$\displaystyle\int_{2C}^{\infty}\mathbf{m}_{\chi}(ds)\leq\frac{\int_{2C}^{\infty}s^{j-1}\mathbf{m}_{\chi}(ds)}{(2C)^{j-1}}\leq\frac{1}{4C^{2}}\left(\frac{1}{2}\right)^{j+1}\longrightarrow
0$

as $j\rightarrow\infty$. Hence we obtain that $\mathbf{m}_{\chi}(ds)$ has
compact support in $(0,\infty)$.

We now compute the density of the measure $\mathbf{m}_{\chi}(ds)$ with respect
to the Lebesgue measure on ${\mathbb{R}}$. Recall that

$\displaystyle\frac{\mathbf{m}_{\chi}(ds)}{ds}=-\lim_{\epsilon\rightarrow
0+}\frac{1}{\pi}\Im\left(\mathrm{St}_{\mathbf{m}_{\chi}}(s+\mathbf{i}\epsilon)\right)$

where $\Im$ denotes the imaginary part of a complex number and
$\mathrm{St}_{\mathbf{m}_{\chi}}$ is the Stieltjes transform of the measure
$\mathbf{m}_{\chi}$, which can be computed as follows: for
$\zeta\in\mathbb{C}\setminus\mathrm{supp}(\mathbf{m}_{\chi})$,

$\displaystyle\mathrm{St}_{\mathbf{m}_{\chi}}(\zeta)=\int_{0}^{\infty}\frac{\mathbf{m}_{\chi}(ds)}{\zeta-s}=\sum_{i=0}^{\infty}\int_{0}^{\infty}\frac{s^{i}\mathbf{m}_{\chi}(ds)}{\zeta^{i+1}}=\sum_{j=1}^{\infty}\frac{1}{\zeta^{j}}\int_{0}^{\infty}e^{-j\kappa}\frac{\partial\mathcal{H}(\chi,\kappa)}{\partial\kappa}d\kappa$

Again by (5.3) we obtain

(5.5)
$\displaystyle\mathrm{St}_{\mathbf{m}_{\chi}}(\zeta)=\sum_{j=1}^{\infty}\frac{1}{\zeta^{j}j\pi\mathbf{i}}\oint_{\mathcal{C}}[\mathcal{G}_{\chi}(w)]^{\frac{j}{n}}\prod_{k\geq
1}\left[\mathcal{F}_{u,v,k}(w)\right]^{\frac{j}{n}}\frac{dw}{w}.$

When the contour $\mathcal{C}$ satisfies the the conditions given as in
Proposition 4.5, under Assumptions 5.2 and 5.3, we can write each integral
into the sum of the residue at 0 and the residues at all the points in
$\mathcal{D}_{\mathcal{\chi}}$. We can compute the residue at 0 as follows:

$\displaystyle S_{0}:=$
$\displaystyle\sum_{j=1}^{\infty}\frac{2}{\zeta^{j}j}[\mathcal{G}_{\chi}(0)]^{\frac{j}{n}}\prod_{k\geq
1}\left[\mathcal{F}_{u,v,k}(0)\right]^{\frac{j}{n}}$
$\displaystyle=-2\prod_{k\geq
1}\log\left(1-\frac{[\mathcal{G}_{\chi}(0)]^{\frac{1}{n}}\prod_{k\geq
1}\left[\mathcal{F}_{u,v,k}(0)\right]^{\frac{1}{n}}}{\zeta}\right).$

where $\left|\frac{[\mathcal{G}_{\chi}(0)]^{\frac{1}{n}}\prod_{k\geq
1}\left[\mathcal{F}_{u,v,k}(0)\right]^{\frac{1}{n}}}{\zeta}\right|<1$ to
ensure the convergence of the Maclaurin series.

######  Lemma 5.5.

The part of (5.5) corresponding to the sum of residues at all the points of
$\mathcal{D}_{\chi}$ can be computed by

$\displaystyle-\frac{1}{\pi\mathbf{i}}\prod_{k\geq
1}\lim_{K\rightarrow\infty}\oint_{\mathcal{C}_{1,K}}\log\left(1-\frac{[\mathcal{G}_{\chi}(w)]^{\frac{1}{n}}\prod_{k\in[K]}\left[\mathcal{F}_{u,v,k}(w)\right]^{\frac{j}{n}}}{\zeta}\right)\frac{dw}{w}$

where

$\displaystyle\sup_{K}\max_{w\in\mathcal{C}_{1,K}}\left|\frac{[\mathcal{G}_{\chi}(w)]^{\frac{1}{n}}\prod_{k\in[K]}\left[\mathcal{F}_{u,v,k}(w)\right]^{\frac{1}{n}}}{\zeta}\right|<1$

to ensure the convergence of Maclaurin series; $\mathcal{C}_{1,K}$ is a simple
closed curve enclosing all the poles of
$\left[\prod_{k\in[K]}\mathcal{F}_{u,v,k}^{(1)}(w)\right]\mathcal{G}_{1,>\chi}(w)\mathcal{G}_{0,>\chi}(w)$,
but no other poles of the integrand with

$\displaystyle\mathcal{F}_{u,v,k}^{(1)}(w):$
$\displaystyle=\mathcal{G}_{1,>V_{0}}(u^{-2k}v^{-2k}w)\mathcal{G}_{0,>V_{0}}(u^{-2k}v^{-2k}w)\mathcal{G}_{1,>V_{0}}(u^{2-2k}v^{-2k}w)\mathcal{G}_{0,>V_{0}}(u^{2-2k}v^{-2k}w);$

###### Proof.

Define

$\displaystyle\mathcal{F}_{u,v,k}^{(2)}(w):$
$\displaystyle=\mathcal{G}_{1,<V_{m}}(u^{2k}v^{2k}w)\mathcal{G}_{0,<V_{m}}(u^{2k}v^{2k}w)\mathcal{G}_{1,<V_{m}}(u^{2k}v^{2k-2}w)\mathcal{G}_{0,<V_{m}}(u^{2k}v^{2k-2}w).$

Then the part of (5.5) corresponding to the sum of residues at all the points
of $\mathcal{D}_{\chi}$ is

$\displaystyle S_{\mathcal{D}_{\chi}}:=$
$\displaystyle\left[\sum_{j=1}^{\infty}\frac{1}{\zeta^{j}j\pi\mathbf{i}}\oint_{\mathcal{C}}[\mathcal{G}_{\chi}(w)]^{\frac{j}{n}}\prod_{k\geq
1}\left[\mathcal{F}_{u,v,k}^{(2)}(w)\right]^{\frac{j}{n}}\lim_{K\rightarrow\infty}\prod_{k\in[K]}\left[\mathcal{F}_{u,v,k}^{(1)}(w)\right]^{\frac{j}{n}}\frac{dw}{w}\right]-S_{0}$

Under Assumptions 5.2 and 5.3, $\mathcal{C}$ is a simple closed curve
enclosing all the poles of $\prod_{k\geq
1}\mathcal{F}_{u,v,k}^{(1)}(w)\mathcal{G}_{1,>\chi}(w)\mathcal{G}_{0,>\chi}(w)$
and 0, but no other poles of the integrand. By the bounded convergence theorem
we obtain that

$\displaystyle S_{\mathcal{D}_{\chi}}:=$
$\displaystyle\left[\sum_{j=1}^{\infty}\frac{1}{\zeta^{j}j\pi\mathbf{i}}\lim_{K\rightarrow\infty}\oint_{\mathcal{C}}[\mathcal{G}_{\chi}(w)]^{\frac{j}{n}}\prod_{k\geq
1}\left[\mathcal{F}_{u,v,k}^{(2)}(w)\right]^{\frac{j}{n}}\prod_{k\in[K]}\left[\mathcal{F}_{u,v,k}^{(1)}(w)\right]^{\frac{j}{n}}\frac{dw}{w}\right]-S_{0}$

Since $\prod_{k\in[K]}\left[\mathcal{F}_{u,v,k}^{(1)}(w)\right]^{\frac{j}{n}}$
has only finitely many poles, we obtain

$\displaystyle S_{\mathcal{D}_{\chi}}:=$
$\displaystyle\sum_{j=1}^{\infty}\frac{1}{\zeta^{j}j\pi\mathbf{i}}\lim_{K\rightarrow\infty}I_{K},$

where

$\displaystyle
I_{K}:=\oint_{\mathcal{C}_{1,K}}[\mathcal{G}_{\chi}(w)]^{\frac{j}{n}}\prod_{k\geq
1}\left[\mathcal{F}_{u,v,k}^{(2)}(w)\right]^{\frac{j}{n}}\prod_{k\in[K]}\left[\mathcal{F}_{u,v,k}^{(1)}(w)\right]^{\frac{j}{n}}\frac{dw}{w}$

As $k\rightarrow\infty$, the poles of $\mathcal{F}_{u,v,k}^{(1)}(w)$ converge
to 0 exponentially fast, we obtain

$\displaystyle\left|I_{K}-\lim_{K\rightarrow\infty}I_{K}\right|\leq
CKj\alpha^{Kj}$

where $C>0$, $\alpha\in(0,1)$ are constants independent of $K$ and $j$. Then
we obtain

$\displaystyle S_{\mathcal{D}_{\chi}}$
$\displaystyle=\lim_{K\rightarrow\infty}\sum_{j=1}^{\infty}\frac{1}{\zeta^{j}j\pi\mathbf{i}}\oint_{\mathcal{C}_{1,K}}[\mathcal{G}_{\chi}(w)]^{\frac{j}{n}}\prod_{k\geq
1}\left[\mathcal{F}_{u,v,k}^{(2)}(w)\right]^{\frac{j}{n}}\prod_{k\in[K]}\left[\mathcal{F}_{u,v,k}^{(1)}(w)\right]^{\frac{j}{n}}\frac{dw}{w}$
$\displaystyle=\lim_{K\rightarrow\infty}\prod_{k\geq
1}\frac{1}{1-u^{k}v^{k}}\left[\sum_{j=1}^{\infty}\frac{1}{\zeta^{j}j\pi\mathbf{i}}\oint_{\mathcal{C}_{1,K}}[\mathcal{G}_{\chi}(w)]^{\frac{j}{n}}\prod_{k\in[K]}\left[\mathcal{F}_{u,v,k}(w)\right]^{\frac{j}{n}}\frac{dw}{w}\right]$

Then the lemma follows. ∎

Hence we have

$\displaystyle\mathrm{St}_{\mathbf{m}_{\chi}}(\zeta)$
$\displaystyle=\prod_{k\geq
1}\left[-2\log\left(1-\frac{[\mathcal{G}_{\chi}(0)]^{\frac{1}{n}}\prod_{k\geq
1}\left[\mathcal{F}_{u,v,k}(0)\right]^{\frac{1}{n}}}{\zeta}\right)\right.$
$\displaystyle\left.-\frac{1}{\pi\mathbf{i}}\lim_{K\rightarrow\infty}\oint_{\mathcal{C}_{1,K}}\log\left(1-\frac{[\mathcal{G}_{\chi}(w)]^{\frac{1}{n}}\prod_{k\in[K]}\left[\mathcal{F}_{u,v,k}(w)\right]^{\frac{1}{n}}}{\zeta}\right)\frac{dw}{w}\right]$

We would like to get rid of the fractal exponent for the simplicity of
computing complex integrals. To that end, we define another function

(5.6) $\displaystyle\Theta_{\chi}(\zeta):$ $\displaystyle=\prod_{k\geq
1}\left[-2\log\left(1-\frac{\mathcal{G}_{\chi}(0)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(0)}{\zeta^{n}}\right)\right.$
$\displaystyle\left.-\frac{1}{\pi\mathbf{i}}\lim_{K\rightarrow\infty}\oint_{\mathcal{C}_{1,K}}\log\left(1-\frac{\mathcal{G}_{\chi}(w)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(w)}{\zeta^{n}}\right)\frac{dw}{w}\right]$

Let $\omega=e^{\frac{2\pi\mathbf{i}}{n}}$, then it is straightforward to check
that
$\Theta_{\chi}(\zeta)=\sum_{i=0}^{n-1}\mathrm{St}_{\mathbf{m}_{\chi}}(\omega^{-i}\zeta)$.
Then we obtain

$\displaystyle\frac{1}{\pi}\lim_{\epsilon\rightarrow
0+}\Im\Theta_{\chi}(s+\mathbf{i}\epsilon)=\frac{1}{\pi}\lim_{\epsilon\rightarrow
0+}\sum_{i=0}^{n-1}\Im\mathrm{St}_{\mathbf{m}_{\chi}}(\omega^{-i}(s+\mathbf{i}\epsilon))$

Since $\mathrm{St}_{\mathbf{m}_{\chi}}(\zeta)$ is continuous in $\zeta$ when
$\zeta\in\mathbb{C}\setminus\mathrm{supp}(\mathbf{m}_{\chi})$,
$\mathrm{supp}(\mathbf{m}_{\chi})\in(0,\infty)$, and
$\mathrm{St}_{\mathbf{m}_{\chi}}(\overline{\zeta})=\overline{\mathrm{St}_{\mathbf{m}_{\chi}}(\zeta)}$,
we obtain that when $s\in\mathrm{supp}(\mathbf{m}_{\chi})$,

(5.7) $\displaystyle\frac{1}{\pi}\lim_{\epsilon\rightarrow
0+}\Im\Theta_{\chi}(s+\mathbf{i}\epsilon)$
$\displaystyle=\frac{1}{\pi}\sum_{i=1}^{n-1}\Im\mathrm{St}_{\mathbf{m}_{\chi}}(\omega^{-i}s)+\frac{1}{\pi}\lim_{\epsilon\rightarrow
0+}\Im\mathrm{St}_{\mathbf{m}_{\chi}}(s+\mathbf{i}\epsilon)$
$\displaystyle=\frac{1}{\pi}\lim_{\epsilon\rightarrow
0+}\Im\mathrm{St}_{\mathbf{m}_{\chi}}(s+\mathbf{i}\epsilon)=-\frac{\mathbf{m}_{\chi}(ds)}{ds}$

Hence by (5.4), to compute the slope of the limiting rescaled height function
in the $\kappa$-direction, it suffices to compute
$-\frac{1}{\pi}\lim_{\epsilon\rightarrow
0+}\Im\Theta_{\chi}(s+\mathbf{i}\epsilon)$ when $s=e^{-\kappa}$.

By (5.6) we obtain

(5.8) $\displaystyle\Theta_{\chi}(\zeta):$
$\displaystyle=-2\log\left(1-\frac{\mathcal{G}_{\chi}(0)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(0)}{\zeta^{n}}\right)-\frac{1}{\pi\mathbf{i}}\lim_{K\rightarrow\infty}\oint_{\mathcal{C}_{1,K}}\log\left(1-\frac{\mathcal{G}_{\chi}(w)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(w)}{\zeta^{n}}\right)d\log
w$ $\displaystyle=-2\log\left(1-\frac{\mathcal{G}_{\chi}(0)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(0)}{\zeta^{n}}\right)-\frac{1}{\pi\mathbf{i}}\lim_{K\rightarrow\infty}\oint_{\mathcal{C}_{1,K}}d\left[\log\left(1-\frac{\mathcal{G}_{\chi}(w)\prod_{k\in
K}\mathcal{F}_{u,v,k}(w)}{\zeta^{n}}\right)\log w\right]$
$\displaystyle-\frac{1}{\pi\mathbf{i}}\lim_{K\rightarrow\infty}\oint_{\mathcal{C}_{1,K}}\frac{\log
w}{\zeta^{n}-\mathcal{G}_{\chi}(w)\prod_{k\in
K}\mathcal{F}_{u,v,k}(w)}\frac{d[\mathcal{G}_{\chi}(w)\prod_{k\in
K}\mathcal{F}_{u,v,k}(w)]}{dw}dw$

To compute the contour integral above, we need to consider the root of the
following equation in $w$:

(5.9)
$\displaystyle\mathcal{G}_{\chi}(w)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(w)=\zeta^{n}.$

Recall that $\mathcal{C}_{1,K}$ is the union of positively oriented simple
closed curves enclosing every poles of
$\mathcal{G}_{1,>\chi}(w)\mathcal{G}_{0,>\chi}(w)\prod_{k\in[K]}\mathcal{F}^{(1)}_{u,v,k}(w)$;
but no other zeros or poles of the integrand. Let $\mathcal{D}_{\chi,K}$ be
the set consisting of all the poles of
$\mathcal{G}_{1,>\chi}(w)\mathcal{G}_{0,>\chi}(w)\prod_{k\in[K]}\mathcal{F}^{(1)}_{u,v,k}(w)$.
Since $\mathcal{D}_{\chi,K}$ is finite, we may decompose

$\displaystyle\mathcal{C}_{1,K}=\cup_{\xi\in\mathcal{D}_{\chi,K}}\mathcal{C}_{\xi};$

where $\mathcal{C}_{\xi}$ is a positively oriented simple closed curve
enclosing $\xi$ but no other zeros or poles of the integrand.

When $\zeta\rightarrow\infty$, zeros of (5.9) will approach poles of
$\mathcal{G}_{\chi}\prod_{k\in[K]}\mathcal{F}_{u,v,k}$. For each
$\xi\in\mathcal{D}_{\chi,K}$, let $w_{\xi,\chi,K}(\zeta)$ be the root of (5.9)
such that $\lim_{\zeta\rightarrow\infty}w_{\xi,\chi,K}(\zeta)=\xi$.

When $|\zeta|$ is sufficiently large, $w_{\xi,\chi,K}(\zeta)$ is enclosed by
$\mathcal{C}_{\xi}$. Enclosed by each $\mathcal{C}_{\xi}$, there is exactly
one zero and one pole for
$1-\frac{\mathcal{G}_{\chi}(w)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(w)}{\zeta^{n}}$,
hence

(5.10)
$\displaystyle\frac{1}{\pi\mathbf{i}}\oint_{\mathcal{C}_{1,K}}d\left[\log\left(1-\frac{\mathcal{G}_{\chi}(w)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(w)}{\zeta^{n}}\right)\log
w\right]=0.$

By computing residues at each $w_{\xi,\chi,K}(\zeta)$ and $\xi$, we obtain

(5.11)
$\displaystyle-\frac{1}{\pi\mathbf{i}}\oint_{\mathcal{C}_{1,K}}\frac{\log
w}{\zeta^{n}-\mathcal{G}_{\chi}(w)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(w)}\frac{d[\mathcal{G}_{\chi}(w)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(w)]}{dw}dw$
$\displaystyle=2\sum_{\xi\in\mathcal{D}_{\chi,K}}\left[\log
w_{\xi,\chi,K}(\zeta)-\log\xi\right].$

######  Lemma 5.6.

Suppose that either Assumption 5.1 holds or both Assumption 5.2 and Assumption
5.3 hold. Then for any $s\in{\mathbb{R}}$ and $K\in{\mathbb{N}}$, the equation
in $w$

(5.12)
$\displaystyle\mathcal{G}_{\chi}(w)\prod_{k\in[K]}[\mathcal{F}_{u,v,k}(w)]=s^{n},$

has at most one pair of complex conjugate roots.

###### Proof.

The lemma follows from the same argument as the proof of Lemma 6.5 in [32].
The idea is that the roots of (5.12) are roots of a polynomial, hence the
total number of roots (counting multiplicities) is the degree $d_{K}$ of the
polynomial. The interlacing assumptions of 5.1 or 5.2 and 5.3 guaranteed the
existence of at least $d_{K}-2$ real roots. ∎

######  Lemma 5.7.

Suppose Assumption 2.6 and (4.6) hold. Suppose that either Assumption 5.1
holds or Assumptions 5.2 and 5.3 hold. Assume that equation (5.12) in $w$ with
$s=e^{-\kappa}$ has exactly one pair of nonreal conjugate roots and
$\mathcal{D}_{\chi,K}\neq\emptyset$.

  1. (1)

If
$[\mathcal{R}_{\chi,1,1}\setminus\mathcal{R}_{\chi,1,2}]\cup[\cup_{k\in[K]}\left([\mathcal{R}_{5,1,k,k}\setminus\mathcal{R}_{5,2,k,k}]\cup[\mathcal{R}_{5,1,k-1,k}\setminus\mathcal{R}_{5,2,k-1,k}]\right)]=\emptyset$,
then

$\displaystyle\mathcal{G}_{\chi}(0)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(0)<e^{-n\kappa}.$

  2. (2)

If
$[\mathcal{R}_{\chi,3,1}\setminus\mathcal{R}_{\chi,3,2}][\cup_{k\in[K]}\left([\mathcal{R}_{7,1,k,k}\setminus\mathcal{R}_{7,2,k,k}]\cup[\mathcal{R}_{7,1,k-1,k}\setminus\mathcal{R}_{7,2,k-1,k}]\right)=\emptyset$,
then

$\displaystyle\mathcal{G}_{\chi}(0)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(0)>e^{-n\kappa}.$

###### Proof.

Same arguments as the proof of Lemma 6.6 in [32]. ∎

######  Proposition 5.8.

Suppose Assumption 2.6 and (4.6) hold. Suppose that either Assumption 5.1
holds or both Assumption 5.2 and Assumption 5.3 hold. Let
$\mathcal{H}(\chi,\kappa)$ be the limit of the rescaled height function of
free boundary dimer coverings on rail yard graphs as $\epsilon\rightarrow 0$,
as obtained in Lemma 2.7. Assume that equation (5.12) in $w$ with
$s=e^{-\kappa}$ has exactly one pair of nonreal conjugate roots and
$\mathcal{D}_{\chi}\neq\emptyset$ and $\mathcal{G}_{\chi}(0)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(0)\neq e^{-n\kappa}$. Then

(5.13)
$\displaystyle\frac{\partial\mathcal{H}(\chi,\kappa)}{\partial\kappa}=2-\lim_{K\rightarrow\infty}\frac{2\mathrm{arg}(\mathbf{w}_{+,K})}{\pi}$

where $\mathbf{w}_{+,K}$ is the unique nonreal root of
$\mathcal{G}_{\chi}(w)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(w)=e^{-n\kappa}$ in
the upper half plane, and the branch of $\mathrm{arg}(\cdot)$ is chosen such
that $\mathrm{arg}(\mathbf{w}_{+,K})\in(0,\pi)$.

###### Proof.

By (5.4) (5.7)(5.8), (5.10), (5.11), we obtain

(5.14) $\displaystyle\frac{\partial\mathcal{H}(\chi,\kappa)}{\partial\kappa}$
$\displaystyle=-\left.\lim_{\epsilon\rightarrow
0+}\frac{1}{\pi}\Im\Theta_{\chi}(s+\mathbf{i}\epsilon)\right|_{s=e^{-\kappa}}$
$\displaystyle=2\mathbf{1}_{\mathcal{G}_{\chi}(0)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(0)>e^{-n\kappa}}-\frac{2}{\pi}\lim_{K\rightarrow\infty}\sum_{\xi\in\mathcal{D}_{\chi,K}}\left[\mathrm{arg}(w_{\xi,\chi,K}(e^{-\kappa}))-\mathrm{arg}(\xi)\right]$
$\displaystyle=\lim_{K\rightarrow\infty}\left(2\mathbf{1}_{\mathcal{G}_{\chi}(0)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(0)>e^{-n\kappa}}-\frac{2}{\pi}\sum_{\xi\in\mathcal{D}_{\chi,K}}\left[\mathrm{arg}(w_{\xi,\chi,K}(e^{-\kappa}))-\mathrm{arg}(\xi)\right]\right)$

where the branch of $\mathrm{arg}$ is chosen to have range $(-\pi,\pi]$. Since
each $\xi$ is a nonzero real number, we have
$\mathrm{arg}(\xi)=\mathbf{1}_{\xi<0}$.

Under the assumption that $\mathcal{D}_{\chi}\neq\emptyset$ and
$\mathcal{G}_{\chi}(0)\prod_{k\geq 1}\mathcal{F}_{u,v,k}(0)\neq e^{-n\kappa}$,
we obtain that $\mathcal{G}_{\chi}(0)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(0)\neq
e^{-n\kappa}$ for all sufficiently large $K$. The following cases might occur:

  1. (1)

If

$\displaystyle[\mathcal{R}_{\chi,3,1}\setminus\mathcal{R}_{\chi,3,2}]\cup$
$\displaystyle[\cup_{k\geq
1}\left([\mathcal{R}_{7,1,k,k}\setminus\mathcal{R}_{7,2,k,k}]\cup[\mathcal{R}_{7,1,k-1,k}\setminus\mathcal{R}_{7,2,k-1,k}]\right)\neq\emptyset;$

then for all sufficiently large $K$,

$\displaystyle
A_{K}:=[\mathcal{R}_{\chi,3,1}\setminus\mathcal{R}_{\chi,3,2}]\cup$
$\displaystyle[\cup_{k\in[K]}\left([\mathcal{R}_{7,1,k,k}\setminus\mathcal{R}_{7,2,k,k}]\cup[\mathcal{R}_{7,1,k-1,k}\setminus\mathcal{R}_{7,2,k-1,k}]\right)\neq\emptyset;$

In this case the number of negative poles in $\mathcal{D}_{\chi,K}$ is the
cardinality $|A_{K}|$ of the set $A_{K}$ above. Under Assumption 5.1 or
Assumptions 5.2 and 5.3, the range of the function
$\mathcal{G}_{\chi}(0)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(0)$ between any two
consecutive points in $A_{K}$ is $(-\infty,\infty)$, hence there are at least
$|A_{K}|-1$ negative real roots in
$\\{w_{\xi,\chi,K}(e^{-\kappa})\\}_{\xi\in\mathcal{D}_{\chi,K}}$. Let

$\displaystyle c_{7,K}:=\max A_{1,K}.$

    1. (a)

If

$\displaystyle E:=[\mathcal{R}_{\chi,1,1}\setminus\mathcal{R}_{\chi,1,2}]\cup$
$\displaystyle[\cup_{k\geq
1}\left([\mathcal{R}_{5,1,k,k}\setminus\mathcal{R}_{5,2,k,k}]\cup[\mathcal{R}_{5,1,k-1,k}\setminus\mathcal{R}_{5,2,k-1,k}]\right)\neq\emptyset$

Then for all sufficiently large $K$,

$\displaystyle
E_{K}:=[\mathcal{R}_{\chi,1,1}\setminus\mathcal{R}_{\chi,1,2}]\cup$
$\displaystyle[\cup_{k\in[K]}\left([\mathcal{R}_{5,1,k,k}\setminus\mathcal{R}_{5,2,k,k}]\cup[\mathcal{R}_{5,1,k-1,k}\setminus\mathcal{R}_{5,2,k-1,k}]\right)\neq\emptyset$

let

$\displaystyle c_{8,K}:=\min E_{K}>0.$

    2. (b)

If $E=\emptyset$ and

$\displaystyle J:=[\mathcal{R}_{\chi,2,1}\setminus\mathcal{R}_{\chi,2,2}]\cup$
$\displaystyle[\cup_{k\geq
1}\left([\mathcal{R}_{6,1,k,k}\setminus\mathcal{R}_{6,2,k,k}]\cup[\mathcal{R}_{6,1,k-1,k}\setminus\mathcal{R}_{6,2,k-1,k}]\right)\neq\emptyset$

Then for all sufficiently large $K$,

$\displaystyle
J_{K}:=[\mathcal{R}_{\chi,2,1}\setminus\mathcal{R}_{\chi,2,2}]\cup$
$\displaystyle[\cup_{k\in[K]}\left([\mathcal{R}_{6,1,k,k}\setminus\mathcal{R}_{6,2,k,k}]\cup[\mathcal{R}_{6,1,k-1,k}\setminus\mathcal{R}_{6,2,k-1,k}]\right)\neq\emptyset.$

Let

$\displaystyle C_{8,K}=\min J_{K}>0.$

    3. (c)

If $E=J=\emptyset$, let

$\displaystyle C_{8,K}=+\infty.$

Then by (4.7)-(4.10), the following cases might occur

    1. (a)

$\mathcal{G}_{\chi}(0)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(0)>e^{-\kappa}$: then
there exists a unique root in
$\\{w_{\xi,\chi,K}(e^{-\kappa})\\}_{\xi\in\mathcal{D}_{\chi,K}}\cap(B_{1},B_{2})$
which is negative; in this case the argument of each negative pole in
$\mathcal{D}_{\chi,K}$ cancels with an argument of a unique negative root in
$\\{w_{\xi,\chi,K}(e^{-\kappa})\\}_{\xi\in\mathcal{D}_{\chi,K}}$;

    2. (b)

$\mathcal{G}_{\chi}(0)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(0)<e^{-\kappa}$: then
there exists no root in
$\\{w_{\xi,\chi,K}(e^{-\kappa})\\}_{\xi\in\mathcal{D}_{\chi,K}}\cap(B_{1},0)$;
in this case there is a unique negative pole in $\mathcal{D}_{\chi,K}$ whose
argument cannot cancel with an argument of a unique negative root in
$\\{w_{\xi,\chi,K}(e^{-\kappa})\\}_{\xi\in\mathcal{R}_{\chi}}$;

in either case we have

$\displaystyle
2\mathbf{1}_{\mathcal{G}_{\chi}(0)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(0)>e^{-n\kappa}}-\frac{2}{\pi}\sum_{\xi\in\mathcal{D}_{\chi,K}}\left[\mathrm{arg}(w_{\xi,\chi,K}(e^{-\kappa}))-\mathrm{arg}(\xi)\right]=2-\frac{2\arg(\mathbf{w}_{+,K})}{\pi}$

Then (5.18) holds.

  2. (2)

$E\neq\emptyset$ and $A=\emptyset$. In this case $A_{K}=\emptyset$ for any
positive integer $K$. There are neither negative poles in
$\mathcal{D}_{\chi,K}$ nor negative real roots in
$\\{w_{\xi,\chi,K}(e^{-\kappa})\\}_{\xi\in\mathcal{D}_{\chi,K}}$. By Lemma
5.7, we have
$\mathcal{G}_{\chi}(0)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(0)>e^{-n\kappa}$.
Then (5.18) follows from (5.14).

∎

######  Lemma 5.9.

Suppose that either Assumption 5.1 holds or both Assumption 5.2 and Assumption
5.3 hold. Then for any $s\in{\mathbb{R}}$, the equation in $w$

(5.15) $\displaystyle\mathcal{G}_{\chi}(w)\prod_{k\geq
1}[\mathcal{F}_{u,v,k}(w)]=s^{n},$

has at most one pair of complex conjugate roots. Moreover, if $\mathbf{w}_{+}$
be the unique root of (5.15) in the upper half plane; then

(5.16) $\displaystyle\mathbf{w}_{+}=\lim_{K\rightarrow\infty}\mathbf{w}_{+,K}$

where $\mathbf{w}_{+,K}$ is given as in Proposition 5.8.

###### Proof.

Note that as $K\rightarrow\infty$,

$\displaystyle\mathcal{G}_{\chi}(w)\prod_{k\in
K}[\mathcal{F}_{u,v,k}(w)]\rightarrow\mathcal{G}_{\chi}(w)\prod_{k\geq
1}[\mathcal{F}_{u,v,k}(w)]$

uniformly on compact subsets of $\mathbb{C}$ that do not intersect
${\mathbb{R}}$. Assume that (5.15) has more than one pair of complex conjugate
roots; Then by the Hurwitz Theorem (see Lemma A.8), when $K$ is sufficiently
large, then equation (5.12) has more than one pair of complex conjugate roots.
But this is a contradiction to Lemma 5.6. The convergence (5.16) again follows
from Lemmas A.8 and 5.6. ∎

Proof of Theorem 2.9. By Lemma A.8, from (5.14) we obtain By (5.4) (5.7)(5.8),
(5.10), (5.11), we obtain

(5.17)
$\displaystyle\frac{\partial\mathcal{H}(\chi,\kappa)}{\partial\kappa}=2\mathbf{1}_{\mathcal{G}_{\chi}(0)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(0)>e^{-n\kappa}}-\frac{2}{\pi}\sum_{\xi\in\mathcal{D}_{\chi}}\left[\mathrm{arg}(w_{\xi,\chi}(e^{-\kappa}))-\mathrm{arg}(\xi)\right]$

where $w_{\xi,\chi}=\lim_{K\rightarrow\infty}w_{\xi,\chi,K}$ is a root of the
1st equation of (2.13) which approaches the pole $\xi$ when
$\kappa\rightarrow-\infty$. It follows that when the first equation of (2.13)
has only real roots, $(\chi,\kappa)$ is always in the frozen region.

When the first equation of (2.13) has a unique pair of nonreal conjugate
roots, by Lemma 5.9 and (5.18) we obtain

(5.18)
$\displaystyle\frac{\partial\mathcal{H}(\chi,\kappa)}{\partial\kappa}=2-\frac{2\mathrm{arg}(\mathbf{w}_{+})}{\pi}.$

where $\mathrm{arg}(\mathbf{w}_{+})\in(0,\pi)$, hence $(\chi,\kappa)$ is in
the liquid region in this case.

Then the frozen boundary is given by the condition that the first equation of
(2.13) has double real roots, which gives the system of equations (2.13).
$\Box$

See (A.12) for the expression of the second equation in (2.13).

##  6\. Height Fluctuations and Gaussian Free Field

In Section 6, we prove that the fluctuations of unrescaled height function
converges to the pull-back of the restriction Gaussian Free Field (GFF) in the
upper half plane to a subset under a diffeomorphism from the liquid region to
the subset of the upper half plane. The main theorem proved in this section is
Theorem 2.10.

###  6.1. Gaussian free field

Let $C_{0}^{\infty}$ be the space of smooth real-valued functions with compact
support in the upper half plane ${\mathbb{H}}$. The _Gaussian free field_
(GFF) $\Xi$ on ${\mathbb{H}}$ with the zero boundary condition is a collection
of Gaussian random variables $\\{\Xi_{f}\\}_{f\in C_{0}^{\infty}}$ indexed by
functions in $C_{0}^{\infty}$, such that the covariance of two Gaussian random
variables $\Xi_{f_{1}}$, $\Xi_{f_{2}}$ is given by

(6.1)
$\mathrm{Cov}(\Xi_{f_{1}},\Xi_{f_{2}})=\int_{{\mathbb{H}}}\int_{{\mathbb{H}}}f_{1}(z)f_{2}(w)G_{{\mathbb{H}}}(z,w)dzd\overline{z}dwd\overline{w},$

where

$G_{{\mathbb{H}}}(z,w):=-\frac{1}{2\pi}\ln\left|\frac{z-w}{z-\overline{w}}\right|,\qquad
z,w\in{\mathbb{H}}$

is the Green’s function of the Laplacian operator on ${\mathbb{H}}$ with the
Dirichlet boundary condition. The Gaussian free field $\Xi$ can also be
considered as a random distribution on $C_{0}^{\infty}$, such that for any
$f\in C_{0}^{\infty}$, we have

$\Xi(f)=\int_{{\mathbb{H}}}f(z)\Xi(z)dz:=\xi_{f}.$

Here $\Xi(f)$ is the Gaussian random variable with respect to $f$, which has
mean 0 and variance given by (6.1)) with $f_{1}$ and $f_{2}$ replaced by $f$.
See [38] for more about the GFF.

###  6.2. $\mathbf{w}_{+}$ as a mapping from $\mathcal{L}$ to $\mathbb{H}$

:

By (4.7)-(4.10) and (4.12)-(4.13), we may write
$\mathcal{G}_{\chi}(w)\prod_{k\geq 1}\mathcal{F}_{u,v,k}(w)$ as the quotient
of two functions $U_{\chi}(w)$ and $R(w)$, such that $U_{\chi}(w)$ depends on
$\chi$ and $R(w)$ is independent of $\chi$. More precisely,

$\displaystyle\mathcal{G}_{\chi}(w)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(w)=\frac{U_{\chi}(w)}{R(w)};$

where

$\displaystyle
U_{\chi}(w)=e^{\phi(\chi)}\frac{\prod_{j\in[n],a_{j}=R}\left(1+e^{-\chi}w\tau_{j}\right)}{\prod_{j\in[n],a_{j}=L}({1-e^{-\chi}w\tau_{j}})}$

and

$\displaystyle\phi(\chi)$
$\displaystyle=\sum_{p\in[m],V_{p-1}>\chi}\sum_{j\in[n],(a_{j},b_{j}(p-1,p))=(L,-)}(V_{p}-V_{p-1})+\sum_{p\in[m],V_{p-1}<\chi<V_{p}}\sum_{j\in[n],(a_{j},b_{j}(p-1,p))=(L,-)}(V_{p}-\chi)$
$\displaystyle-\sum_{p\in[m],V_{p-1}>\chi}\sum_{j\in[n],(a_{j},b_{j}(p-1,p))=(R,-)}(V_{p}-V_{p-1})-\sum_{p\in[m],V_{p-1}<\chi<V_{p}}\sum_{j\in[n],(a_{j},b_{j}(p-1,p))=(R,-)}(V_{p}-\chi)$
(6.2) $\displaystyle R(w)=\frac{A_{1}\cdot A_{2}}{\prod_{k\geq
1}\mathcal{F}_{u,v,k}(w)}$ $\displaystyle A_{1}$
$\displaystyle=\prod_{j=1}^{n}\prod_{(p\in[m],V_{p}>\chi)}\left(1-e^{-V_{p}}w\tau_{j}\right)^{-\mathbf{1}_{\mathcal{E}_{j,p,>,1}}}\prod_{(p\in[m],V_{p-1}>\chi)}\left(1-we^{-V_{p-1}}\tau_{j}\right)^{\mathbf{1}_{\mathcal{E}_{j,p,>,1}}}$
$\displaystyle\times\prod_{(p\in[m],V_{p-1}<\chi)}\left(1-we^{-V_{p-1}}\tau_{j}\right)^{-\mathbf{1}_{\mathcal{E}_{j,p,<,1}}}\prod_{(p\in[m],V_{p}<\chi)}\left(1-e^{-V_{p}}w\tau_{j}\right)^{\mathbf{1}_{\mathcal{E}_{j,p,<,1}}}$
$\displaystyle A_{2}$
$\displaystyle=\prod_{j=1}^{n}\prod_{(p\in[m],V_{p}>\chi)}\left(1+e^{-V_{p}}w\tau_{j}\right)^{\mathbf{1}_{\mathcal{E}_{j,p,>,0}}}\prod_{(p\in[m],V_{p-1}>\chi)}\left({1+we^{-V_{p-1}}\tau_{j}}\right)^{-\mathbf{1}_{\mathcal{E}_{j,p,>,0}}}$
$\displaystyle\times\prod_{(p\in[m],V_{p}<\chi)}\left(1+e^{-V_{p}}w\tau_{j}\right)^{-\mathbf{1}_{\mathcal{E}_{j,p,<,0}}}\prod_{(p\in[m],V_{p-1}<\chi)}\left({1+we^{-V_{p-1}}\tau_{j}}\right)^{\mathbf{1}_{\mathcal{E}_{j,p,<,0}}}$

For $j\in[n]$ and $p\in[m]$ we use $b_{j}(p-1,p)=+$ to denote that for each
$\epsilon>0$, and each integer $k$ such that $\epsilon
k\in\left(V_{p-1},V_{p}\right)$ and $[(k-j)\mod n]=0$, $b_{k}^{(\epsilon)}=+$.
Then

$\displaystyle A_{1}$
$\displaystyle=\prod_{j\in[n]:a_{j}=L}\left[\left(1-e^{-V_{0}}w\tau_{j}\right)^{-\mathbf{1}_{b_{j}(0,1)=+}}\left(1-e^{-V_{m}}w\tau_{j}\right)^{-\mathbf{1}_{b_{j}(m-1,m)=-}}\right]$
(6.3)
$\displaystyle\times\left[\prod_{p=1}^{m-1}\left(1-e^{-V_{p}}\tau_{j}w\right)^{-\mathbf{1}_{b_{j}(p,p+1)=+}+\mathbf{1}_{b_{j}(p-1,p)=+}}\right]$

and

$\displaystyle A_{2}$
$\displaystyle=\prod_{j\in[n]:a_{j}=R}\left[\left(1+e^{-V_{0}}w\tau_{j}\right)^{\mathbf{1}_{b_{j}(0,1)=+}}\left(1+e^{-V_{m}}w\tau_{j}\right)^{\mathbf{1}_{b_{j}(m-1,m)=-}}\right]$
(6.4)
$\displaystyle\times\left[\prod_{p=1}^{m-1}\left(1+e^{-V_{p}}\tau_{j}w\right)^{\mathbf{1}_{b_{j}(p,p+1)=+}-\mathbf{1}_{b_{j}(p-1,p)=+}}\right]$

We shall always use $[\cdot]^{\frac{1}{n}}$ to denote the branch which takes
positive real values on the positive real line. Define

$\displaystyle
U(w_{*},z_{*})=\left[\frac{\prod_{j\in[n],a_{j}=R}\left(1+w_{*}\tau_{j}\right)}{\prod_{j\in[n],a_{j}=L}({1-w_{*}\tau_{j}})}\right]^{\frac{1}{n}}-z_{*}$

Hence we have

(6.5) $\displaystyle\mathcal{G}_{\chi}(w)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(w)=e^{-n\kappa}$

if and only if

(6.6) $\displaystyle\begin{cases}U(w_{*},z_{*})=0\\\ [R(w)]^{\frac{1}{n}}=z\\\
(w_{*},z_{*})=(e^{-\chi}w,e^{-\kappa-\frac{\phi(\chi)}{n}}z)\end{cases}$

######  Lemma 6.1.

Let

$\displaystyle n_{R}:=|\\{j\in[n]:a_{j}=R\\}|;\qquad
n_{L}:=|\\{j\in[n]:a_{j}=L\\}|;$

For any $(\alpha,\theta)$ such that

(6.7)
$\displaystyle\theta\in(0,\pi),\qquad\alpha\in\left(0,\frac{n_{L}\pi+(n_{R}-n_{L})\theta}{n}\right),$

there exists a unique pair $(w_{*},z_{*})$ such that $\arg z_{*}=\alpha$,
$\arg w_{*}=\theta$ and $U(w_{*},z_{*})=0$.

###### Proof.

See Lemma 7.1 of [32]. ∎

######  Proposition 6.2.

For each $(\chi,\kappa)\in\mathcal{L}$, let $\mathbf{w}_{+}(\chi,\kappa)$ be
the unique root of (6.5) in the upper half plane ${\mathbb{H}}$. Then
$\mathbf{w}_{+}:\mathcal{L}\rightarrow\mathbf{w}_{+}(\mathcal{L})\subset{\mathbb{H}}$
is a diffeomorphism.

###### Proof.

We first show that $\mathbf{w}_{+}$ is a bijection between $\mathcal{L}$ and
$\mathbf{w}_{+}(\mathcal{L})$. For any
$w\in\mathbf{w}_{+}(\mathcal{L})\subset{\mathbb{H}}$, let
$z=[R(w)]^{\frac{1}{n}}$. Let

$\displaystyle\theta:=\arg w\in(0,\pi);\qquad\alpha:=\arg z$

By Lemma 6.1, if

$\displaystyle\alpha\in\left(0,\frac{n_{L}\pi+(n_{R}-n_{L})\theta}{n}\right)$

there exists a unique pair $(w_{*},z_{*})$, such that $\arg z_{*}=\alpha$,
$\arg w_{*}=\theta$ and $U(w_{*},z_{*})=0$.

Let

$\displaystyle\chi:=\log\left(\frac{w}{w_{*}}\right);\qquad\kappa:=\log\left(\frac{z}{z_{*}}\right)-\frac{\phi(\chi)}{n}$

where the branch of the $\log(\cdot)$ is chosen such that it takes real values
on the positive real axis. Then we deduce that $\mathbf{w}_{+}$ is a
bijection. From the process we see that both the mapping $\mathbf{w}_{+}$ and
its inverse are differentiable. Then the proposition follows. ∎

######  Lemma 6.3.

Suppose that either Assumption 5.1 holds or both Assumptions 5.2 and 5.3 hold.
Let $K\geq 1$. One of the following two conditions holds:

  1. (1)

(6.8) $\displaystyle\lim_{x\rightarrow
c_{3}}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=+\infty;\qquad\lim_{x\rightarrow
c_{6}}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=+\infty;$
(6.9) $\displaystyle\lim_{x\rightarrow
c_{8}}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=-\infty;\qquad\lim_{x\rightarrow
c_{2}}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=-\infty.$
(6.10)
$\displaystyle\infty>\min_{x\in(c_{3},c_{6})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)>\max_{x\in(c_{8},c_{2})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)>-\infty.$

  2. (2)

(6.11) $\displaystyle\lim_{x\rightarrow
c_{3}}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=-\infty;\qquad\lim_{x\rightarrow
c_{6}}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=-\infty;$
(6.12) $\displaystyle\lim_{x\rightarrow
c_{8}}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=+\infty;\qquad\lim_{x\rightarrow
c_{2}}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=+\infty.$
(6.13)
$\displaystyle\infty>\min_{x\in(c_{8},c_{2})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)>\max_{x\in(c_{3},c_{6})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)>-\infty.$

where $c_{3}=c_{3,K}$, $c_{2}=c_{2,K}$ for all $K$, $c_{3,K}$, $c_{2},K$,
$c_{6}$, $c_{5}$ are given as in Assumption 5.1.

###### Proof.

Note that under Assumption 5.1, we may write

(6.14)
$\displaystyle\mathcal{G}_{\chi}(w)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(w)=C_{K}\frac{B_{K}(w)}{T_{K}(w)}$

where

$\displaystyle\lim_{K\rightarrow\infty}C_{K}=C>0$

and

$\displaystyle B_{K}(w)=\prod_{f_{i,K}\in D_{2,K}\cup D_{4,K}}(w-f_{i,K})$
$\displaystyle T_{K}(w)=\prod_{g_{i,K}\in D_{1,K}\cup D_{3,K}}(w-g_{i,K})$

For $i\in[4]$, let $D_{i,K}$ be given as in Assumption 5.1. We can decompose
(6.14) into partial fractions as follows:

$\displaystyle(\ref{qgf})=C_{K}\left(1+\sum_{g_{i,K}\in D_{1,K}\cup
D_{3,K}}\frac{1}{w-g_{i,K}}\frac{B_{K}(g_{i,K})}{T_{k}^{\prime}(g_{i,K})}\right)$

Under Assumption 5.1, we obtain

  * •

It is straightforward to check that for each $g_{i,k}\in D_{1,K}\cup D_{3,K}$,
either

$\displaystyle\lim_{x\rightarrow
g_{i,K+}}(\ref{qgf})=-\infty;\qquad\lim_{x\rightarrow
g_{i+1,K-}}(\ref{qgf})=+\infty.$

or

$\displaystyle\lim_{x\rightarrow
g_{i,K+}}(\ref{qgf})=+\infty;\qquad\lim_{x\rightarrow
g_{i+1,K-}}(\ref{qgf})=-\infty.$

  * •

If one of the following two conditions holds

    1. (1)

$g_{i,K}<g_{i+1,K}$ are two consecutive points in $D_{1,K}$; or

    2. (2)

$g_{i,K}<g_{i+1,K}$ are two consecutive points in $D_{3,K}$.

In case (1), there are exactly one point in $D_{2,K}$ and no point in
$D_{4,K}$ between $g_{i,K}$ and $g_{i+1,K}$; in case (2), there are exactly
one point in $D_{4,K}$ and no point in $D_{2,K}$ between $g_{i,K}$ and
$g_{i+1,K}$; hence in either case we have

$\displaystyle\mathrm{sgn}\left(\frac{B_{K}(g_{i,K}+)}{T_{K}^{\prime}(g_{i,K}+)}\right)=-\mathrm{sgn}\left(\frac{B_{K}(g_{i+1,K}-)}{T_{K}^{\prime}(g_{i+1,K}-)}\right)$

Therefore either

$\displaystyle\lim_{x\rightarrow
g_{i,K-}}(\ref{qgf})=-\infty;\qquad\lim_{x\rightarrow
g_{i,K-}}(\ref{qgf})=+\infty.$

or

$\displaystyle\lim_{x\rightarrow
g_{i,K-}}(\ref{qgf})=+\infty;\qquad\lim_{x\rightarrow
g_{i,K-}}(\ref{qgf})=-\infty.$

  * •

If $g_{i,K}<g_{i+1,K}$ are two consecutive points in $D_{1,K}\cup D_{3,K}$;
and one of the following two conditions holds

    1. (1)

$g_{i,K}\in D_{1,K}$ and $g_{i+1,K}\in D_{3,K}$;

    2. (2)

$g_{i,K}\in D_{3,K}$ and $g_{i+1,K}\in D_{1,K}$;

In case (1), there are exactly one point in $D_{2,K}$ and one point in
$D_{4,K}$ between $g_{i,K}$ and $g_{i+1,K}$; in case (2), there are exactly no
point in $D_{4,K}$ and no point in $D_{2,K}$ between $g_{i,K}$ and
$g_{i+1,K}$; hence in either case we have

$\displaystyle\mathrm{sgn}\left(\frac{B_{K}(g_{i,K}+)}{T_{K}^{\prime}(g_{i,K}+)}\right)=\mathrm{sgn}\left(\frac{B_{K}(g_{i+1,K}-)}{T_{K}^{\prime}(g_{i+1,K}-)}\right)$

Therefore either

$\displaystyle\lim_{x\rightarrow
g_{i,K-}}(\ref{qgf})=-\infty;\qquad\lim_{x\rightarrow
g_{i,K-}}(\ref{qgf})=-\infty.$

or

$\displaystyle\lim_{x\rightarrow
g_{i,K-}}(\ref{qgf})=+\infty;\qquad\lim_{x\rightarrow
g_{i,K-}}(\ref{qgf})=+\infty.$

Then (6.8) and (6.9) or (6.11) and (6.12) follows.

We now show that if (6.8) and (6.9) hold, then (6.10) holds. We can show that
if (6.11) and (6.12) hold, then (6.13) holds similarly.

Suppose that (6.8) and (6.9) hold, and

$\displaystyle\min_{x\in(c_{3},c_{6})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)\leq\max_{x\in(c_{8},c_{2})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)$

then we can find

$\displaystyle
s\in\left[\min_{x\in(c_{3},c_{6})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x),\max_{x\in(c_{8},c_{2})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)\right]$

such that the number of real roots of the equation

$\displaystyle\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=s$

is at least $|D_{1,k}\cup D_{3,K}|+2$ (counting multiplicities), which is a
contradiction to the fact that the equation has exactly $|D_{1,k}\cup
D_{3,K}|$ roots in the complex plane (counting multiplicities). Then the lemma
follows. ∎

######  Lemma 6.4.

Suppose that either Assumption 5.1 holds or both Assumptions 5.2 and 5.3 hold.
Let $\chi\in{\mathbb{R}}$.

  * •

If case (1) in Lemma 6.3 holds, let

$\displaystyle
x_{m,K}=\mathrm{argmax}_{x\in(c_{8},c_{2})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)$
$\displaystyle
x_{M,K}=\mathrm{argmin}_{x\in(c_{3},c_{6})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)$
$\displaystyle
m_{K}=\mathrm{max}_{x\in(c_{8},c_{2})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)$
$\displaystyle
M_{K}=\mathrm{min}_{x\in(c_{3},c_{6})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)$

  * •

If case (2) in Lemma 6.3 holds, let

$\displaystyle
x_{m,K}=\mathrm{argmax}_{x\in(c_{3},c_{6})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)$
$\displaystyle
x_{M,K}=\mathrm{argmin}_{x\in(c_{8},c_{2})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)$
$\displaystyle
m_{k}=\mathrm{max}_{x\in(c_{3},c_{6})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)$
$\displaystyle
M_{K}=\mathrm{argmin}_{x\in(c_{8},c_{2})}\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)$

Then both $x_{m,K}$ and $x_{M,K}$ are well defined and unique. Moreover,
$x_{m,K}$ is the unique double real root of

$\displaystyle\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=m_{K}$

and $x_{M,K}$ is the unique double real root of

$\displaystyle\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=M_{K}$

and for any $s\in(-\infty,m_{K})\cup(M_{k},\infty)$, the equation

$\displaystyle\mathcal{G}_{\chi}(x)\prod_{k\in[K]}\mathcal{F}_{u,v,k}(x)=s$

has no non-real roots.

###### Proof.

The lemma follows from the proof of Lemma 6.3. ∎

######  Lemma 6.5.

Suppose that either Assumption 5.1 holds or both Assumptions 5.2 and 5.3 hold.
Consider the vertical line $\chi=\chi_{0}$. Let $\kappa_{0,m}$ (resp.
$\kappa_{0,M}$) be the minimal (resp. maximal) ordinate of points in the
intersection of the frozen boundary with $\chi=\chi_{0}$. Let $x_{m}$ be the
unique nonzero double real root of

$\displaystyle\mathcal{G}_{\chi_{0}}(w)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(w)=e^{-n\kappa_{0,m}}$

and let $x_{M}$ be the unique nonzero double real root of

$\displaystyle\mathcal{G}_{\chi_{0}}(w)\prod_{k\geq
1}\mathcal{F}_{u,v,k}(w)=e^{-n\kappa_{0,M}}$

then one of the following two conditions hold.

  * •

$x_{m}\in(c_{3},c_{6})$, and $x_{M}\in(c_{8},c_{2})$; or

  * •

$x_{m}\in(c_{8},c_{2})$, and $x_{M}\in(c_{3},c_{6})$.

###### Proof.

The lemma follows from Lemma 6.4 by letting $K\rightarrow\infty$. ∎

Proof of Theorem 2.10. Same arguments as the proof of Theorem 7.7 in [32].
$\Box$

##  7\. Sampling Algorithm

Now we describe an algorithm to sample double-free boundary dimer coverings on
rail-yard graphs. Algorithms to sample one free boundary, one empty boundary
Schur processes can be found in [3]. The idea is based on repeatedly commuting
and reflecting the Bosonic operators with respect to the left and right
boundary, inspired by the computation of partition functions of double-sided
free boundary steep tilings in [10]. Since the partition function includes an
infinite product, the sampling algorithm here to sample the double-sided free
boundary Schur processes is not an exact sampling, but the sample obtained by
this algorithm has the distribution an approximation of the Gibbs measure, if
we run the algorithm for sufficiently many steps.

First we review a few algorithms in [3]. The goal of algorithm 7.1 is, given 3
partitions $\lambda,\mu,\kappa\in{\mathbb{Y}}$ and a geometric random variable
$G\in{\mathbb{N}}$ satisfying $\kappa\prec\lambda$ and $\kappa\prec\mu$,
sample a partition $\nu$ such that $\mu\prec\nu$, $\lambda\prec\nu$,
$|\nu|+|\kappa|=|\lambda|+|\mu|+G$, and for fixed $(\lambda,\mu)$, the mapping
$(\kappa,G)\rightarrow\nu$ is a bijection.

######  Algorithm 7.1.

def $HH(\lambda,\mu,\kappa,\xi)$

  * •

sample $G\sim\mathrm{Geom}(\xi)$;

  * •

construct a partition $\nu\in{\mathbb{Y}}$ as follows:

$\displaystyle\nu_{i}=\begin{cases}\max\\{\lambda_{1},\mu_{1}\\}+G&\mathrm{if}\
i=1\\\
\max\\{\lambda_{i},\mu_{i}\\}+\min\\{\lambda_{i-1},\mu_{i-1}\\}-\kappa_{i-1}&\mathrm{if}\
i>1.\end{cases}$

  * •

return $\nu$.

Here $\mathrm{Geom}(\xi)$ is a geometric random variable such that
$P(\mathrm{Geom}(\xi)=k)=\xi^{k}(1-\xi)$ for $k\geq 0$.

The goal of algorithm 7.2 is, given 3 partitions
$\lambda,\mu,\kappa\in{\mathbb{Y}}$ and a Bernoulli random variable
$B\in\\{0,1\\}$ satisfying $\kappa\prec^{\prime}\lambda$ and $\kappa\prec\mu$,
sample a partition $\nu$ such that $\mu\prec^{\prime}\nu$, $\lambda\prec\nu$,
$|\nu|+|\kappa|=|\lambda|+|\mu|+B$, and for fixed $(\lambda,\mu)$, the mapping
$(\kappa,B)\rightarrow\nu$ is a bijection.

######  Algorithm 7.2.

def $HV(\lambda,\mu,\kappa,\xi)$

  * •

sample $B\sim\mathrm{Bern}\left(\frac{\xi}{1+\xi}\right)$;

  * •

for $i=1\ldots\max\\{l(\lambda),l(\mu)\\}+1$,

    * –

if $\lambda_{i}\leq\mu_{i}<\lambda_{i-1}$, then
$\nu_{i}=\max\\{\lambda_{i},\mu_{i}\\}+B$, else
$\nu_{i}=\max\\{\lambda_{i},\mu_{i}\\}$;

    * –

if $\mu_{i+1}<\lambda_{i}\leq\mu_{i}$, then
$B=\min\\{\lambda_{i},\mu_{i}\\}-\kappa_{i}$.

  * •

return $\nu$.

Here $\mathrm{Bern}(p)$ is a Bernoulli random variable taking value 1 with
probability $p$.

Besides Algorithms 7.1 and 7.2, we now introduce two other algorithms.

The goal of algorithm 7.3 is, given 3 partitions
$\lambda,\mu,\kappa\in{\mathbb{Y}}$ satisfying $\mu\prec\kappa\prec\lambda$,
sample a partition $\nu$ such that $\mu\prec\nu\prec\lambda$,
$|\nu|+|\kappa|=|\lambda|+|\mu|$, and for fixed $(\lambda,\mu)$, the mapping
$\kappa\rightarrow\nu$ is a bijection.

######  Algorithm 7.3.

def $AA(\lambda,\mu,\kappa)$

  * •

construct a partition $\nu\in{\mathbb{Y}}$ as follows:
$\nu_{i}=\min\\{\lambda_{i},\mu_{i-1}\\}+\max\\{\lambda_{i+1},\mu_{i}\\}-\kappa_{i}$.

  * •

return $\nu$.

The goal of algorithm 7.4 is, given 3 partitions
$\lambda,\mu,\kappa\in{\mathbb{Y}}$ satisfying
$\mu\prec^{\prime}\kappa\prec\lambda$, sample a partition $\nu$ such that
$\mu\prec\nu\prec^{\prime}\lambda$, $|\nu|+|\kappa|=|\lambda|+|\mu|$, and for
fixed $(\lambda,\mu)$, the mapping $\kappa\rightarrow\nu$ is a bijection.

######  Algorithm 7.4.

def $AB(\lambda,\mu,\kappa)$

  * •

for $1\leq i\leq\max\\{l(\lambda),l(\mu)\\}+1$;

    * –

If $\lambda_{i+1}\leq\mu_{i}<\lambda_{i}$,
$\delta=\kappa_{i}-\max\\{\lambda_{i+1},\mu_{i}\\}$; else $\delta=0$.

    * –

if $\mu_{i}<\lambda_{i}\leq\mu_{i-1}$,
$\nu_{i}=\min\\{\lambda_{i},\mu_{i-1}\\}-\delta$; else,
$\nu_{i}=\min\\{\lambda_{i},\mu_{i-1}\\}$.

  * •

return $\nu$.

######  Lemma 7.5.

Given 3 partitions $\lambda,\mu,\kappa\in{\mathbb{Y}}$ satisfying
$\mu\prec^{\prime}\kappa\prec\lambda$, Algorithm 7.4 returns a partition $\nu$
such that $\mu\prec\nu\prec^{\prime}\lambda$,
$|\nu|+|\kappa|=|\lambda|+|\mu|$, and for fixed $(\lambda,\mu)$, the mapping
$\kappa\rightarrow\nu$ is a bijection.

###### Proof.

Note that the condition $\mu\prec^{\prime}\kappa\prec\lambda$ gives

$\displaystyle\max\\{\lambda_{i+1},\mu_{i}\\}\leq\kappa_{i}\leq\min\\{\lambda_{i},\mu_{i}+1\\}.$

Then condition $\mu\prec\nu\prec^{\prime}\lambda$ gives

(7.1)
$\displaystyle\max\\{\lambda_{i}-1,\mu_{i}\\}\leq\nu_{i}\leq\min\\{\lambda_{i},\mu_{i-1}\\}.$

Hence we obtain that $\min\\{\lambda_{i},\mu_{i-1}\\}-\nu_{i}=0$ unless

(7.2) $\displaystyle\mu_{i}<\lambda_{i}\leq\mu_{i-1},$

in which case $\min\\{\lambda_{i},\mu_{i-1}\\}-\nu_{i}$ may take the value 1
as well. Let $i_{1}<i_{2}<\ldots<i_{r}$ be the $i$’s such that (7.2) holds.
Similarly $\kappa_{i}-\max\\{\lambda_{i+1},\mu_{i}\\}=0$ unless

(7.3) $\displaystyle\lambda_{i+1}\leq\mu_{i}<\lambda_{i}$

in which case $\kappa_{i}-\max\\{\lambda_{i+1},\mu_{i}\\}=0$ may take the
value 1 as well. Let $j_{1}<j_{2}<\ldots<j_{s}$ be the $i$’s such that (7.3)
holds. We claim that $s=r$ and

$\displaystyle j_{1}<i_{1}\leq j_{2}<i_{2}\ldots\leq j_{r}<i_{r}.$

Indeed, the mapping $\gamma$ defined by

$\displaystyle\gamma(j):=\min\\{i>j:\mu_{i}<\lambda_{i}\\}$

gives a bijection from $\\{j_{1},\ldots,j_{r}\\}$ to
$\\{i_{1},\ldots,i_{r}\\}$. Then it follows that
$|\nu|+|\kappa|=|\lambda|+|\mu|$. (7.1), and therefore
$\mu\prec\nu\prec^{\prime}\lambda$ can also be checked from the construction
of $\nu$. ∎

Let $K\in{\mathbb{N}}$ be a fixed large positive integer. Let
$m_{1},\ldots,m_{K}$ be sampled according to independent geometric
distributions such that

$\displaystyle m_{k}\sim\mathrm{Geom}((uv)^{k}),\ \forall k\in[K].$

Then we construct a partition $\lambda\in{\mathbb{Y}}$ whose maximal part is
at most $K$ satisfying

$\displaystyle m_{k}(\lambda)=m_{k},\ \forall k\in[K]$

where $m_{k}(\lambda)$ is the number of parts in $\lambda$ equal to $k$, as in
(A.5).

Let

$\displaystyle N_{-}:=\\{j\in[l..r],b_{j}=-\\};\qquad
N_{+}:=\\{j\in[l..r],b_{j}=+\\};$ $\displaystyle N:=N_{-}+N_{+}.$

This also gives an increasing bijection $\psi_{-}$ from $[N_{-}]$ to
$\\{j\in[l..r],b_{j}=-\\}$, and an increasing bijection $\psi_{+}$ from
$[N_{+}]$ to $\\{j\in[l..r],b_{j}=+\\}$.

To each point in the set

$\displaystyle
I:=\\{(i,-NK),(-NK,j):i\in[-N_{+}-NK..-NK],j\in[-N_{-}-NK..-NK]\\}$

associate a partition

$\displaystyle\lambda^{(i,j)}=\lambda;\ \forall(i,j)\in I.$

Let

$\displaystyle s:=-Nr+q+1;\qquad r\in[K],\ q\in[0..N-1]$

Assume we have obtained partitions

$\displaystyle\lambda^{(i,s-1)},\forall i\in[-N_{+}+s-1..s-1];\ \mathrm{and}\
\lambda^{(s-1,j)},\forall j\in[-N_{-}+s-1..s-1].$

for some $s\in[K]$. Now we sample partitions

$\displaystyle\lambda^{(i,s)},\forall i\in[-N_{+}+s..s];\ \mathrm{and}\
\lambda^{(s,j)},\forall j\in[-N_{-}+s..s].$

in the following way:

  1. (1)

If $q<N_{-}$:

    1. (a)

When $i$ decreases from $-N_{-}+s-1+q$ to $-N_{-}+s$, let $j:=i+N_{-}-s$.
Update the partition $\lambda^{(s-1,i)}$ to $\tilde{\lambda}^{(s-1,i)}$ as
follows:

      1. (i)

Let $\tilde{\lambda}^{(s-1,-N_{-}+s-1+q)}=\lambda^{(s-1,-N_{-}+s-1+q)}$;

      2. (ii)

If $a_{\psi_{-}(N_{-}-q)}=L$ and $a_{\psi_{-}(N_{-}-j)}=L$, let

$\displaystyle\tilde{\lambda}^{(s-1,i)}=AA(\tilde{\lambda}^{(s-1,i+1)},\lambda^{(s-1,i-1)},\lambda^{(s-1,i)})$

      3. (iii)

If $a_{\psi_{-}(N_{-}-q)}=L$ and $a_{\psi_{-}(N_{-}-j)}=R$, let

$\displaystyle\tilde{\lambda}^{(s-1,i)}=AB(\tilde{\lambda}^{(s-1,i+1)},\lambda^{(s-1,i-1)},\lambda^{(s-1,i)})$

      4. (iv)

If $a_{\psi_{-}(N_{-}-q)}=R$ and $a_{\psi_{-}(N_{-}-j)}=L$, let

$\displaystyle\tilde{\lambda}^{(s-1,i)}=[AB([\tilde{\lambda}^{(s-1,i+1)}]^{\prime},[\lambda^{(s-1,i-1)}]^{\prime},[\lambda^{(s-1,i)}]^{\prime})]^{\prime}$

      5. (v)

If $a_{\psi_{-}(N_{-}-q)}=R$ and $a_{\psi_{-}(N_{-}-j)}=R$, let

$\displaystyle\tilde{\lambda}^{(s-1,i)}=[AA([\tilde{\lambda}^{(s-1,i+1)}]^{\prime},[\lambda^{(s-1,i-1)}]^{\prime},[\lambda^{(s-1,i)}]^{\prime})]^{\prime}$

After all these updates, we still use $\lambda^{(s-1,i)}$ to denote the
updated partitions $\tilde{\lambda}^{(s-1,i)}$.

    2. (b)

Sample the partition $\lambda^{(s,-N_{-}+s)}$ as follows:

      1. (i)

If $a_{\psi_{-}(N_{-}-q)}=L$, let

$\displaystyle\lambda^{(s,-N_{-}+s)}=$ $\displaystyle
HH(\lambda^{(s-1,-N_{-}+s)},\lambda^{(s-1,-N_{-}+s)},\lambda^{(s-1,-N_{-}+s-1)},u^{2r}v^{2r-1}x_{\psi_{-}(N_{-}-q)}).$

      2. (ii)

If $a_{\psi_{-}(N_{-}-q)}=R$, let

$\displaystyle\lambda^{(s,-N_{-}+s)}=$
$\displaystyle[HH([\lambda^{(s-1,-N_{-}+s)}]^{\prime},[\lambda^{(s-1,-N_{-}+s)}]^{\prime},[\lambda^{(s-1,-N_{-}+s-1)}]^{\prime},u^{2r}v^{2r-1}x_{\psi_{-}(N_{-}-q)})]^{\prime}.$

    3. (c)

For $t\in[-N_{-}+s+1..-N_{-}+s+q]$ sample the partition $\lambda^{(s,t)}$ as
follows

      1. (i)

If $a_{\psi_{-}(-t+s+1)}=a_{\psi_{-}(N_{-}-q)}=L$, let

$\displaystyle\lambda^{(s,t)}=HH(\lambda^{(s-1,t)},\lambda^{(s,t-1)},\lambda^{(s-1,t-1)},u^{4r-2}v^{4r-4}x_{\psi_{-}(-t+s+1)}x_{\psi_{-}(N_{-}-q)})$

      2. (ii)

If $a_{\psi_{-}(-t+s+1)}=a_{\psi_{-}(N_{-}-q)}=R$, let

$\displaystyle\lambda^{(s,t)}=[HH([\lambda^{(s-1,t)}]^{\prime},[\lambda^{(s,t-1)}]^{\prime},[\lambda^{(s-1,t-1)}]^{\prime},u^{4r-2}v^{4r-4}x_{\psi_{-}(-t+s+1)}x_{\psi_{-}(N_{-}-q)})]^{\prime}$

      3. (iii)

If $a_{\psi_{-}(-t+s+1)}=L$, and $a_{\psi_{-}(N_{-}-q)}=R$, let

$\displaystyle\lambda^{(s,t)}=HV(\lambda^{(s-1,t)},\lambda^{(s,t-1)},\lambda^{(s-1,t-1)},u^{4r-2}v^{4r-4}x_{\psi_{-}(-t+s+1)}x_{\psi_{-}(N_{-}-q)})$

      4. (iv)

If $a_{\psi_{-}(-t+s+1)}=R$, and $a_{\psi_{-}(N_{-}-q)}=L$, let

$\displaystyle\lambda^{(s,t)}=[HV([\lambda^{(s-1,t)}]^{\prime},[\lambda^{(s,t-1)}]^{\prime},[\lambda^{(s-1,t-1)}]^{\prime},u^{4r-2}v^{4r-4}x_{\psi_{-}(-t+s+1)}x_{\psi_{-}(N_{-}-q)})]^{\prime}$

    4. (d)

For $t\in[-N_{-}+s+q+1..s-1]$ sample the partition $\lambda^{(s,t)}$ as
follows

      1. (i)

If $a_{\psi_{-}(-t+s)}=a_{\psi_{-}(N_{-}-q)}=L$, let

$\displaystyle\lambda^{(s,t)}=HH(\lambda^{(s-1,t)},\lambda^{(s,t-1)},\lambda^{(s-1,t-1)},u^{4r}v^{4r-2}x_{\psi_{-}(-t+s)}x_{\psi_{-}(N_{-}-q)})$

      2. (ii)

If $a_{\psi_{-}(-t+s)}=a_{\psi_{-}(N_{-}-q)}=R$, let

$\displaystyle\lambda^{(s,t)}=[HH([\lambda^{(s-1,t)}]^{\prime},[\lambda^{(s,t-1)}]^{\prime},[\lambda^{(s-1,t-1)}]^{\prime},u^{4r}v^{4r-2}x_{\psi_{-}(-t+s)}x_{\psi_{-}(N_{-}-q)})]^{\prime}$

      3. (iii)

If $a_{\psi_{-}(-t+s)}=L$, and $a_{\psi_{-}(N_{-}-q)}=R$, let

$\displaystyle\lambda^{(s,t)}=HV(\lambda^{(s-1,t)},\lambda^{(s,t-1)},\lambda^{(s-1,t-1)},u^{4r}v^{4r-2}x_{\psi_{-}(-t+s)}x_{\psi_{-}(N_{-}-q)})$

      4. (iv)

If $a_{\psi_{-}(-t+s)}=R$, and $a_{\psi_{-}(N_{-}-q)}=L$, let

$\displaystyle\lambda^{(s,t)}=[HV([\lambda^{(s-1,t)}]^{\prime},[\lambda^{(s,t-1)}]^{\prime},[\lambda^{(s-1,t-1)}]^{\prime},u^{4r}v^{4r-2}x_{\psi_{-}(-t+s)}x_{\psi_{-}(N_{-}-q)})]^{\prime}$

    5. (e)

When $i$ decreases from $s-1$ to $-N_{+}+s$, let $j:=i+N_{+}-s+1$. Update the
partition $\lambda^{(i,s-1)}$ to $\tilde{\lambda}^{(i,s-1)}$ as follows:

      1. (i)

Let $\tilde{\lambda}^{(s,s-1)}=\lambda^{(s,s-1)}$;

      2. (ii)

If $a_{\psi_{-}(N_{-}-q)}=L$ and $a_{\psi_{+}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(i,s-1)}=AA(\tilde{\lambda}^{(i+1,s-1)},\lambda^{(i-1,s-1)},\lambda^{(i,s-1)})$

      3. (iii)

If $a_{\psi_{-}(N_{-}-q)}=L$ and $a_{\psi_{+}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(i,s-1)}=AB(\tilde{\lambda}^{(i+1,s-1)},\lambda^{(i-1,s-1)},\lambda^{(i,s-1)})$

      4. (iv)

If $a_{\psi_{-}(N_{-}-q)}=R$ and $a_{\psi_{+}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(i,s-1)}=[AB([\tilde{\lambda}^{(i+1,s-1)}]^{\prime},[\lambda^{(i-1,s-1)}]^{\prime},[\lambda^{(i,s-1)}]^{\prime})]^{\prime}$

      5. (v)

If $a_{\psi_{-}(N_{-}-q)}=R$ and $a_{\psi_{+}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(i,s-1)}=[AA([\tilde{\lambda}^{(i+1,s-1)}]^{\prime},[\lambda^{(i-1,s-1)}]^{\prime},[\lambda^{(i,s-1)}]^{\prime})]^{\prime}$

After all these updates, we still use $\lambda^{(i,s-1)}$ to denote the
updated partitions $\tilde{\lambda}^{(i,s-1)}$.

    6. (f)

Sample the partition $\lambda^{(-N_{+}+s,s)}$ as follows:

      1. (i)

If $a_{\psi_{-}(N_{-}-q)}=L$, let

$\displaystyle\lambda^{(-N_{+}+s,s)}=$ $\displaystyle
HH(\lambda^{(-N_{+}+s,s-1)},\lambda^{(-N_{+}+s,s-1)},\lambda^{(-N_{+}+s-1,s-1)},u^{2r-1}v^{2r-2}x_{\psi_{-}(N_{-}-q)}).$

      2. (ii)

If $a_{\psi_{-}(N_{-}-q)}=R$, let

$\displaystyle\lambda^{(-N_{+}+s,s)}=$
$\displaystyle[HH([\lambda^{(-N_{+}+s,s-1)}]^{\prime},[\lambda^{(,-N_{+}+s,s-1)}]^{\prime},[\lambda^{(-N_{+}+s-1,s-1)}]^{\prime},u^{2r-1}v^{2r-2}x_{\psi_{-}(N_{-}-q)})]^{\prime}.$

    7. (g)

For $t\in[-N_{+}+s+1..s]$ sample the partition $\lambda^{(t,s)}$ as follows

      1. (i)

If $a_{\psi_{+}(t+N_{+}-s)}=a_{\psi_{-}(N_{-}-q)}=L$, let

$\displaystyle\lambda^{(t,s)}=HH(\lambda^{(t-1,s)},\lambda^{(t,s-1)},\lambda^{(t-1,s-1)},u^{4r-2}v^{4r-2}x_{\psi_{+}(t+N_{+}-s)}x_{\psi_{-}(N_{-}-q)})$

      2. (ii)

If $a_{\psi_{+}(t+N_{+}-s)}=a_{\psi_{-}(N_{-}-q)}=R$, let

$\displaystyle\lambda^{(t,s)}=[HH([\lambda^{(t-1,s)}]^{\prime},[\lambda^{(t,s-1)}]^{\prime},[\lambda^{(t-1,s-1)}]^{\prime},u^{4r-2}v^{4r-2}x_{\psi_{+}(t+N_{+}-s)}x_{\psi_{-}(N_{-}-q)})]^{\prime}$

      3. (iii)

If $a_{\psi_{+}(t+N_{+}-s)}=L$, and $a_{\psi_{-}(N_{-}-q)}=R$, let

$\displaystyle\lambda^{(t,s)}=HV(\lambda^{(t-1,s)},\lambda^{(t,s-1)},\lambda^{(t-1,s-1)},u^{4r-2}v^{4r-2}x_{\psi_{+}(t+N_{+}-s)}x_{\psi_{-}(N_{-}-q)})$

      4. (iv)

If $a_{\psi_{+}(-t+N_{+}-s)}=R$, and $a_{\psi_{-}(N_{-}-q)}=L$, let

$\displaystyle\lambda^{(t,s)}=[HV([\lambda^{(t-1,s)}]^{\prime},[\lambda^{(t,s-1)}]^{\prime},[\lambda^{(t-1,s-1)}]^{\prime},u^{4r-2}v^{4r-2}x_{\psi_{+}(t+N_{+}-s)}x_{\psi_{-}(N_{-}-q)})]^{\prime}$

    8. (h)

When $i$ decreases from $s-1$ to $-N_{-}+s+q+1$, let $j:=s-i$. Update the
partition $\lambda^{(s,i)}$ to $\tilde{\lambda}^{(s,i)}$ as follows:

      1. (i)

Let $\tilde{\lambda}^{(s,s)}=\lambda^{(s,s)}$;

      2. (ii)

If $a_{\psi_{-}(N_{-}-q)}=L$ and $a_{\psi_{-}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(s,i)}=AA(\tilde{\lambda}^{(s,i+1)},\lambda^{(s,i-1)},\lambda^{(s,i)})$

      3. (iii)

If $a_{\psi_{-}(N_{-}-q)}=L$ and $a_{\psi_{-}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(s,i)}=AB(\tilde{\lambda}^{(s,i+1)},\lambda^{(s,i-1)},\lambda^{(s,i)})$

      4. (iv)

If $a_{\psi_{-}(N_{-}-q)}=R$ and $a_{\psi_{-}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(s,i)}=[AB([\tilde{\lambda}^{(s,i+1)}]^{\prime},[\lambda^{(s,i-1)}]^{\prime},[\lambda^{(s,i)}]^{\prime})]^{\prime}$

      5. (v)

If $a_{\psi_{-}(N_{-}-q)}=R$ and $a_{\psi_{-}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(s,i)}=[AA([\tilde{\lambda}^{(s,i+1)}]^{\prime},[\lambda^{(s,i-1)}]^{\prime},[\lambda^{(s,i)}]^{\prime})]^{\prime}$

After all these updates, we still use $\lambda^{(s,i)}$ to denote the updated
partitions $\tilde{\lambda}^{(s,i)}$.

  2. (2)

if $q\geq N_{-}$ and $r>1$:

    1. (a)

When $i$ decreases from $s-1-q+N_{-}-1$ to $-N_{+}+s$, let $j:=i+N_{+}-s+1$.
Update the partition $\lambda^{(i,s-1)}$ to $\tilde{\lambda}^{(i,s-1)}$ as
follows:

      1. (i)

Let $\tilde{\lambda}^{(-N_{+}+s+q,s-1)}=\lambda^{(-N_{+}+s+q,s-1)}$;

      2. (ii)

If $a_{\psi_{+}(N-q)}=L$ and $a_{\psi_{+}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(i,s-1)}=AA(\tilde{\lambda}^{(i+1,s-1)},\lambda^{(i-1,s-1)},\lambda^{(i,s-1)})$

      3. (iii)

If $a_{\psi_{+}(N-q)}=L$ and $a_{\psi_{+}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(i,s-1)}=AB(\tilde{\lambda}^{(i+1,s-1)},\lambda^{(i-1,s-1)},\lambda^{(i,s-1)})$

      4. (iv)

If $a_{\psi_{+}(N-q)}=R$ and $a_{\psi_{+}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(i,s-1)}=[AB([\tilde{\lambda}^{(i+1,s-1)}]^{\prime},[\lambda^{(i-1,s-1)}]^{\prime},[\lambda^{(i,s-1)}]^{\prime})]^{\prime}$

      5. (v)

If $a_{\psi_{+}(N-q)}=R$ and $a_{\psi_{+}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(i,s-1)}=[AA([\tilde{\lambda}^{(i+1,s-1)}]^{\prime},[\lambda^{(i-1,s-1)}]^{\prime},[\lambda^{(i,s-1)}]^{\prime})]^{\prime}$

After all these updates, we still use $\lambda^{(i,s-1)}$ to denote the
updated partitions $\tilde{\lambda}^{(i,s-1)}$.

    2. (b)

Sample the partition $\lambda^{(-N_{+}+s,s)}$ as follows:

      1. (i)

If $a_{\psi_{+}(N-q)}=L$, let

$\displaystyle\lambda^{(-N_{+}+s,s)}=$ $\displaystyle
HH(\lambda^{(-N_{+}+s,s-1)},\lambda^{(-N_{-}+s,s-1)},\lambda^{(-N_{-}+s-1,s-1)},u^{2r-1}v^{2r}x_{\psi_{+}(N-q)}).$

      2. (ii)

If $a_{\psi_{+}(N-q)}=R$, let

$\displaystyle\lambda^{(-N_{+}+s,s)}=$
$\displaystyle[HH([\lambda^{(-N_{-}+s,s-1)}]^{\prime},[\lambda^{(-N_{+}+s,s-1)}]^{\prime},[\lambda^{(-N_{-}+s-1,s-1)}]^{\prime},u^{2r-1}v^{2r}x_{\psi_{+}(N-q)})]^{\prime}.$

    3. (c)

For $t\in[-N_{+}+s+1..s+N_{-}-q-1]$ sample the partition $\lambda^{(t,s)}$ as
follows

      1. (i)

If $a_{\psi_{+}(t-s+N_{+})}=a_{\psi_{+}(N-q)}=L$, let

$\displaystyle\lambda^{(t,s)}=HH(\lambda^{(t-1,s)},\lambda^{(t,s-1)},\lambda^{(t-1,s-1)},u^{4r-4}v^{4r-2}x_{\psi_{+}(t-s+N_{+})}x_{\psi_{+}(N-q)})$

      2. (ii)

If $a_{\psi_{+}(t-s+N_{+})}=a_{\psi_{+}(N-q)}=R$, let

$\displaystyle\lambda^{(t,s)}=[HH([\lambda^{(t-1,s)}]^{\prime},[\lambda^{(t,s-1)}]^{\prime},[\lambda^{(t-1,s-1)}]^{\prime},u^{4r-4}v^{4r-2}x_{\psi_{+}(t-s+N_{+})}x_{\psi_{+}(N-q)})]^{\prime}$

      3. (iii)

If $a_{\psi_{+}(t-s+N^{+})}=R$, and $a_{\psi_{+}(N-q)}=L$, let

$\displaystyle\lambda^{(t,s)}=HV(\lambda^{(t,s-1)},\lambda^{(t-1,s)},\lambda^{(t-1,s-1)},u^{4r-4}v^{4r-2}x_{\psi_{+}(t-s+N^{+})}x_{\psi_{+}(N-q)})$

      4. (iv)

If $a_{\psi_{+}(t-s+N^{+})}=L$, and $a_{\psi_{+}(N-q)}=R$, let

$\displaystyle\lambda^{(t,s)}=[HV([\lambda^{(t,s-1)}]^{\prime},[\lambda^{(t-1,s)}]^{\prime},[\lambda^{(t-1,s-1)}]^{\prime},u^{4r-4}v^{4r-2}x_{\psi_{+}(t-s+N^{+})}x_{\psi_{+}(N-q)})]^{\prime}$

    4. (d)

For $t\in[s+N_{-}-q..s-1]$ sample the partition $\lambda^{(t,s)}$ as follows

      1. (i)

If $a_{\psi_{+}(t-s+N_{+}+1)}=a_{\psi_{+}(N-q)}=L$, let

$\displaystyle\lambda^{(t,s)}=HH(\lambda^{(t-1,s)},\lambda^{(t,s-1)},\lambda^{(t-1,s-1)},u^{4r-2}v^{4r}x_{\psi_{+}(t-s+N_{+}+1)}x_{\psi_{+}(N-q)})$

      2. (ii)

If $a_{\psi_{+}(t-s+N_{+}+1)}=a_{\psi_{+}(N-q)}=R$, let

$\displaystyle\lambda^{(t,s)}=[HH([\lambda^{(t-1,s)}]^{\prime},[\lambda^{(t,s-1)}]^{\prime},[\lambda^{t-1,s-1)}]^{\prime},u^{4r-2}v^{4r}x_{\psi_{+}(t-s+N_{+}+1)}x_{\psi_{+}(N-q)})]^{\prime}$

      3. (iii)

If $a_{\psi_{+}(t-s+N_{+}+1)}=R$, and $a_{\psi_{+}(N-q)}=L$, let

$\displaystyle\lambda^{(t,s)}=HV(\lambda^{(t,s-1)},\lambda^{(t-1,s)},\lambda^{(t-1,s-1)},u^{4r-2}v^{4r}x_{\psi_{+}(t-s+N_{+}+1)}x_{\psi_{+}(N-q)})$

      4. (iv)

If $a_{\psi_{+}(t-s+N_{+}+1}=L$, and $a_{\psi_{+}(N-q)}=R$, let

$\displaystyle\lambda^{(t,s)}=[HV([\lambda^{(t,s-1)}]^{\prime},[\lambda^{(t-1,s)}]^{\prime},[\lambda^{(t-1,s-1)}]^{\prime},u^{4r-2}v^{4r}x_{\psi_{+}(t-s+N_{+}+1)}x_{\psi_{+}(N-q)})]^{\prime}$

    5. (e)

When $i$ decreases from $s-1$ to $-N_{-}+s$, let $j:=s-i$. Update the
partition $\lambda^{(s-1,i)}$ to $\tilde{\lambda}^{(s-1,i)}$ as follows:

      1. (i)

Let $\tilde{\lambda}^{(s-1,s)}=\lambda^{(s-1,s)}$;

      2. (ii)

If $a_{\psi_{+}(N-q)}=L$ and $a_{\psi_{-}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(s-1,i)}=AA(\tilde{\lambda}^{(s-1,i+1)},\lambda^{(s-1,i-1)},\lambda^{(s,i)})$

      3. (iii)

If $a_{\psi_{+}(N-q)}=L$ and $a_{\psi_{-}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(s-1,i)}=AB(\tilde{\lambda}^{(s-1,i+1)},\lambda^{(s-1,i-1)},\lambda^{(s-1,i)})$

      4. (iv)

If $a_{\psi_{+}(N-q)}=R$ and $a_{\psi_{-}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(s-1,i)}=[AB([\tilde{\lambda}^{(s-1,i+1)}]^{\prime},[\lambda^{(s-1,i-1)}]^{\prime},[\lambda^{(s-1,i)}]^{\prime})]^{\prime}$

      5. (v)

If $a_{\psi_{+}(N-q)}=R$ and $a_{\psi_{-}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(s-1,i)}=[AA([\tilde{\lambda}^{(s-1,i+1)}]^{\prime},[\lambda^{(s-1,i-1)}]^{\prime},[\lambda^{(s-1,i)}]^{\prime})]^{\prime}$

After all these updates, we still use $\lambda^{(s-1,i)}$ to denote the
updated partitions $\tilde{\lambda}^{(s-1,i)}$.

    6. (f)

Sample the partition $\lambda^{(s,-N_{-}+s)}$ as follows:

      1. (i)

If $a_{\psi_{+}(N-q)}=L$, let

$\displaystyle\lambda^{(s,-N_{-}+s)}=$ $\displaystyle
HH(\lambda^{(s-1,-N_{-}+s)},\lambda^{(s-1,-N_{-}+s)},\lambda^{(s-1,-N_{-}+s-1)},u^{2r-2}v^{2r-1}x_{\psi_{+}(N-q)}).$

      2. (ii)

If $a_{\psi_{+}(N-q)}=R$, let

$\displaystyle\lambda^{(s,-N_{-}+s)}=$
$\displaystyle[HH([\lambda^{(s-1,-N_{-}+s)}]^{\prime},[\lambda^{(s-1,-N_{-}+s)}]^{\prime},[\lambda^{(s-1,-N_{-}+s-1)}]^{\prime},u^{2r-2}v^{2r-1}x_{\psi_{+}(N-q)})]^{\prime}.$

    7. (g)

For $t\in[-N_{-}+s+1..s]$, sample the partition $\lambda^{(s,t)}$ as follows

      1. (i)

If $a_{\psi_{-}(s+1-t)}=a_{\psi_{+}(N-q)}=L$, let

$\displaystyle\lambda^{(s,t)}=HH(\lambda^{(s,t-1)},\lambda^{(s-1,t)},\lambda^{(s-1,t-1)},u^{4r-4}v^{4r-4}x_{\psi_{-}(s+1-t)}x_{\psi_{+}(N-q)})$

      2. (ii)

If $a_{\psi_{-}(s+1-t)}=a_{\psi_{+}(q+1-N_{-})}=R$, let

$\displaystyle\lambda^{(s,t)}=[HH([\lambda^{(s,t-1)}]^{\prime},[\lambda^{(s-1,t)}]^{\prime},[\lambda^{(s-1,t-1)}]^{\prime},u^{4r-4}v^{4r-4}x_{\psi_{-}(s+1-t)}x_{\psi_{+}(N-q)})]^{\prime}$

      3. (iii)

If $a_{\psi_{-}(s+1-t)}=R$, and $a_{\psi_{+}(N-q)}=L$, let

$\displaystyle\lambda^{(s,t)}=HV(\lambda^{(s-1,t)},\lambda^{(s,t-1)},\lambda^{(s-1,t-1)},u^{4r-4}v^{4r-4}x_{\psi_{-}(s+1-t)}x_{\psi_{+}(N-q)})$

      4. (iv)

If $a_{\psi_{-}(s+1-t)}=L$, and $a_{\psi_{+}(N-q)}=R$, let

$\displaystyle\lambda^{(s,t)}=[HV([\lambda^{(s-1,t)}]^{\prime},[\lambda^{(s,t-1)}]^{\prime},[\lambda^{(s-1,t-1)}]^{\prime},u^{4r-4}v^{4r-4}x_{\psi_{-}(s+1-t)}x_{\psi_{+}(N-q)})]^{\prime}$

    8. (h)

When $i$ decreases from $s-1$ to $s-q+N_{-}$, let $j:=N_{+}-s+i+1$. Update the
partition $\lambda^{(i,s)}$ to $\tilde{\lambda}^{(i,s)}$ as follows:

      1. (i)

Let $\tilde{\lambda}^{(s,s)}=\lambda^{(s,s)}$;

      2. (ii)

If $a_{\psi_{+}(N-q)}=L$ and $a_{\psi_{+}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(i,s)}=AA(\tilde{\lambda}^{(i+1,s)},\lambda^{(i-1,s)},\lambda^{(i,s)})$

      3. (iii)

If $a_{\psi_{+}(N-q)}=L$ and $a_{\psi_{+}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(i,s)}=AB(\tilde{\lambda}^{(i+1,s)},\lambda^{(i-1,s)},\lambda^{(i,s)})$

      4. (iv)

If $a_{\psi_{+}(N-q)}=R$ and $a_{\psi_{+}(j)}=L$, let

$\displaystyle\tilde{\lambda}^{(i,s)}=[AB([\tilde{\lambda}^{(i+1,s)}]^{\prime},[\lambda^{(i-1,s)}]^{\prime},[\lambda^{(i,s)}]^{\prime})]^{\prime}$

      5. (v)

If $a_{\psi_{+}(N-q)}=R$ and $a_{\psi_{+}(j)}=R$, let

$\displaystyle\tilde{\lambda}^{(i,s)}=[AA([\tilde{\lambda}^{(i+1,s)}]^{\prime},[\lambda^{(i-1,s)}]^{\prime},[\lambda^{(i,s)}]^{\prime})]^{\prime}$

After all these updates, we still use $\lambda^{(i,s)}$ to denote the updated
partitions $\tilde{\lambda}^{(i,s)}$.

  3. (3)

if $q\geq N_{-}$ and $r=1$:

    1. (a)

When $i$ decreases from $s+N_{-}-q-2$ to $-N_{+}+s$, Update the partition
$\lambda^{(i,s-1)}$ to $\tilde{\lambda}^{(i,s-1)}$ as in Step (2a) above.

    2. (b)

Sample the partition $\lambda^{(-N_{+}+s,s)}$ as in Step (2b) above.

    3. (c)

For $t\in[-N_{+}+s+1..s+N_{-}-q-1]$, sample the partition $\lambda^{(t,s)}$ as
in Step (2c).

    4. (d)

For $t\in[s+N_{-}-q..s-1]$, let $\pi_{t}$ be the part of the boundary of of
all the integral points $(i,j)$, where $\lambda^{(i,j)}$ has been sampled,
starting from $(-N_{+}+s,s)$, ending in $(s-1,-N_{-}+s-1)$, and winding
clockwise. Let $i_{t}\in{\mathbb{Z}}$ be the vertical coordinate of the
$(t+N_{+}-s-1)$th unit horizontal segment of $\pi_{t}$ and let $z$ be the
right endpoint of of the $(t+N_{+}-s-1)$th unit horizontal segment of
$\pi_{t}$, then following $z$, along $\pi_{t}$ there is a vertical edge $zw$.
The following cases might occur:

      1. (i)

If following $w$, $\pi_{t}$ has a horizontal unit, sample the partition
$\lambda^{(t+1,i_{t})}$ as follows

        1. (A)

If $a_{\psi_{+}(t-s+N_{+}+1)}=a_{\psi_{+}(q+1-N_{-})}=L$, let

$\displaystyle\lambda^{(t+1,i_{t})}=HH(\lambda^{(t,i_{t})},\lambda^{(t+1,i_{t}-1)},\lambda^{(t,i_{t}-1)},u^{4r-4}v^{4r-2}x_{\psi_{+}(t-s+N_{+}+1)}x_{\psi_{+}(q+1-N_{-})})$

        2. (B)

If $a_{\psi_{+}(t-s+N_{+}+1)}=a_{\psi_{+}(q+1-N_{-})}=R$, let

$\displaystyle\lambda^{(t+1,i_{t})}=[HH([\lambda^{(t,i_{t})}]^{\prime},[\lambda^{(t+1,i_{t}-1)}]^{\prime},[\lambda^{(t,i_{t}-1)}]^{\prime},u^{4r-4}v^{4r-2}x_{\psi_{+}(t-s+N_{+}+1)}x_{\psi_{+}(q+1-N_{-})})]^{\prime}$

        3. (C)

If $a_{\psi_{+}(t-s+N_{+}+1)}=L$, and $a_{\psi_{+}(q+1-N_{-})}=R$, let

$\displaystyle\lambda^{(t+1,i_{t})}=HV(\lambda^{(t+1,i_{t}-1)},\lambda^{(t,i_{t})},\lambda^{(t,i_{t}-1)},u^{4r-4}v^{4r-2}x_{\psi_{+}(t-s+N_{+}+1)}x_{\psi_{+}(q+1-N_{-})})$

        4. (D)

If $a_{\psi_{+}(t-s+N^{+})}=R$, and $a_{\psi_{+}(q+1-N_{-})}=L$, let

$\displaystyle\lambda^{(t+1,i_{t})}=[HV([\lambda^{(t+1,i_{t}-1)}]^{\prime},[\lambda^{(t,i_{t})}]^{\prime},[\lambda^{(t,i_{t}-1)}]^{\prime},u^{4r-4}v^{4r-2}x_{\psi_{+}(t-s+N_{+}+1)}x_{\psi_{+}(q+1-N_{-})})]^{\prime}$

      2. (ii)

If following $w$, $\pi_{t}$ has a vertical unit $ww_{1}$. Assume $ww_{1}$ is
the $h_{ww_{1}}$th vertical unit along $\pi_{t}$, then update the partition
$\lambda^{(t,i_{t}-1)}$ to $\tilde{\lambda}^{(t,i_{t}-1)}$ as follows

        1. (A)

Let $\tilde{\lambda}^{(t,i_{t})}=\lambda^{(t,i_{t})}$

        2. (B)

If $a_{\psi_{+}(q+1-N_{-})}=a_{\psi_{-}(h_{ww_{1}}-1)}=L$, then

$\displaystyle\tilde{\lambda}^{(t,i_{t}-1)}=AA(\tilde{\lambda}^{(t,i_{t})},\lambda^{(t,i_{t}-2)},\lambda^{(t,i_{t}-1)})$

        3. (C)

If $a_{\psi_{+}(q+1-N_{-})}=L,a_{\psi_{-}(h_{ww_{1}}-1)}=R$, then

$\displaystyle\tilde{\lambda}^{(t,i_{t}-1)}=AB(\tilde{\lambda}^{(t,i_{t})},\lambda^{(t,i_{t}-2)},\lambda^{(t,i_{t}-1)})$

        4. (D)

If $a_{\psi_{+}(q+1-N_{-})}=R,a_{\psi_{-}(h_{ww_{1}}-1)}=L$, then

$\displaystyle\tilde{\lambda}^{(t,i_{t}-1)}=[AB([\tilde{\lambda}^{(t,i_{t})}]^{\prime},[\lambda^{(t,i_{t}-2)}]^{\prime},[\lambda^{(t,i_{t}-1)}]^{\prime})]^{\prime}$

        5. (E)

If $a_{\psi_{+}(q+1-N_{-})}=a_{\psi_{-}(h_{ww_{1}}-1)}=R$, then

$\displaystyle\tilde{\lambda}^{(t,i_{t}-1)}=[AA([\tilde{\lambda}^{(t,i_{t})}]^{\prime},[\lambda^{(t,i_{t}-2)}]^{\prime},[\lambda^{(t,i_{t}-1)}]^{\prime})]^{\prime}$

        6. (F)

continue similar updates in (B) until the next edge along $\pi_{t}$ is
horizontal. Then we still use $\lambda$’s to denoted the updated partitions
$\tilde{\lambda}$’s, and increase $t$ by 1 and continue the step (3d) until
$t=s$ then move to (3di).

    5. (e)

Sample the partition $\lambda^{(s,-N_{-}+s)}$ as in Step (2f).

    6. (f)

for $t\in[-N_{-}+s+1..s]$, let $\sigma_{t}$ be the part of the boundary of all
the integral points $(i,j)$, where $\lambda^{(i,j)}$ has been sampled,
starting from $(s,-N_{-}+s)$, ending in $(-N_{+}+s,s)$ and winding
counterclockwise. Let $j_{t}\in{\mathbb{Z}}$ be the horizontal coordinate of
the $(t+N_{-}-s-1)$th unit vertical segment of $\sigma_{t}$; and let $z$ be
the upper endpoint of the $(t+N_{-}-s-1)$th unit vertical segment of
$\sigma_{t}$. Then following $z$ there is a horizontal unit $zw$ along
$\sigma_{t}$. Let

$\displaystyle\eta:=|\\{i\in[l..r]:b_{i}=-,i>\psi_{+}(q+1-N)\\}|$

For $t\in[-N_{-}+s+1..-N_{-}+s+\eta]$, the following cases might occur

      1. (i)

If following $w$ along $\sigma_{t}$, $ww_{1}$ is a vertical unit segment,
sample the partition $\lambda^{(j_{t},t)}$ as follows

        1. (A)

If $a_{\psi_{-}(t+N_{-}-s)}=a_{\psi_{+}(q+1-N_{-})}=L$, let

$\displaystyle\lambda^{(j_{t},t)}=HH(\lambda^{(j_{t},t-1)},\lambda^{(j_{t}-1,t)},\lambda^{(j_{t}-1,t-1)},u^{4r-4}v^{4r-4}x_{\psi_{-}(t+N_{-}-s)}x_{\psi_{+}(q+1-N_{-})})$

        2. (B)

If $a_{\psi_{-}(t+N_{-}-s)}=a_{\psi_{+}(q+1-N_{-})}=R$, let

$\displaystyle\lambda^{(j_{t},t)}=[HH([\lambda^{(j_{t},t-1)}]^{\prime},[\lambda^{(j_{t}-1,t)}]^{\prime},[\lambda^{(j_{t}-1,t-1)}]^{\prime},u^{4r-4}v^{4r-4}x_{\psi_{-}(t+N_{-}-s)}x_{\psi_{+}(q+1-N_{-})})]^{\prime}$

        3. (C)

If $a_{\psi_{-}(t+N_{-}-s)}=L$, and $a_{\psi_{-}(q+1-N_{-})}=R$, let

$\displaystyle\lambda^{(j_{t},t)}=HV(\lambda^{(j_{t}-1,t)},\lambda^{(j_{t},t-1)},\lambda^{(j_{t}-1,t-1)},u^{4r-4}v^{4r-4}x_{\psi_{-}(t+N_{-}-s)}x_{\psi_{+}(q+1-N_{-})})$

        4. (D)

If $a_{\psi_{+}(-t+N_{+}-s)}=R$, and $a_{\psi_{-}(N_{-}-q)}=L$, let

$\displaystyle\lambda^{(j_{t},t)}=[HV([\lambda^{(j_{t}-1,t)}]^{\prime},[\lambda^{(j_{t},t-1)}]^{\prime},[\lambda^{(j_{t}-1,t-1)}]^{\prime},u^{4r-4}v^{4r-4}x_{\psi_{-}(t+N_{-}-s)}x_{\psi_{+}(q+1-N_{-})})]^{\prime}$

      2. (ii)

If following $w$ along $\sigma_{t}$, $ww_{1}$ is a horizontal unit segment.
Assume $ww_{1}$ is the $q_{ww_{1}}$th horizontal unit along $-\sigma_{t}$
($-\sigma_{t}$ is the path obtained from $\sigma_{t}$ by reversing the
orientation) update the partition $\lambda^{(j_{t}-1,t)}$ to
$\tilde{\lambda}^{(j_{t}-1,t)}$ as follows

        1. (A)

If $a_{\psi_{+}(q+1-N_{-})}=a_{\psi_{+}(h_{ww_{1}}+1)}=L$, then

$\displaystyle\tilde{\lambda}^{(j_{t}-1,t)}=AA(\tilde{\lambda}^{(j_{t},t)},\lambda^{(j_{t}-2,t)},\lambda^{(j_{t}-1,t)})$

        2. (B)

If $a_{\psi_{+}(q+1-N_{-})}=L,a_{\psi_{+}(h_{ww_{1}}+1)}=R$, then

$\displaystyle\tilde{\lambda}^{(j_{t}-1,t)}=AB(\tilde{\lambda}^{(j_{t},t)},\lambda^{(j_{t}-2,t)},\lambda^{(j_{t}-1,t)})$

        3. (C)

If $a_{\psi_{+}(q+1-N_{-})}=R,a_{\psi_{+}(h_{ww_{1}}+1)}=L$, then

$\displaystyle\tilde{\lambda}^{(j_{t}-1,t)}=[AB([\tilde{\lambda}^{(j_{t},t)}]^{\prime},[\lambda^{(,j_{t}-2,t)}]^{\prime},[\lambda^{(j_{t}-1,t)}]^{\prime})]^{\prime}$

        4. (D)

If $a_{\psi_{+}(q+1-N_{-})}=a_{\psi_{+}(h_{ww_{1}}+1)}=R$, then

$\displaystyle\tilde{\lambda}^{(j_{t}-1,t)}=[AA([\tilde{\lambda}^{(j_{t},t)}]^{\prime},[\lambda^{(j_{t}-2,t)}]^{\prime},[\lambda^{(j_{t}-1,t)}]^{\prime})]^{\prime}$

        5. (E)

continue similar updates in (B) until the next edge along $\pi_{t}$ is
horizontal. Then we still use $\lambda$’s to denoted the updated partitions
$\tilde{\lambda}$’s, and increase $t$ by 1 and continue the step (3fi) until
$t>s$.

Consider the part $\mathcal{D}$ of the boundary of all points
$(i,j)\in{\mathbb{Z}}^{2}$ such that $\lambda^{(i,j)}$ has been sampled,
starting from $(-N_{+},0)$ and ending in $(0,-N_{-})$ and winding clockwise,
the sequence of partitions $\lambda^{(i,j)}$ along all the points of this part
of (oriented) boundary gives a sample of sequence of partitions corresponding
to double-free-boundary dimer coverings on the rail yard graph.

######  Theorem 7.6.

Assume $\mathcal{D}=(z_{0},\ldots,z_{N})$. Then

  1. (1)

the ordered sequence of partitions

(7.4) $\displaystyle(\lambda^{z_{0}},\ldots,\lambda^{z_{N}})$

have distribution proportional to the product of weights of present edges in
the dimer covering, conditional on

     * •

$\lambda_{1}^{z_{0}}\leq K$ and $\lambda_{1}^{z_{N}}\leq K$; and

     * •

$\lambda^{(-N_{+}-NK,-NK)}=\lambda^{(-N_{+}-NK+1,-NK)}=\ldots=\lambda^{(-NK,-NK)}=\lambda^{(-NK,-NK-1)}=\ldots=\lambda^{(-NK,-NK-
N_{-})}$;

  2. (2)

as $K\rightarrow\infty$, the distribution of (7.4) converges to the Gibbs
measure of double-sided free boundary dimer configurations.

###### Proof.

It is straightforward to check Part (1) of the theorem from the sampling
process. For Part (2), note that the partition function of double-sided free
boundary conditions that do not satisfy the condition of Part (1) if of order
$(uv)^{K}$ if the partition function of all the configurations, which tends to
0 as $K\rightarrow\infty$. ∎

##  8\. Example

######  Example 8.1.

Let

$\displaystyle V_{0}=0,\ V_{1}=1,\ m=1,\ n=4$

and

$\displaystyle(a_{1},b_{1})=(L,-),\ (a_{2},b_{2})=(L,+),\
(a_{3},b_{3})=(R,+),\ (a_{4},b_{4})=(R,-),\ $

and

$\displaystyle\tau_{1}=\tau_{4}=1,\ \tau_{2}=\tau_{3}=0.3,\ u=v=0.1$

We have

(8.1) $\displaystyle g_{k}(w):$
$\displaystyle=\frac{1}{w-(0.01)^{2k}e}-\frac{1}{w-(0.01)^{2k}}+\frac{1}{w-(0.1)^{4k-2}e}-\frac{1}{w-(0.1)^{4k-2}}$
$\displaystyle+\frac{1}{w-(0.01)^{-2k}[0.3]^{-1}}-\frac{1}{w-(0.01)^{-2k}e[0.3]^{-1}}$
$\displaystyle+\frac{1}{w-(0.1)^{2-4k}[0.3]^{-1}}-\frac{1}{w-(0.1)^{2-4k}e[0.3]^{-1}}$
$\displaystyle+\frac{1}{w+(0.01)^{2k}}-\frac{1}{w+(0.01)^{2k}e}+\frac{1}{w+(0.1)^{4k-2}}-\frac{1}{w+(0.1)^{4k-2}e}$
$\displaystyle+\frac{1}{w+(0.01)^{-2k}e[0.3]^{-1}}-\frac{1}{w+(0.01)^{-2k}[0.3]^{-1}}$
$\displaystyle+\frac{1}{w+(0.1)^{2-4k}e[0.3]^{-1}}-\frac{1}{w+(0.1)^{2-4k}[0.3]^{-1}}$

The second equation in (2.13) gives

(8.2) $\displaystyle 0$
$\displaystyle=\frac{1}{w-e}-\frac{1}{w-e^{\chi}}+\frac{1}{w-[0.3]^{-1}}-\frac{1}{w-e^{\chi}[0.3]^{-1}}$
$\displaystyle+\frac{1}{w+e^{\chi}}-\frac{1}{w+e}+\frac{1}{w+e^{\chi}[0.3]^{-1}}-\frac{1}{w+[0.3]^{-1}}+\sum_{k=1}^{\infty}g_{k}(w)$

Moreover,

$\displaystyle\mathcal{G}_{\chi}(w)$
$\displaystyle=\left(\frac{w-e}{w-e^{\chi}}\right)\left(\frac{1-0.3w}{1-0.3e^{-\chi}w}\right)\left(\frac{w+e^{\chi}}{w+e}\right)\left(\frac{1+0.3e^{-\chi}w}{1+0.3w}\right)$
$\displaystyle\mathcal{F}_{u,v,k}(w)$
$\displaystyle=\left(\frac{w-(0.01)^{2k}e}{w-(0.01)^{2k}}\right)\left(\frac{w-(0.1)^{4k-2}e}{w-(0.1)^{4k-2}}\right)$
$\displaystyle\left(\frac{1-[0.3]w(0.01)^{2k}}{1-[0.3]e^{-1}w(0.01)^{2k}}\right)\left(\frac{1-[0.3]w(0.1)^{4k-2}}{1-[0.3]e^{-1}w(0.1)^{4k-2}}\right)$
$\displaystyle\left(\frac{w+(0.01)^{2k}}{w+(0.01)^{2k}e}\right)\left(\frac{w+(0.1)^{4k-2}}{w+(0.1)^{4k-2}e}\right)$
$\displaystyle\left(\frac{1+[0.3]e^{-1}(0.01)^{2k}w}{1+[0.3](0.01)^{2k}w}\right)\left(\frac{1+[0.3]e^{-1}(0.1)^{4k-2}w}{1+[0.3](0.1)^{4k-2}w}\right)$

See Figure for an approximation of the frozen boundary with $K=10$; See Figure
for a simulation of the configurations with $K=3$, $N=70$ using the algorithm
in Section 8.

Figure 8.1. Approximation of the frozen boundary in Example 8.1 with $K=10$.
Figure 8.2. Approximation of the frozen boundary in Example 8.1 with $K=3$,
$N=70$

##  Appendix A Technical Results

In Appendix A we review facts about Macdonald polynomials and include some
known technical results used in this paper.

Let $\mathcal{A}$ be a ${\mathbb{Z}}_{\geq 0}$ graded algebra. The topological
completion $\overline{\mathcal{A}}$ of $\mathcal{A}$ is defined to be the
algebra of all formal series

$\displaystyle a=\sum_{k=0}^{\infty}a_{k},\
a_{k}\in\mathcal{A},\mathrm{deg}(a_{k})=k.$

For $a\in\overline{\mathcal{A}}$, the lower degree $\mathrm{ldeg}(a)$ of $a$
is defined as follows

(A.1) $\displaystyle\mathrm{ldeg}(a)=\max\\{K:a_{k}=0,\ \forall k<K\\}$

Let $\Lambda_{X}$ be the algebra of symmetric functions on a countable set of
variables $X=(x_{1},\ldots,x_{n},\ldots)$. A specialization

$\displaystyle\rho:\Lambda_{X}\rightarrow\mathbb{C}$

is an algebra homomorphism. Define the power sum symmetric function on
countable many variables by

(A.2) $\displaystyle p_{n}(X):=\begin{cases}1&\mathrm{if}\ n=0\\\
\sum_{x_{i}\in X}x_{i}^{n}&\mathrm{if}\ n\geq 1.\end{cases}$

Then $\Lambda_{X}$ is generated by these power sum symmetric functions. Hence
a specialization $\rho$ is uniquely determined by the values
$\\{\rho(p_{n}(X))\\}_{n=0}^{\infty}$; see [5].

For symmetric functions $f\in\Lambda_{X}$, we write $f(\rho):=\rho(f)$. Then a
specialization $\rho$ is also uniquely determined by its generating function
$H(\rho;t)$ defined as follows:

$\displaystyle H(\rho;t):=\exp\left(\sum_{n\geq
1}\frac{p_{n}(\rho)t^{n}}{n}\right)$

Let $\rho_{1},\rho_{2}$ be two specializations and $c\in\mathbb{C}$. Define
$\rho_{1}\cup\rho_{2}$, $c\rho$, $c*\rho$ be new specializations with
generating functions given by

$\displaystyle H(\rho_{1}\cup\rho_{2};t):=H(\rho_{1},t)H(\rho_{2},t);\qquad
H(c*\rho;t):=[H(\rho,t)]^{c};\qquad$ $\displaystyle H(c\rho;t):=H(\rho,ct).$

respectively. Equivalently for all $n\geq 1$,

$\displaystyle
p_{n}(\rho_{1}\cup\rho_{2})=p_{n}(\rho_{1})+p_{n}(\rho_{2});\qquad$
$\displaystyle p_{n}(c*\rho)=cp_{n}(\rho);\qquad$ $\displaystyle
p_{n}(c\rho)=c^{n}p_{n}(\rho).$

Define

(A.3) $\displaystyle H(\rho_{1};\rho_{2})$ $\displaystyle=$
$\displaystyle\sum_{\lambda\in{\mathbb{Y}}}s_{\lambda}(\rho_{1})s_{\lambda}(\rho_{2})=\exp\left(\sum_{n\geq
1}\frac{p_{n}(\rho_{1})p_{n}(\rho_{2})}{n}\right)$ (A.4)
$\displaystyle\tilde{H}(\rho):$ $\displaystyle=$
$\displaystyle\sum_{\lambda\in{\mathbb{Y}}}s_{\lambda}(\rho)=\exp\left(\sum_{n\geq
1}\frac{p_{2n-1}(\rho)}{2n-1}+\frac{p_{n}(\rho)^{2}}{2n}\right)$

In particular, when $\rho_{2}$ is the specialization of a single variable $t$,
$H(\rho_{1};\rho_{2})=H(\rho_{1},t)$. Moreover,

$\displaystyle
H(\rho;\rho_{1}\cup\rho_{2})=H(\rho;\rho_{1})H(\rho;\rho_{2});\qquad\widetilde{H}(\rho_{1}\cup\rho_{2})=\widetilde{H}(\rho_{1})\widetilde{H}(\rho_{2})H(\rho_{1};\rho_{2}).$

For each pair of parameters $q,t\in(0,1)$ and each partition
$\lambda\in{\mathbb{Y}}$, define the normalized Macdonald polynomial to be a
symmetric polynomial as follows:

$\displaystyle
P_{\lambda}(\cdot;q,t)=m_{\lambda}+\sum_{\mu<\lambda}u_{\lambda,\mu}(q,t)m_{\mu}$

where $u_{\lambda,\mu}(q,t)\in{\mathbb{R}}$ and $m_{\mu}$ are the monomial
symmetric polynomials; see [33]. It is known that
$\\{P_{\lambda}(X,q,t)\\}_{\lambda\in{\mathbb{Y}}}$ forms a linear basis for
$\Lambda_{X}$. More generally, for each specialization $\rho$,
$\\{P_{\lambda}(\rho,q,t)\\}_{\lambda\in{\mathbb{Y}}}$ can also be obtained
from $p_{n}(\rho)$, as explained above.

Let $\lambda\in{\mathbb{Y}}$, define

$\displaystyle
p_{\lambda}(\rho)=\prod_{i\in{\mathbb{N}}}p_{\lambda_{i}}(\rho).$

Then $\\{p_{\lambda}(X)\\}_{\lambda\in{\mathbb{Y}}}$ is a linear basis for
$\Lambda_{X}$.

For each fixed pair of parameters $q,t\in(0,1)$ and
$\lambda,\mu\in{\mathbb{Y}}$ define the scalar product
$\langle\cdot,\cdot\rangle:\Lambda_{X}\times\Lambda_{X}\rightarrow{\mathbb{R}}$
as a bilinear map such that:

(A.5) $\displaystyle\langle
p_{\lambda},p_{\mu}\rangle=\delta_{\lambda\mu}\left[\prod_{i=1}^{l(\lambda)}\frac{1-q^{\lambda_{i}}}{1-t^{\lambda_{i}}}\right]\left[\prod_{j=1}^{\infty}j^{m_{j}(\lambda)}(m_{j}(\lambda))!\right]$

where $\delta_{\lambda\mu}=1$ if and only if $\lambda=\mu$, and
$m_{j}(\lambda)$ is the number of parts in $\lambda$ equal to $j$. Note that
the scalar product defined above depends on parameters $(q,t)$. However, when
$q=t$ (A.5) does not depend on $q$ or $t$.

Define $Q_{\lambda}(X;q,t)$ to be a multiple of $P_{\lambda}(X,q,t)$ such that

$\displaystyle\langle P_{\lambda}(X,q,t),Q_{\lambda}(X,q,t)\rangle=1$

For $\lambda,\mu\in{\mathbb{Y}}$, define the skew MacDonald symmetric
functions by

$\displaystyle P_{\lambda}(X,Y;q,t)$ $\displaystyle=$
$\displaystyle\sum_{\mu\in{\mathbb{Y}}}P_{\lambda/\mu}(X;q,t)P_{\mu}(Y;q,t);$
$\displaystyle Q_{\lambda}(X,Y;q,t)$ $\displaystyle=$
$\displaystyle\sum_{\mu\in{\mathbb{Y}}}Q_{\lambda/\mu}(X;q,t)Q_{\mu}(Y;q,t);$

It is known that (see Remarks 1. on Page 346 of [33]) for a single variable
$x$

$\displaystyle
P_{\lambda/\mu}(x)=\delta_{\lambda\prec\mu}\psi_{\lambda/\mu}(q,t)x^{|\lambda|-|\mu|};\qquad
Q_{\lambda/\mu}(x)=\delta_{\mu\prec\lambda}\phi_{\lambda/\mu}(q,t)x^{|\lambda|-|\mu|}$

where $\psi_{\lambda/\mu}(q,t)$ and $\phi_{\lambda/\mu}(q,t)$ are independent
of $x$ and furthermore

$\displaystyle\left.\psi_{\lambda/\mu}(q,t)\right|_{q=t}=\left.\phi_{\lambda/\mu}(q,t)\right|_{q=t}=1$

Let $\lambda,\mu\in{\mathbb{Y}}$, we write $\lambda\subseteq\mu$ if

  * •

$\lambda_{i}\leq\mu_{i}$, for all $i\in{\mathbb{N}}$.

######  Lemma A.1.

Let $X$ be a countable set of variables, then

$\displaystyle P_{\lambda/\mu}(X;q,t)=Q_{\lambda/\mu}(X;q.t)=0$

unless $\mu\subseteq\lambda$. When $\mu\subseteq\lambda$,
$Q_{\lambda/\mu}(X;q.t)$ and $P_{\lambda/\mu}(X;q,t)$ are homogeneous in $X$
of degree $|\lambda|-|\mu|$.

###### Proof.

See (7.7) on Page 344 of [33]. ∎

In particular, when $q=t$,

$\displaystyle P_{\lambda}(X;t,t)=Q_{\lambda}(X;t,t)=s_{\lambda}(X);$ (A.6)
$\displaystyle
P_{\lambda/\mu}(X;t,t)=Q_{\lambda/\mu}(X;t,t)=s_{\lambda/\mu}(X).$

See (4.14) on Page 324 of [33].

######  Definition A.2.

Let $k\in\mathbb{Z}^{+}$ and $q,t\in\mathbb{R}^{+}$ be parameters. Let
$D_{-k,X}$ be an operator acting on symmetric functions $\Lambda_{X}$. For any
analytic symmetric function $F(X)$ satisfying

$\displaystyle
F(X)=\sum_{\lambda\in{\mathbb{Y}}}c_{\lambda}P_{\lambda}(X;q,t),$

where $c_{\lambda}$’s are complex coefficients, define
$D_{-k,X}F\in\Lambda_{X}$ to be

(A.7) $\displaystyle
D_{-k,X;q,t}F(X)=\sum_{\lambda\in{\mathbb{Y}}}c_{\lambda}\left\\{(1-t^{-k})\left[\sum_{i=1}^{l(\lambda)}(q^{\lambda_{i}}t^{-i+1})^{k}\right]+t^{-kl(\lambda)}\right\\}P_{\lambda}(X;q,t).$

Let $W=(w_{1},\ldots,w_{k})$ be an ordered set of variables. Define

(A.8) $\displaystyle
D(W;q,t)=\frac{(-1)^{k-1}}{(2\pi\mathbf{i})^{k}}\frac{\sum_{i=1}^{k}\frac{w_{k}t^{k-i}}{w_{i}q^{k-i}}}{\left(1-\frac{tw_{2}}{qw_{1}}\right)\ldots\left(1-\frac{tw_{k}}{qw_{k-1}}\right)}\prod_{i<j}\frac{(1-\frac{w_{i}}{w_{j}})(1-\frac{qw_{i}}{tw_{j}})}{\left(1-\frac{w_{i}}{tw_{j}}\right)\left(1-\frac{qw_{i}}{w_{j}}\right)}\prod_{i=1}^{k}\frac{dw_{i}}{w_{i}}$

Define

(A.9) $\displaystyle L(W,X;q,t)=\prod_{i=1}^{k}\prod_{x_{j}\in
X}\frac{w_{i}-\frac{qx_{j}}{t}}{w_{i}-qx_{j}}$

The following proposition is a slightly more general form of Proposition 4.10
of [17].

######  Proposition A.3.

Assume one of the following two conditions holds

  1. (1)

$q\in(0,1)$ and $t\in(0,1)$; or

  2. (2)

$q\in(1,\infty)$ and $t\in(1,\infty)$.

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a function analytic in a
neighborhood of 0, and $f(0)\neq 0$. Let $g:\mathbb{C}\rightarrow\mathbb{C}$
be a function analytic in a neighborhood of 0, and

$\displaystyle g(z)=\frac{f(z)}{f(q^{-1}z)};$

for $z$ in a small neighborhood of $0$. Then

(A.10) $\displaystyle D_{-k,X;q,t}\left(\prod_{x_{i}\in X}f(x_{i})\right)$
$\displaystyle=$ $\displaystyle\left(\prod_{x_{i}\in
X}f(x_{i})\right)\oint\cdots\oint
D(W;q,t)L(W,X;q,t)\left(\prod_{i=1}^{k}g(w_{i})\right)$

where the contours of the integral satisfy the following conditions

  * •

all the contours are in the neighborhood of $0$ such that both $f$ and $g$ are
analytic;

  * •

each contour enclose $0$ and $\\{qx_{i}\\}_{x_{i}\in X}$;

  * •

If case (1) holds, $|w_{i}|\leq|tw_{i+1}|$ for all $i\in[k-1]$;

  * •

If case (2) holds, $|w_{i}|\leq\left|\frac{1}{q}w_{i+1}\right|$ for all
$i\in[k-1]$;

$L(W,X;q,t)$ is given by (A.9), and $D(W;q,t)$ is given by (A.8).

###### Proof.

See Proposition A.2 of [32]. ∎

######  Lemma A.4.

Let $(a,q)_{\infty}=\prod_{r=0}^{\infty}(1-aq^{r})$ and

(A.11)
$\displaystyle\Pi(X,Y;q,t):=\frac{(tx_{i}y_{j};q)_{\infty}}{(x_{i}y_{j};q)_{\infty}},\quad\Pi^{\prime}(X,Y):=\prod_{i,j}(1+x_{i}y_{j}).$

Then

$\displaystyle\sum_{\lambda\in{\mathbb{Y}}}P_{\lambda}(X;q,t)Q_{\lambda}(Y;q,t)$
$\displaystyle=\sum_{\lambda\in{\mathbb{Y}}}P_{\lambda^{\prime}}(X;q,t)Q_{\lambda^{\prime}}(Y;q,t)=\Pi(X,Y;q,t)$
$\displaystyle\sum_{\lambda\in{\mathbb{Y}}}P_{\lambda}(X;q,t)P_{\lambda^{\prime}}(Y;t,q)$
$\displaystyle=\sum_{\lambda\in{\mathbb{Y}}}Q_{\lambda}(X;q,t)Q_{\lambda^{\prime}}(Y;t,q)=\Pi^{\prime}(X,Y)$

In particular, when $q=t$ we obtain the Cauchy identities for Schur
polynomials.

$\displaystyle\sum_{\lambda\in{\mathbb{Y}}}s_{\lambda}(X)s_{\lambda}(Y)=\prod_{i,j}\frac{1}{1-x_{i}y_{j}};$
$\displaystyle\sum_{\lambda\in{\mathbb{Y}}}s_{\lambda}(X)s_{\lambda^{\prime}}(Y)=\prod_{i,j}(1+x_{i}y_{j}).$

###### Proof.

See (2.5), (4.13), and (5.4) in Section VI of [33]. ∎

######  Lemma A.5.

Let $\Pi$, $\Pi^{\prime}$, and $L$ be as in (A.11) and (A.9), then

$\displaystyle\Pi(X,Y;q,t)=\exp\left(\sum_{n=1}^{\infty}\frac{1-t^{n}}{1-q^{n}}\frac{1}{n}p_{n}(X)p_{n}(Y)\right),$
$\displaystyle\Pi^{\prime}(X,Y)=\exp\left(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}p_{n}(X)p_{n}(Y)\right),$
$\displaystyle
L(X,Y;q,t)=\exp\left(\sum_{n=1}^{\infty}\frac{1-t^{-n}}{n}p_{n}(qX^{-1})p_{n}(Y)\right).$

###### Proof.

The first identity follows from Page 310 of [33]. The other two follow from

$\displaystyle\Pi^{\prime}(X,Y)$
$\displaystyle=\left[\Pi(-X,Y;0,0)\right]^{-1},$ $\displaystyle L(X,Y;q,t)$
$\displaystyle=\Pi(qX^{-1},Y;0,t^{-1}).$

∎

######  Lemma A.6.

(Lemma 5.7 of [2]) Let $\theta\in(0,\pi)$, and $\xi>0$. Define

$\displaystyle
R_{\epsilon,\theta,\xi}:=\\{w\in\mathbb{C}:\mathrm{dist}(w,[1,\infty))\leq\xi\\}\cap\\{w\in\mathbb{C}:|\arg(w-(1-\epsilon))|\leq\theta\\}.$

Let $\alpha>0$ and suppose $N(\epsilon)\in{\mathbb{Z}}>0$ such that
$\limsup_{\epsilon\rightarrow 0}\epsilon N(\epsilon)>0$ as
$\epsilon\rightarrow 0$. Then for any fixed $\theta\in(0,\pi),\xi>0$, we have

$\displaystyle\frac{(z;e^{-\epsilon})_{N(\epsilon)}}{(e^{-\epsilon\alpha}z;e^{-\epsilon})_{N(\epsilon)}}=\left(\frac{1-z}{1-e^{-\epsilon
N(\epsilon)}z}\right)^{\alpha}\exp\left(O\left(\frac{\epsilon\min\\{|z|,|z|^{2}\\}}{|1-z|}\right)\right)$

uniformly for $z\in\mathbb{C}\setminus R_{\epsilon,\theta,\xi}$ and $\epsilon$
arbitrarily small.

######  Lemma A.7.

(Corollary A.2 in [17]) Let $d,h,k$ be positive integers. Let $f$, $g_{1}$,…,
$g_{d}$ be meromorphic functions with possible poles at ${z_{1},...,z_{h}}$.
Then for $k\geq 2$,

$\displaystyle\frac{1}{(2\pi\mathbf{i})^{k}}\oint\ldots\oint\frac{1}{(v_{2}-v_{1})\cdots\ldots\cdot(v_{k}-v_{k-1})}\prod_{j=1}^{d}\left(\sum_{i=1}^{k}g_{j}(v_{i})\right)\prod_{i=1}^{k}f(v_{i})dv_{i}$
$\displaystyle=\frac{k^{d-1}}{2\pi\mathbf{i}}\oint
f(v)^{k}\prod_{j=1}^{d}g_{j}(v)dv.$

where the contours contain $\\{z_{1},...,z_{h}\\}$ and on the left side we
require that the $v_{i}$-contour is contained in the $v_{j}$-contour whenever
$i<j$.

######  Lemma A.8.

(Hurwitz’s theorem)Let $\\{f_{k}\\}$ be a sequence of holomorphic functions on
a connected open set $G$ that converge uniformly on compact subsets of G to a
holomorphic function $f$ which is not constantly zero on G. If f has a zero of
order $m$ at $z_{0}$ then for every small enough $\rho>0$ and for sufficiently
large $k\in{\mathbb{N}}$ (depending on $\rho$), $f_{k}$ has precisely $m$
zeroes in the disk defined by $|z-z_{0}|<\rho$, including multiplicity.
Furthermore, these zeroes converge to $z_{0}$ as $k\rightarrow\infty$.

###### Proof.

See Page 178 of [1], proof of Theorem 2. ∎

The second equation in (2.13) gives

(A.12) $\displaystyle 0$
$\displaystyle=\sum_{\begin{subarray}{c}(p\in[m],V_{p}>\chi)\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,>,1}}=1\end{subarray}}\frac{1}{w-e^{V_{p}}\tau_{j}^{-1}}-\frac{1}{w-e^{\max\\{V_{p-1},\chi\\}}\tau_{j}^{-1}}+\sum_{\begin{subarray}{c}(p\in[m],V_{p-1}<\chi)\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,<,1}}=1\end{subarray}}\frac{1}{w-e^{V_{p-1}}\tau_{j}^{-1}}-\frac{1}{w-e^{\min\\{V_{p},\chi\\}}\tau_{j}^{-1}}$
$\displaystyle+\sum_{\begin{subarray}{c}(p\in[m],V_{p}>\chi)\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,>,0}}=1\end{subarray}}\frac{1}{w+e^{\max\\{V_{p-1},\chi\\}}\tau_{j}^{-1}}-\frac{1}{w+e^{V_{p}}\tau_{j}^{-1}}+\sum_{\begin{subarray}{c}(p\in[m],V_{p-1}<\chi)\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,<,0}}=1\end{subarray}}\frac{1}{w+e^{\min\\{V_{p},\chi\\}}\tau_{j}^{-1}}-\frac{1}{w+e^{V_{p-1}}\tau_{j}^{-1}}$
$\displaystyle+\sum_{k=1}^{\infty}g_{k}(w)$

where

(A.13) $\displaystyle g_{k}(w):$
$\displaystyle=\sum_{\begin{subarray}{c}(p\in[m])\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,>,1}}=1\end{subarray}}\frac{1}{w-u^{2k}v^{2k}e^{V_{p}}\tau_{j}^{-1}}-\frac{1}{w-u^{2k}v^{2k}e^{V_{p-1}}\tau_{j}^{-1}}$
$\displaystyle+\sum_{\begin{subarray}{c}(p\in[m])\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,>,1}}=1\end{subarray}}\frac{1}{w-u^{2(k-1)}v^{2k}e^{V_{p}}\tau_{j}^{-1}}-\frac{1}{w-u^{2(k-1)}v^{2k}e^{V_{p-1}}\tau_{j}^{-1}}$
$\displaystyle+\sum_{\begin{subarray}{c}(p\in[m])\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,<,1}}=1\end{subarray}}\frac{1}{w-u^{-2k}v^{-2k}e^{V_{p-1}}\tau_{j}^{-1}}-\frac{1}{w-u^{-2k}v^{-2k}e^{V_{p}}\tau_{j}^{-1}}$
$\displaystyle+\sum_{\begin{subarray}{c}(p\in[m])\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,<,1}}=1\end{subarray}}\frac{1}{w-u^{-2k}v^{-2(k-1)}e^{V_{p-1}}\tau_{j}^{-1}}-\frac{1}{w-u^{-2k}v^{-2(k-1)}e^{V_{p}}\tau_{j}^{-1}}$
$\displaystyle+\sum_{\begin{subarray}{c}(p\in[m])\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,>,0}}=1\end{subarray}}\frac{1}{w+u^{2k}v^{2k}e^{V_{p-1}}\tau_{j}^{-1}}-\frac{1}{w+u^{2k}v^{2k}e^{V_{p}}\tau_{j}^{-1}}$
$\displaystyle+\sum_{\begin{subarray}{c}(p\in[m])\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,>,0}}=1\end{subarray}}\frac{1}{w+u^{2(k-1)}v^{2k}e^{V_{p-1}}\tau_{j}^{-1}}-\frac{1}{w+u^{2(k-1)}v^{2k}e^{V_{p}}\tau_{j}^{-1}}$
$\displaystyle+\sum_{\begin{subarray}{c}(p\in[m])\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,<,0}}=1\end{subarray}}\frac{1}{w+u^{-2k}v^{-2k}e^{V_{p}}\tau_{j}^{-1}}-\frac{1}{w+u^{-2k}v^{-2k}e^{V_{p-1}}\tau_{j}^{-1}}$
$\displaystyle+\sum_{\begin{subarray}{c}(p\in[m])\\\
j\in[n]:\mathbf{1}_{\mathcal{E}_{j,p,<,0}}=1\end{subarray}}\frac{1}{w+u^{-2k}v^{-2(k-1)}e^{V_{p}}\tau_{j}^{-1}}-\frac{1}{w+u^{-2k}v^{-2(k-1)}e^{V_{p-1}}\tau_{j}^{-1}}$

Acknowledgements. ZL acknowledges support from National Science Foundation
grant DMS1608896 and Simons Foundation grant 638143. ZL thanks Mirjana Vuletić
for helpful discussions.

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