$\mathbf{1}(\text{condition})$ is 1 or 0 depending on whether the condition is true or false). Crucially, we have (for nonnegative integral $x$) $$\sum_{y=1}^x g(y;\vec{\gamma},\vec{\delta})= f(x;\vec{\gamma},\vec{\delta}).\label{eqn.integration}$$ We are now only thinking of ratios of height 1. Therefore, by **(P0)**, the values of $f$ are either 0 or 1. Hence, the values of $g$ are either 0,1 or -1. It is clear that $g$ is periodic as well, of period $P$.

Because Question \[q.hook\] is stronger than Question \[q.alt\], we can assume all the properties **(P0)**-**(P3)** for the functions $f$ and $g$. These requirements come from considering partitions with just one row.

We now consider partitions of hook shape only. Take $\mu = (1+a,1,\cdots,1)$, with $l$ repetitions of 1 at the end ($a$ stands for arm, $l$ for leg). This partition has hooks of length $1,2,\cdots,l$, $1,2,\cdots,a$, and $a+l+1$. Therefore, Condition  for finding a $\mu$ reduces to $$\sum_{i=1}^a g(i) +\sum_{i=1}^l g(i) +g(a+l+1) = f(a)+f(l)+f(a+l+1)-f(a+l) < 0.
\label{eqn.conditionf}$$ Given the possible values of $f$, we are left with no choice. We must have $$\label{eqn.setup}
f(a)=f(l)=f(a+l+1)=0 \text{ and }f(a+l)=1.$$

We will soon need Kneser’s theorem, a generalization of the Cauchy-Davenport theorem. Let $G$ be an abelian group. If $A \subseteq G$, define the *stabilizer* of 