juggling of factorials. Substituting this into the above expression for c(x) produces, after further manipulation,

Equating coefficients yields the desired formula for C_n.

Catalan number - Second proof

This proof depends on a trick due to D. André, which is now more generally known as the reflection principle (not to be confused with the Schwarz reflection theorem in complex analysis). It is most easily expressed in terms of the "monotonic paths which do not cross the diagonal" problem (see above).

Suppose we are given a monotonic path in an n × n grid that does cross the diagonal. Find the first edge in the path that lies above the diagonal, and flip the portion of the path occurring after that edge, along a line parallel to the diagonal. (In terms of Dyck words, we are starting with a sequence of n X's and n Y's which is not a Dyck word, and exchanging all X's with Y's after the first Y that violates the Dyck condition.) The resulting path is a monotonic path in an (n − 1) × (n + 1) grid. Figure 1 illustrates this procedure; the green portion of the path is the portion being flipped.

Since every monotonic path in the (n − 1) × (n + 1) grid must cross the diagonal at some point, every such path can be obtained in this fashion in precisely one way. The number of these paths is equal to

Therefore, to calculate the number of monotonic n × n paths which do not cross the diagonal, we need to subtract this from the total number of monotonic n × n paths, so we finally obtain

which is the nth Catalan number C_n.

Catalan number - Third proof

The following bijective proof, while being more involved than the previous one, provides a more natural explanation for the term n + 1 appearing in the denominator of the formula for C_n.

Suppose we are given a monotonic path, which may happen to cross the diagonal. The exceedance of the path is defined to be the number of pairs of edges which lie above the diagonal. For example, in Figure 2, the edges lying above the diagonal are marked in red, so the exceedance of the path is 5.

Now, if we are given a monotonic path whose exceedance is not zero, then we may apply the following algorithm to construct a new path whose exceedance is one less than the one we started with.

• Starting from the bottom left, follow the path until it first travels above the diagonal.
• Continue to follow the path until it touches the diagonal again. Denote by X the first such edge that is reached.
• Swap the portion of the path occurring before X with the portion occurring after X.

The following example should make this clearer. In Figure 3, the black circle indicates the point where the path first crosses the diagonal. The black edge is X, and we swap the red portion with the green portion to make a new path, shown in the second diagram.

Notice that the exceedance has dropped from three to two. In fact, the algorithm will cause the exceedance to decrease by one, for any path that we feed it.

It is also not difficult to see that this process is reversible: given any path P whose exceedance is less than n, there is exactly one path which yields P when the algorithm is applied to it.

This implies that the number of paths of exceedance n is equal to the number of paths of exceedance n − 1, which is equal to the number of paths of exceedance n − 2, and so on, down to zero. In other words, we have split up the set of all monotonic paths into n + 1 equally sized classes, corresponding to the possible exceedances between 0 and n. Since there are

monotonic paths, we obtain the desired formula

Figure 4 illustrates the situation for n = 3. Each of the 20 possible monotonic paths appears somewhere in the table. The first column shows all paths of exceedance three, which lie entirely above the diagonal. The columns to the right show the result of successive applications of the algorithm, with the exceedance decreasing one unit at a time. Since there are five rows, C₃ = 5.

Adapted from the Wikipedia article "Proof of the formula", under the G.N U Free Docmentation License. Please also see http://en.wikipedia.org/wiki