# An Explicit Algorithm of Rigged Configuration Bijection for the Adjoint
Crystal of Type $G_{2}^{(1)}$

Toya Hiroshima  OCAMI, Osaka City University, 3-3-138 Sugimoto, Sumiyoshi-ku,
Osaka 558-8585, Japan<EMAIL_ADDRESS>
###### Abstract.

We construct an explicit algorithm of the static-preserving bijection between
the rigged configurations and the highest weight paths of the form
$(B^{2,1})^{\otimes L}$ in the $G_{2}^{(1)}$ adjoint crystals.

###### Key words and phrases:

crystal, rigged configuration, quantum group

###### 2010 Mathematics Subject Classification:

Primary 05A19, 17B37; Secondary 05E10, 17B25, 81R10

###### Contents

  1. 1 Introduction
  2. 2 Affine Algebra $G_{2}^{(1)}$ and the KR Crystal
    1. 2.1 Affine algebra $G_{2}^{(1)}$
    2. 2.2 KR crystal
  3. 3 Rigged Configuration and the Bijection
    1. 3.1 Rigged configuration
    2. 3.2 The bijection from RCs to paths.
  4. 4 The Bijection
    1. 4.1 Algorithm $\delta_{\theta}$
    2. 4.2 Inverse algorithm $\tilde{\delta}_{\theta}$
  5. 5 Proof of Theorem 3.1
  6. 6 Verifications of the Rules of Forbidden Box Marking

##  1\. Introduction

Kerov, Kirillov, and Reshetikhin introduced a combinatorial object, called a
rigged configuration, through Bethe Ansatz analysis of the isotropic
Heisenberg spin chain [12]. They constructed a bijection between rigged
configurations and the semistandard Young tableaux [12, 13], through which a
bijection between rigged configurations and highest weight element of a tensor
product of Killirov-Reshetikhin (KR) crystals or highest weight paths was
formulated. The rigged configuration possess a natural statistic and it
coincides with the charge introduced by Lascoux and
Sch$\ddot{\mathrm{u}}$tzenberger [16]. On the other hand, the highest weight
path carries the statistic called energy by Nakayashiki and Yamada [22]. The
bijection, which is called the rigged configuration bijection, is a bijection
such that the charge of a rigged configuration is sent to the energy of the
corresponding path.

The bijection of Kerov, Kirillov, and Reshetikhin is a bijection from rigged
configurations to the paths of the form $(B^{1,1})^{\otimes L}$ in type
$A_{n}^{(1)}$ where $B^{r,s}$ denotes the KR crystal. Until now, this has been
extended in more general setting and in various types [3, 14, 25, 26, 27, 28,
23, 24, 30, 31, 34, 35, 32]. For nonexceptional types, the generalization of
the rigged configuration bijection has been established [29]. On the other
hand, there remain unsolved problems in exception types. This paper concerns
one of these problems. For the $G_{2}^{(1)}$ adjoint crystals, the explicit
algorithm of the rigged configuration bijection $\Phi$ is not known as pointed
out in [35] though the crystal structure is very simple (see Fig. 2.2). In
this paper, we construct a map $\Phi$ from rigged configurations to highest
weight elements of $(B^{2,1})^{\otimes L}$ by executing a fundamental
procedure $\delta_{\theta}$ repeatedly. Our result provides an alternate but
direct proof of $X=M$ conjecture of [4, 5] in our setting, which has been
proved previously by Naoi [20].

This paper is organized as follows. In Section 2, we give the necessary
background on KR crystals and paths for $G_{2}^{(1)}$. Rigged configurations
and the bijection $\Phi$ are described in Section 3, where we state our main
theorem (Theorem 3.1). In Section 4, we provide an explicit description of the
bijection $\delta_{\theta}$ as well as its inverse $\tilde{\delta}_{\theta}$
for $G_{2}^{(1)}$ adjoint crystals. Section 5 is devoted to the proof of
Theorem 3.1. In Section 6, we explain some forbidden rules in the algorithm.

##  2\. Affine Algebra $G_{2}^{(1)}$ and the KR Crystal

###  2.1. Affine algebra $G_{2}^{(1)}$

We consider in this paper the exceptional untwisted affine algebra
$G_{2}^{(1)}$. The Dynkin diagram is depicted in Figure 2.1. We follow [7] for
the labeling of the Dynkin nodes. Let $I$ be the index set of the Dynkin nodes
and let $\alpha_{i}$, $\alpha_{i}^{\vee}$, $\Lambda_{i}$ $(i\in I)$ be simple
roots, simple coroots, fundamental weights, respectively. Following the
notation in [7] we denote the projection of $\Lambda_{i}$ onto the weight
space of $G_{2}$ by $\bar{\Lambda}_{i}$ ($i\in I_{0}:=I\backslash{0}$) and set
$\bar{P}=\bigoplus_{i\in I_{0}}\mathbb{Z}\bar{\Lambda}_{i}$,
$\bar{P}^{+}=\bigoplus_{i\in I_{0}}\mathbb{Z}_{\geq 0}\bar{\Lambda}_{i}$. Let
$(A_{i,j})_{i,j\in I}$ stand for the Cartan matrix for $G_{2}^{(1)}$. The
canonical pairing $\left<\;,\;\right>:P^{\vee}\times P\rightarrow\mathbb{Z}$
is given by $\left<\alpha_{i}^{\vee},\alpha_{j}\right>=A_{i,j}$.

$<$$1$$2$$0$ Figure 2.1. The Dynkin diagram for $G_{2}^{(1)}$ in the Bourbaki
labeling.

###  2.2. KR crystal

Let $\mathfrak{g}$ be any affine algebra and $U_{q}^{\prime}(\mathfrak{g})$
the corresponding quantized enveloping algebra without the degree operator.
Among finite-dimensional $U_{q}^{\prime}(\mathfrak{g})$-modules there is a
distinguished family called Kirillov-Reshetikhin (KR) modules [6, 15, 21]. One
of the remarkable properties of KR modules is that they admit crystal bases
introduced by Kashiwara [8, 9]. Such a crystal base is called a KR crystal.
The KR crystal is indexed by
$(a,i)\in\mathcal{H}_{0}:=I_{0}\times\mathbb{Z}_{>0}$ and denoted by
$B^{a,i}$. For exceptional types the KR crystal is known to exist when the KR
module is irreducible or the index $a$ is adjacent to $0$ [11].

The KR crystal we are interested in in this paper is $G_{2}^{(1)}$ adjoint
crystal $B^{2,1}$, which is a level 1 perfect crystal and is constructed in
[1]. The crystal structure is depicted in Figure 2.2. See also [17, 18] (note
the difference indexing of the Dynkin diagram). Here vertices in the graph
signify elements of $B^{2,1}$ and $b\stackrel{{\scriptstyle
i}}{{\longrightarrow}}b^{\prime}$ stands for $f_{i}b=b^{\prime}$ or
equivalently $b=e_{i}b^{\prime}$. we adopt the anti-Kashiwara convention for
the tensor products of crystals as in [2]. Namely, if $B_{1}$ and $B_{2}$ are
crystals, then for $b_{1}\otimes b_{2}\in B_{1}\otimes B_{2}$ the action of
$e_{i}$ is defined as

$e_{i}(b_{1}\otimes b_{2})=\begin{cases}e_{i}b_{1}\otimes
b_{2}&if\;\varphi_{i}(b_{2})<\varepsilon_{i}(b_{1}),\\\ b_{1}\otimes
e_{i}b_{2}&if\;\varphi_{i}(b_{2})\geq\varepsilon_{i}(b_{1}),\end{cases}$

where $\varepsilon_{i}(b)=\max\left\\{k\geq
0\mathrel{}\middle|\mathrel{}e_{i}^{k}b\in B\right\\}$ and
$\varphi_{i}(b)=\max\left\\{k\geq 0\mathrel{}\middle|\mathrel{}f_{i}^{k}b\in
B\right\\}$.

In what follow in this paper $B=B^{2,1}$. The set of _classically restricted
paths_ in $B^{\otimes L}$ of weight $\lambda\in\bar{P}^{+}$ is by definition

$\mathcal{P}(\lambda,L)=\left\\{b\in B^{\otimes
L}\mathrel{}\middle|\mathrel{}\mathrm{wt}(b)=\lambda\text{ and }e_{i}b=0\text{
for all }i\in I_{0}\right\\}.$

One may check that the following are equivalent for $b=b_{1}\otimes
b_{2}\otimes\cdots\otimes b_{L}\in B^{\otimes L}$ and $\lambda\in\bar{P}^{+}$.

  * (1)

$b$ is a classically restricted path of weight $\lambda\in\bar{P}^{+}$.

  * (2)

$b_{2}\otimes\cdots\otimes b_{L}$ is a classically restricted path of weight
$\lambda-\mathrm{wt}(b_{1})$ and
$\varepsilon_{i}\leq\left<\alpha_{i}^{\vee},\lambda-\mathrm{wt}(b_{1})\right>$
for all $i\in I_{0}$.

The weight function $\mathrm{wt}:B\rightarrow\bar{P}$ is given by
$\mathrm{wt}(b)=\sum_{i\in
I_{0}}(\varphi_{i}(b)-\varepsilon_{i}(b))\bar{\Lambda}_{i}$. The weight
function $\mathrm{wt}:B^{\otimes L}\rightarrow\bar{P}$ is defined by
$\mathrm{wt}(b_{1}\otimes\cdots\otimes
b_{L})=\sum_{j=1}^{L}\mathrm{wt}(b_{j})$.

In Figure 2.2, we show the crystal subgraph for $B_{0}$, which is obtained by
ignoring the $0$-arrow from the crystal graph for $B^{2,1}$. All the
$0$-arrows are listed below.

$10$$2$$11$$3$$12$$4$$13$$6$$0$$0$$0$$0$

and

$14$$\emptyset$$1$.$0$$0$

$3$$2$$1$$4$$6$$5$$8$$7$$10$$9$$11$$14$$13$$12$$1$$2$$1$$1$$2$$2$$1$$2$$1$$1$$2$$1$$2$$1$
Figure 2.2. Crystal graph of $B_{0}$ for $B^{2,1}$.

The _energy function_ $D:B^{\otimes L}\rightarrow\mathbb{Z}$ gives the grading
on $B^{\otimes L}$. In our case where a path is an element of the tensor
product of a single KR crystal it takes a simple form. Due to the existence of
the universal $R$-matrix and the fact that $B\otimes B$ is connected, by [10]
there is a unique (up to global additive constant) function $H:B\otimes
B\rightarrow\mathbb{Z}$ called the _local energy function_ , such that

$H(e_{i}(b\otimes b^{\prime}))=\begin{cases}H(b\otimes b^{\prime})-1&\text{ if
}i=0\text{ and }e_{0}(b\otimes b^{\prime})=e_{0}b\otimes b^{\prime},\\\
H(b\otimes b^{\prime})+1&\text{ if }i=0\text{ and }e_{0}(b\otimes
b^{\prime})=b\otimes e_{0}b^{\prime},\\\ H(b\otimes
b^{\prime})&\text{otherwise}.\end{cases}$

We normalize $H$ by the condition $H(\framebox{$1$}\otimes\framebox{$1$})=0$.

The crystal graph of $B_{0}\otimes B_{0}$ decomposes into five connected
components as

(2.1) $B_{0}\otimes B_{0}=B(2\bar{\Lambda}_{2})\oplus
B(3\bar{\Lambda}_{1})\oplus B(2\bar{\Lambda}_{1})\oplus
B(\bar{\Lambda}_{2})\oplus B(0),$

where $B(\lambda)$ stands for the highest weight $G_{2}$ crystal of weight
$\lambda$ and the highest vector are given by
$\framebox{$1$}\otimes\framebox{$1$}$ ($\lambda=2\bar{\Lambda}_{2}$),
$\framebox{$2$}\otimes\framebox{$1$}$ ($\lambda=3\bar{\Lambda}_{1}$),
$\framebox{$5$}\otimes\framebox{$1$}$ ($\lambda=2\bar{\Lambda}_{1}$),
$\framebox{$8$}\otimes\framebox{$1$}$ ($\lambda=\bar{\Lambda}_{2}$), and
$\framebox{$14$}\otimes\framebox{$1$}$ ($\lambda=0$). $H$ is constant on each
component, and takes the value $H(\framebox{$1$}\otimes\framebox{$1$})=0$,
$H(\framebox{$2$}\otimes\framebox{$1$})=-1$, and $H=-2$ for the rest. It is
easily verified that
$H(\framebox{$\emptyset$}\otimes\framebox{$\emptyset$})=-2$ and
$H(\framebox{$\emptyset$}\otimes\framebox{$i$})=H(\framebox{$i$}\otimes\framebox{$\emptyset$})=-1$
$(1\leq i\leq 14)$. With this $H$, the energy function $D$ is defined by

(2.2) $D(b_{1}\otimes\cdots\otimes b_{L})=\sum_{j=1}^{L}jH(b_{j}\otimes
b_{j+1}),$

where $b_{L+1}=\framebox{$1$}$.

##  3\. Rigged Configuration and the Bijection

###  3.1. Rigged configuration

Fix $\lambda\in\bar{P}^{+}$ and a matrix
$\boldsymbol{L}=(L_{i}^{(a)})_{(a,i)\in\mathcal{H}_{0}}$ of nonnegative
integers. Let $\nu=(m_{i}^{(a)})_{(a,i)\in\mathcal{H}_{0}}$. Say that $\nu$ is
a $\lambda$-configuration if it satisfies

(3.1)
$\sum_{(a,i)\in\mathcal{H}_{0}}im_{i}^{(a)}\alpha_{a}=\sum_{(a,i)\in\mathcal{H}_{0}}iL_{i}^{(a)}\bar{\Lambda}_{a}-\lambda$

and $p_{i}^{(a)}\geq 0$ for all $(a,i)\in\mathcal{H}_{0}$, where

(3.2) $p_{i}^{(a)}=\sum_{j\in\mathbb{Z}_{>0}}L_{j}^{(a)}\min(i,j)-\sum_{b\in
I_{0}}\frac{A_{ab}}{\gamma_{b}}\sum_{j\in\mathbb{Z}_{>0}}\min(\gamma_{a}i,\gamma_{b}j)m_{j}^{(b)}$

is the _vacancy numbers_ and $\gamma_{a}$ ($a\in I_{0}$) are scaling factors
which are given by $\gamma_{1}=1$ and $\gamma_{2}=3$ for $G_{2}$.

Let $\nu$ be an admissible configuration. We identify $\nu$ with a sequence of
partitions $\\{\nu^{(a)}\\}_{a\in I_{0}}$ such that
$\nu^{(a)}=(1^{m_{1}^{(a)}}2^{m_{2}^{(a)}}\cdots)$. Let
$J=\\{J^{(a,i)}\\}_{(a,i)\in\mathcal{H}_{0}}$ be a double sequence of
partitions. Then a rigged configuration (RC) is a pair $(\nu,J)$ subject to
the restriction Eq. (3.1) and the requirement that $J^{(a,i)}$ be a partition
contained in a $m_{i}^{(a)}\times p_{i}^{(a)}$ rectangle. The set of rigged
configurations for fixed $\lambda$ and $\boldsymbol{L}$ is denoted by
$\mathrm{RC}(\lambda,\boldsymbol{L})$.

Define the _charge_ of a configuration $\nu$ by

$\displaystyle c(\nu)=$ $\displaystyle\frac{1}{2}\sum_{a,b\in
I_{0}}\frac{A_{ab}}{\gamma_{b}}\sum_{i,j\in\mathbb{Z}_{>0}}\min(\gamma_{a}i,\gamma_{b}j)m_{i}^{(a)}m_{j}^{(b)}$
$\displaystyle-\sum_{a\in
I_{0},j,k\in\mathbb{Z}_{>0}}L_{j}^{(a)}\min(j,k)m_{k}^{(a)}.$

To obtain the charge of a rigged configuration $(\nu,J)$, we add all of the
riggings to $c(\nu)$;

$c(\nu,J)=c(\nu)+|J|,$

where $|J|=\sum_{(a,i)\in\mathcal{H}_{0}}|J^{(a,i)}|$.

We now set

(3.3) $L_{i}^{(a)}=L\delta_{a,2}\delta_{i,1}\quad((a,i)\in\mathcal{H}_{0}),$

which corresponds to considering paths in $(B^{2,1})^{\otimes L}$. The set
$\mathrm{RC}(\lambda,\boldsymbol{L})$ with the restriction Eq. (3.3) is
denoted by $\mathrm{RC}(\lambda,L)$. The vacancy numbers $p_{i}^{(a)}$ and the
charge $c(\nu)$ take the following forms.

(3.4) $\displaystyle p_{i}^{(1)}$
$\displaystyle=-2\sum_{j\in\mathbb{Z}_{>0}}\min(i,j)m_{j}^{(1)}+\sum_{j\in\mathbb{Z}_{>0}}\min(i,3j)m_{j}^{(2)},$
(3.5) $\displaystyle p_{i}^{(2)}$
$\displaystyle=L+\sum_{j\in\mathbb{Z}_{>0}}\min(3i,j)m_{j}^{(1)}-2\sum_{j\in\mathbb{Z}_{>0}}\min(i,j)m_{j}^{(2)},$

and

(3.6) $\displaystyle c(\nu)=$
$\displaystyle\sum_{i,j\in\mathbb{Z}_{>0}}\min(i,j)m_{i}^{(1)}m_{j}^{(1)}-\sum_{i,j\in\mathbb{Z}_{>0}}\min(i,3j)m_{i}^{(1)}m_{j}^{(2)}$
$\displaystyle+\sum_{i,j\in\mathbb{Z}_{>0}}\min(i,j)m_{i}^{(2)}m_{j}^{(2)}-L\sum_{j\in\mathbb{Z}_{>0}}m_{j}^{(2)}.$

###  3.2. The bijection from RCs to paths.

We now describe the bijection
$\Phi:\mathrm{RC}(\lambda,L)\rightarrow\mathcal{P}(\lambda,L)$. Let
$(\nu,J)\in\mathrm{RC}(\lambda,L)$. We shall define a map
$\gamma:\mathrm{RC}(\lambda,L)\rightarrow B$ which associates to $(\nu,J)$ an
element of $B$. Define by $\mathrm{RC}_{b}(\lambda,L)$ the elements of
$\mathrm{RC}(\lambda,L)$ such that $\gamma(\nu,L)=b$. We shall define a
bijection
$\delta_{\theta}:\mathrm{RC}_{b}(\lambda,L)\rightarrow\mathrm{RC}(\lambda-\mathrm{wt}(b),L-1)$.
The disjoint union of these bijections then defines a bijection
$\delta_{\theta}:\mathrm{RC}(\lambda,L)\rightarrow\bigsqcup_{b\in
B}\mathrm{RC}(\lambda-\mathrm{wt}(b),L-1)$.

The bijection $\Phi$ is defined recursively as follows. For $b\in B$ let
$\mathcal{P}_{b}(\lambda,L)$ be the set of paths in $B^{\otimes L}$ that have
$b$ as leftmost tensor factor. For $L=0$ the bijection $\Phi$ sends the empty
rigged configuration (the only element of the set $\mathrm{RC}(\lambda,L)$) to
the empty path (the only element of $\mathcal{P}(\lambda,L)$). Otherwise
assume that $\Phi$ has been defined for $B^{\otimes(L-1)}$ and define it for
$B^{\otimes L}$ by the commutative diagram

$\begin{CD}\mathrm{RC}_{b}(\lambda,L)@>{\Phi}>{}>\mathcal{P}_{b}(\lambda,L)\\\
@V{\delta_{\theta}}V{}V@V{}V{}V\\\
\mathrm{RC}(\lambda-\mathrm{wt}(b),L-1)@>{\Phi}>{}>\mathcal{P}(\lambda-\mathrm{wt}(b),L-1)\end{CD}$

where the right-hand side vertical map removes the leftmost tensor factor $b$.

Here follows the main theorem of our paper.

######  Theorem 3.1.

$\Phi:\mathrm{RC}(\lambda,L)\rightarrow\mathcal{P}(\lambda,L)$ is a bijection
such that

(3.7) $c(\nu,J)=D(\Phi(\nu,J))\text{ for all
}(\nu,J)\in\mathrm{RC}(\lambda,L).$

##  4\. The Bijection

In this section, for $(\nu,J)\in\mathrm{RC}(\lambda,L)$, an algorithm is given
which defines $b=\gamma(\nu,J)$, the new smaller rigged configuration
$(\tilde{\nu},\tilde{J})=\delta(\nu,J)$ such that
$(\tilde{\nu},\tilde{J})\in\mathrm{RC}(\rho,L-1)$ where
$\rho=\lambda-\mathrm{wt}(b)$, and the new vacancy numbers
$\tilde{p}_{i}^{(a)}$ in terms of the old $p_{i}^{(a)}$. Before describing the
algorithm, we give an example of $\Phi$, where we fix notation and give
several definitions.

######  Example 4.1.

The algorithm $\Phi$ for an RC of $L=4$ and $\lambda=2\bar{\Lambda}_{1}$ is
depicted at each step $\delta_{\theta}$ below, where partitions $\nu^{(1)}$
and $\nu^{(2)}$ are illustrated as the left and the right Young diagrams in
the RC. A row in a Young diagram is called a _string_.

$\downarrow$$0$$2$$0$$1$$[3]$$[2]$$[5]$$1$$0$$1$$0$$0$$0$$[1]$$[4]$$\longrightarrow$$7$$\downarrow$$1$$1$$[8]$$[7]$$[4]$$[3]$$[2]$$1$$1$$1$$0$$0$$[1]$$[6]$$[5]$

$\longrightarrow$$12$$\emptyset$$0$$0$$[1]$$\longrightarrow$$2$$\emptyset$$\emptyset$$\longrightarrow$$1$$\emptyset$$\emptyset$

so that the first RC corresponds to the path
$\framebox{$7$}\otimes\framebox{$12$}\otimes\framebox{$2$}\otimes\framebox{$1$}$.
The energy of this path is computed to be $-8$. The computation uses Eq. (2.2)
and Eq. (5.16). The charge of the first RC is computed to be $-8$ using Eq.
(3.6).

The down arrows marked in Example 4.1 are called _delimiters_ , which are
marked at each end of the consecutive sequence of three boxes of a string in
$\nu^{(1)}$. The numbers marked on the left of strings are vacancy numbers,
which are common for the strings of the same length. The numbers marked on the
right of strings are riggings. We follow the convention that riggings in
strings of the same length are sorted in an increasing order (from the top).
We call a string of length $i$ in $\nu^{(a)}$ _singular_ , _q-singular_ , and
_qq-singular_ if its rigging is equal to $p_{i}^{(a)}$, $p_{i}^{(a)}-1$, and
$p_{i}^{(a)}-2$, respectively. Here, “q-singular” is the abbreviation of
“quasi singular” firstly introduced in [19] and commonly used in the
literature [29, 35]. The “qq-singular” strings are firstly introduced in this
paper. A singular or q-singular string is written as a singular/q-singular
string. The other cases are similar. We say that the string which is not
singular is _q-singular at best_. A string of _qq-singular at best_ is defined
similarly. A string which is qq-singular at best but not qq-singular is called
_qqq-singular at best_. The boxes marked by $[n]$ are deleted at each step
$\delta_{\theta}$. The number $n$ indicates the order of marking process. We
call this process the _box marking_ and say that the box or the string is
marked by $[n]$. We also say that $[n]$ is marked in a box or a string when
the box or the string is marked by $[n]$. The string marked by $[n]$ is called
the _selected $i_{n}$-string_ or simply the _$i_{n}$ -string_. The length of
the $i_{n}$-string is also denoted by $i_{n}$. Therefore, the $i_{5}$-string
in $\nu^{(1)}$ in the first RC is also a 6-string. In the first RC in Example
4.1, the $i_{2}$-string coincide with the $i_{3}$-string. We write
$i_{2}=i_{3}$ in this case. The length of the $i_{5}$-string is larger than
that of the $i_{3}$-string. We write $i_{3}<i_{5}$ in this case. This notation
is generalized in an obvious manner. Strings in $\nu^{(1)}$ are classified by
their types.

$\downarrow$$\downarrow$$\downarrow$

Strings of the first, the second, and the third types are called _type-0_ ,
_type-I_ , and _type-II_ , respectively. A string of type-0 or type-I is
written as a string of type-0/I. The other cases are similar. The _effective
length_ of a string of length $i$ in $\nu^{(1)}$ is defined by
$i^{eff}:=\lceil i/3\rceil$. For example, $i_{3}^{eff}=1$ and $i_{5}^{eff}=2$
in the first RC in Example 4.1. The real length $i$ is recovered as
$3i^{eff}$, $3i^{eff}-1$, or $3i^{eff}-2$ if the string is type-0, type-I, or
type-II, respectively. We form a new RC by removing marked boxes, adjusting
the values of riggings of the box-deleted strings, and keeping the other the
same.

###  4.1. Algorithm $\delta_{\theta}$

The algorithm consists of box marking and adjusting the values of riggings of
the box-deleted strings. We begin with the algorithm of box marking.

(BM-1):

Marking $[1]$.

Find the singular string of minimum length in $\nu^{(2)}$. If such a string
exists, then mark the rightmost box of the string by $[1]$. If not, then
terminate the algorithm and return $\framebox{$1$}$. When $[1]$ is marked, we
prescribe that boxes of length $(\leq 3(i_{1}-1))$ in $\nu^{(1)}$ cannot be
marked.

(BM-2):

Marking $[2]$.

Find the strings of effective length $i_{1}$ in $\nu^{(1)}$. If such strings
exist, then mark the rightmost box of one of such strings by [2] according to
the following preferential rule.

| type-0 | type-I | type-II
---|---|---|---
singular | 1 | 2 | 3
q-singular | 4 | 5 |
qq-singular | 6 |  |

For example, firstly find a singular string of length $3i_{1}$. If such a
string exists, then mark the string by $[2]$. If such strings do not exist,
then find the singular string of length $3i_{1}-1$, etc. If $[2]$ cannot be
marked in a string of effective length $i_{1}$, then find the string of
minimum length $(\geq 3i_{1}+1)$ which is singular/q-singular/qq-singular. In
this search, the type-II (resp. type-I) qq-singular string must be ignored if
there exists a type-I (resp. type-0) singular string of the same effective
length (see Example 4.2). We prescribe that the type-II qq-singular string of
length ($\geq 3i_{1}+1$) is ignored if there exists a type-0 singular string
of the same effective length. We also prescribe that the type-I q-singular
(resp. qq-singular) string of length ($\geq 3i_{1}+1$) is ignored if there
exists a type-0 singular (resp. q-singular) string of the same effective
length (see Example 4.3). If the search is successful, then mark the rightmost
box of the founded string by $[2]$. If not, then delete the marked box and
return $\framebox{$2$}$. If [2] is marked, then we prescribe that boxes of
north and northwest of the box marked by [2] cannot be marked. This rule is
also applied in any box marking in $\nu^{(1)}$ (not in $\nu^{(2)}$).

######  Example 4.2.

$L=6$ and $\lambda=2\bar{\Lambda}_{1}+\bar{\Lambda}_{2}$.

$\downarrow$$2$$2$$0$$2$$[4]$$[3]$$[2]$$0$$1$$0$$0$$0$$0$$0$$0$$[1]$$[5]$$\longrightarrow$$8$$\downarrow$$3$$1$$3$$0$$[4]$$[3]$$[2]$$1$$1$$1$$0$$0$$0$$0$$[1]$$[5]$$\longrightarrow$$8$

which corresponds to the path
$\framebox{$8$}\otimes\framebox{$8$}\otimes\framebox{$1$}\otimes\framebox{$5$}\otimes\framebox{$8$}\otimes\framebox{$1$}$.
The charge of the first RC and the energy of this path coincide, which is
computed to be $-19$. If we selected the qq-singular string for the
$i_{2}$-string in the first RC, then we would have

$\downarrow$$2$$2$$0$$2$$[2]$$[3]$$0$$1$$0$$0$$0$$0$$0$$0$$[1]$$\longrightarrow$$4$$\downarrow$$1$$1$$1$$0$$[2]$$1$$0$$0$$0$$0$$0$$0$$[1]$$\longrightarrow$$3$

which corresponds to the path
$\framebox{$4$}\otimes\framebox{$3$}\otimes\framebox{$\emptyset$}\otimes\framebox{$5$}\otimes\framebox{$8$}\otimes\framebox{$1$}$
whose energy is computed to be $-20$.

######  Example 4.3.

$L=6$ and $\lambda=2\bar{\Lambda}_{2}+\bar{\Lambda}_{2}$ as in Example 4.2.

$\downarrow$$2$$2$$0$$1$$[2]$$[3]$$0$$1$$1$$0$$0$$0$$1$$0$$[1]$$[4]$$\longrightarrow$$5$$\downarrow$$-1$$0$$1$$2$

The second RC is not admissible.

(BM-3):

Marking $[3]$.

  * (1):

$i_{2}=3i_{1}-2$.

Find the q-singular string of length $3i_{1}$ ignoring q-singular strings of
length $3i_{1}-1$ even if they exist. If such s string of length $3i_{1}$
exists, then mark the rightmost box of the founded string by $[3]$. If such a
string does not exist, then find the string (singular or q-singular) of length
$(\geq 3i_{1}+1)$. If such a string exists, then mark the rightmost box of the
founded string by $[3]$. If not, then delete the marked boxes and return
$\framebox{$3$}$.

  * (2):

$i_{2}=3i_{1}-1$.

    * (a):

The selected $i_{2}$-string is singular.

Mark the box on the left of $[2]$ by $[3]$.

    * (b):

The selected $i_{2}$-string is q-singular.

Find the singular/q-singular string of minimum length $(\geq 3i_{1}+1)$. If
such a string exists, then mark the rightmost box of the founded string by
$[3]$. If not, then delete the marked boxes and return $\framebox{$3$}$.

  * (3):

$i_{2}=3i_{1}$.

    * (a):

The selected $i_{2}$-string is q-singular.

Mark the box on the left of $[2]$ by $[3]$.

    * (b):

The selected $i_{2}$-string is qq-singular.

Find the singular/q-singular string of minimum length $(\geq i_{2}+1)$. If
such a string exists, then mark the rightmost box of the founded string by
[3]. If not, then delete the marked boxes and return $\framebox{$3$}$.

  * (4):

$i_{2}\geq 3i_{1}+1$.

    * (a):

The selected $i_{2}$-string is singular/q-singular.

Mark the box on the left of [2] by [3].

    * (b):

The selected $i_{2}$-string is qq-singular.

Follow the same as in case (3-b).

In cases (3) and (4), if the selected $i_{2}$-string is qq-singular of type-0
and there exists a singular string of length $i_{2}+1$, then discard the
previous $i_{2}$-string selection and do the following box marking.

$\downarrow$$[3]$$[2]$

######  Example 4.4.

$L=5$ and $\lambda=\bar{\Lambda}_{1}+2\bar{\Lambda}_{2}$.

$\downarrow$$3$$1$$1$$1$$[4]$$[3]$$[2]$$1$$1$$0$$0$$0$$0$$[1]$$\longrightarrow$$6$$0$$4$$0$$1$$[2]$$0$$0$$0$$0$$0$$[1]$$\longrightarrow$$3$

which corresponds to the path
$\framebox{$6$}\otimes\framebox{$3$}\otimes\framebox{$3$}\otimes\framebox{$5$}\otimes\framebox{$1$}$.
The charge of the first RC and the energy of this path coincide, which is
computed to be $-14$. If we overrode the above rule, then we would have

$\downarrow$$3$$1$$1$$1$$[2]$$[3]$$1$$1$$0$$0$$0$$0$$[1]$$\longrightarrow$$4$$0$$2$$0$$1$$1$$0$$0$$0$$0$$\longrightarrow$$1$

which corresponds to the path
$\framebox{$4$}\otimes\framebox{$1$}\otimes\framebox{$7$}\otimes\framebox{$5$}\otimes\framebox{$1$}$.
However, the energy of this path is computed to be $-15$.

(BM-4):

Marking $[4]$.

  * (1):

The selected $i_{3}$-string is singular.

The $i_{3}$-string is possibly marked by [2].

    * (a):

The selected $i_{3}$-string is type-0.

Mark the box on the left of $[3]$ by $[4]$.

    * (b):

The selected $i_{3}$-string is type-I.

If $i_{3}^{eff}>i_{1}$, then mark the box on the left of $[3]$ by $[4]$.

If $i_{3}^{eff}=i_{1}$, i.e. $i_{3}=3i_{1}-1$, then the $i_{3}$-string is
marked by [2]. Find the singular string of minimum length $(\geq i_{3}^{eff})$
in $\nu^{(2)}$. If such a string exists, then let $l_{4}^{(2)}$ be the length
of the founded string. If not, then set $l_{4}^{(2)}=\infty$. Find the
singular string in $\nu^{(1)}$ of minimum length $(\geq i_{3}+1)$. Note that
the singular strings of length $3i_{1}$ do not exists in this case. If such a
string exists, then let $l_{4}^{(1)}$ be the effective length of the founded
string. If not, then set $l_{4}^{(1)}=\infty$. If $l_{4}^{(1)}\geq
l_{4}^{(2)}$, then mark the rightmost box of the founded singular string of
length $l_{4}^{(2)}$ in $\nu^{(2)}$ by [4]. If $l_{4}^{(1)}<l_{4}^{(2)}$, then
mark the rightmost box of the founded singular string of effective length
$l_{4}^{(1)}$ in $\nu^{(1)}$ by $[4]$. If $l_{4}^{(1)}=l_{4}^{(2)}=\infty$,
then delete the marked boxes and return $\framebox{$4$}$.

    * (c):

The selected $i_{3}$-string is type-II.

Find the singular string in $\nu^{(2)}$ of length $i_{3}^{eff}-1$ (the
effective length reduction). If such a string exists, then mark the rightmost
box of the founded string by $[4]$ (see Examples 4.5 and 4.6). If not, then
find the singular string in $\nu^{(2)}$ of length $i_{3}^{eff}$. If such a
string exists, then mark the rightmost box of the founded string by $[4]$. If
not, then mark the box on the left of $[3]$ by $[4]$.

  * (2):

The selected $i_{3}$-string is q-singular.

Follow the same as in (1-b).

If the selected $i_{4}$-string is type-0 and the selected $i_{3}$-string is
type-I with the same effective length, then discard the $i_{3}$-string
selection and do the following box marking.

$\downarrow$$[4]$$[3]$

This rule guarantees the bijection when the q-singular $i_{5}$-string is in
$\nu^{(2)}$ with $i_{4}^{eff}=i_{5}$ and when the singular $i_{5}$-string is
in $\nu^{(2)}$ with $i_{4}^{eff}=i_{5}-1$ or $i_{4}^{eff}=i_{5}$ (see Examples
4.7 and 4.8).

######  Example 4.5.

$L=3$ and $\lambda=\bar{\Lambda}_{1}+\bar{\Lambda}_{2}$. The only admissible
configuration $\nu$ is

$\downarrow$$1$$0$

which has the charge $c(\nu)=-5$. The first two steps of $\Phi$ are depicted
as

$\downarrow$$1$$1$$[6]$$[5]$$[3]$$[2]$$0$$0$$0$$0$$[1]$$[4]$$\longrightarrow$$9$$\emptyset$$0$$0$$[1]$$\longrightarrow$$2$

which corresponds to the path
$\framebox{$9$}\otimes\framebox{$2$}\otimes\framebox{$1$}$ whose energy is
$-4$ and

$\downarrow$$1$$0$$[3]$$[2]$$0$$0$$0$$0$$[1]$$\longrightarrow$$4$$0$$0$$[3]$$[2]$$0$$0$$0$$[1]$$[4]$$\longrightarrow$$5$

which corresponds to the path
$\framebox{$4$}\otimes\framebox{$5$}\otimes\framebox{$1$}$ whose energy is
$-5$.

######  Example 4.6.

$L=4$ and $\lambda=2\bar{\Lambda}_{1}+\bar{\Lambda}_{2}$.

$\downarrow$$0$$2$$0$$2$$[2]$$[6]$$[5]$$[3]$$0$$0$$0$$0$$0$$[1]$$[4]$$\longrightarrow$$9$$0$$0$$[2]$$0$$0$$0$$[1]$$\longrightarrow$$3$

which corresponds to the path
$\framebox{$9$}\otimes\framebox{$3$}\otimes\framebox{$2$}\otimes\framebox{$1$}$
whose energy is $-7$.

######  Example 4.7.

$L=6$ and $\lambda=\bar{\Lambda}_{1}+\bar{\Lambda}_{2}$.

$\downarrow$$2$$1$$1$$1$$0$$1$$[2]$$[4]$$[3]$$0$$1$$0$$0$$0$$0$$0$$0$$0$$[1]$$[5]$$\longrightarrow$$8$$\downarrow$$0$$1$$0$$0$$1$$0$$[2]$$[6]$$[5]$$[3]$$0$$1$$0$$0$$0$$0$$0$$0$$[1]$$[4]$$\longrightarrow$$9$

which corresponds to the path
$\framebox{$8$}\otimes\framebox{$9$}\otimes\framebox{$3$}\otimes\framebox{$5$}\otimes\framebox{$8$}\otimes\framebox{$1$}$.
The charge of the first RC and the energy of this path coincide, which is
computed to be $-22$. If we selected the q-singular string for the
$i_{3}$-string in the first RC, then we would have

$\downarrow$$2$$1$$1$$1$$0$$1$$[2]$$[3]$$[4]$$0$$1$$0$$0$$0$$0$$0$$0$$0$$[1]$$[5]$$\longrightarrow$$8$$\downarrow$$0$$1$$0$$0$$0$$0$$[2]$$[3]$$[4]$$0$$1$$0$$0$$0$$0$$0$$0$$[1]$$[5]$$\longrightarrow$$8$

which corresponds to the path
$\framebox{$8$}\otimes\framebox{$8$}\otimes\framebox{$7$}\otimes\framebox{$4$}\otimes\framebox{$5$}\otimes\framebox{$1$}$.
However, the energy of this path is computed to be $-23$.

######  Example 4.8.

$L=7$ and $\lambda=3\bar{\Lambda}_{1}+\bar{\Lambda}_{2}$.

$\downarrow$$1$$1$$0$$1$$0$$0$$[2]$$[4]$$[3]$$0$$1$$0$$0$$0$$0$$0$$0$$1$$[1]$$[6]$$[5]$$\longrightarrow$$10$$\downarrow$$2$$0$$2$$0$$[6]$$[5]$$[3]$$[2]$$0$$0$$0$$0$$0$$0$$0$$[1]$$[4]$$\longrightarrow$$9$

which corresponds to the path
$\framebox{$10$}\otimes\framebox{$9$}\otimes\framebox{$3$}\otimes\framebox{$1$}\otimes\framebox{$9$}\otimes\framebox{$2$}\otimes\framebox{$1$}$.
The charge of the first RC and the energy of this path coincide, which is
computed to be $-24$. If we selected the q-singular string for the
$i_{3}$-string in the first RC, then we would have

$\downarrow$$1$$1$$0$$1$$0$$0$$[2]$$[3]$$[4]$$0$$1$$0$$0$$0$$0$$0$$0$$1$$[1]$$[6]$$[5]$$\longrightarrow$$10$$\downarrow$$2$$0$$1$$0$$[3]$$[2]$$[4]$$0$$0$$0$$0$$0$$0$$0$$[1]$$\longrightarrow$$6$

which corresponds to the path
$\framebox{$10$}\otimes\framebox{$6$}\otimes\framebox{$5$}\otimes\framebox{$1$}\otimes\framebox{$9$}\otimes\framebox{$2$}\otimes\framebox{$1$}$.
However, the energy of this path is computed to be $-25$.

(BM-5):

Marking $[5]$.

  * (1):

The selected $i_{4}$-string is in $\nu^{(1)}$.

Find the singular/q-singular string of minimum length $(\geq i_{4}^{eff})$ in
$\nu^{(2)}$. Suppose that such a string exists. If the founded string is
singular and the length is $1$, then delete the marked boxes and return
$\framebox{$\emptyset$}$. In this case, the box marking of $\nu^{(1)}$ is

$\uparrow$$[4]$$[3]$$[2]$$3$

If the founded string is q-singular or singular with length $(\geq 2)$, then
the rightmost box of the founded string by $[5]$. If not, then delete the
marked boxes and return $\framebox{$6$}$.

  * (2):

The selected $i_{4}$-string is in $\nu^{(2)}$.

Find the singular or q-singular string of effective length $(\geq i_{4})$ in
$\nu^{(1)}$. If such a string exists and it is not a type-I/II q-singular
string of effective length $i_{4}$, then mark the rightmost box of the founded
string by $[5]$. If such a string does not exist, then delete the marked boxes
and return $\framebox{$5$}$.

######  Example 4.9.

$L=4$ and $\lambda=3\Lambda_{1}$.

$\downarrow$$0$$2$$0$$1$$[2]$$[3]$$0$$0$$0$$0$$0$$0$$[1]$$[4]$$\longrightarrow$$5$

which corresponds to the path
$\framebox{$5$}\otimes\framebox{$9$}\otimes\framebox{$2$}\otimes\framebox{$1$}$.
The charge of the first RC and the energy of this path coincide, which is
computed to be $-8$. For the rest steps, see Example 4.5. If [5] were marked
on the left of [3], then we would have

$\downarrow$$0$$2$$0$$1$$[2]$$[5]$$[3]$$0$$0$$0$$0$$0$$0$$[1]$$[4]$$\longrightarrow$$7$$3$$3$$[4]$$[3]$$[2]$$0$$0$$0$$0$$[1]$$[5]$$\longrightarrow$$\emptyset$

which corresponds to the path
$\framebox{$7$}\otimes\framebox{$\emptyset$}\otimes\framebox{$2$}\otimes\framebox{$1$}$.
However, the energy of this path is computed to be $-6$.

(BM-6):

Marking $[6]$.

  * (1):

The selected $i_{5}$-string is q-singular.

Find the singular string of minimum length $(\geq i_{5}+1)$. If such a string
exists, then mark the rightmost box of the founded string by $[6]$. If such a
string does not exist, then delete the marked boxes and return
$\framebox{$7$}$ when the selected $i_{5}$-string is in $\nu^{(1)}$ and
$\framebox{$8$}$ when the selected $i_{5}$-string is in $\nu^{(2)}$.

  * (2):

The selected $i_{5}$-string is singular.

Mark the box on the left of $[5]$ by $[6]$.

According to the marking rule of [4] the box marking such that the selected
$i_{4}$-string in $\nu^{(1)}$ is type-II with $i_{4}^{eff}=i_{6}$ cannot
occur.

(BM-7):

Marking $[7]$.

  * (1):

The selected $i_{6}$-string is in $\nu^{(1)}$.

Find the singular string of minimum length $(\geq i_{6}^{eff})$. If such a
string exists, then mark the rightmost unmarked box of the founded string by
$[7]$. If not, then delete the marked boxes and return $\framebox{$9$}$. Note
that it is possible $[7]$ is marked on the left of $[4]$.

  * (2):

The selected $i_{6}$-string is in $\nu^{(2)}$.

Find the selected $i_{4}$-string with $i_{4}^{eff}=i_{6}$, which is type-0/I.
If the search is successful, then mark [7] on the left of [4]. If not, then
follow the box marking rule (BM-2) with replacing $i_{1}$ by $i_{6}$. If the
search is not successful, then delete the marked boxes and return
$\framebox{$10$}$.

(BM-8):

Marking $[8]$.

  * (1):

The selected $i_{7}$-string is in $\nu^{(1)}$.

Find the selected $i_{7}$-string which is also marked by [4]. If such a string
exists, then mark [8] on the left of [7]. If the selected string is type-I and
the rightmost box of the selected string is marked by [3], then prohibit the
further box marking of this string (see Example 4.10). We say that this string
is _inactivated_. If the selected $i_{7}$-string is not marked by [4], then
follow the box marking rule (BM-3) with replacing $i_{1}$ by $i_{6}$ and
$i_{2}$ by $i_{7}$ except that [9] cannot be marked in $\nu^{(2)}$. If the
search is not successful, then delete the marked boxes and return
$\framebox{$11$}$.

######  Example 4.10.

$L=5$ and $\lambda=2\bar{\Lambda}_{1}$.

$\downarrow$$\downarrow$$3$$1$$1$$1$$[2]$$[8]$$[7]$$[4]$$[3]$$1$$0$$0$$1$$0$$0$$0$$0$$[1]$$[6]$$[5]$$\longrightarrow$$12$$\downarrow$$0$$1$$0$$1$$[3]$$[2]$$[6]$$[5]$$1$$0$$1$$0$$0$$0$$[1]$$[7]$$[4]$

which corresponds to the path
$\framebox{$12$}\otimes\framebox{$11$}\otimes\framebox{$1$}\otimes\framebox{$5$}\otimes\framebox{$1$}$.
The charge of the first RC and the energy of this path coincide, which is
computed to be $-13$. If [9] were marked on the left of [8], then the first
return would be $13$. However,
$\lambda-\mathrm{wt}(\framebox{$13$})=2\bar{\Lambda}_{1}-(-3\bar{\Lambda}_{1}+\bar{\Lambda}_{2})=5\bar{\Lambda_{1}}-\bar{\Lambda_{2}}$,
which is not dominant.

  * (2):

The selected $i_{7}$-string is in $\nu^{(2)}$.

Find the singular/q-singular string of effective length $i_{7}$ in
$\nu^{(1)}$, which is possibly marked by [6], according to the following
preferential rule.

| type-0 | type-I
---|---|---
singular | 1 | 2
q-singular | 3 |

If such a string exists, then mark the rightmost unmarked box of the founded
string by $[8]$. If not, then find the singular or q-singular string of
minimum length whose effective length $(\geq i_{7}+1)$. If the search is
successful, then mark the rightmost box of the unmarked box of the founded
string by $[8]$. If not, then delete the marked boxes and return
$\framebox{$11$}$. If the selected $i_{8}$-string is type-I and the rightmost
box of this string is marked by [5] or [6], then prohibit the further box
marking of this string (see Examples 4.11 and 4.12). We also say that this
string is inactivated.

######  Example 4.11.

$L=5$ and $\lambda=2\bar{\Lambda}_{1}$ as in Example 4.10.

$\downarrow$$\downarrow$$3$$1$$2$$1$$[3]$$[2]$$[8]$$[6]$$[5]$$1$$0$$0$$1$$0$$0$$0$$0$$[1]$$[4]$$[7]$$\longrightarrow$$12$$\downarrow$$0$$2$$0$$2$$[2]$$[8]$$[7]$$[4]$$[3]$$0$$0$$0$$0$$0$$0$$[1]$$[6]$$[5]$

which corresponds to the path
$\framebox{$12$}\otimes\framebox{$12$}\otimes\framebox{$3$}\otimes\framebox{$2$}\otimes\framebox{$1$}$.
The charge of the first RC and the energy of this path coincide, which is
computed to be $-12$. By the same reasoning as in Example 4.10, [9] cannot be
marked on the left of [8] in the first RC.

######  Example 4.12.

$L=5$ and $\lambda=2\bar{\Lambda}_{1}$ as in Example 4.10.

$\downarrow$$\downarrow$$3$$1$$2$$1$$[5]$$[3]$$[2]$$[8]$$[6]$$1$$0$$0$$1$$1$$0$$0$$0$$[1]$$[4]$$[7]$$\longrightarrow$$12$$\downarrow$$3$$2$$[5]$$[3]$$[2]$$1$$0$$0$$0$$0$$[1]$$[4]$

which corresponds to the path
$\framebox{$12$}\otimes\framebox{$7$}\otimes\framebox{$\emptyset$}\otimes\framebox{$2$}\otimes\framebox{$1$}$.
The charge of the first RC and the energy of this path coincide, which is
computed to be $-11$. By the same reasoning as in Example 4.10, [9] cannot be
marked on the left of [8] in the first RC.

(BM-9):

Marking $[9]$.

Find the active singular string that is not inactivated of length $(\geq
i_{8})$ in $\nu^{(1)}$. If the founded string whose length is denoted by
$l_{9}^{(1)}$ is type-I and the rightmost box of the founded string is marked
by $[2]$, then discard this selection and find the string of length
$(>l_{9}^{(1)})$. If the founded string whose length is denoted by
$l_{9}^{(1)}$ is type-II and the rightmost box of the founded string is marked
by $[3]$, then discard this selection and find the string of length
$(>l_{9}^{(1)})$. The last two rules prohibit the box marking in a single
string across two delimiters. If the string which can be marked by $[9]$ is
founded, then mark the rightmost box of the unmarked box of the founded string
by $[9]$. If not, then delete the marked boxes and return $\framebox{$12$}$.

If the selected $i_{9}$-string is type-0 and the selected $i_{8}$-string is
type-I with the same effective length, then discard the $i_{8}$-string
selection and do the following box marking.

$\downarrow$$[9]$$[8]$

This rule guarantees the bijection when [10] is marked in a singular string
with $i_{9}^{eff}=i_{10}$ in $\nu^{(2)}$.

(BM-10):

Marking $[10]$.

Find the singular string of minimum length $(\geq i_{9}^{eff})$ in
$\nu^{(2)}$. If such a string exists, then mark the rightmost unmarked box of
the founded string by $[10]$ and delete the marked boxes returning
$\framebox{$14$}$. If not, then delete the marked boxes and return
$\framebox{$13$}$.

There are several exceptional rules for box marking.

######  Example 4.13.

$L=4$ and $\lambda=2\bar{\Lambda}_{2}$.

According to the box marking rules prescribed above, we have

$\downarrow$$3$$0$$1$$0$$[2]$$[7]$$[4]$$[3]$$0$$1$$0$$0$$0$$0$$0$$[1]$$[6]$$[5]$$\longrightarrow$$11$

However,
$\lambda-\mathrm{wt}(\framebox{$11$})=2\bar{\Lambda}_{2}-(\bar{\Lambda}_{1}-\bar{\Lambda}_{2})=-\bar{\Lambda}_{1}+3\bar{\Lambda}_{2}$,
which is not dominant. For the above RC, we must do the following box marking.

$\downarrow$$3$$0$$1$$0$$[9]$$[8]$$[7]$$[4]$$[3]$$[2]$$0$$1$$0$$0$$0$$0$$0$$[1]$$[6]$$[5]$$\longrightarrow$$13$

The first string in $\nu^{(1)}$ is called a _boomerang string_. Here we
summarize the rules of finding boomerang strings, which have a common feature;
the box marking restarts in the boomerang string leading to the larger box
marking.

(BS-1):

The selected $i_{2}$-string is type-II q-singular.

The type-0/I singular string of effective length $i_{2}^{eff}$ under the
condition that $\nu^{(2)}$ has a singular string of length $i_{2}^{eff}$.

(BS-2):

The selected $i_{2}$-string is type-0/I q-singular or type-I qq-singular.

The following singular strings are boomerang strings for the $i_{2}$-string.
If there exist boomerang strings more than two, then select the string of
minimum length.

  * (1):

The type-II singular string of effective length $i_{2}^{eff}+1$ under the
condition that $\nu^{(2)}$ has a singular string of length $i_{2}^{eff}$ or
$i_{2}^{eff}+1$.

  * (2):

The type-0/I singular string of effective length $i_{2}^{eff}+1$ under the
condition that $\nu^{(2)}$ has a singular string of length $i_{2}^{eff}+1$.

(BS-3):

The selected $i_{2}$-string is type-0 qq-singular.

The type-0/I singular string of effective length $i_{2}^{eff}+1$ is a
boomerang string if $\nu^{(2)}$ has a singular string of length
$i_{2}^{eff}+1$. See Example 4.13. Note that we always select the type-II
singular string of effective length $i_{2}^{eff}+1$ if it exists. See Example
4.4.

(BS-4):

The selected $i_{3}$-string with $i_{2}<i_{3}$ is type-II q-singular.

The type-0/I singular string of effective length $i_{3}^{eff}$ is a boomerang
string for the $i_{3}$-string if $\nu^{(2)}$ has a singular string of length
$i_{3}^{eff}$.

(BS-5):

The selected $i_{3}$-string with $i_{2}<i_{3}$ is type-0/I q-singular.

The singular string of any type and of effective length $i_{3}^{eff}+1$ is a
boomerang string for the $i_{3}$-string if $\nu^{(2)}$ has a singular string
of length $i_{3}^{eff}+1$.

(BS-6):

The selected $i_{5}$-string in $\nu^{(1)}$ with $i_{3}<i_{5}$ is q-singular
and type-I/II.

The singular string of length ($>i_{5}$) and of the effective length
$i_{5}^{eff}$ is a boomerang string for the $i_{5}$-string if $\nu^{(2)}$ has
a singular string of length $i_{5}^{eff}$.

Here, we summarize the rule of adjusting the values of riggings in
$\tilde{\nu}^{(a)}$ $(a=1,2)$. The fundamental rule is that we set the box-
deleted string in $\tilde{\nu}^{(a)}$ $(a=1,2)$ to be singular and that we
keep the riggings of the box-undeleted strings the same.

(RA-1):

When $[6]$ is marked in $\nu^{(1)}$ or $\nu^{(2)}$ with $i_{5}<i_{6}$, we set
the box-deleted $(i_{6}-n)$-string to be q-singular where $n\geq 1$. See
Example 4.12.

(RA-2):

When $[3]$ is marked in $\nu^{(1)}$ with $i_{2}<i_{3}$ and if [4] is not
marked in $\nu^{(1)}$ and the $i_{3}$-string is not marked by [5], we do as
follows according to the type of the selected $i_{3}$-string.

  * (1):

type-II.

We set the box-deleted $(i_{3}-1)$-string to be q-singular unconditionally.

  * (2):

type-0/I.

If [4] is marked in $\nu^{(2)}$ with $i_{3}^{eff}=i_{4}$, we set the box-
deleted $(i_{3}-1)$-string to be singular. Otherwise, we set the box-deleted
$(i_{3}-1)$-string to be q-singular.

(RA-3):

When $[4]$ is marked in $\nu^{(1)}$ with $i_{3}<i_{4}$ and the selected
$i_{4}$-string is marked by $[4]$ only, we do as follows according to the type
of the selected $i_{4}$-string.

  * (1):

type-0.

If $[5]$ is marked in a q-singular string in $\nu^{(2)}$ with
$i_{4}^{eff}=i_{5}$ or $[5]$ and $[6]$ are marked in the same singular string
in $\nu^{(2)}$ with $i_{4}^{eff}=i_{6}-1$, then we set the box-deleted
$(i_{4}-1)$-string to be singular. Otherwise, we set the box-deleted
$(i_{4}-1)$-string to be qq-singular.

  * (2):

type-I.

If $[5]$ is not marked in $\nu^{(2)}$, then we set the box-deleted
$(i_{4}-1)$-string to be qq-singular. If $[5]$ is marked in a q-singular
string in $\nu^{(2)}$ with $i_{4}^{eff}=i_{5}$ or [5] and [6] are marked in
the same singular string in $\nu^{(2)}$ with $i_{4}^{eff}=i_{6}-1$, then we
set the box-deleted $(i_{4}-1)$-string to be q-singular. Otherwise, we set the
box-deleted $(i_{4}-1)$-string to be qq-singular.

  * (3):

type-II.

We set the box-deleted $(i_{4}-1)$-string to be qq-singular unconditionally.

In either case, if the selected $i_{3}$-string is marked by [3] only, we set
the box-deleted $(i_{3}-1)$-string to be q-singular.

(RA-4):

When [4] is marked in $\nu^{(1)}$ with $i_{2}<i_{3}=i_{4}$ and the selected
$i_{4}$-string is not marked by [7], we do as follows according to the type of
the selected $i_{4}$-string.

  * (1):

type-0.

If [5] is marked in a q-singular string in $\nu^{(2)}$ with
$i_{4}^{eff}=i_{5}$ or [5] and [6] are marked in the same singular string in
$\nu^{(2)}$ with $i_{4}^{eff}=i_{6}-1$, then we set the box-deleted
$(i_{4}-2)$-string to be singular. Otherwise, we set the box-deleted
$(i_{4}-2)$-string to be q-singular.

  * (2):

type-I/II.

We set the box-deleted $(i_{4}-2)$-string to be q-singular unconditionally.

(RA-5):

When [8] is marked in $\nu^{(1)}$ but [9] is not and the selected
$i_{8}$-string is marked by [8] only, we set the box-deleted
$(i_{8}-1)$-string in $\tilde{\nu}^{(1)}$ to be q-singular whether [7] is
marked in $\nu^{(1)}$ or $\nu^{(2)}$.

(RA-6):

When [9] is marked in $\nu^{(1)}$ and the rightmost box of the selected
$i_{9}$-string is marked by [8], i.e., $i_{8}=i_{9}$, we do as follows
according to the type of the selected $i_{9}$-string whether [7] is marked in
$\nu^{(1)}$ or $\nu^{(2)}$.

  * (1):

type-0.

If [10] is marked in $\nu^{(2)}$ with $i_{9}^{eff}=i_{10}$, then we set the
box-deleted $(i_{9}-2)$-string to be singular. Otherwise, we set this string
to be q-singular.

  * (2):

type-I/II.

We set the box-deleted $(i_{9}-2)$-string to be q-singular unconditionally.

(RA-7):

When [9] are marked in $\nu^{(1)}$ with $i_{8}<i_{9}$, we do as follows
according to the type of the $i_{9}$-string whether [7] is marked in
$\nu^{(1)}$ or $\nu^{(2)}$.

  * (1):

type-0.

If [10] is marked in $\nu^{(2)}$ with $i_{9}^{eff}=i_{10}$, then we set the
box-deleted $(i_{9}-1)$-string to be singular. Otherwise, we set this string
to be qq-singular.

  * (2):

type-I.

If [10] is marked in $\nu^{(2)}$ with $i_{9}^{eff}=i_{10}$, then we set the
box-deleted $(i_{9}-1)$-string to be q-singular. Otherwise, we set this string
to be qq-singular.

  * (3):

type-II.

We set the box-deleted $(i_{9}-1)$-string to be qq-singular unconditionally.

In either case, if the selected $i_{8}$-string is marked by [8] only, we set
the box-deleted $(i_{8}-1)$-string to be q-singular.

Let $A$ be a statement, then $\chi(A)=1$ if $A$ is true and $\chi(A)=0$ if $A$
is false. The changes of vacancy numbers are defined by $\Delta
p_{i}^{(a)}:=\tilde{p}_{i}^{(a)}-p_{i}^{(a)}$. Here we compute the changes of
vacancy numbers according to the return $b$ of $\delta_{\theta}$.

(VC-1):

$b=\framebox{$1$}$.

From Eqs. (3.4) and (3.5), we have $\Delta p_{i}^{(1)}=0$ $(i\geq 1)$ and
$\Delta p_{i}^{(2)}=-1$ $(i\geq 1)$.

(VC-2):

$b=\framebox{$2$}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting

$\displaystyle\min(i,3i_{1})(m_{i_{1}}^{(2)}-1)+\min(i,3(i_{1}-1))(m_{i_{1}-1}^{(2)}+1)$
$\displaystyle-\min(i,3i_{1})m_{i_{1}}^{(2)}-\min(i,3(i_{1}-1))m_{i_{1}-1}^{(2)}$
$\displaystyle=-\min(i,3i_{1})+\min(i,3(i_{1}-1))$ $\displaystyle=-3\chi(i\geq
3i_{1})-2\chi(i=3i_{1}-1)-\chi(i=3i_{1}-2)$

on $\Delta p_{i}^{(1)}$ in (VC-1) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting

$\displaystyle-$ $\displaystyle
2\min(i,i_{1})(m_{i_{1}}^{(2)}-1)-2\min(i,i_{1}-1)(m_{i_{1}-1}^{(2)}+1)$
$\displaystyle+2\min(i,i_{1})m_{i_{1}}^{(2)}+2\min(i,i_{1}-1)m_{i_{1}-1}^{(2)}$
$\displaystyle=2\min(i,i_{1})-2\min(i,i_{1}-1)=2\chi(i\geq i_{1})$

on $\Delta p_{i}^{(2)}$ in (VC-1). The changed $\Delta p_{i}^{(1)}$ and
$\Delta p_{i}^{(2)}$ are listed below.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-3$ | $(i\geq 3i_{1})$ | 1 | $(i\geq i_{1})$
$-2$ | $(i=3i_{1}-1)$ |  |
$-1$ | $(i=3i_{1}-2)$ |  |

(VC-3):

$b=\framebox{$3$}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting $2\chi(i\geq i_{2})$ on $\Delta p_{i}^{(1)}$ in (VC-2) and $\Delta
p_{i}^{(2)}$ are obtained by overwriting $-\chi(3i\geq i_{2})$ on $\Delta
p_{i}^{(2)}$ in (VC-2).

  * (1):

$i_{2}\leq 3i_{1}-1$.

$\Delta p_{i}^{(1)}$ |  $i_{2}$:I |  $i_{2}$:II |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $-1$ | $-1$ | $(i\geq 3i_{1})$ | 0 | $(i\geq i_{1})$
| $0$ | $0$ | $(i=3i_{1}-1)$ |  |
|  | $1$ | $(i=3i_{1}-2)$ |  |

where $i_{2}$:I (resp. $i_{2}$:II) stands for that the $i_{2}$-string is
type-I (resp. type-II).

  * (2):

$i_{2}\geq 3i_{1}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-1$ | $(i\geq i_{2})$ | 0 | $(i\geq i_{2}^{eff})$

(VC-4):

$b=\framebox{$4$}$.

  * (1):

$i_{2}=i_{3}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting $4\chi(i\geq i_{3})+2\chi(i=i_{3}-1)$ on $\Delta p_{i}^{(1)}$ in
(VC-2) and $\Delta p_{i}^{(2)}$ are obtained by overwriting $-2\chi(3i\geq
i_{3})-\chi(3i=i_{3}-1)$ on $\Delta p_{i}^{(2)}$ in (VC-2).

    * (a):

$i_{3}^{eff}=i_{1}$.

$\Delta p_{i}^{(1)}$ |  $i_{3}$:0 |  $i_{3}$:I |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $1$ | $1$ | $(i\geq 3i_{1})$ | $-1$ | $(i\geq i_{1})$
| $0$ | $2$ | $(i=3i_{1}-1)$ |  |
| $-1$ | $1$ | $(i=3i_{1}-2)$ |  |

    * (b):

$i_{3}^{eff}>i_{1}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |  $i_{3}$:0/I |  $i_{3}$:II |
---|---|---|---|---|---
$1$ | $(i\geq i_{3})$ |  | $-1$ | $-1$ | $(i\geq i_{3}^{eff})$
$-1$ | $(i=i_{3}-1)$ |  | $1$ | $0$ | $(i=i_{3}^{eff}-1)$

where $i_{3}$:0/I (resp. $i_{3}$:II) stands for that the $i_{3}$-string is
type-0/I (resp. type-II).

  * (2):

$i_{2}<i_{3}$.

In this case, $i_{3}\geq 3i_{1}$. The changes of vacancy numbers $\Delta
p_{i}^{(1)}$ are obtained by overwriting $2\chi(i\geq i_{3})$ on $\Delta
p_{i}^{(1)}$ in (VC-3) and $\Delta p_{i}^{(2)}$ are obtained by overwriting
$-\chi(3i\geq i_{3})$ on $\Delta p_{i}^{(2)}$ in (VC-3).

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq i_{3})$ | $-1$ | $(i\geq i_{3}^{eff})$

(VC-5):

$b=\framebox{$5$}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting

$-3\chi(i\geq 3i_{4})-2\chi(i=3i_{4}-1)-\chi(i=3i_{4}-2)$

on $\Delta p_{i}^{(1)}$ in (VC-4) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting $2\chi(i\geq i_{4})$ on $\Delta p_{i}^{(2)}$ in (VC-4).

  * (1):

$i_{2}=i_{3}$.

    * (a):

$i_{3}=3i_{4}-1$.

$\Delta p_{i}^{(1)}$ | $i_{1}=i_{4}$ | $i_{1}<i_{4}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $-2$ | $-2$ | $(i\geq 3i_{4})$ | $1$ | $(i\geq i_{1})$
| $0$ | $-1$ | $(i=3i_{4}-1)$ |  |
| $0$ | $-2$ | $(i=3i_{4}-2)$ |  |

Note that “$i\geq i_{1}$” in $\Delta p_{i}^{(2)}$ is not a typo.

    * (b):

otherwise.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-2$ | $(i\geq 3i_{4})$ | $1$ | $(i\geq i_{4})$
$-1$ | $(i=3i_{4}-1)$ |  |
$0$ | $(i=3i_{4}-2)$ |  |

Note that $i_{3}^{eff}<i_{4}$ when the $i_{3}$-string is type-0.

  * (2):

$i_{2}<i_{3}$.

    * (a):

$i_{3}^{eff}<i_{4}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-2$ | $(i\geq 3i_{4})$ | $1$ | $(i\geq i_{4})$
$-1$ | $(i=3i_{4}-1)$ |  |
$0$ | $(i=3i_{4}-2)$ |  |

    * (b):

$i_{3}=3i_{4}-1$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-2$ | $(i\geq 3i_{4})$ | $1$ | $(i\geq i_{4})$
$-1$ | $(i=3i_{4}-1)$ |  |
$-2$ | $(i=3i_{4}-2)$ |  |

    * (c):

$i_{3}=3i_{4}-2$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-2$ | $(i\geq 3i_{4})$ | $1$ | $(i\geq i_{4})$
$-1$ | $(i=3i_{4}-1)$ |  |
$0$ | $(i=3i_{4}-2)$ |  |

Note that the case when $i_{3}=3i_{4}$ cannot occur.

(VC-6):

$b=\framebox{$6$}$.

  * (1):

$i_{2}=i_{3}=i_{4}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting

$6\chi(i\geq i_{4})+4\chi(i=i_{4}-1)+2\chi(i=i_{4}-2)$

on $\Delta p_{i}^{(1)}$ in (VC-2) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting

$-3\chi(3i\geq i_{4})-2\chi(3i=i_{4}-1)-\chi(3i=i_{4}-2)$

on $\Delta p_{i}^{(2)}$ in (VC-2).

    * (a):

The $i_{4}$-string is type-0 with $i_{4}^{eff}=i_{1}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$3$ | $(i\geq i_{4})$ | $-2$ | $(i\geq i_{4}^{eff})$
$2$ | $(i=i_{4}-1)$ |  |
$1$ | $(i=i_{4}-2)$ |  |

    * (b):

The $i_{4}$-string is type-0 with $i_{4}^{eff}>i_{1}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$3$ | $(i\geq i_{4})$ | $-2$ | $(i\geq i_{4}^{eff})$
$1$ | $(i=i_{4}-1)$ | $1$ | $(i_{1}\leq i\leq i_{4}^{eff}-1)$
$-1$ | $(i=i_{4}-2)$ |  |

    * (c):

The $i_{4}$-string is type-I.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$3$ | $(i\geq i_{4})$ | $-2$ | $(i\geq i_{4}^{eff})$
$1$ | $(i=i_{4}-1)$ | $0$ | $(i=i_{4}^{eff}-1)$
$-1$ | $(i=i_{4}-2)$ |  |

    * (d):

The $i_{4}$-string is type-II.

$\Delta p_{i}^{(1)}$ | $i_{4}^{eff}=i_{1}+1$ | $i_{4}^{eff}>i_{1}+1$ |
---|---|---|---
| $3$ | $3$ | $(i\geq i_{4})$
| $1$ | $1$ | $(i=i_{4}-1)$
| $0$ | $-1$ | $(i=i_{4}-2)$

$\Delta p_{i}^{(2)}$ |
---|---
$-2$ | $(i\geq i_{4}^{eff})$
$-1$ | $(i=i_{4}^{eff}-1)$

  * (2):

$i_{2}<i_{3}=i_{4}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting $4\chi(i\geq i_{4})+2\chi(i=i_{4}-1)$ on $\Delta p_{i}^{(1)}$ in
(VC-3) and $\Delta p_{i}^{(2)}$ are obtained by overwriting $-2\chi(3i\geq
i_{4})-\chi(3i=i_{4}-1)$ on $\Delta p_{i}^{(2)}$ in (VC-3).

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |  $i_{4}$:0/I |  $i_{4}$:II |
---|---|---|---|---|---
$3$ | $(i\geq i_{4})$ |  | $-2$ | $-2$ | $(i\geq i_{4}^{eff})$
$1$ | $(i=i_{4}-1)$ |  | $0$ | $-1$ | $(i=i_{4}^{eff}-1)$

  * (3):

$i_{3}<i_{4}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting $2\chi(i\geq i_{4})$ on $\Delta p_{i}^{(1)}$ in (VC-4) and $\Delta
p_{i}^{(2)}$ are obtained by overwriting $-\chi(3i\geq i_{4})$ on $\Delta
p_{i}^{(2)}$ in (VC-4).

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$3$ | $(i\geq i_{4})$ | $-2$ | $(i\geq i_{4}^{eff})$

(VC-7):

$b=\framebox{$7$}$.

  * (1):

$i_{2}=i_{3}=i_{5}$.

In this case, $i_{5}=3i_{4}$. The changes of vacancy numbers $\Delta
p_{i}^{(1)}$ are obtained by overwriting

$\displaystyle 6\chi(i\geq i_{5})+4\chi(i=i_{5}-1)+2\chi(i=i_{5}-2)$
$\displaystyle-3\chi(i\geq 3i_{4})-2\chi(i=3i_{4}-1)-\chi(i=3i_{4}-2)$
$\displaystyle=$ $\displaystyle 3\chi(i\geq
i_{5})+2\chi(i=i_{5}-1)+\chi(i=i_{5}-2)$

on $\Delta p_{i}^{(1)}$ in (VC-2) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting

$\displaystyle-3\chi(3i\geq
i_{5})-2\chi(3i=i_{5}-1)-\chi(3i=i_{5}-2)+2\chi(i\geq i_{4})$ $\displaystyle=$
$\displaystyle-\chi(i\geq i_{4})$

on $\Delta p_{i}^{(2)}$ in (VC-2).

$\Delta p_{i}^{(1)}$ | $i_{1}=i_{4}$ | $i_{1}<i_{4}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $0$ | $0$ | $(i\geq i_{5})$ | $0$ | $(i\geq i_{4})$
| $0$ | $-1$ | $(i=i_{5}-1)$ |  |
| $0$ | $-2$ | $(i=i_{5}-2)$ |  |

  * (2):

$i_{2}<i_{3}=i_{5}$.

In this case $i_{5}=3i_{4}$. The changes of vacancy numbers $\Delta
p_{i}^{(1)}$ are obtained by overwriting

$\displaystyle 4\chi(i\geq i_{5})+2\chi(i=i_{5}-1)$ $\displaystyle-3\chi(i\geq
3i_{4})-2\chi(i=3i_{4}-1)-\chi(i=3i_{4}-2)$ $\displaystyle=$
$\displaystyle\chi(i\geq i_{5})-\chi(i=i_{5}-2)$

on $\Delta p_{i}^{(1)}$ in (VC-3) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting

$-2\chi(3i\geq i_{5})-\chi(3i=i_{5}-1)+2\chi(i\geq i_{4})=0$

on $\Delta p_{i}^{(2)}$ in (VC-3). The changes of vacancy numbers coincide
with those in case (1).

  * (3):

$i_{3}<i_{5}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting $2\chi(i\geq i_{5})$ on $\Delta p_{i}^{(1)}$ in (VC-5) and the
changes of vacancy numbers $\Delta p_{i}^{(2)}$ are obtained by overwriting
$-\chi(3i\geq i_{5})$ on $\Delta p_{i}^{(2)}$ in (VC-5).

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$0$ | $(i\geq i_{5})$ | $0$ | $(i\geq i_{5}^{eff})$

(VC-8):

$b=\framebox{$8$}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting

$-3\chi(i\geq 3i_{5})-2\chi(i=3i_{5}-1)-\chi(i=3i_{5}-2)$

on $\Delta p_{i}^{(1)}$ in (VC-6) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting $2\chi(i\geq i_{5})$ on $\Delta p_{i}^{(2)}$ in (VC-6).

  * (1):

$i_{4}^{eff}<i_{5}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$0$ | $(i\geq 3i_{5})$ | $0$ | $(i\geq i_{5})$
$1$ | $(i=3i_{5}-1)$ |  |
$2$ | $(i=3i_{5}-2)$ |  |

  * (2):

$i_{2}=i_{3}=i_{4}$ and $i_{4}^{eff}=i_{5}$.

In this case, the $i_{4}$-string is type-0/I.

$\Delta p_{i}^{(1)}$ | $i_{1}=i_{5}$ | $i_{1}<i_{5}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $0$ | $0$ | $(i\geq 3i_{5})$ | $0$ | $(i\geq i_{5})$
| $0$ | $-1$ | $(i=3i_{5}-1)$ |  |
| $0$ | $-2$ | $(i=3i_{5}-2)$ |  |

  * (3):

$i_{2}<i_{3}=i_{4}$ and $i_{4}^{eff}=i_{5}$.

$\Delta p_{i}^{(1)}$ |  $i_{4}$:0 |  $i_{4}$:I |  $i_{4}$:II |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---|---
| $0$ | $0$ | $0$ | $(i\geq 3i_{5})$ | $0$ | $(i\geq i_{5})$
| $-1$ | $1$ | $1$ | $(i=3i_{5}-1)$ |  |
| $-2$ | $0$ | $2$ | $(i=3i_{5}-2)$ |  |

  * (4):

$i_{3}<i_{4}$ and $i_{4}^{eff}=i_{5}$.

$\Delta p_{i}^{(1)}$ |  $i_{4}$:0 |  $i_{4}$:I |  $i_{4}$:II |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---|---
| $0$ | $0$ | $0$ | $(i\geq 3i_{5})$ | $0$ | $(i\geq i_{5})$
| $-1$ | $1$ | $1$ | $(i=3i_{5}-1)$ |  |
| $0$ | $0$ | $2$ | $(i=3i_{5}-2)$ |  |

Note that $i_{3}\leq 3i_{5}-2$ when $i_{4}=3i_{5}$ (see (BM-4)).

(VC-9):

$b=\framebox{$9$}$.

  * (1):

$i_{2}=i_{3}=i_{5}=i_{6}$.

In this case, the $i_{6}$-string is type-II and $i_{6}^{eff}=i_{4}+1$. The
changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by overwriting

$\displaystyle 8\chi(i\geq
i_{6})+6\chi(i=i_{6}-1)+4\chi(i=i_{6}-2)+2\chi(i=i_{6}-3)$
$\displaystyle-3\chi(i\geq 3i_{4})-2\chi(i=3i_{4}-1)-\chi(i=3i_{4}-2)$
$\displaystyle=$ $\displaystyle 5\chi(i\geq
i_{6})+3\chi(i=i_{6}-1)+2\chi(i=i_{6}-2)+\chi(i=i_{6}-3)$

on $\Delta p_{i}^{(1)}$ in (VC-2) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting

$\displaystyle-4\chi(3i\geq i_{6})-3\chi(3i=i_{6}-1)-2\chi(3i=i_{6}-2)$
$\displaystyle-\chi(3i=i_{6}-3)+2\chi(i\geq i_{4})=-2\chi(i\geq
i_{4}+1)-\chi(i=i_{4})$

on $\Delta p_{i}^{(2)}$ in (VC-2).

$\Delta p_{i}^{(1)}$ | $i_{1}=i_{4}$ | $i_{1}<i_{4}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $2$ | $2$ | $(i\geq i_{6})$ | $-1$ | $(i\geq i_{4}+1)$
| $0$ | $0$ | $(i=i_{6}-1)$ | $0$ | $(i=i_{4})$
| $0$ | $-1$ | $(i=i_{6}-2)$ |  |
| $0$ | $-2$ | $(i=i_{6}-3)$ |  |

  * (2):

$i_{2}<i_{3}=i_{5}=i_{6}$.

In this case, the $i_{6}$-string is type-II and $i_{6}^{eff}=i_{4}+1$. The
changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by overwriting

$\displaystyle 6\chi(i\geq i_{6})+4\chi(i=i_{6}-1)+2\chi(i=i_{6}-2)$
$\displaystyle-3\chi(i\geq 3i_{4})-2\chi(i=3i_{4}-1)-\chi(i=3i_{4}-2)$
$\displaystyle=$ $\displaystyle 3\chi(i\geq
i_{6})+\chi(i=i_{6}-1)-\chi(i=i_{6}-3)$

on $\Delta p_{i}^{(1)}$ in (VC-3) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting

$\displaystyle-3\chi(3i\geq i_{6})-2\chi(3i=i_{6}-1)-\chi(3i=i_{6}-2)$
$\displaystyle+2\chi(i\geq i_{4})=-\chi(i\geq i_{4}+1)$

on $\Delta p_{i}^{(2)}$ in (VC-3). The changes of vacancy numbers coincide
with those in case (1).

  * (3):

$i_{3}<i_{5}=i_{6}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting $4\chi(i\geq i_{6})+2\chi(i=i_{6}-1)$ on $\Delta p_{i}^{(1)}$ in
(VC-5) and $\Delta p_{i}^{(2)}$ are obtained by overwriting $-2\chi(3i\geq
i_{6})-\chi(3i=i_{6}-1)$ on $\Delta p_{i}^{(2)}$ in (VC-5).

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |  $i_{6}$:0/I |  $i_{6}$:II |
---|---|---|---|---|---
$2$ | $(i\geq i_{6})$ |  | $-1$ | $-1$ | $(i\geq i_{6}^{eff})$
$0$ | $(i=i_{6}-1)$ |  | $1$ | $0$ | $(i=i_{6}^{eff}-1)$

  * (4):

$i_{5}<i_{6}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting $2\chi(i\geq i_{6})$ on $\Delta p_{i}^{(1)}$ in (VC-7) and $\Delta
p_{i}^{(2)}$ are obtained by overwriting $-\chi(3i\geq i_{6})$ on $\Delta
p_{i}^{(2)}$ in (VC-7).

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$2$ | $(i\geq i_{6})$ | $-1$ | $(i\geq i_{6}^{eff})$

(VC-10):

$b=\framebox{$10$}$.

  * (1):

$i_{5}=i_{6}$.

In this case $i_{6}\geq i_{4}^{eff}+1$. The changes of vacancy numbers $\Delta
p_{i}^{(1)}$ are obtained by overwriting

$-6\chi(i\geq 3i_{6})-5\chi(i=3i_{6}-1)-\cdots-\chi(i=3i_{6}-5)$

on $\Delta p_{i}^{(1)}$ in (VC-6) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting $4\chi(i\geq i_{6})+2\chi(i=i_{6}-1)$ on $\Delta p_{i}^{(2)}$ in
(VC-6).

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-3$ | $(i\geq 3i_{6})$ | $2$ | $(i\geq i_{6})$
$-2$ | $(i=3i_{6}-1)$ | $0$ | $(i=i_{6}-1)$
$-1$ | $(i=3i_{6}-2)$ |  |
$0$ | $(i=3i_{6}-3)$ |  |

We omit $\Delta p_{i}^{(1)}$ $(3i_{6}-5\leq i\leq 3i_{6}-4)$.

  * (2):

$i_{5}<i_{6}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting

$-3\chi(i\geq 3i_{6})-2\chi(i=3i_{6}-1)-\chi(i=3i_{6}-2)$

on $\Delta p_{i}^{(1)}$ in (VC-8) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting $2\chi(i\geq i_{6})$ on $\Delta p_{i}^{(2)}$ in (VC-8).

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-3$ | $(i\geq 3i_{6})$ | $2$ | $(i\geq i_{6})$
$-2$ | $(i=3i_{6}-1)$ | $0$ | $(i=i_{6}-1)$
$-1$ | $(i=3i_{6}-2)$ |  |

(VC-11):

$b=\framebox{$11$}$.

  * (1):

The $i_{7}$-string is in $\nu^{(1)}$.

In this case [7] is marked in the rightmost box of the $i_{7}$-string. The
changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by overwriting
$2\chi(i\geq i_{7})$ on $\Delta p_{i}^{(1)}$ in (VC-10) and $\Delta
p_{i}^{(2)}$ are obtained by overwriting $-\chi(3i\geq i_{7})$ on $\Delta
p_{i}^{(2)}$ in (VC-10).

    * (a):

$i_{7}^{eff}=i_{6}$.

$\Delta p_{i}^{(1)}$ |  $i_{7}$:0 |  $i_{7}$:I |  $i_{7}$:II |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---|---
| $-1$ | $-1$ | $-1$ | $(i\geq 3i_{6})$ | $1$ | $(i\geq i_{6})$
|  | $0$ | $0$ | $(i\geq 3i_{6}-1)$ |  |
|  |  | $1$ | $(i\geq 3i_{6}-2)$ |  |

    * (b):

$i_{7}^{eff}>i_{6}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-1$ | $(i\geq i_{7})$ | $1$ | $(i\geq i_{7}^{eff})$

  * (2):

The $i_{7}$-string is in $\nu^{(2)}$ and $i_{4}=i_{7}$.

We have

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-1$ | $(i\geq 3i_{7})$ | $1$ | $(i\geq i_{7})$
$0$ | $(i=3i_{7}-1)$ | $0$ | $(i=i_{7}-1)$
$1$ | $(i=3i_{7}-2)$ |  |
$0$ | $(i=3i_{7}-3)$ |  |

for all possible box marking patterns of $\nu^{(1)}$,

$\downarrow$$[6]$$[5]$$[3]$$[2]$(a)$\downarrow$$[6]$$[5]$$[3]$(b)$\downarrow$$[6]$$[5]$(c)

where the $i_{6}$-string is always type-II. We omit the unmarked strings. In
case (a), $\Delta p_{i}^{(1)}$ $(3i_{7}-5\leq i\leq 3i_{7}-4)$ are obtained by
overwriting $2\chi(i=3i_{7}-4)+\chi(i=3i_{7}-5)$ on $\Delta p_{i}^{(1)}$ in
(VC-2) and $\Delta p_{i}^{(1)}$ $(i\leq 3i_{7}-6)$ and $\Delta p_{i}^{(2)}$
$(i\leq i_{7}-2)$ are the same as in (VC-2). In case (b), $\Delta p_{i}^{(1)}$
$(3i_{7}-5\leq i\leq 3i_{7}-4)$ are obtained by overwriting
$-\chi(i=3i_{7}-5)$ on $\Delta p_{i}^{(1)}$ in (VC-3) and $\Delta p_{i}^{(1)}$
$(i\leq 3i_{7}-6)$ and $\Delta p_{i}^{(2)}$ $(i\leq i_{7}-2)$ are the same as
in (VC-3). In case (c), $\Delta p_{i}^{(1)}$ $(3i_{7}-5\leq i\leq 3i_{7}-4)$
are obtained by overwriting $-2\chi(i=3i_{7}-4)-\chi(i=3i_{7}-5)$ on $\Delta
p_{i}^{(1)}$ in (VC-4) and $\Delta p_{i}^{(1)}$ $(i\leq 3i_{7}-6)$ and $\Delta
p_{i}^{(2)}$ $(i\leq i_{7}-2)$ are the same as in (VC-4).

  * (3):

The $i_{7}$-string is in $\nu^{(2)}$ and $i_{4}<i_{7}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-1$ | $(i\geq 3i_{7})$ | $1$ | $(i\geq i_{7})$
$0$ | $(i=3i_{7}-1)$ |  |
$1$ | $(i=3i_{7}-2)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq 3i_{7}-3)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{7}-1)$ are the same as in (VC-9).

(VC-12):

$b=\framebox{$12$}$.

  * (1):

The selected $i_{7}$-string is in $\nu^{(1)}$ and $i_{4}=i_{7}=i_{8}$.

In this case $i_{5}=i_{6}$. Possible box marking patterns in $\nu^{(1)}$ are
following two (see (BM-8)).

$\downarrow$$[8]$$[7]$$[4]$$[3]$$[2]$(a)$\downarrow$$[8]$$[7]$$[4]$$[3]$(b)

$\Delta p_{i}^{(1)}$ | $i_{6}=i_{1}+1$ | $i_{6}>i_{1}+1$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $1$ | $1$ | $(i\geq 3i_{6})$ | $0$ | $(i\geq i_{6}-1)$
| $2$ | $2$ | $(i=3i_{6}-1)$ |  |
| $1$ | $1$ | $(i=3i_{6}-2)$ |  |
| $0$ | $0$ | $(i=3i_{6}-3)$ |  |
| $0$ | $-1$ | $(i=3i_{6}-4)$ |  |
| $0$ | $-2$ | $(i=3i_{6}-5)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq 3i_{6}-6)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{6}-2)$ are the same as in (VC-2) (case (a))
and (VC-3) (case (b)).

  * (2):

The selected $i_{7}$-string is in $\nu^{(1)}$ and $i_{4}<i_{7}=i_{8}$.

    * (a):

$i_{8}^{eff}=i_{6}$.

$\Delta p_{i}^{(1)}$ |  $i_{8}$:0 |  $i_{8}$:I |  $i_{8}$:II |
---|---|---|---|---
| $1$ | $1$ | $1$ | $(i\geq 3i_{6})$
| $0$ | $2$ | $2$ | $(i=3i_{6}-1)$
|  | $1$ | $3$ | $(i=3i_{6}-2)$
|  |  | $2$ | $(i=3i_{6}-3)$

$\Delta p_{i}^{(2)}$ |  $i_{8}$:0/I |  $i_{8}$:II |
---|---|---|---
| $0$ | $0$ | $(i\geq i_{6})$
| $0$ | $-1$ | $(i=i_{6}-1)$

    * (b):

$i_{8}^{eff}>i_{6}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |  $i_{8}$:0/I |  $i_{8}$:II |
---|---|---|---|---|---
$1$ | $(i\geq i_{8})$ |  | $0$ | $0$ | $(i\geq i_{8}^{eff})$
$2$ | $(i=i_{8}-1)$ |  | $0$ | $-1$ | $(i=i_{8}^{eff}-1)$

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{8}-2)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{8}^{eff}-2)$ are the same as in (VC-10).

  * (3):

The selected $i_{7}$-string is in $\nu^{(1)}$ and $i_{7}<i_{8}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq i_{8})$ | $0$ | $(i\geq i_{8}^{eff})$

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{8}-1)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{8}^{eff}-1)$ are the same as in (VC-11) (1).

  * (4):

The selected $i_{7}$-string is in $\nu^{(2)}$ and $i_{4}=i_{7}$.

In this case $i_{8}^{eff}=i_{7}$. The only possible box marking in $\nu^{(1)}$
is

$\downarrow$$[8]$$[6]$$[5]$

$\Delta p_{i}^{(1)}$ | $i_{7}=i_{1}+1$ | $i_{7}>i_{1}+1$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $1$ | $1$ | $(i\geq 3i_{7})$ | $0$ | $(i\geq i_{7}-1)$
| $2$ | $2$ | $(i=3i_{7}-1)$ |  |
| $1$ | $1$ | $(i=3i_{7}-2)$ |  |
| $0$ | $0$ | $(i=3i_{7}-3)$ |  |
| $0$ | $-1$ | $(i=3i_{7}-4)$ |  |
| $0$ | $0$ | $(i=3i_{7}-5)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq 3i_{7}-6)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{7}-2)$ are the same as in (VC-4).

  * (5):

The selected $i_{7}$-string is in $\nu^{(2)}$ and $i_{4}<i_{7}$.

    * (a):

The box marking in $\nu^{(1)}$ is

$\downarrow$$[8]$$[6]$$[5]$

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq 3i_{7})$ | $0$ | $(i\geq i_{7}-1)$
$2$ | $(i=3i_{7}-1)$ |  |
$1$ | $(i=3i_{7}-2)$ |  |
$0$ | $(i=3i_{7}-3)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq 3i_{7}-4)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{7}-2)$ are the same as in (VC-5).

    * (b):

The box marking in $\nu^{(1)}$ is

$\downarrow$$[8]$$[6]$

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq 3i_{7})$ | $-$ |
$2$ | $(i=3i_{7}-1)$ |  |
$1$ | $(i=3i_{7}-2)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq 3i_{7}-3)$ and
$\Delta p_{i}^{(2)}$ are the same as in (VC-7).

    * (c):

$i_{8}^{eff}>i_{7}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq i_{8})$ | $0$ | $(i\geq i_{8}^{eff})$

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{8}-1)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{8}^{eff}-1)$ are the same as in (VC-11) (2) or
(3).

(VC-13):

$b=\framebox{$13$}$.

  * (1):

The selected $i_{7}$-string is in $\nu^{(2)}$ with $i_{4}=i_{7}$.

The only possible box marking of the selected $i_{9}$-string is

$\downarrow$$[9]$$[8]$$[6]$$[5]$

$\Delta p_{i}^{(1)}$ | $i_{7}=i_{1}+1$ | $i_{7}>i_{1}+1$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $3$ | $3$ | $(i\geq i_{9})$ | $-1$ | $(i\geq i_{7})$
| $2$ | $2$ | $(i=i_{9}-1)$ | $0$ | $(i=i_{7}-1)$
| $1$ | $1$ | $(i=i_{9}-2)$ |  |
| $0$ | $0$ | $(i=i_{9}-3)$ |  |
| $0$ | $-1$ | $(i=i_{9}-4)$ |  |
| $0$ | $0$ | $(i=i_{9}-5)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{9}-6)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{7}-2)$ are the same as in (VC-4).

  * (2):

The selected $i_{7}$-string is in $\nu^{(2)}$ with $i_{4}<i_{7}$ and the box
marking in $\nu^{(1)}$ is either

$\downarrow$$[9]$$[8]$$[6]$$[5]$(a)or$\downarrow$$[9]$$[8]$$[6]$(b)

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$3$ | $(i\geq i_{9})$ | $-1$ | $(i\geq i_{7})$
$2$ | $(i=i_{9}-1)$ | $0$ | $(i=i_{7}-1)$
$1$ | $(i=i_{9}-2)$ |  |
$0$ | $(i=i_{9}-3)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{9}-4)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{7}-2)$ are the same as in (VC-5) (case (a))
and (VC-7) (case (b)).

  * (3):

The selected $i_{6}$-string is in $\nu^{(2)}$ with $i_{5}=i_{6}$ and the box
marking in $\nu^{(1)}$ is either

$\downarrow$$[9]$$[8]$$[7]$$[4]$$[3]$$[2]$(a)or$\downarrow$$[9]$$[8]$$[7]$$[4]$$[3]$(b)

$\Delta p_{i}^{(1)}$ | $i_{6}=i_{1}+1$ | $i_{6}>i_{1}+1$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $3$ | $3$ | $(i\geq i_{9})$ | $-1$ | $(i\geq i_{6})$
| $2$ | $2$ | $(i=i_{9}-1)$ | $0$ | $(i=i_{6}-1)$
| $1$ | $1$ | $(i=i_{9}-2)$ |  |
| $0$ | $0$ | $(i=i_{9}-3)$ |  |
| $0$ | $-1$ | $(i=i_{9}-4)$ |  |
| $0$ | $-2$ | $(i=i_{9}-5)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{9}-6)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{6}-2)$ are the same as in (VC-2) (case (a))
and (VC-3) (case (b)).

  * (4):

The rightmost box of the $i_{9}$-string is marked by [7].

$\Delta p_{i}^{(1)}$ | $i_{9}^{eff}=i_{6}$ | $i_{9}^{eff}>i_{6}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---|---|---
| $3$ | $3$ | $(i\geq i_{9})$ | $-1$ | $(i\geq i_{9}^{eff})$
| $2$ | $1$ | $(i=i_{9}-1)$ |  |
| $1$ | $-1$ | $(i=i_{9}-2)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{9}-3)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{9}^{eff}-1)$ are the same as in (VC-10).

  * (5):

The rightmost box of the $i_{9}$-string is marked by [8].

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |  $i_{9}$:0/I |  $i_{9}$:II |
---|---|---|---|---|---
$3$ | $(i\geq i_{9})$ |  | $-1$ | $-1$ | $(i\geq i_{9}^{eff})$
$1$ | $(i=i_{9}-1)$ |  | $1$ | $0$ | $(i=i_{9}^{eff}-1)$

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{9}-2)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{9}^{eff}-2)$ are the same as in (VC-11).

  * (6):

The $i_{9}$-string is marked by [9] only.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$3$ | $(i\geq i_{9})$ | $-1$ | $(i\geq i_{9}^{eff})$

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{9}-1)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{9}^{eff}-1)$ are the same as in (VC-12).

(VC-14):

$b=\framebox{$14$}$.

  * (1):

$i_{1}<i_{10}$.

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by
overwriting

$-3\chi(i\geq 3i_{10})-2\chi(i=3i_{10}-1)-\chi(i=3i_{10}-2)$

on $\Delta p_{i}^{(1)}$ in (VC-13) and $\Delta p_{i}^{(2)}$ are obtained by
overwriting $2\chi(i\geq i_{10})$ on $\Delta p_{i}^{(2)}$ in (VC-13).

  * (2):

$i_{1}=i_{10}$.

The only possible box marking of $\nu^{(1)}$ (left) and $\nu^{(2)}$ (right) is

$\downarrow$$\uparrow$$[9]$$[8]$$[7]$$[4]$$[3]$$[2]$$3i_{1}$$\uparrow$$[10]$$[6]$$[1]$$[5]$$i_{1}$

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
0 | $(i\geq 1)$ | $1$ | $(i\geq i_{1})$
|  | 0 | $(i=i_{1}-1)$
|  | $-1$ | $(1\leq i\leq i_{1}-2)$

(VC-$\emptyset$):

$b=\framebox{$\emptyset$}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$0$ | $(i\geq 1)$ | $0$ | $(i\geq 1)$

###  4.2. Inverse algorithm $\tilde{\delta}_{\theta}$

The reader may skip this subsection as we do not use it in the following
sections. This is provided just for completeness.

For a given RC $(\tilde{\nu},\tilde{J})$ and $b\in B$ the inverse algorithm
$\tilde{\delta}_{\theta}$ of $\delta_{\theta}$ is described in this
subsection. The inverse algorithm consists of adding boxes to strings in
$\tilde{\nu}$ and adjusting the numbers of riggings. We omit the rule of
adjusting the numbers of riggings as it is obvious by the algorithm of
$\delta_{\theta}$. We follow the convention that riggings in strings of the
same length are sorted in an decreasing order (from the top). We assume that
$\tilde{\nu}^{(a)}$ has singular strings of length zero with zero vacancy (and
therefore zero rigging). In the following, strings of length $l$ under some
conditions are selected and the $l$-string is referred to the selected string
unless stated otherwise. We denote by $[l]_{eff}$ the effective length of the
(selected or unselected) string of length $l$ in $\tilde{\nu}^{(1)}$.

Case 1. $b=\framebox{$\emptyset$}$.

Add three boxes to the string of length zero in $\tilde{\nu}^{(1)}$ and add
one box to two strings of length zero in $\tilde{\nu}^{(2)}$.

Case 2. $b=\framebox{$2$}$.

Find the singular string of maximum length in $\tilde{\nu}^{(2)}$ and add one
box in this string. This corresponds to the box marking in this string in
$\nu^{(2)}$ by $[1]$. The augmented string is called the box-added
$i_{1}$-string. This terminology is generalized in an obvious manner.

Case 3. $b=\framebox{$3$}$.

Find the singular string of maximum length in $\tilde{\nu}^{(1)}$, add one box
in this string, and follow the algorithm of Case 2 such that the length of the
box-added $i_{1}$-string does not exceed the effective length of the box-added
$i_{2}$-string.

Case 4. $b=\framebox{$4$}$.

Find the singular/q-singular string of maximum length $l^{(1)}$ in
$\tilde{\nu}^{(1)}$. We choose the singular string when there exist singular
and q-singular strings of the same length.

If the $l^{(1)}$-string is singular, then add two boxes to this string and
follow the algorithm of Case 2. If the $l^{(1)}$-string is q-singular, then
add one box to this string and follow the algorithm of Case 3.

Case 5. $b=\framebox{$5$}$.

Find the singular string of maximum length $l^{(2)}$ in $\tilde{\nu}^{(2)}$
and add one box in this string. Then find singular/q-singular string of
maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$. We choose the singular string
when there exist singular and q-singular strings of the same length and do the
following cases.

  * (1)

The $l^{(1)}$-string is singular of length $3l^{(2)}+1$.

Add one box to the $l^{(1)}$-string and follow the algorithm, of Case 3. This
corresponds to the following box marking in $\nu^{(1)}$.

$[3]$$\downarrow$

with $i_{3}^{eff}=i_{4}$.

  * (2)

The $l^{(1)}$-string is singular of length $3l^{(2)}$ or $3l^{(2)}-1$.

Add two boxes to the $l^{(1)}$-string and follow the algorithm of Case 2. This
corresponds to the following box marking in $\nu^{(1)}$

$\downarrow$$[3]$$[2]$or$\downarrow$$[3]$$[2]$

with $i_{3}^{eff}=i_{4}$.

  * (3)

The $l^{(1)}$-string is q-singular of length $3l^{(2)}$.

Add one box to the founded string and follow the algorithm of Case 3. This
corresponds to the following box marking in $\nu^{(1)}$

$[3]$$\downarrow$

with $i_{3}^{eff}=i_{4}$.

If the box adding not successful, then follow the algorithm of Case 4. This
corresponds to the box marking in $\nu$ with $i_{3}^{eff}<i_{4}$.

Case 6. $b=\framebox{$6$}$.

Find the singular/q-singular/qq-singular string of maximum length $l^{(1)}$ in
$\tilde{\nu}^{(1)}$. We choose the singular string when there exists singular
and q/qq-singular strings of the same length. Similarly, we choose the
q-singular string when there exists q-singular and qq-singular strings of the
same length. That is, we choose the string of the same length by the following
preferential rule

singular | q-singular | qq-singular
---|---|---
1 | 2 | 3

In addition, we impose the following preferential rule for the
$l^{(1)}$-string. That is, if the $l^{(1)}$-string is qq-singular and there
exists a singular string of length $l^{(1)}-1$ in $\tilde{\nu}^{(1)}$, then
reset the $l^{(1)}$-string to be the $(l^{(1)}-1)$-string.

The box adding goes as follows. If the $l^{(1)}$-string is singular (resp.
q-singular), then add three (resp. two) boxes to the $l^{(1)}$-string and
follow the algorithm of Case 2 (resp. Case 3). If the $l^{(1)}$ is qq-
singular, then add one box to the $l^{(1)}$-string and follow the algorithm of
Case 4.

Case 7. $b=\framebox{$7$}$.

Find the singular string of maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$
and do as follows according to the type of the $l^{(1)}$-string.

  * (1)

type-0.

Find the singular string of length $[l^{(1)}]_{eff}$ in $\tilde{\nu}^{(2)}$.
If such a string exists, then add three boxes to the $l^{(1)}$-string in
$\tilde{\nu}^{(1)}$ and one box to the founded string in $\tilde{\nu}^{(2)}$
and follow the algorithm of Case 2. This corresponds to the following box
marking in $\nu^{(1)}$

$[5]$$[3]$$[2]$$\downarrow$

with $i_{5}^{eff}=i_{4}$.

  * (2)

type-I.

Find the singular string of length $[l^{(1)}]_{eff}-1$ in $\tilde{\nu}^{(2)}$.
If such a string exists, then add one box to the $l^{(1)}$-string in
$\tilde{\nu}^{(1)}$ and follow the algorithm of Case 5 where the length of the
box-added $i_{4}$-string in $\tilde{\nu}^{(2)}$ is $[l^{(1)}]_{eff}$.

  * (3)

type-II.

Find the singular string of length $[l^{(1)}]_{eff}-1$ in $\tilde{\nu}^{(2)}$.
If such a string exists, then add two boxes to the $l^{(1)}$-string in
$\tilde{\nu}^{(1)}$, one box to the founded string in $\tilde{\nu}^{(2)}$ and
follow the algorithm of Case 3. This corresponds to the following box marking
in $\nu^{(1)}$

$[5]$$[3]$$\downarrow$

with $i_{5}^{eff}=i_{4}$.

If the box adding is not successful, then add one box to the $l^{(1)}$-string
in $\tilde{\nu}^{(1)}$ and follow the algorithm of Case 5. This corresponds to
the box marking in $\nu$ with $i_{5}^{eff}>i_{4}$.

Case 8. $b=\framebox{$8$}$.

Find the singular string of maximum length $l^{(2)}$ in $\tilde{\nu}^{(2)}$
and add one box to this string. The resulting string is the box-added
$i_{5}$-string. Then, find the string of maximum length $l^{(1)}$ in
$\tilde{\nu}^{(1)}$ such that it is

  * (1)

a singular string of type-I/II (resp. type-0) with $[l^{(1)}]_{eff}=l^{(2)}+1$
(resp. $[l^{(1)}]_{eff}=l^{(2)}$) or

  * (2)

a q-singular string of type-II (resp. type-0) with $[l^{(1)}]_{eff}=l^{(2)}+1$
(resp. $[l^{(1)}]_{eff}=l^{(2)}$) or

  * (3)

a singular string of type-I/II with $[l^{(1)}]_{eff}=l^{(2)}$ or

  * (4)

a q-singular string of type-I with with $[l^{(1)}]_{eff}=l^{(2)}$ oir

  * (5)

a qq-singular string of type-0 with $[l^{(1)}]_{eff}=l^{(2)}$.

We choose the string of the same length by the following preferential rule

singular | q-singular | qq-singular
---|---|---
1 | 2 | 3

and do as follows according to the type of the $l^{(1)}$-string. All
correspond to the box marking in $\nu$ with $i_{4}^{eff}=i_{5}$.

  * (1)

type-0.

If the $l^{(1)}$-string is singular (resp. q-singular), then add three (resp.
two) boxes to this string and follow the algorithm of Case 2 (resp. Case 3).
If the $l^{(1)}$-string is qq-singular, then add one box to this string and
follow the algorithm of Case 4.

  * (2)

type-I.

If the $l^{(1)}$-string is singular with with $[l^{(1)}]_{eff}=l^{(2)}+1$,
then add one box to this string and follow the algorithm of Case 4 ignoring
the q-singular string of length $l^{(1)}-1$ in $\tilde{\nu}^{(1)}$. This is
because the following box marking in $\nu^{(1)}$

$\downarrow$$[3]$$[4]$

with $i_{4}^{eff}=i_{5}$ is forbidden. If the selected string is singular
(resp. q-singular) with $[l^{(1)}]_{eff}=l^{(2)}$, then add three (resp. two)
boxes to this string and follow the algorithm of Case 2 (resp. Case 3).

  * (3)

type-II.

If the $l^{(1)}$-string is singular (resp. q-singular) with
$[l^{(1)}]_{eff}=l^{(2)}+1$, then add two (resp. one) box(es) to this string
and follow the algorithm of Case 3 (resp. Case 4). If the $l^{(1)}$-string is
singular with $[l^{(1)}]_{eff}=l^{(2)}$, then add three boxes to this string
and follow the algorithm of Case 2.

If the box adding is not successful, then follow the algorithm of Case 6 such
that the effective length of the box-added $i_{4}$-string is smaller than
$l^{(2)}+1$, which corresponds to the box marking in $\nu$ with
$i_{4}^{eff}<i_{5}$.

Case 9. $b=\framebox{$9$}$.

Find the singular string of maximum length $l^{(2)}$ in $\tilde{\nu}^{(2)}$.
Then, find the singular/q-singular string of maximum length $l^{(1)}$ in
$\tilde{\nu}^{(1)}$. We choose the singular string when there exist singular
and q-singular strings of the same length. In addition, we impose the
following preferential rule for the $l^{(1)}$-string.

  * •

If the $l^{(1)}$-string in $\tilde{\nu}^{(1)}$ is type-I q-singular with
$[l^{(1)}]_{eff}=l^{(2)}+1$ and there exists a type-II singular string of the
same effective length in $\tilde{\nu}^{(1)}$, then reset the $l^{(1)}$-string
to be the founded type-II singular string.

  * •

If the $l^{(1)}$-string is type-I/II q-singular string with
$[l^{(1)}]_{eff}=l^{(2)}+1$ and there exists a type-0 singular string of
effective length $l^{(2)}$, then reset the $l^{(1)}$-string to be the founded
type-0 singular string.

The box adding goes as follows. If the $l^{(1)}$-string is q-singular, then
add one box to this string and follow the algorithm of Case 7. If the
$l^{(1)}$-string is singular, then do as follows according to the type of the
$l^{(1)}$-string.

  * (1)

type-0.

If $[l^{(1)}]_{eff}=l^{(2)}$, then add four boxes to this string, one box to
the $l^{(2)}$-string, and follow the algorithm of Case 2.

  * (2)

type-II.

If $[l^{(1)}]_{eff}=l^{(2)}+1$, then add three boxes to this string, one box
to the $l^{(2)}$-string, and follow the algorithm of Case 3.

  * (3)

type-I.

If $[l^{(1)}]_{eff}=l^{(2)}$, then add two boxes to this string, one box to
the $l^{(2)}$-string, and follow the algorithm of Case 4 ignoring the
q-singular string of length $l^{(1)}-1$ in $\tilde{\nu}^{(1)}$. This is due to
the rule of (BS-5).

The corresponding box marking in $\nu^{(1)}$ is one of the following three.

$\downarrow$$[6]$$[5]$$[3]$$[2]$or$\downarrow$$[6]$$[5]$$[3]$or$\downarrow$$[6]$$[5]$

where $i_{6}^{eff}=i_{4}$ in the left two and $i_{6}^{eff}=i_{4}+1$ in the
rightmost.

If the box adding is not successful, then add two boxes to the
$l^{(1)}$-string and follow the algorithm of Case 5.

Case 10. $b=\framebox{$10$}$.

Find the singular/q-singular string of maximum length $l^{(2)}$ in
$\tilde{\nu}^{(2)}$. We choose the singular string when there exist singular
and q-singular strings of the same length.

If the $l^{(2)}$-string is q-singular, then add one box to this string and
follow the algorithm of Case 8.

If the $l^{(2)}$-string is singular, then add two boxes to the
$l^{(2)}$-string and find the string of maximum length $l^{(1)}$ in
$\tilde{\nu}^{(1)}$ such that it is

  * (1)

a singular string of type-I/II with $[l^{(1)}]_{eff}=l^{(2)}+1$ or a singular
string (of any type) with $[l^{(1)}]_{eff}=l^{(2)}$ or

  * (2)

a q-singular string of type-II (resp. type-0/I) with
$[l^{(1)}]_{eff}=l^{(2)}+1$ (resp. $[l^{(1)}]_{eff}=l^{(2)}$) or

  * (3)

a qq-singular string of type-0 with $[l^{(1)}]_{eff}=l^{(2)}$.

In addition, we impose the following preferential rule for the
$l^{(1)}$-string.

  * •

If the $l^{(1)}$-string is type-II q-singular and there exists a type-0
singular string of effective length $[l^{(1)}]_{eff}-1$, then reset the
$l^{(1)}$-string to be the founded singular string.

  * •

If the $l^{(1)}$-string is qq-singular and there exists a type-I singular
string of the same effective length, then reset the $l^{(1)}$-string to be the
founded singular string.

The box adding goes as follows.

  * (1)

The $l^{(1)}$-string is singular.

Do the following according to the type of the $l^{(1)}$-string.

    * (a)

type-0.

Add three boxes to this string and follow the algorithm of Case 2.

    * (b)

type-I.

If $[l^{(1)}]_{eff}=l^{(2)}$, then add three boxes to this string and follow
the algorithm of Case 2. If $[l^{(1)}]_{eff}=l^{(2)}+1$, then add one box to
this string and follow the algorithm of Case 4.

    * (c)

type-II.

If $[l^{(1)}]_{eff}=l^{(2)}$, then add three boxes to this string and follow
the algorithm of Case 2. If $[l^{(1)}]_{eff}=l^{(2)}+1$, then add two boxes to
this string and follow the algorithm of Case 3.

  * (2)

The $l^{(1)}$-string is q-singular. If the $l^{(1)}$-string is type-0/I (resp.
type-II), then add two (resp. one) box(es) to this string and follow the
algorithm of Case 3 (resp. Case 4).

  * (3)

The $l^{(1)}$-string is qq-singular.

Add one box to this string and follow the algorithm of Case 4.

If the box adding is not successful, then follow the algorithm of Case 6 such
that the effective length of the box-added $i_{4}$-string does not exceed
$l^{(2)}$.

Case 11. $b=\framebox{$11$}$.

Find the singular string of maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$.
Find the singular string of maximum length $l_{s}^{(2)}$ in
$\tilde{\nu}^{(2)}$. Find the q-singular string of maximum length
$l_{q}^{(2)}$ satisfying $[l^{(1)}+1]_{eff}\geq l_{q}^{(2)}$. If such a
q-singular string does not exist, then set $l_{q}^{(2)}=0$. If
$l_{q}^{(2)}>l_{s}^{(2)}$, then add one box to the $l^{(1)}$-string in
$\tilde{\nu}^{(1)}$, add one box to the $l_{q}^{(2)}$-string in
$\tilde{\nu}^{(2)}$ and follow the algorithm of Case 8. This corresponds to
the following box marking in $\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$[7]$$[6]$

If $l_{s}^{(2)}>l_{q}^{(2)}$, then do as follows. If $[l^{(1)}+1]_{eff}\geq
l_{s}^{(2)}$, the add one box to the $l^{(1)}$-string in $\tilde{\nu}^{(1)}$,
add two boxes to the $l_{s}^{(2)}$ singular string in $\tilde{\nu}^{(2)}$ and
follow the algorithm of Case 6. This corresponds to the following box marking
in $\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$[7]$$[6]$$[5]$

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l^{(2)}$ in $\tilde{\nu}^{(2)}$. Find the singular
string of maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$. If the
$l^{(1)}$-string is type-0, $[l^{(1)}+4]_{eff}=l^{(2)}+2$, and there are no
q-singular strings of length $l$ ($l^{(1)}+1\leq l\leq l^{(1)}+3$), then add
four boxes to the $l^{(1)}$-string, add two boxes to the $l^{(2)}$-string, and
follow the algorithm of Case 2. This corresponds to the following box marking
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[6]$$[5]$$[3]$$[2]$$[7]$$[4]$

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Find the
singular string of maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$. If
$l^{(1)}$-string is type-0, $[l^{(1)}+4]_{eff}\leq l^{(2)}+1$, and there are
no q-singular strings of length $l$ ($l^{(1)}+1\leq l\leq 3l_{1}^{(2)}$), then
find the singular string of length $l_{2}^{(2)}$ in $\tilde{\nu}^{(2)}$
satisfying $l_{2}^{(2)}+1=[l^{(1)}+4]_{eff}-1$. If such a singular string
exists, then add four boxes to the $l^{(1)}$-string, add one box the
$l_{1}^{(2)}$\- and $l_{2}^{(2)}$-string and follow the algorithm of Case 2.
This corresponds to the following box marking in $\nu^{(1)}$ (left) and
$\nu^{(2)}$ (right)

$\downarrow$$[6]$$[5]$$[3]$$[2]$$[4]$$[7]$

where $i_{6}^{eff}=i_{4}+1\leq i_{7}$.

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Find the
singular string of maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$. If the
$l^{(1)}$-string is type-II, $[l^{(1)}+3]_{eff}\leq l_{1}^{(2)}+1$, and there
are no q-singular strings of length $l$ ($l^{(1)}+1\leq l\leq 3l_{1}^{(2)}$),
then find the singular string of length $l_{2}^{(2)}$ in $\tilde{\nu}^{(2)}$
satisfying $l_{2}^{(2)}+1=[l^{(1)}+3]_{eff}-1$. If such a singular string
exists, then add three boxes to the $l^{(1)}$-string, add one box the
$l_{1}^{(2)}$\- and $l_{2}^{(2)}$-string and follow the algorithm of Case 3.
This corresponds to the following box marking in $\nu^{(1)}$ (left) and
$\nu^{(2)}$ (right)

$\downarrow$$[6]$$[5]$$[3]$$[4]$$[7]$

where $i_{6}^{eff}=i_{4}+1\leq i_{7}$.

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l^{(2)}$ in $\tilde{\nu}^{(2)}$. Find the singular
string of maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$. If the
$l^{(1)}$-string is type-I and $[l^{(1)}+2]_{eff}=l^{(2)}+2$, then add two
boxes to the $l^{(1)}$\- and $l^{(2)}$-strings and follow the algorithm of
Case 4 ignoring the q-singular string of length $l^{(1)}-1$ in
$\tilde{\nu}^{(1)}$. The box adding in this case corresponds to the following
box marking in $\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[6]$$[5]$$[7]$$[4]$

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l^{(2)}$ in $\tilde{\nu}^{(2)}$. Find the singular
string of maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$. If the
$l^{(1)}$-string is type-I and $[l^{(1)}+2]_{eff}=l^{(2)}+1$, then add two
boxes to the $l^{(1)}$-string and one box to the $l^{(2)}$-strings and follow
the algorithm of Case 5. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right)

$\downarrow$$[6]$$[5]$$[7]$

where $i_{6}^{eff}=i_{7}$.

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l^{(2)}$ in $\tilde{\nu}^{(2)}$. Find the q-singular
string of maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$. If the
$l^{(1)}$-string is type-0 and $[l^{(1)}+1]_{eff}=l^{(2)}+1$, then add one box
to the $l^{(1)}$\- and the $l^{(2)}$-strings and follow the algorithm of Case
7. This corresponds to the following box marking in $\nu^{(1)}$ (left) and
$\nu^{(2)}$ (right)

$\downarrow$$[6]$$[7]$

where $i_{6}^{eff}=i_{7}$.

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l^{(2)}$ in $\tilde{\nu}^{(2)}$. Then, find the
string of maximum length $l^{(1)}$ such that the $l^{(1)}$-string is singular
with $[l^{(1)}+2]_{eff}<l^{(2)}+1$ or the $l^{(1)}$-string is q-singular with
$[l^{(1)}+1]_{eff}<l^{(2)}+1$. If the $l^{(1)}$-string is singular, then add
two boxes to the $l^{(1)}$-string and one box to the $l^{(2)}$-strings and
follow the algorithm of Case 5. If the $l^{(1)}$-string is q-singular, then
add one box to the $l^{(1)}$\- and the $l^{(2)}$-strings and follow the
algorithm of Case 7. The corresponding box marking in $\nu^{(1)}$ is

$[6]$$[5]$or$[6]$

with $i_{6}^{eff}<i_{7}$.

Case 12. $b=\framebox{$12$}$.

Find the singular/q-singular string of maximum length $l^{(1)}$ in
$\tilde{\nu}^{(1)}$. We choose the singular string when there exist singular
and q-singular strings of the same length. In addition, we impose the
following preferential rules.

  * •

If the $l^{(1)}$-string is q-singular (of any type), then find the type-0
singular string of type-0 of length $l^{(1)\prime}$ whose effective length
$[l^{(1)}]_{eff}-1$ in $\tilde{\nu}^{(1)}$ such that there exists a singular
string of length $l^{(2)}$ satisfying $[l^{(1)\prime}+5]_{eff}=l^{(2)}+2$. If
such an $l^{(1)\prime}$-string exists, then reset the $l^{(1)}$-string to be
$l^{(1)\prime}$-string.

  * •

If the $l^{(1)}$-string is q-singular of type I, then find the type-II of
length $l^{(1)\prime}=l^{(1)}-1$ such that there exists a singular string of
length $l^{(2)}$ satisfying $[l^{(1)\prime}+4]_{eff}=l^{(2)}+2$. If such an
$l^{(1)\prime}$-string exists, then reset the $l^{(1)}$-string to be
$l^{(1)\prime}$-string.

The box adding goes as follows.

If the $l^{(1)}$-string is singular of type-0, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+5]_{eff}=l^{(2)}+2$. If such a string exists, then add five boxes to
the singular $l^{(1)}$-string, two boxes to the $l^{(2)}$-string, and follow
the algorithm of Case 2. This corresponds to the following box marking
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[8]$$[7]$$[4]$$[3]$$[2]$$[6]$$[5]$

If the $l^{(1)}$-string is singular of type-II, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+4]_{eff}=l^{(2)}+2$. If such a string exists, then add four boxes to
the singular $l^{(1)}$-string, two boxes to the $l^{(2)}$-string, and follow
the algorithm of Case 3. This corresponds to the following box marking
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[8]$$[7]$$[4]$$[3]$$[6]$$[5]$

If the $l^{(1)}$-string is singular of type-I, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+3]_{eff}=l^{(2)}+1$. If such a string exists, then add three boxes
to the singular $l^{(1)}$-string, one box to the $l^{(2)}$-string, and follow
the algorithm of Case 5. This corresponds to the following box marking
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[8]$$[6]$$[5]$$[7]$

If the $l^{(1)}$-string is singular of type-I, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+3]_{eff}=l^{(2)}+2$. If such a string exists, then add three boxes
to the singular $l^{(1)}$-string, two boxes to the $l^{(2)}$-string, and
follow the algorithm of Case 4 ignoring the q-singular string of length
$l^{(1)}-1$ in $\tilde{\nu}^{(1)}$. This corresponds to the following box
marking $\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[8]$$[6]$$[5]$$[7]$$[4]$

If the $l^{(1)}$-string is singular of type-0/II, then find a singular (resp.
q-singular) string of length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+2]_{eff}=l^{(2)}+2$ (resp. $[l^{(1)}+2]_{eff}=l^{(2)}+1$). If such a
string exists, then add two boxes to the $l^{(1)}$-string and follow the
algorithm of Case 10 such that $i_{8}^{eff}=i_{6}$. This corresponds to the
following box marking in $\nu^{(1)}$.

$[8]$$[7]$

If the $l^{(1)}$-string is q-singular of type-0, then find a singular string
of length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+2]_{eff}=l^{(2)}+1$. If such a string exists, then add two boxes to
the $l^{(1)}$-string, one box to the $l^{(2)}$-string, and follow the
algorithm of Case 7. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[8]$$[6]$$[7]$

If the $l^{(1)}$-string is q-singular of type-I/II, then find a singular
string of length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+2]_{eff}=l^{(2)}+2$ or $[l^{(1)}+2]_{eff}=l^{(2)}+1$. If such a
string exists, then add two boxes to the $l^{(1)}$-string and follow the
algorithm of Case 11. This corresponds to the following box marking in
$\nu^{(2)}$.

$[7]$$[4]$or$[7]$

with $i_{8}^{eff}=i_{7}$ and the $i_{8}$-string in $\nu^{(1)}$ is marked by
[8] only.

If the box adding so far is not successful, then do as follows. If the
$l^{(1)}$-string is singular, then add two boxes to the $l^{(1)}$-string and
follow the algorithm of Case 10 such that $i_{8}^{eff}>i_{6}$. If the
$l^{(1)}$-string is q-singular, then add one box to the $l^{(1)}$-string and
follow the algorithm of Case 11.

Case 13. $b=\framebox{$13$}$.

Find the singular/q-singular/qq-singular string of maximum length $l^{(1)}$ in
$\tilde{\nu}^{(1)}$ We choose the string of the same length by the following
preferential rule

singular | q-singular | qq-singular
---|---|---
1 | 2 | 3

In addition, we impose the following preferential rules.

  * •

If the $l^{(1)}$-string is q-singular/qq-singular of any type, then find a
type-0 singular string of length $l^{(1)\prime}$ in $\tilde{\nu}^{(1)}$ with
$[l^{(1)\prime}]_{eff}=[l^{(1)}]_{eff}-1$ such that there exists a singular
string of length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)\prime}+6]_{eff}=l^{(2)}+2$. If such an $l^{(1)\prime}$-string exists,
then reset the $l^{(1)}$-string to be the $l^{(1)\prime}$-string.

  * •

If the $l^{(1)}$-string is qq-singular of type-II, then find a singular string
of length $l^{(1)\prime}=l^{(1)}-4$ in $\tilde{\nu}^{(1)}$ such that there
exists a singular string of length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)\prime}+6]_{eff}=l^{(2)}+2$. If such an $l^{(1)\prime}$-string exists,
then reset the $l^{(1)}$-string to be the $l^{(1)\prime}$-string.

  * •

If the $l^{(1)}$-string is q-singular of type-I, then find a singular string
of length $l^{(1)\prime}=l^{(1)}-1$ in $\tilde{\nu}^{(1)}$ such that there
exists a singular string of length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)\prime}+5]_{eff}=l^{(2)}+2$. If such an $l^{(1)\prime}$-string exists,
then reset the $l^{(1)}$-string to be the $l^{(1)\prime}$-string.

The box adding goes as follows.

If the $l^{(1)}$-string is qq-singular, then add one box to the
$l^{(1)}$-string and follow the algorithm of Case 12.

If the $l^{(1)}$-string is singular of type-0, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+6]_{eff}=l^{(2)}+2$. If such a string exists, then add six boxes to
the $l^{(1)}$-string, two boxes to the $l^{(2)}$-string, and follow the
algorithm of Case 2. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[7]$$[4]$$[3]$$[2]$$[6]$$[5]$

If the $l^{(1)}$-string is singular of type-II, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+5]_{eff}=l^{(2)}+2$. If such a string exists, then add five boxes to
the $l^{(1)}$-string, two boxes to the $l^{(2)}$-string, and follow the
algorithm of Case 3. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[7]$$[4]$$[3]$$[6]$$[5]$

If the $l^{(1)}$-string is singular of type-I, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+4]_{eff}=l^{(2)}+1$. If such a string exists, then add four boxes to
the $l^{(1)}$-string, one box to the $l^{(2)}$-string, and follow the
algorithm of Case 5. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[6]$$[5]$$[7]$

If the $l^{(1)}$-string is singular of type-I, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+4]_{eff}=l^{(2)}+2$ . If such a string exists, then add four boxes
to the $l^{(1)}$-string, two boxes to the $l^{(2)}$-string, and follow the
algorithm of Case 4 ignoring the q-singular string of length $l^{(1)}-1$ in
$\tilde{\nu}^{(1)}$. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[6]$$[5]$$[7]$$[4]$

If the box adding is not successful, then do as follows. If the
$l^{(1)}$-string is singular of type-0, then find a singular string of length
$l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying $[l^{(1)}+3]_{eff}=l^{(2)}+2$ or a
q-singular string of length $l^{(2)}$ satisfying $[l^{(1)}+3]_{eff}=l^{(2)}+1$
. If such a string exists, then add three boxes to the $l^{(1)}$-string, and
follow the algorithm of Case 10. This corresponds to the following box marking
in $\nu^{(1)}$ (left) and $\nu^{(2)}$ (right)

$\downarrow$$[9]$$[8]$$[7]$$x$

where $x$ is $[6]$$[5]$ or $[6]$ with $i_{9}^{eff}=i_{6}$.

If the box adding is not successful, then do as follows. If the
$l^{(1)}$-string is singular of any type, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying $[l^{(1)}+3]_{eff}-1\geq
l^{(2)}+2$ or a q-singular string of length $l^{(2)}$ satisfying
$[l^{(1)}+3]_{eff}-1\geq l^{(2)}+1$. If such a string exists, then add three
boxes to the $l^{(1)}$-string, and follow the algorithm of Case 10. This
corresponds to the box marking $\nu^{(1)}$ and $\nu^{(2)}$ depicted above but
with $i_{9}^{eff}>i_{6}$. If the $l^{(1)}$-string is q-singular of type-0,
then find a singular string of length $l^{(2)}$ in $\tilde{\nu}^{(2)}$
satisfying $[l^{(1)}+3]_{eff}=l^{(2)}+1$. If such a string exists, then add
three boxes to the $l^{(1)}$-string, one box to the $l^{(2)}$-string, and
follow the algorithm of Case 7. This corresponds to the following box marking
in $\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[6]$$[7]$

If the $l^{(1)}$-string is q-singular of type-II, then find a singular string
of length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)}+2]_{eff}=l^{(2)}+1$ or $[l^{(1)}+2]_{eff}=l^{(2)}+2$. If such a
string exists, then add two boxes to the $l^{(1)}$-string and follow the
algorithm of Case 11 such that the length of the box-added $i_{7}$-string is
equal to $[l^{(1)}+2]_{eff}$. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right)

$\downarrow$$[9]$$[8]$$x$

where $x$ is $[7]$$[4]$ or $[7]$ with $i_{9}^{eff}=i_{7}$.

If the box adding so far is not successful, then do as follows. If the
$l^{(1)}$-string is q-singular of any type, then find a singular string of
length $l^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying $[l^{(1)}+2]_{eff}-1\geq
l^{(2)}+1$. If such a string exists, then add two boxes to the
$l^{(1)}$-string and follow the algorithm of Case 11 such that the length of
the box-added $i_{7}$-string is smaller than $[l^{(1)}+2]_{eff}$.

Case 14. $b=\framebox{$14$}$.

Find two longest singular strings of length $l_{1}^{(2)}$ and
$l_{2}^{(2)}(\leq l_{1}^{(2)})$ in $\tilde{\nu}^{(2)}$. If
$l_{1}^{(2)}=l_{2}^{(2)}$, then find the singular string of length $l^{(1)}$
in $\tilde{\nu}^{(1)}$ satisfying $[l^{(1)}+6]_{eff}=l_{1}^{(2)}$. If such a
string exists, then add six boxes to the $l^{(1)}$-string, two boxes to the
$l_{1}^{(2)}$\- and $l_{2}^{(2)}$-strings, and terminate the box adding. This
corresponds to the following box marking in $\nu^{(1)}$ (left) and $\nu^{(2)}$
(right).

$\downarrow$$[9]$$[8]$$[7]$$[4]$$[3]$$[2]$$[10]$$[6]$$[1]$$[5]$

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the
singular string of length $l_{2}^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$l_{2}^{(2)}=l_{1}^{(2)}-1$ and the type-0 singular string of length $l^{(1)}$
in $\tilde{\nu}^{(1)}$ satisfying $[l^{(1)}+6]_{eff}=l_{1}^{(2)}+1$. If such
strings exist, then add six boxes to the $l^{(1)}$-string, one box to the
$l_{1}^{(2)}$-string, two boxes to the $l_{2}^{(2)}$-string, and follow the
algorithm of Case 2. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[7]$$[4]$$[3]$$[2]$$[6]$$[5]$$[10]$

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the
singular string of length $l_{2}^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$l_{2}^{(2)}=l_{1}^{(2)}-1$ and the type-II singular string of length
$l^{(1)}$ in $\tilde{\nu}^{(1)}$ satisfying $[l^{(1)}+5]_{eff}=l_{1}^{(2)}+1$.
If such strings exist, then add five boxes to the $l^{(1)}$-string, one box to
the $l_{1}^{(2)}$-string, two boxes to the $l_{2}^{(2)}$-string, and follow
the algorithm of Case 3. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[7]$$[4]$$[3]$$[6]$$[5]$$[10]$

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the
singular string of length $l_{2}^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$l_{2}^{(2)}=l_{1}^{(2)}-1$ and the type-I singular string of length $l^{(1)}$
in $\tilde{\nu}^{(1)}$ satisfying $[l^{(1)}+4]_{eff}=l_{1}^{(2)}+1$. If such
strings exist, then add four boxes to the $l^{(1)}$-string, one box to the
$l_{1}^{(2)}$-string, two boxes to the $l_{2}^{(2)}$-string, and follow the
algorithm of Case 4 ignoring the q-singular string of length $l^{(1)}-1$ in
$\tilde{\nu}^{(1)}$. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[6]$$[5]$$[7]$$[4]$$[10]$

In this case, if there exists a type-II singular string of length
$l^{(1)\prime}=l^{(1)}-1$ in $\tilde{\nu}^{(1)}$, then discard the box adding
above, reset the $l^{(1)}$-string to be the $l^{(1)\prime}$-string, and follow
the previous box marking.

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the
singular string of length $l_{2}^{(2)}$ in $\tilde{\nu}^{(2)}$ satisfying
$l_{2}^{(2)}=l_{1}^{(2)}$ and the type-I singular string of length $l^{(1)}$
in $\tilde{\nu}^{(1)}$ satisfying $[l^{(1)}+4]_{eff}=l_{1}^{(2)}+1$. If such
strings exist, then add four boxes to the $l^{(1)}$-string, one box to the
$l_{1}^{(2)}$\- and $l_{2}^{(2)}$-strings, and follow the algorithm of Case 5.
This corresponds to the following box marking in $\nu^{(1)}$ (left) and
$\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[6]$$[5]$$[7]$$[10]$

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the
singular string of length $l^{(1)}$ in $\tilde{\nu}^{(1)}$ satisfying
$[l^{(1)}+3]_{eff}=l_{1}^{(2)}+1$. If such strings exist, then add three boxes
to the $l^{(1)}$-string, one box to the $l_{1}^{(2)}$-string, and follow the
algorithm of Case 10. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right)

$[9]$$[8]$$[7]$$[10]$

with $i_{9}^{eff}=i_{10}$. When the box-added $i_{9}$-string is type-0, the
length of the box-added $i_{6}$-string is smaller than or equal to
$i_{9}^{eff}$ and when the box-added $i_{9}$-string is type-I/II, the length
of the box-added $i_{6}$-string is strictly smaller than $i_{9}^{eff}$.

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the
type-II singular string of length $l^{(1)}$ in $\tilde{\nu}^{(1)}$ satisfying
$[l^{(1)}+2]_{eff}=l_{1}^{(2)}+1$. If such strings exist, then add two boxes
to the $l^{(1)}$-string, one box to the $l_{1}^{(2)}$-string, and follow the
algorithm of Case 11. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right)

$\downarrow$$[9]$$[8]$$[10]$

with $i_{9}^{eff}=i_{10}$.

If the box adding is not successful, then do as follows. Find the singular
string of maximum length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the
type-I singular string of length $l^{(1)}$ in $\tilde{\nu}^{(1)}$ satisfying
$[l^{(1)}+1]_{eff}=l_{1}^{(2)}+1$. If such strings exist, then add one boxes
to the $l^{(1)}$-string, one box to the $l_{1}^{(2)}$-string, and follow the
algorithm of Case 12. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right)

$\downarrow$$[9]$$[10]$

with $i_{9}^{eff}=i_{10}$. In this case, if there exists a type-II singular
string of length $l^{(1)\prime}=l^{(1)}-1$ in $\tilde{\nu}^{(1)}$, then
discard the box adding above, reset the $l^{(1)}$-string to be the
$l^{(1)\prime}$-string, and follow the previous box marking.

If the box adding so far is not successful, do as follows. Let $l_{1}^{(2)}$
and $l_{2}^{(2)}(\leq l_{1}^{(2)})$ be the two largest length of singular
strings in $\tilde{\nu}^{(2)}$. Find the q-singular/qq-singular string of
maximum length $l^{(1)}$ in $\tilde{\nu}^{(1)}$ such that it is

  * (1)

a q-singular string of type-0 with
$[l^{(1)}+3]_{eff}=l_{1}^{(2)}+1=l_{2}^{(2)}+1$ or

  * (2)

a q-singular string of type-0/I with $[l^{(1)}+2]_{eff}=l_{1}^{(2)}+1$ or

  * (3)

a q-singular string of type-II with $[l^{(1)}+1]_{eff}=l_{1}^{(2)}+1$ or

  * (4)

a qq-singular string of type-0 with $[l^{(1)}+1]_{eff}=l_{1}^{(2)}+1$.

We choose the q-singular string when there exist q-singular and qq-singular
strings of the same length. In addition, we impose the following preferential
rules.

  * •

If the $l^{(1)}$-string is q-singular/qq-singular of type-0 or q-singular of
type-I, then find a type-0 singular string of length $l^{(1)\prime}$ in
$\tilde{\nu}^{(1)}$ with $[l^{(1)\prime}]_{eff}=[l^{(1)}]_{eff}-1$ such that
there exist singular strings of length $l_{1}^{(2)}$ and $l_{2}^{(2)}$ in
$\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)\prime}+6]_{eff}=l_{1}^{(2)}+1=l_{2}^{(2)}+2$. If such an
$l^{(1)\prime}$-string exists, then reset the $l^{(1)}$-string to be the
$l^{(1)\prime}$-string.

  * •

If the $l^{(1)}$-string is q-singular of type-II, then find a string of length
$l^{(1)\prime}=l^{(1)}-4$ in $\tilde{\nu}^{(1)}$ such that there exist
singular strings of length $l_{1}^{(2)}$ and $l_{2}^{(2)}$ in
$\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)\prime}+6]_{eff}=l_{1}^{(2)}+1=l_{2}^{(2)}+2$. If such an
$l^{(1)\prime}$-string exists, then reset the $l^{(1)}$-string to be the
$l^{(1)\prime}$-string.

  * •

If the $l^{(1)}$-string is q-singular of type-0, then find a string of length
$l^{(1)\prime}=l^{(1)}-2$ in $\tilde{\nu}^{(1)}$ such that there exist
singular strings of length $l_{1}^{(2)}$ and $l_{2}^{(2)}$ in
$\tilde{\nu}^{(2)}$ satisfying
$[l^{(1)\prime}+5]_{eff}=l_{1}^{(2)}+1=l_{2}^{(2)}+2$. If such an
$l^{(1)\prime}$-string exists, then reset the $l^{(1)}$-string to be the
$l^{(1)\prime}$-string.

If the box adding is not successful, then do as follows. If the
$l^{(1)}$-string is singular, then do as before.

Find two longest singular strings of length $l_{1}^{(2)}$ and $l_{2}^{(2)}$.
If $l_{1}^{(2)}=l_{2}^{(2)}$, then find the type-0 q-singular string of length
$l^{(1)}(\geq 3)$ in $\tilde{\nu}^{(1)}$ satisfying
$[l^{(1)}+3]_{eff}=l_{1}^{(2)}+1$. If such a string exists, then add three
boxes to the $l^{(1)}$-string, one box to the $l_{1}^{(2)}$\- and
$l_{2}^{(2)}$-strings, and follow the algorithm of Case 7. This corresponds to
the following box marking in $\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[8]$$[6]$$[7]$$[10]$

If the box adding is not successful, then do as follows. Find the singular
strings of length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the type-0/I
q-singular string of length $l^{(1)}$ in $\tilde{\nu}^{(1)}$ satisfying
$[l^{(1)}+2]_{eff}=l_{1}^{(2)}+1$. If such a string exists, then add two boxes
to the $l^{(1)}$-string, one box to the $l_{1}^{(2)}$-string, and follow the
algorithm of Case 11. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$[9]$$[8]$$[10]$

with $i_{9}^{eff}=i_{10}$.

If the box adding is not successful, then do as follows. Find the singular
strings of length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the type-II
q-singular string of length $l^{(1)}$ in $\tilde{\nu}^{(1)}$ satisfying
$[l^{(1)}+1]_{eff}=l_{1}^{(2)}+1$. If such a string exists, then add one box
to the $l^{(1)}$-string, one box to the $l_{1}^{(2)}$-string, and follow the
algorithm of Case 12. This corresponds to the following box marking in
$\nu^{(1)}$ (left) and $\nu^{(2)}$ (right).

$\downarrow$$[9]$$[10]$

with $i_{9}^{eff}=i_{10}$.

If the box adding is not successful, then do as follows. Find the singular
strings of length $l_{1}^{(2)}$ in $\tilde{\nu}^{(2)}$. Then find the type-0
qq-singular string of length $l^{(1)}(\geq 4)$ in $\tilde{\nu}^{(1)}$
satisfying $[l^{(1)}+1]_{eff}=l_{1}^{(2)}+1$. If such a string exists, then
add one box to the $l^{(1)}$-string, one box to the $l_{1}^{(2)}$-string, and
follow the algorithm of Case 12. This corresponds to the following box marking
in $\nu^{(1)}$ (left) and $\nu^{(2)}$ (right)

$\downarrow$$[9]$$[10]$

with $i_{9}^{eff}=i_{10}$.

If the box adding is not successful, then do as follows. Find the singular
string of maximum length in $\tilde{\nu}^{(2)}$. Then add one box to this
string and follow the algorithm of Case 13. This corresponds to the box
marking in $\nu$ depicted above but with $i_{9}^{eff}<i_{10}$.

##  5\. Proof of Theorem 3.1

Theorem 3.1 is proved in this section. The following notation is used. Let
$(\nu,J)\in\mathrm{RC}(\lambda,L)$, $b=\gamma(\nu,J)$,
$\rho=\lambda-\mathrm{wt}(b)$, and
$(\tilde{\nu},\tilde{J})=\delta_{\theta}(\nu,J)$. For
$(\nu,J)\in\mathrm{RC}(\lambda,L)$, we define $\Delta
c(\nu,J)=c(\nu,J)-c(\delta_{\theta}(\nu,J))$. The following lemma is
essentially the same as [26, Lemma 5.1].

######  Lemma 5.1.

To prove that Eq. (3.7) holds, it suffices to show that it holds for $L=1$ and
that for $L\geq 2$ with $\Phi(\nu,J)=b_{1}\otimes\cdots\otimes b_{L}$, we have

(5.1) $\Delta c(\nu,J)=-\alpha_{1}^{(2)}+\chi(b_{1}=\emptyset)$

and

(5.2) $H(b_{1}\otimes
b_{2})=\tilde{\alpha}_{1}^{(2)}-\alpha_{1}^{(2)}+\chi(b_{1}=\emptyset)-\chi(b_{2}=\emptyset),$

where $\alpha_{1}^{(2)}$ and $\tilde{\alpha}_{1}^{(2)}$ are the lengths of the
first columns in $\nu^{(2)}$ and $\tilde{\nu}^{(2)}$ respectively, and
$\delta_{\theta}(\nu,J)=(\tilde{\nu},\tilde{J})$.

There are five things that must be verified:

  * (I)

$\rho$ is dominant.

  * (II)

$(\tilde{\nu},\tilde{J})\in\mathrm{RC}(\rho,L-1)$.

  * (III)

$b$ can be appended to $(\tilde{\nu},\tilde{J})$.

  * (IV)

(5.1) in Lemma 5.1 holds.

  * (V)

(5.2) in Lemma 5.1 holds.

Parts (I) and (II) show that $\delta_{\theta}$ is well-defined. Part (III)
shows that $\delta_{\theta}$ has an inverse. Parts (IV) and (V) suffice to
prove that $\Phi$ preserves statistics. We omit the proof of (III) as it is
very similar to the proof of well-definedness.

We need several preliminary lemmas on the convexity and nonnegativity of the
vacancy numbers $p_{i}^{(a)}$.

######  Lemma 5.2.

For large $i$, we have

$p_{i}^{(a)}=\lambda_{a}$

where $\lambda_{a}$ is defined by $\lambda=\sum_{a\in
I_{0}}\lambda_{a}\bar{\Lambda}_{a}$.

###### Proof.

This follows from the formula for the vacancy number Eq. (3.2) and the
constraint Eq. (3.1). ∎

Direct calculations show that

(5.3) $\displaystyle-p_{3i-2}^{(1)}+2p_{3i-1}^{(1)}-p_{3i}^{(1)}$
$\displaystyle=-2m_{3i-1}^{(1)},$ (5.4) $\displaystyle-
p_{3i-1}^{(1)}+2p_{3i}^{(1)}-p_{3i+1}^{(1)}$
$\displaystyle=-2m_{3i}^{(1)}+m_{i}^{(2)},$ (5.5) $\displaystyle-
p_{3i}^{(1)}+2p_{3i+1}^{(1)}-p_{3i+2}^{(1)}$ $\displaystyle=-2m_{3i+1}^{(1)},$

and

(5.6) $\displaystyle-p_{i-1}^{(2)}+2p_{i}^{(2)}-p_{i+1}^{(2)}=$ $\displaystyle
3m_{3i}^{(1)}+2(m_{3i-1}^{(1)}+m_{3i+1}^{(1)})+m_{3i-2}^{(1)}+m_{3i+2}^{(1)}$
$\displaystyle-$ $\displaystyle 2m_{i}^{(2)}.$

In particular, these equations imply the convexity condition

(5.7) $p_{i}^{(a)}\geq\frac{1}{2}(p_{i-1}^{(a)}+p_{i+1}^{(a)})\quad\text{if
}m_{i}^{(a)}=0.$

The following two lemmas follow immediately from Lemma 5.2 and the convexity
condition Eq. (5.7).

######  Lemma 5.3.

Let $\nu$ be a configuration and let $n$ be $0$ or $1$. The following are
equivalent:

  * (1)

$p_{i}^{(a)}\geq n$ for all $i\in\mathbb{Z}_{>0}$, $a\in I_{0}$;

  * (2)

$p_{i}^{(a)}\geq n$ for all $i\in\mathbb{Z}_{>0}$, $a\in I_{0}$ such that
$m_{i}^{(a)}>0$.

######  Lemma 5.4.

Let $\nu$ be a configuration and let $n$ be $2$ or $3$. Let
$p_{i_{0}}^{(a)}\geq n$. The following are equivalent:

  * (1)

$p_{i}^{(a)}\geq n$ for all $i\geq i_{0}$, $a\in I_{0}$;

  * (2)

$p_{i}^{(a)}\geq n$ for all $i\geq i_{0}$, $a\in I_{0}$ such that
$m_{i}^{(a)}>0$.

######  Lemma 5.5.

Suppose that

$p_{i}^{(a)}\geq 1,\ldots,p_{j}^{(a)}\geq 1\quad(i<j)$

with $m_{i}^{(a)}=\cdots=m_{j}^{(a)}=0$. If $p_{k}^{(a)}=1$ for some $k$
$(i\leq k\leq j)$, then

$p_{i}^{(a)}=1,\ldots,p_{j}^{(a)}=1\quad(i<j).$

###### Proof.

This follows immediately from the convexity condition Eq. (5.7). ∎

###### Proof of (I).

Here we show $\rho=\lambda-\mathrm{wt}(b)$ is dominant with $l\geq 1$. Let us
assume that $\nu^{(1)}\neq\emptyset$ and $\nu^{(2)}\neq\emptyset$ and let
$l^{(a)}$ be the largest part in $\nu^{(a)}$ $(a=1,2)$. In particular,
$m_{l^{(1)}}^{(1)}\neq 0$ and $m_{l^{(2)}}^{(2)}\neq 0$. The proof for the
case when $\nu^{(1)}=\emptyset$ or $\nu^{(2)}=\emptyset$ is much simpler and
we omit the details.

Case 0. $b=\framebox{$\emptyset$}$.

Since $\mathrm{wt}(\framebox{$\emptyset$})=0$, there is nothing to prove.

Case 1. $b=\framebox{$1$}$.

Since $\mathrm{wt}(\framebox{$1$})=\bar{\Lambda}_{2}$, we must show that
$\lambda_{2}\geq 1$. Suppose that $\lambda_{2}=0$ so that
$p_{\infty}^{(2)}=0$. The convexity condition implies $p_{l^{(2)}}^{(2)}=0$ so
that $\nu^{(2)}$ has a singular string of length $l^{(2)}$, which contradicts
$b=\framebox{$1$}$. Hence $\lambda_{2}\geq 1$.

Case 2. $b=\framebox{$2$}$.

Since $\mathrm{wt}(\framebox{$2$})=3\bar{\Lambda}_{1}-\bar{\Lambda}_{2}$, we
must show that $\lambda_{1}\geq 3$. We first suppose that

$\lambda_{1}=-2\sum_{j=1}^{l^{(1)}}jm_{j}^{(1)}+3\sum_{j=1}^{l^{(2)}}jm_{j}^{(2)}=0.$

Then

$p_{l^{(1)}}^{(1)}=-2\sum_{j=1}^{l^{(1)}}jm_{j}^{(1)}+\sum_{j=1}^{l^{(2)}}\min(l^{(1)},3j)m_{j}^{(2)}=\sum_{j=1}^{l^{(2)}}\left(\min(l^{(1)},3j)-3j\right)m_{j}^{(2)}.$

If $l^{(1)}<3l^{(2)}$, then $p_{l^{(1)}}^{(1)}<0$. Hence $l^{(1)}\geq
3l^{(2)}$, which yields $p_{l^{(1)}}^{(1)}=0$. This implies that $\nu^{(1)}$
has a singular string of effective length $(\geq l^{(2)})$, which contradicts
$b=\framebox{$2$}$. Secondly, we suppose that $\lambda_{1}=1$. Then

$p_{l^{(1)}}^{(1)}=1+\sum_{j=1}^{l^{(2)}}\left(\min(l^{(1)},3j)-3j\right)m_{j}^{(2)}.$

If $m_{l^{(2)}}^{(2)}=1$, then $p_{l^{(1)}}^{(1)}<0$ when $l^{(1)}\leq
3l^{(2)}-2$. If $m_{l^{(2)}}^{(2)}\geq 2$, then $p_{l^{(1)}}^{(1)}<0$ when
$l^{(1)}\leq 3l^{(2)}-1$. Hence $l^{(1)}\geq 3l^{(2)}-1$ if
$m_{l^{(2)}}^{(2)}=1$ and $l^{(1)}\geq 3l^{(2)}$ if $m_{l^{(2)}}^{(2)}\geq 2$.
Simple calculations yield two possible cases:

  * (1)

$m_{l^{(2)}}^{(2)}=1$, $l^{(1)}=3l^{(2)}-1$, and $p_{l^{(1)}}^{(1)}=0$.

  * (2)

$m_{l^{(2)}}^{(2)}\geq 1$, $l^{(1)}\geq 3l^{(2)}$, and $p_{l^{(1)}}^{(1)}=1$.

Case (1) contradicts $b=\framebox{$2$}$ because $\nu^{(1)}$ has a type-I
singular string of effective length $l^{(2)}$. Case (2) also contradicts
$b=\framebox{$2$}$ because $\nu^{(1)}$ has a string (singular or q-singular)
of length $(\geq 3l^{(2)})$. Hence $\lambda_{1}\geq 2$. Thirdly, we suppose
that $\lambda_{1}=2$. By the similar calculations as above yield four possible
cases:

  * (1)

$m_{l^{(2)}}^{(2)}=1$, $l^{(1)}=3l^{(2)}-2$, and $p_{l^{(1)}}^{(1)}=0$.

  * (2)

$m_{l^{(2)}}^{(2)}=1$, $l^{(1)}=3l^{(2)}-1$, and $p_{l^{(1)}}^{(1)}=1$.

  * (3)

$m_{l^{(2)}}^{(2)}=2$, $l^{(1)}=3l^{(2)}-1$, and $p_{l^{(1)}}^{(1)}=0$.

  * (4)

$m_{l^{(2)}}^{(2)}\geq 1$, $l^{(1)}\geq 3l^{(2)}$, and $p_{l^{(1)}}^{(1)}=2$.

Case (1) (resp. (3)) implies that $\nu^{(1)}$ has a type-II (resp. I) singular
string of effective length $l^{(2)}$. Case (2) implies that $\nu^{(1)}$ has a
type-I string (singular or q-singular) of effective length $l^{(2)}$. Case (4)
implies that $\nu^{(1)}$ has a string (singular or q-singular or qq-singular)
of length $(\geq 3l^{(2)})$. All contradict $b=\framebox{$2$}$ by the
algorithm of $\delta_{\theta}$. Hence, $\lambda_{1}\geq 3$.

Case 3. $b=\framebox{$3$}$.

In this case, after $[2]$ was marked in $\nu^{(1)}$ the box marking has
terminated. Since $\mathrm{wt}(\framebox{$3$})=\bar{\Lambda}_{1}$, we must
show that $\lambda_{1}\geq 1$. Suppose that $\lambda_{1}=0$. As in Case 2, it
follows that $\nu^{(1)}$ has a singular string of length ($\leq 3i_{1}$)
contradicting $b=\framebox{$3$}$. Hence $\lambda_{1}\geq 1$. Note that even if
[2] was marked in this singular string, [3] can also be marked in this string.

Case 4. $b=\framebox{$4$}$.

In this case, after $[3]$ was marked in $\nu^{(1)}$ the box marking has
terminated. Since
$\mathrm{wt}(\framebox{$4$})=-\bar{\Lambda}_{1}+\bar{\Lambda}_{2}$, we must
show that $\lambda_{2}\geq 1$. We first suppose that

$\lambda_{2}=L+\sum_{j=1}^{l^{(1)}}jm_{j}^{(1)}-2\sum_{j=1}^{l^{(2)}}jm_{j}^{(2)}=0.$

Then we have $p_{l^{(2)}+1}^{(2)}$ by convexity and

$p_{l^{(2)}}^{(2)}=\sum_{j=1}^{l^{(1)}}\left(\min(3l^{(2)},j)-j\right)m_{j}^{(1)}.$

If $l^{(1)}>3l^{(2)}$, then $p_{l^{(2)}}^{(2)}<0$. Hence $l^{(1)}\leq
3l^{(2)}$, which yields $p_{l^{(2)}}^{(2)}=0$ so that all $l^{(2)}$-strings in
$\nu^{(2)}$ are singular. If [3] is marked in a string of of effective length
$(<l^{(2)})$, then [4] can be marked in a singular string of length $l^{(2)}$
in $\nu^{(2)}$. This is a contradiction. Note that [1] must have been marked
in a singular string of length $(<l^{(2)})$ in this case. Therefore, it
suffices to consider the case when [3] had been marked in a string of
effective length $l^{(2)}$ in $\nu^{(1)}$.

  * (1)

$l^{(1)}=3l^{(2)}$.

We show that $m_{l^{(2)}}^{(2)}\geq 2$. Suppose that $m_{l^{(2)}}^{(2)}=1$.
From Eq. (5.6), we have

(5.8)
$-p_{l^{(2)}-1}^{(2)}=3m_{3l^{(2)}}^{(1)}+2m_{3l^{(2)}-1}^{(1)}+m_{3l^{(2)}-2}^{(1)}-2.$

However, this does not hold because $m_{3l^{(2)}}^{(1)}=m_{l^{(1)}}^{(1)}>0$.
Hence $m_{l^{(2)}}^{(2)}\geq 2$ so that [4] can be marked in an
$l^{(2)}$-string in $\nu^{(2)}$ even if [1] is already marked in another
$l^{(2)}$-string in $\nu^{(2)}$, which contradicts $b=\framebox{$4$}$.

  * (2)

$l^{(1)}=3l^{(2)}-1$.

Suppose that $m_{l^{(2)}}^{(2)}=1$. Equation (5.8) still holds, from which we
have that $p_{l^{(2)}-1}^{(2)}=0$ because
$m_{l~{}{(1)}}^{(1)}=m_{3l^{(2)}-1}^{(1)}>0$. Suppose that $m_{j}^{(2)}\neq 0$
for some $j(\leq l^{(2)}-1)$. Then, $p_{j}^{(2)}\geq 1$ as otherwise [1] would
have been marked in a singular string of length ($\leq l^{(2)}-2$)
contradicting $b=\framebox{$4$}$. However, $p_{j}^{(2)}\geq 1$ implies that
$p_{l^{(2)}-1}^{(2)}\geq 1$ by convexity, which in turn contradicts Eq. (5.8).
Therefore it suffices to consider the case when $m_{j}^{(2)}=0$ $(j\leq
l^{(2)}-1)$. We show that $m_{l^{(2)}}^{(2)}\geq 2$. Suppose that
$m_{l^{(2)}}^{(2)}=1$. Then from Eq. (3.4), we have $p_{l^{(1)}}^{(1)}<0$ so
that $m_{l^{(2)}}^{(2)}\geq 2$, which implies that [4] can be marked in an
$l^{(2)}$-string in $\nu^{(2)}$ even if [1] is already marked in another
$l^{(2)}$-string in $\nu^{(2)}$, which contradicts $b=\framebox{$4$}$.

  * (3)

$l^{(1)}=3l^{(2)}-2$.

In this case, [1] been marked in a singular string of length ($\leq
l^{(2)}-1$) in $\nu^{(2)}$. Therefore, [4] can be marked in an
$l^{(2)}$-string in $\nu^{(2)}$, which contradicts $b=\framebox{$4$}$.

All cases (a), (b), and (c) contradict $b=\framebox{$4$}$. Hence
$\lambda_{2}\geq 1$.

Case 5. $b=\framebox{$5$}$.

In this case, after $[4]$ was marked in $\nu^{(2)}$ the box marking has
terminated. Since
$\mathrm{wt}(\framebox{$5$})=2\bar{\Lambda}_{1}-\bar{\Lambda}_{2}$, we must
show that $\lambda_{1}\geq 2$. We first suppose that $\lambda_{1}=0$. Then we
have $l^{(1)}\geq 3l^{(2)}$ and $p_{l^{(1)}}^{(1)}=0$ as in Case 2. Note that
all $l^{(1)}$-strings in $\nu^{(1)}$ are singular.

  * (1)

$l^{(1)}=3l^{(2)}$.

Note that $[3]$ is not marked in an $l^{(1)}$-string (singular) in
$\nu^{(1)}$. Otherwise, $[4]$ would be marked in the same string because it is
singular. Therefore, $[5]$ can be marked in an $l^{(1)}$-string in
$\nu^{(1)}$.

  * (2)

$l^{(1)}=3l^{(2)}+1$.

Note that $[3]$ can be marked in an $l^{(1)}$-string in $\nu^{(1)}$. In this
case, the effective length is reduced by one ($l^{(2)}+1\rightarrow l^{(2)}$)
and [4] is marked in an $l^{(2)}$-string in $\nu^{(2)}$ followed by the box
marking of an $l^{(1)}$-string in $\nu^{(1)}$ by $[5]$.

  * (3)

$l^{(1)}\geq 3l^{(2)}+2$.

$[3]$ is not marked in the $l^{(1)}$-string. Otherwise, $[4]$ could not be
marked in $\nu^{(2)}$ because the effective length of $l^{(1)}$-strings is
larger than $l^{(2)}$. It is obvious that $[5]$ can be marked in the
$l^{(1)}$-string.

In three cases above, $[5]$ can be marked in the $l^{(1)}$-string
contradicting $b=\framebox{$5$}$. Next, we suppose that $\lambda_{1}=1$. As in
Case 2, we have two possible cases:

  * (1)

$m_{l^{(2)}}^{(2)}=1$, $l^{(1)}=3l^{(2)}-1$, and $p_{l^{(1)}}^{(1)}=0$.

  * (2)

$m_{l^{(2)}}^{(2)}\geq 1$, $l^{(1)}\geq 3l^{(2)}$, and $p_{l^{(1)}}^{(1)}=1$.

Note that $l^{(1)}$-strings in $\nu^{(1)}$ are singular or q-singular. Here we
assume that $[3]$ is marked in an $l^{(1)}$-string as otherwise $[5]$ could be
marked in this string contradicting $b=\framebox{$5$}$. Note that case (1)
must be excluded. This is shown as follows. If $[1]$ were marked in a singular
row of length $l^{(2)}$, then $[4]$ could not be marked in $\nu^{(2)}$ because
$m_{l^{(2)}}^{(2)}=1$. If $[1]$ were marked in a singular row of length
$(<l^{(2)})$, then $[4]$ would be marked on the left of the box marked by
$[3]$ in the $l^{(1)}$-string. Thus, case (b) survives. If one of the
$l^{(1)}$-strings is singular, the same arguments as those in the case of
$\lambda_{1}=0$ hold and lead to a contradiction. Therefore all the
$l^{(1)}$-strings must be q-singular so that $[4]$ is not marked in an
$l^{(1)}$-string. This implies $[5]$ can be marked in an $l^{(1)}$-string
contradicting $b=\framebox{$5$}$. Hence we have $\lambda_{1}\geq 2$.

Case 6. $b=\framebox{$6$}$.

In this case, after $[4]$ was marked in $\nu^{(1)}$ the box marking has
terminated. Since
$\mathrm{wt}(\framebox{$6$})=-3\bar{\Lambda}_{1}+2\bar{\Lambda}_{2}$, we must
show that $\lambda_{2}\geq 2$. We first suppose that $\lambda_{2}=0$. As in
Case 4 this results in a contradiction. Secondly we suppose that
$\lambda_{2}=1$. Then $p_{l^{(2)}+1}^{(2)}=1$ by convexity and

$p_{l^{(2)}}^{(2)}=1+\sum_{j=1}^{l^{(1)}}\left(\min(3l^{(2)},j)-j\right)m_{j}^{(1)}.$

We have following two possible cases in a similar fashion as in Case 2:.

  * (1)

$m_{l^{(1)}}^{(1)}=1$, $l^{(1)}=3l^{(2)}+1$, and $p_{l^{(2)}}^{(2)}=0$.

  * (2)

$m_{l^{(1)}}^{(1)}\geq 1$, $l^{(1)}\leq 3l^{(2)}$, and $p_{l^{(2)}}^{(2)}=1$.

Let us show that we have a contradiction in both cases.

In case (1), we show that $m_{l^{(2)}}^{(2)}\geq 2$, which in turn leads to a
contradiction. Suppose that $m_{l^{(2)}}^{(2)}=1$. From Eq. (5.6), we have

$-p_{l^{(2)}-1}^{(2)}-1=3m_{3l^{(2)}}^{(1)}+2(m_{3l^{(2)}+1}^{(1)}+m_{3l^{(2)}-1}^{(1)})+m_{3l^{(2)}-2}^{(1)}-2.$

However, this does not hold because
$m_{l^{)1)}}^{(1)}=m_{3l^{(2)}+1}^{(1)}>0$. Therefore, $m_{l^{(2)}}^{(2)}\geq
2$. If the $l^{(1)}$-string is q-singular at best, then [4] must have been
marked in a string of length ($\leq 3l^{(2)}$) so that [5] can be marked in
one of the $l^{(2)}$-strings in $\nu^{(2)}$ even if [5] cannot be marked in a
string of length ($\leq l^{(2)}-1$). If the $l^{(1)}$-string is singular and
[3] was marked in a string of length ($\leq 3l^{(2)}$), then [4] would not be
the $l^{(1)}$-string in $\nu^{(1)}$ because the effective length of the
$l^{(1)}$-string is larger than $l^{(2)}$ so that [4] must have been marked in
a string of length ($\leq 3l^{(2)}$) in $\nu^{(1)}$ which leads to a
contradiction as before. Thus, it remains to consider the case when the
$l^{(1)}$-string is singular and [3] has been marked in this string. We have
two possible ways of box marking in $\nu^{(1)}$.

$\downarrow$$\uparrow$$[3]$$3l^{(2)}+1$or$\downarrow$$\uparrow$$[3]$$[2]$$3l^{(2)}+1$

In both cases, the effective length is reduced by one ($l^{(2)}+1\rightarrow
l^{(2)}$) and $[4]$ is marked in a singular string of length $l^{(2)}$ in
$\nu^{(2)}$, which is also a contradiction. Hence, $\lambda_{2}\geq 2$.

In case (2), we also show that $m_{l^{(2)}}^{(2)}\geq 2$ leading to a
contradiction. Suppose that $m_{l^{(2)}}^{(2)}=1$. From Eq. (5.6), we have

$-p_{l^{(2)}-1}^{(2)}+1=3m_{3l^{(2)}}^{(1)}+2m_{3l^{(2)}-1}^{(1)}+m_{3l^{(2)}-2}^{(1)}-2\geq
1,$

which forces that $p_{l^{(2)}-1}^{(2)}=0$, which implies that
$m_{l^{(2)}-1}^{(2)}\neq 0$ by convexity. Therefore [1] must be marked in a