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I am reading Proofs and Confirmations, the history behind the alternating sign matrix conjecture, regarding counting $n \times n$ alternating sign matrices. In the introduction, it is written that the people (Mills Robbins and Rumsey) behind the ASM conjecture added an extra parameter (the position of the 1 in the first row), and by using this refined count, managed to guess a beautiful formula.

There are other examples where refining the family reveals beautiful extra structure. My current personal favorite is counting crossings in perfect matchings. $$ T_n(q) := \sum_{M \in PM(n)} q^{crossings(M)} = \frac{1}{(1-q)^n} \sum_{j=-n}^{n} (-1)^j q^{\frac{j(j-1)}{2}} \binom{2n}{n+j} $$ see A067311. The formula on the right hand side is not that nice, but by refining crossings by gathering matchings according to which vertices are starting vertices, one can easily prove that $$ T_n(q) = \sum_{a \in Area(n)} \prod_{i=1}^n [a_i+1]_q $$ where the sum is now over Catalan$(n)$ objects, namely area sequences of Dyck paths. This example IMHO illustrates that refining a problem can reveal nicer properties. (From this formula, one can apply general theory by Flajolet and immediately get a nice continued fraction expansion).

Another famous example would perhaps be the recent proof of the shuffle conjecture, which really relies on the refined conjecture (the compositional shuffle conjecture).

Question: What are some legendary examples where refining the problem made a significant impact on the solution? Your favorite example?

Willie Wong
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Per Alexandersson
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    Certainly not legendary, but my personal favorite discovery (in particular, because it appeared to me in a dream): consider all paths with north and east steps in a fixed region delimited by two such paths. Then there are as many paths with $b$ horizontal steps on the bottom and $t$ horizontal steps on the top boundary as there are paths with $b$ and $t$ switched, see https://arxiv.org/abs/1305.2206. The crucial insight was that the multiset of $y$-coordinates of the horizontal steps which do not lie on the boundary could be fixed. – Martin Rubey Jul 12 '19 at 16:45
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    Also not legendary, but in order to count the number of reduced decompositions of the longest element $w_0$ of $S_n$, it was necessary to compute their generating function according to descent set (http://math.mit.edu/~rstan/pubs/pubfiles/56.pdf). – Richard Stanley Jul 12 '19 at 18:55
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    Nt at all legendary, but here is an example of "adding extra variables" https://mathoverflow.net/a/29179/6101 – Pietro Majer Jul 12 '19 at 19:36
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    Related question: Generalizing a problem to make it easier – Wolfgang Jul 13 '19 at 05:24
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    the method of generating functions in general is about adding extra parameters to simplify the counting (this seems to be the most legendary example :-)) – Dima Pasechnik Jul 13 '19 at 06:59

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Something like this is ubiquitous in lattice path enumeration problems, especially for problems like the enumeration of lattice paths restricted to the quarter plane (a focus of a tremendous amount of research in the past 20 years). The basic idea is that instead of studying the generating function $Q(n)$ of lattice paths of length $n$ starting from the origin, you introduce auxiliary parameters $x$ and $y$ and study the generating function $Q(n,x,y)$ of lattice paths of length $n$ starting from the origin and ending at $(x,y)$. This lets you write a functional equation satisfied by $Q(n,x,y)$. This basic idea is the starting point for the so-called "kernel method" (see e.g. https://arxiv.org/abs/0810.4387) which has been tremendously successful at resolving these enumeration problems.

Sam Hopkins
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