It is rather well-known that the number of up-down walks that start at $0$ and stay nonnegative until ending at $0$ after $2n$ steps is the $n$th Catalan number, $C_n=\frac{1}{n+1}\binom{2n}{n}$. Such paths, in which we either go up by $1$ or down by $1$ at each step and never fall below the $x$-axis, are called *Dyck paths*. An example for $n=5$ is shown below.

What is not so well-known is that:

The number of up-down walks of length $2n$ that stay nonnegative, but do not necessarily return to $0$ at the end, is simply $\binom{2n}{n}$.

This fact came up in my research a few days ago, and I figured that this had to be a trivial theorem that was easy to prove. Several hours of failed attempts later, I found myself googling “positive walks central binomial coefficients” and came across a wonderful recent paper of M. van Leeuwen.

He describes a bijective way of proving this fact by matching the walks with the recurrent (returning-to-zero) walks that are not necessarily positive. Clearly there are $\binom{2n}{n}$ such walks of the latter sort, since they can be thought of as walks from one corner to the opposite corner on a rotated $n\times n$ grid.

To construct the bijection, consider any recurrent walk from $0$ to $2n$. If the walk is already positive, simply map it to itself. If the walk becomes negative at any point, consider the first times the walk reaches $-1,-2,\ldots,-d$, where $-d$ is the minimum value of the walk. The step just before each of these positions must have been a down-step, and by simply changing each of these $d$ down-steps to an up-step, we end up with a positive walk of length $2n$.

To see that this is bijective, notice that if we had flipped $d$ of the down-steps of the recurrent walk, the last step of the new walk ends at height $2d$. So, to reverse the process, let $2d$ be the ending height of a given positive walk, and we wish to flip $d$ of the up-steps back to down-steps. But the ones we had flipped must be the steps following the last visits to heights $0,1,2,\ldots,d-1$ in the positive walk, by the above construction, so we simply find these visits and flip down the steps following each of them. Beautiful!

Hi Maria,

Very nice! This reminds me very much of the following problem: given a positive integer n, construct a bijection from the n-subsets of [2n+1] to the (n+1)-subsets of [2n+1] such that every n-subset is contained in its image.

Steven

I’ve seen this bijection rediscovered and reproven so many times now that I start to wonder where it appeared first. The earliest mention I know is in Christian Stump’s 2008 PhD thesis, Proposition 2.33., but clearly it must have been around way longer.

Interesting, glad to hear the alternate reference. I think it’s often re-proven because you usually don’t see it mentioned in the standard combinatorics texts. Hopefully a combinatorics text will come out that includes it, so that we can have a standard reference for it.