Number of balanced sequences

Question

A sequence of length $2n$ that contains only $1, -1$ (every time I say "sequence", I mean that these properties hold) is said to be balanced if all the $2n$ partial sums are non-negtive.

I'm requested to calculate the number of balanced sequences (given $n$).

I already know the answer, by using another result I've proven in the past: the number of balanced sequences with sum $2m$ for $0\leq m\leq n$ is $\binom{2n}{n+m}-\binom{2n}{n+m+1}$. So when we sum this expression over all the possible values for $m$, we get the right answer to my first question: $\binom{2n}{n}$. (Note that the sum has to be even).

Now, this is exactly the number of sequences with sum $0$, so my question is if there is a better way to see it, i.e:

Is there a natural bijection from the set of balanced sequences to the set of sequences with sum $0$?

My attempt: Let $\left\{a_i \right \}_{i=1}^{2n}$ a balanced sequence, and let $2m$ be it's sum. Let $k_0\geq 1$ be the first index that satisifies $\sum_{i=1}^{k_0}a_i=m$ (it exists!) and define a new sequence: $\bar{a}_i=\left\{\begin{matrix} a_i & 1\leq i \leq k_0 \\ -a_i & i>k_0 \end{matrix}\right.$ The new sequence has a sum $0$, but is it a bijection?

I'd like to hear your suggestions. Thanks!

Answer 1

The proof of the following proposition gives you the desired bijection:

Proposition. For $\ell\in\Bbb Z^+$, the number of zero-sum sequences of length $2n$ whose minimum partial sum is $-\ell$ is equal to the number of balanced sequences of length $2n$ whose sum is $2\ell$.

Proof. Let $\alpha=\langle a_1,\ldots,a_{2n}\rangle$ be a zero-sum sequence, and for $k=1,\ldots,2n$ let $s_k$ be the partial sum $s_k=\sum_{i=1}^ka_i$. Suppose that $\min\{s_k:k=1,\ldots,2n\}=-\ell$ for some $\ell\in\Bbb Z^+$. For $i=1,\ldots,\ell$ let $k_i$ be minimal such that $s_{k_i}=-i$; clearly $1\le s_1<\ldots<s_\ell<2n-\ell$, and $a_{k_i}=-1$ for $i=1,\ldots,\ell$.

Let $\hat\alpha=\langle\hat a_1,\ldots,\hat a_{2n}\rangle$ be the sequence obtained from $\alpha$ by replacing each $a_{k_i}$ for $i=1,\ldots,\ell$ with $1$, and let $\hat s_k=\sum_{i=1}^k\hat a_i$ for $k=1,\ldots,2n$. Clearly $\hat s_{2n}=s_{2n}+2\ell=2\ell$. Moreover, it’s not hard to check that $\hat s_k\ge 0$ for $k=1,\ldots,2n$. In fact, $\hat s_k\ge 0$ for $1\le k<k_1$, $\hat s_k\ge 1$ for $k_1\le k<k_2$, $\hat s_k\ge 2$ for $k_2\le k<k_3$, and so on.

Now let $\alpha=\langle a_1,\ldots,a_{2n}\rangle$ be a balanced sequence, define the partial sums $s_k$ as before, and suppose that $s_{2n}=2\ell$. For $i=1,\ldots,\ell$ let $k_i$ be minimal such that $s_{k_i}=i$; clearly $1\le s_1<\ldots<s_\ell<2n-\ell$, and $a_{k_i}=1$ for $i=1,\ldots,\ell$.

Let $\bar\alpha=\langle\bar a_1,\ldots,\bar a_{2n}\rangle$ be the sequence obtained from $\alpha$ be replacing each $a_{k_i}$ for $i=1,\ldots,\ell$ with $-1$, and let $\bar s_k=\sum_{i=1}^k\bar a_i$. Clearly $\bar s_{2n}=s_{2n}-2\ell=0$, and it’s not hard to check that for $i=1,\ldots,\ell$, $k_i$ is minimal such that $\bar s_{k_i}=-i$. Thus, $\hat{\bar\alpha}=\alpha$, and the maps $\alpha\mapsto\hat\alpha$ and $\alpha\mapsto\bar\alpha$ are inverses of each other and bijections between the set of zero-sum sequences with minimum partial sum $-\ell$ and the set of balanced sequences with sum $2\ell$. $\dashv$