[**The srank Conjecture on Schur’s $Q$-Functions**]{}

William Y. C. Chen$^{1}$, Donna Q. J. Dou$^2$,\
Robert L. Tang$^3$ and Arthur L. B. Yang$^{4}$\
Center for Combinatorics, LPMC-TJKLC\
Nankai University, Tianjin 300071, P. R. China\
$^{1}$[chen@nankai.edu.cn]{}, $^{2}$[qjdou@cfc.nankai.edu.cn]{}, $^{3}$[tangling@cfc.nankai.edu.cn]{}, $^{4}$[yang@nankai.edu.cn]{}

**Abstract.** We show that the shifted rank, or srank, of any partition $\lambda$ with distinct parts equals the lowest degree of the terms appearing in the expansion of Schur’s $Q_{\lambda}$ function in terms of power sum symmetric functions. This gives an affirmative answer to a conjecture of Clifford. As pointed out by Clifford, the notion of the srank can be naturally extended to a skew partition $\lambda/\mu$ as the minimum number of bars among the corresponding skew bar tableaux. While the srank conjecture is not valid for skew partitions, we give an algorithm to compute the srank.

**MSC2000 Subject Classification:** 05E05, 20C25

Introduction
============

The main objective of this paper is to answer two open problems raised by Clifford [@cliff2005] on sranks of partitions with distinct parts, skew partitions and Schur’s $Q$-functions. For any partition $\lambda$ with distinct parts, we give a proof of Clifford’s srank conjecture that the lowest degree of the terms in the power sum expansion of Schur’s $Q$-function $Q_{\lambda}$ is equal to the number of bars in a minimal bar tableaux of shape $\lambda$. Clifford [@cliff2003; @cliff2005] also proposed an open problem of determining the minimum number of bars among bar tableaux of a skew shape $\lambda/\mu$. As noted by Clifford [@cliff2003], this minimum number can be naturally regarded as the shifted rank, or srank, of $\lambda/\mu$, denoted $\mathrm{srank}(\lambda/\mu)$. For a skew bar tableau, we present an algorithm to generate a skew bar tableau without increasing the number of bars. This algorithm eventually leads to a bar tableau with the minimum number of bars.

Schur’s $Q$-functions arise in the study of the projective representations of symmetric groups [@schur1911], see also, Hoffman and Humphreys [@hofhum1992], Humphreys [@humphr1986], J$\rm{\acute{o}}$zefiak [@jozef1989], Morris [@morri1962; @morri1979] and Nazarov [@nazar1988]. Shifted tableaux are closely related to Schur’s $Q$-functions analogous to the role of ordinary tableaux to the Schur functions. Sagan [@sagan1987] and Worley [@worley1984] have independently developed a combinatorial theory of shifted tableaux, which includes shifted versions of the Robinson-Schensted-Knuth correspondence, Knuth’s equivalence relations, Schützenberger’s jeu de taquin, etc. The connections between this combinatorial theory of shifted tableaux and the theory of projective representations of the symmetric groups are further explored by Stembridge [@stemb1989].

Clifford [@cliff2005] studied the srank of shifted diagrams for partitions with distinct parts. Recall that the rank of an ordinary partition is defined as the number of boxes on the main diagonal of the corresponding Young diagram. Nazarov and Tarasov [@naztar2002] found an important generalization of the rank of an ordinary partition to a skew partition in their study of tensor products of Yangian modules. A general theory of border strip decompositions and border strip tableaux of skew partitions is developed by Stanley [@stanl2002], and it has been shown that the rank of a skew partition is the least number of strips to construct a minimal border strip decomposition of the skew diagram. Motivated by Stanley’s theorem, Clifford [@cliff2005] generalized the rank of a partition to the rank of a shifted partition, called srank, in terms of the minimal bar tableaux.

On the other hand, Clifford has noticed that the srank is closely related to Schur’s $Q$-function, as suggested by the work of Stanley [@stanl2002] on the rank of a partition. Stanley introduced a degree operator by taking the degree of the power sum symmetric function $p_{\mu}$ as the number of nonzero parts of the indexing partition $\mu$. Furthermore, Clifford and Stanley [@clista2004] defined the bottom Schur functions to be the sum of the lowest degree terms in the expansion of the Schur functions in terms of the power sums. In [@cliff2005] Clifford studied the lowest degree terms in the expansion of Schur’s $Q$-functions in terms of power sum symmetric functions and conjectured that the lowest degree of the Schur’s $Q$-function $Q_{\lambda}$ is equal to the srank of $\lambda$. Our first result is a proof of this conjecture.

However, in general, the lowest degree of the terms, which appear in the expansion of the skew Schur’s $Q$-function $Q_{\lambda/\mu}$ in terms of the power sums, is not equal to the srank of the shifted skew diagram of $\lambda/\mu$. This is different from the case for ordinary skew partitions and skew Schur functions. Instead, we will take an algorithmic approach to the computation of the srank of a skew partition. It would be interesting to find an algebraic interpretation in terms of Schur’s $Q$-functions.

Shifted diagrams and bar tableaux {#sect2}
=================================

Throughout this paper we will adopt the notation and terminology on partitions and symmetric functions in [@macdon1995]. A *partition* $\lambda$ is a weakly decreasing sequence of positive integers $\lambda_1\geq \lambda_2\geq \ldots\geq
\lambda_k$, denoted $\lambda=(\lambda_1, \lambda_2, \ldots,
\lambda_k)$, and $k$ is called the *length* of $\lambda$, denoted $\ell(\lambda)$. For convenience we may add sufficient 0’s at the end of $\lambda$ if necessary. If $\sum_{i=1}^k\lambda_i=n$, we say that $\lambda$ is a partition of the integer $n$, denoted $\lambda\vdash n$. For each partition $\lambda$ there exists a geometric representation, known as the Young diagram, which is an array of squares in the plane justified from the top and left corner with $\ell(\lambda)$ rows and $\lambda_i$ squares in the $i$-th row. A partition is said to be *odd* (resp. even) if it has an odd (resp. even) number of even parts. Let $\mathcal{P}^o(n)$ denote the set of all partitions of $n$ with only odd parts. We will call a partition *strict* if all its parts are distinct. Let $\mathcal
{D}(n)$ denote the set of all strict partitions of $n$. For each partition $\lambda\in \mathcal {D}(n)$, let $S(\lambda)$ be the shifted diagram of $\lambda$, which is obtained from the Young diagram by shifting the $i$-th row $(i-1)$ squares to the right for each $i>1$. For instance, Figure \[shifted diagram\] illustrates the shifted diagram of shape $(8,7,5,3,1)$.

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Given two partitions $\lambda$ and $\mu$, if for each $i$ we have $\lambda_i\geq \mu_i$, then the skew partition $\lambda/\mu$ is defined to be the diagram obtained from the diagram of $\lambda$ by removing the diagram of $\mu$ at the top-left corner. Similarly, the skew shifted diagram $S(\lambda/\mu)$ is defined as the set-theoretic difference of $S(\lambda)$ and $S(\mu)$.

Now we recall the definitions of bars and bar tableaux as given in Hoffman and Humphreys [@hofhum1992]. Let $\lambda\in \mathcal
{D}(n)$ be a partition with length $\ell(\lambda)=k$. Fixing an odd positive integer $r$, three subsets $I_{+}, I_{0}, I_{-}$ of integers between $1$ and $k$ are defined as follows: $$\begin{aligned}
I_{+}& = &\{i: \lambda_{j+1}<\lambda_i-r<\lambda_j\: \mbox{for
some }
j\leq k,\: \mbox {taking}\:\lambda_{k+1}=0\},\\[5pt]
I_{0} & = & \{i: \lambda_i=r\},\\[5pt]
I_{-} & = & \{i: r-\lambda_{i}=\lambda_j \:\mbox{for some} \:j\:
\mbox{with} \:i<j\leq k\}.\end{aligned}$$

Let $I(\lambda,r)=I_{+}\cup I_{0}\cup I_{-}$. For each $i\in
I(\lambda,r)$, we define a new strict partition $\lambda(i,r)$ of $\mathcal {D}(n-r)$ in the following way:

-   If $i\in I_{+}$, then $\lambda_i>r$, and let $\lambda(i,r)$ be the partition obtained from $\lambda$ by removing $\lambda_i$ and inserting $\lambda_i-r$ between $\lambda_j$ and $\lambda_{j+1}$.

-   If $i\in I_{0}$, let $\lambda(i,r)$ be the partition obtained from $\lambda$ by removing $\lambda_i$.

-   If $i\in I_{-}$, then let $\lambda(i,r)$ be the partition obtained from $\lambda$ by removing both $\lambda_i$ and $\lambda_j$.

Meanwhile, for each $i\in I(\lambda,r)$, the associated $r$-bar is given as follows:

-   If $i\in I_{+}$, the $r$-bar consists of the rightmost $r$ squares in the $i$-th row of $S(\lambda)$, and we say that the $r$-bar is of Type $1$.

-   If $i\in I_{0}$, the $r$-bar consists of all the squares of the $i$-th row of $S(\lambda)$, and we say that the $r$-bar is of Type $2$.

-   If $i\in I_{-}$, the $r$-bar consists of all the squares of the $i$-th and $j$-th rows, and we say that the $r$-bar is of Type $3$.

For example, as shown in Figure \[bar tableau\], the squares filled with $6$ are a $7$-bar of Type $1$, the squares filled with $4$ are a $3$-bar of Type $2$, and the squares filled with $3$ are a $7$-bar of Type $3$.

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A *bar tableau* of shape $\lambda$ is an array of positive integers of shape $S(\lambda)$ subject to the following conditions:

-   It is weakly increasing in every row;

-   The number of parts equal to $i$ is odd for each positive integer $i$;

-   Each positive integer $i$ can appear in at most two rows, and if $i$ appears in two rows, then these two rows must begin with $i$;

-   The composition obtained by removing all squares filled with integers larger than some $i$ has distinct parts.

We say that a bar tableau $T$ is of type $\rho=(\rho_1,\rho_2,\ldots)$ if the total number of $i$’s appearing in $T$ is $\rho_i$. For example, the bar tableau in Figure \[weight\] is of type $(3,1,1,1)$. For a bar tableau $T$ of shape $\lambda$, we define its weight $wt(T)$ recursively by the following procedure. If $T$ is empty, let $wt(T)=1$. Let $\varepsilon(\lambda)$ denote the parity of the partition $\lambda$, i.e., $\varepsilon(\lambda)=0$ if $\lambda$ has an even number of even parts; otherwise, $\varepsilon(\lambda)=1$. Suppose that the largest numbers in $T$ form an $r$-bar, which is associated with an index $i\in I(\lambda, r)$. Let $j$ be the integer that occurrs in the definitions of $I_{+}$ and $I_{-}$. Let $T'$ be the bar tableau of shape $\lambda(i, r)$ obtained from $T$ by removing this $r$-bar. Now, let $$wt(T)=n_i\, wt(T'),$$ where $$n_i=\left\{\begin{array}{cc}
(-1)^{j-i}2^{1-\varepsilon(\lambda)},&
\mbox{if}\  i\in I_{+},\\[6pt]
(-1)^{\ell(\lambda)-i},& \mbox{if}\  i\in I_{0},\\[6pt]
(-1)^{j-i+\lambda_i}2^{1-\varepsilon(\lambda)},& \mbox{if}\  i\in
I_{-}.
\end{array}
\right.$$ For example, the weight of the bar tableau $T$ in Figure \[weight\] equals $$wt(T)=(-1)^{1-1}2^{1-0}\cdot(-1)^{1-1}2^{1-1}\cdot(-1)^{2-2}
\cdot(-1)^{1-1}=2.$$

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The following lemma will be used in Section 3 to determine whether certain terms will vanish in the power sum expansion of Schur’s $Q$-functions indexed by partitions with two distinct parts.

\[vanishbar\] Let $\lambda=(\lambda_1,\lambda_2)$ be a strict partition with the two parts $\lambda_1$ and $\lambda_2$ having the same parity. Given an partition $\sigma=(\sigma_1,\sigma_2)\in
\mathcal{P}^o(|\lambda|)$, if $\sigma_2<\lambda_2$, then among all bar tableaux of shape $\lambda$ there exist only two bar tableaux of type $\sigma$, say $T_1$ and $T_2$, and furthermore, we have $wt(T_1)+wt(T_2)=0$.

Suppose that both $\lambda_1$ and $\lambda_2$ are even. The case when $\lambda_1$ and $\lambda_2$ are odd numbers can be proved similarly. Note that $\sigma_2<\lambda_{2}<\lambda_{1}$. By putting $2$’s in the last $\sigma_2$ squares of the second row and then filling the remaining squares in the diagram with $1$’s, we obtain one tableau $T_1$. By putting $2$’s in the last $\sigma_2$ squares of the first row and then filling the remaining squares with $1$’s, we obtain another tableau $T_2$. Clearly, both $T_1$ and $T_2$ are bar tableaux of shape $\lambda$ and type $\sigma$, and they are the only two such bar tableaux. We notice that $$wt(T_1)=(-1)^{2-2}2^{1-0}\cdot (-1)^{2-1+\lambda_1} 2^{1-1}=-2.$$ While, for the weight of $T_2$, there are two cases to consider. If $\lambda_1-\sigma_2>\lambda_2$, then $$wt(T_2)=(-1)^{1-1}2^{1-0}\cdot
(-1)^{2-1+\lambda_1-\sigma_2}2^{1-1}=2.$$ If $\lambda_1-\sigma_2<\lambda_2$, then $$wt(T_2)=(-1)^{2-1}2^{1-0}\cdot (-1)^{2-1+\lambda_2}2^{1-1}=2.$$ Thus we have $wt(T_2)=2$ in either case, so the relation $wt(T_1)+wt(T_2)=0$ holds.

For example, taking $\lambda=(8,6)$ and $\sigma=(11,3)$, the two bar tableaux $T_1$ and $T_2$ in the above lemma are depicted as in Figure \[2-bar tableaux1\].

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Clifford gave a natural generalization of bar tableaux to skew shapes [@cliff2005]. Formally, a *skew bar tableau* of shape $\lambda/\mu$ is an assignment of nonnegative integers to the squares of $S(\lambda)$ such that in addition to the above four conditions (1)-(4) we further impose the condition that

-   the partition obtained by removing all squares filled with positive integers and reordering the remaining rows is $\mu$.

For example, taking the skew partition $(8,6,5,4,1)/(8,2,1)$, Figure \[skew bar tableau\] is a skew bar tableau of such shape.

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A bar tableau of shape $\lambda$ is said to be *minimal* if there does not exist a bar tableau with fewer bars. Motivated by Stanley’s results in [@stanl2002], Clifford defined the srank of a shifted partition $S(\lambda)$, denoted ${\rm srank}(\lambda)$, as the number of bars in a minimal bar tableau of shape $\lambda$ [@cliff2005]. Clifford also gave the following formula for ${\rm
srank}(\lambda)$.

\[min bar\] Given a strict partition $\lambda$, let $o$ be the number of odd parts of $\lambda$, and let $e$ be the number of even parts. Then ${\rm srank}(\lambda)=\max(o,e+(\ell(\lambda) \ \mathrm{mod}\  2))$.

Next we consider the number of bars in a minimal skew bar tableau of shape $\lambda/\mu$. Note that the squares filled with $0$’s in the skew bar tableau give rise to a shifted diagram of shape $\mu$ by reordering the rows. Let $o_ r$ (resp. $e_r$) be the number of nonempty rows of odd (resp. even) length with blank squares, and let $o_s$ (resp. $e_s$) be the number of rows of $\lambda$ with some squares filled with $0$’s and an odd (resp. even) number of blank squares. It is obvious that the number of bars in a minimal skew bar tableau is greater than or equal to $$o_s+2e_s+\max(o_r,e_r+((e_r+o_r)\ \mathrm{mod}\ 2)).$$ In fact the above quantity has been considered by Clifford [[@cliff2003]]{}. Observe that this quantity depends on the positions of the 0’s.

It should be remarked that a legal bar tableau of shape $\lambda/\mu$ may not exist once the positions of $0$’s are fixed. One open problem proposed by Clifford [@cliff2003] is to find a characterization of ${\rm srank}(\lambda/\mu)$. In Section 5 we will give an algorithm to compute the srank of a skew shape.

Clifford’s conjecture {#sect3}
=====================

In this section, we aim to show that the lowest degree of the power sum expansion of a Schur’s $Q$-function $Q_{\lambda}$ equals ${\rm
srank}(\lambda)$. Let us recall relevant terminology on Schur’s $Q$-functions. Let $x=(x_1,x_2,\ldots)$ be an infinite sequence of independent indeterminates. We define the symmetric functions $q_k=q_k(x)$ in $x_1,x_2,\ldots$ for all integers $k$ by the following expansion of the formal power series in $t$: $$\prod_{i\geq 1}\frac{1+x_it}{1-x_it}=\sum_{k}q_{k}(x)t^k.$$ In particular, $q_k=0$ for $k<0$ and $q_0=1$. It immediately follows that $$\label{eq-def}
\sum_{i+j=n}(-1)^iq_iq_j=0,$$ for all $n\geq 1$. Let $Q_{(a)}=q_a$ and $$Q_{(a,b)}=q_aq_b+2\sum_{m=1}^b(-1)^m q_{a+m}q_{b-m}.$$ From we see that $Q_{(a,b)}=-Q_{(b,a)}$ and thus $Q_{(a,a)}=0$ for any $a,b$. In general, for any strict partition $\lambda$, the symmetric function $Q_{\lambda}$ is defined by the recurrence relations: $$\begin{aligned}
Q_{(\lambda_1,\ldots,\lambda_{2k+1})}&=& \sum_{m=1}^{2k+1}
(-1)^{m+1}
q_{\lambda_m}Q_{(\lambda_1,\ldots,\hat{\lambda}_m,\ldots,\lambda_{2k+1})},\\[5pt]
Q_{(\lambda_1,\ldots,\lambda_{2k})}&=& \sum_{m=2}^{2k} (-1)^{m}
Q_{(\lambda_1,\lambda_m)}Q_{(\lambda_2,\ldots,\hat{\lambda}_m,\ldots,\lambda_{2k})},\end{aligned}$$ where $\hat{}$ stands for a missing entry.

It was known that $Q_{\lambda}$ can be also defined as the specialization at $t=-1$ of the Hall-Littlewood functions associated with $\lambda$ [@macdon1995]. Originally, these $Q_{\lambda}$ symmetric functions were introduced in order to express irreducible projective characters of the symmetric groups [@schur1911]. Note that the irreducible projective representations of $S_n$ are in one-to-one correspondence with partitions of $n$ with distinct parts, see [@jozef1989; @stemb1988; @stemb1989]. For any $\lambda\in \mathcal{D}(n)$, let $\langle\lambda\rangle$ denote the character of the irreducible projective or spin representation indexed by $\lambda$. Morris [@morri1965] has found a combinatorial rule for calculating the characters, which is the projective analogue of the Murnaghan-Nakayama rule. In terms of bar tableaux, Morris’s theorem reads as follows:

\[mnrule\] Let $\lambda\in \mathcal{D}(n)$ and $\pi\in
\mathcal{P}^o(n)$. Then $$\label{mnruleeq}
\langle\lambda\rangle(\pi)=\sum_{T}wt(T)$$ where the sum ranges over all bar tableaux of shape $\lambda$ and type $\pi$.

The above theorem for projective characters implies the following formula, which will be used later in the proof of Lemma \[len2\].

\[2odd\] Let $\lambda$ be a strict partition of length $2$. Suppose that the two parts $\lambda_1,\lambda_2$ are both odd. Then we have $$\langle\lambda\rangle(\lambda)=-1.$$

Let $T$ be the bar tableau obtained by filling the last $\lambda_2$ squares in the first row of $S(\lambda)$ with $2$’s and the remaining squares with $1$’s, and let $T'$ be the bar tableau obtained by filling the first row of $S(\lambda)$ with $1$’s and the second row with $2$’s. Clearly, $T$ and $T'$ are of the same type $\lambda$. Let us first consider the weight of $T$. If $\lambda_1-\lambda_2<\lambda_2$, then $$wt(T)=(-1)^{2-1} 2^{1-0}\cdot (-1)^{2-1+\lambda_2}2^{1-1}=-2.$$ If $\lambda_1-\lambda_2>\lambda_2$, then $$wt(T)=(-1)^{1-1} 2^{1-0}\cdot
(-1)^{2-1+\lambda_1-\lambda_2}2^{1-1}=-2.$$ In both cases, the weight of $T'$ equals $$wt(T')=(-1)^{2-2}\cdot (-1)^{1-1}=1.$$ Since there are only two bar tableaux, $T$ and $T'$, of type $\lambda$, the corollary immediately follows from Theorem \[mnrule\].

Let $p_k(x)$ denote the $k$-th power sum symmetric functions, i.e., $p_k(x)=\sum_{i\geq 1}x_i^k$. For any partition $\lambda=(\lambda_1,\lambda_2,\cdots)$, let $p_{\lambda}=p_{\lambda_1}p_{\lambda_2}\cdots$. The fundamental connection between $Q_{\lambda}$ symmetric functions and the projective representations of the symmetric group is as follows.

\[conn\] Let $\lambda\in \mathcal{D}(n)$. Then we have $$Q_{\lambda}=\sum_{\pi\in \mathcal{P}^o(n)}
2^{[\ell(\lambda)+\ell(\pi)+\varepsilon(\lambda)]/2}
\langle\lambda\rangle(\pi)\frac{p_{\pi}}{z_{\pi}},$$ where $$z_{\pi}=1^{m_1}m_1!\cdot 2^{m_2}m_2!\cdot \cdots, \quad \mbox{if $\pi=\langle 1^{m_1}2^{m_2}\cdots \rangle$.}$$

Stanley [@stanl2002] introduced a degree operator on symmetric functions by defining $\deg(p_i)=1$, and so $\deg(p_{\nu})=\ell(\nu)$. Clifford [@cliff2005] applied this operator to Schur’s $Q$-functions and obtained the following lower bound from Theorem \[conn\].

\[atleast\] The terms of the lowest degree in $Q_{\lambda}$ have degree at least ${\rm srank}(\lambda)$.

The following conjecture is proposed by Clifford:

The terms of the lowest degree in $Q_{\lambda}$ have degree ${\rm
srank}(\lambda)$.

Our proof of the above conjecture depends on the Pfaffian formula for Schur’s $Q$-functions. Given a skew-symmetric matrix $A=(a_{i,j})$ of even size $2n\times 2n$, the *Pfaffian* of $A$, denoted [Pf]{}(A), is defined by $${\rm Pf}(A)=\sum_{\pi}(-1)^{{\rm cr}(\pi)} a_{i_1j_1}\cdots a_{i_nj_n},$$ where the sum ranges over all set partitions $\pi$ of $\{1,2,\cdots,
2n\}$ into two element blocks $i_k<j_k$ and $cr(\pi)$ is the number of crossings of $\pi$, i.e., the number of pairs $h<k$ for which $i_h<i_k<j_h<j_k$.

\[pfexp\] Given a strict partition $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_{2n})$ satisfying $\lambda_1>\ldots>\lambda_{2n}\geq 0$, let $M_{\lambda}=(Q_{(\lambda_i,\lambda_j)})$. Then we have $$Q_{\lambda}={\rm Pf}(M_{\lambda}).$$

We first prove that Clifford’s conjecture holds for strict partitions of length less than three. The proof for the general case relies on this special case.

\[len2\] Let $\lambda$ be a strict partition of length $\ell(\lambda)<3$. Then the terms of the lowest degree in $Q_{\lambda}$ have degree ${\rm srank}(\lambda)$.

In view of Theorem \[mnrule\] and Theorem \[conn\], if there exists a unique bar tableau of shape $\lambda$ and type $\pi$, then the coefficient of $p_{\pi}$ is nonzero in the expansion of $Q_{\lambda}$. There are five cases to consider.

-   $\ell(\lambda)=1$ and $\lambda_1$ is odd. Clearly, we have ${\rm srank}(\lambda)=1$. Note that there exists a unique bar tableau $T$ of shape $\lambda$ and of type $\lambda$ with all squares of $S(\lambda)$ filled with $1$’s. Therefore, the coefficient of $p_{\lambda}$ in the power sum expansion of $Q_{\lambda}$ is nonzero and the lowest degree of $Q_{\lambda}$ is $1$.

-   $\ell(\lambda)=1$ and $\lambda_1$ is even. We see that ${\rm srank}(\lambda)=2$. Since the bars are all of odd size, there does not exist any bar tableau of shape $\lambda$ and of type $\lambda$. But there is a unique bar tableau $T$ of shape $\lambda$ and of type $(\lambda_1-1,1)$, which is obtained by filling the rightmost square of $S(\lambda)$ with $2$ and the remaining squares with $1$’s. So the coefficient of $p_{(\lambda_1-1,1)}$ in the power sum expansion of $Q_{\lambda}$ is nonzero and the terms of the lowest degree in $Q_{\lambda}$ have degree $2$.

-   $\ell(\lambda)=2$ and the two parts $\lambda_1,\lambda_2$ have different parity. In this case, we have ${\rm srank}(\lambda)=1$. Note that there exists a unique bar tableau $T$ of shape $\lambda$ and of type $(\lambda_1+\lambda_2)$, which is obtained by filling all the squares of $S(\lambda)$ with $1$’s. Thus, the coefficient of $p_{\lambda_1+\lambda_2}$ in the power sum expansion of $Q_{\lambda}$ is nonzero and the terms of lowest degree in $Q_{\lambda}$ have degree $1$.

-   $\ell(\lambda)=2$ and the two parts $\lambda_1,\lambda_2$ are both even. It is easy to see that ${\rm
    srank}(\lambda)=2$. Since there exists a unique bar tableau $T$ of shape $\lambda$ and of type $(\lambda_1-1,\lambda_2+1)$, which is obtained by filling the rightmost $\lambda_2+1$ squares in the first row of $S(\lambda)$ with $2$’s and the remaining squares with $1$’s, the coefficient of $p_{(\lambda_1-1,\lambda_2+1)}$ in the power sum expansion of $Q_{\lambda}$ is nonzero; hence the lowest degree of $Q_{\lambda}$ is equal to $2$.

-   $\ell(\lambda)=2$ and the two parts $\lambda_1,\lambda_2$ are both odd. In this case, we have ${\rm
    srank}(\lambda)=2$. By Corollary \[2odd\], the coefficient of $p_{\lambda}$ in the power sum expansion of $Q_{\lambda}$ is nonzero, and therefore the terms of the lowest degree in $Q_{\lambda}$ have degree $2$.

This completes the proof.

Given a strict partition $\lambda$, we consider the Pfaffian expansion of $Q_{\lambda}$ as shown in Theorem \[pfexp\]. To prove Clifford’s conjecture, we need to determine which terms may appear in the expansion of $Q_{\lambda}$ in terms of power sum symmetric functions. Suppose that the Pfaffian expansion of $Q_{\lambda}$ is as follows: $$\label{q-expand}
{\rm Pf}(M_{\lambda})=\sum_{\pi}(-1)^{{\rm cr}(\pi)}
Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})},$$ where the sum ranges over all set partitions $\pi$ of $\{1,2,\cdots,
2m\}$ into two element blocks $\{(\pi_1,\pi_2),\ldots,(\pi_{2m-1},\pi_{2m})\}$ with $\pi_1<\pi_3<\cdots<\pi_{2m-1}$ and $\pi_{2k-1}<\pi_{2k}$ for any $k$. For the above expansion of $Q_{\lambda}$, the following two lemmas will be used to choose certain lowest degree terms in the power sum expansion of $Q_{(\lambda_i,\lambda_j)}$ in the matrix $M_\lambda$.

\[lemma1\] Suppose that $\lambda$ has both odd parts and even parts. Let $\lambda_{i_1}$ (resp. $\lambda_{j_1}$) be the largest odd (resp. even) part of $\lambda$. If the power sum symmetric function $p_{\lambda_{i_1}+\lambda_{j_1}}$ appears in the terms of lowest degree originated from the product $Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$ as in the expansion , then we have $(\pi_1,\pi_2)=(i_1,j_1)$.

Without loss of generality, we may assume that $\lambda_{i_1}>
\lambda_{j_1}$. By Lemma \[len2\], the term $p_{\lambda_{i_1}+\lambda_{j_1}}$ appears in $Q_{(\lambda_{i_1},
\lambda_{j_1})}$ with nonzero coefficients. Since $\lambda_{i_1},
\lambda_{j_1}$ are the largest odd and even parts, $p_{\lambda_{i_1}+\lambda_{j_1}}$ does not appear as a factor of any term of the lowest degree in the expansion of $Q_{(\lambda_{i_k},
\lambda_{j_k})}$, where $\lambda_{i_k}$ and $\lambda_{j_k}$ have different parity. Meanwhile, if $\lambda_{i_k}$ and $\lambda_{j_k}$ have the same parity, then we consider the bar tableaux of shape $(\lambda_{i_k}, \lambda_{j_k})$ and of type $(\lambda_{i_1}+\lambda_{j_1}, \lambda_{i_k}+
\lambda_{j_k}-\lambda_{i_1}-\lambda_{j_1})$. Observe that $\lambda_{i_k}+
\lambda_{j_k}-\lambda_{i_1}-\lambda_{j_1}<\lambda_{j_k}$. Since the lowest degree of $Q_{(\lambda_{i_k}, \lambda_{j_k})}$ is $2$, from Lemma \[vanishbar\] it follows that $p_{\lambda_{i_1}+\lambda_{j_1}}$ can not be a factor of any term of lowest degree in the power sum expansion of $Q_{(\lambda_{i_k},
\lambda_{j_k})}$. This completes the proof.

\[lemma2\] Suppose that $\lambda$ only has even parts. Let $\lambda_1,
\lambda_2$ be the two largest parts of $\lambda$ (allowing $\lambda_2=0$). If the power sums $p_{\lambda_1-1}p_{\lambda_2+1}$ appears in the terms of the lowest degree given by the product $Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$ as in , then we have $(\pi_1,\pi_2)=(1,2)$.

From Case (4) of the proof of Lemma \[len2\] it follows that $p_{\lambda_1-1}p_{\lambda_2+1}$ appears as a term of the lowest degree in the power sum expansion of $Q_{(\lambda_1,\lambda_2)}$. We next consider the power sum expansion of any other $Q_{(\lambda_i,\lambda_j)}$. First, we consider the case when $\lambda_i+\lambda_j>\lambda_2+1$ and $\lambda_i \leq\lambda_2$. Since $\lambda_i+\lambda_j-(\lambda_2+1)<\lambda_j$, by Lemma \[vanishbar\], the term $p_{\lambda_2+1}$ is not a factor of any term of the lowest degree in the power sum expansion of $Q_{(\lambda_i,\lambda_j)}$. Now we are left with the case when $\lambda_i+\lambda_j>\lambda_1-1$ and $\lambda_i\leq \lambda_1-2$. Since $\lambda_i+\lambda_j-(\lambda_1-1)<\lambda_j$, by Lemma \[vanishbar\] the term $p_{\lambda_1-1}$ does not appear as a factor in the terms of the lowest degree of $Q_{(\lambda_i,\lambda_j)}$. So we have shown that if either $p_{\lambda_2+1}$ or $p_{\lambda_1-1}$ appears as a factor of some lowest degree term for $Q_{(\lambda_i,\lambda_j)}$, then we deduce that $\lambda_i=\lambda_1$. Moreover, if both $p_{\lambda_1-1}$ and $p_{\lambda_2+1}$ are factors of the lowest degree terms in the power sum expansion of $Q_{(\lambda_1,\lambda_j)}$, then we have $\lambda_j=\lambda_2$. The proof is complete.

We now present the main result of this paper.

For any $\lambda\in\mathcal{D}(n)$, the terms of the lowest degree in $Q_\lambda$ have degree ${\rm srank}(\lambda)$.

We write the strict partition $\lambda$ in the form $(\lambda_1,\lambda_2,\ldots,\lambda_{2m})$, where $\lambda_1>\ldots>\lambda_{2m}\geq 0$. Suppose that the partition $\lambda$ has $o$ odd parts and $e$ even parts (including $0$ as a part). For the sake of presentation, let $(\lambda_{i_1},\lambda_{i_2},\ldots,\lambda_{i_o})$ denote the sequence of odd parts in decreasing order, and let $(\lambda_{j_1},\lambda_{j_2},\ldots,\lambda_{j_e})$ denote the sequence of even parts in decreasing order.

We first consider the case $o\geq e$. In this case, it will be shown that ${\rm srank}(\lambda)=o$. By Theorem \[min bar\], if $\lambda_{2m}>0$, i.e., $\ell(\lambda)=2m$, then we have $${\rm srank}(\lambda)=\max(o,e+0)=o.$$ If $\lambda_{2m}=0$, i.e., $\ell(\lambda)=2m-1$, then we still have $${\rm srank}(\lambda)=\max(o,(e-1)+1)=o.$$

Let $$A=p_{\lambda_{i_1}+\lambda_{j_1}}\cdots
p_{\lambda_{i_e}+\lambda_{j_e}}p_{\lambda_{i_{e+1}}}p_{\lambda_{i_{e+2}}}\cdots
p_{\lambda_{i_o}}.$$ We claim that $A$ appears as a term of the lowest degree in the power sum expansion of $Q_{\lambda}$. For this purpose, we need to determine those matchings $\pi$ of $\{1,2,\ldots,2m\}$ in , for which the power sum expansion of the product $Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$ contains $A$ as a term of the lowest degree.

By Lemma \[lemma1\], if the $p_{\lambda_{i_1}+\lambda_{j_1}}$ appears as a factor in the lowest degree terms of the power sum expansion of $Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$, then we have $\{\pi_1,\pi_2\}=\{i_1,j_1\}$. Iterating this argument, we see that if $p_{\lambda_{i_1}+\lambda_{j_1}}\cdots
p_{\lambda_{i_e}+\lambda_{j_e}}$ appears as a factor in the lowest degree terms of $Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$, then we have $$\{\pi_1,\pi_2\}=\{i_1,j_1\},\ldots,\{\pi_{2e-1},\pi_{2e}\}=\{i_e,j_e\}.$$ It remains to determine the ordered pairs $$\{(\pi_{2e+1},\pi_{2e+2}),\ldots,(\pi_{2m-1},\pi_{2m})\}.$$ By the same argument as in Case (5) of the proof of Lemma \[len2\], for any $e+1\leq k<l\leq o$, the term $p_{\lambda_{i_{k}}}p_{\lambda_{i_{l}}}$ appears as a term of the lowest degree in the power sum expansion of $Q_{(\lambda_{i_k},\lambda_{i_l})}$. Moreover, if the power sum symmetric function $p_{\lambda_{i_{e+1}}}p_{\lambda_{i_{e+2}}}\cdots
p_{\lambda_{i_o}}$ appears as a term of the lowest degree in the power sum expansion of the product $Q_{(\lambda_{\pi_{2e+1}},\lambda_{\pi_{2e+2}})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$, then the composition of the pairs $\{(\pi_{2e+1},\pi_{2e+2}),\ldots,(\pi_{2m-1},\pi_{2m})\}$ could be any matching of $\{1,2,\ldots,2m\}/\{i_1,j_1,\ldots,i_e,j_e\}$.

To summarize, there are $(2(m-e)-1)!!$ matchings $\pi$ such that $A$ appears as a term of the lowest degree in the power sum expansion of the product $Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$. Combining Corollary \[2odd\] and Theorem \[conn\], we find that the coefficient of $p_{\lambda_{i_k}}p_{\lambda_{i_l}}$ $(e+1\leq k<l\leq o)$ in the power sum expansion of $Q_{(\lambda_{i_k}, \lambda_{i_l})}$ is $-\frac{4}{\lambda_{i_k}\lambda_{i_l}}$. It follows that the coefficient of $A$ in the expansion of the product $Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}}, \lambda_{\pi_{2m}})}$ is independent of the choice of $\pi$. Since $(2(m-e)-1)!!$ is an odd number, the term $A$ will not vanish in the expansion of $Q_{\lambda}$. Note that the degree of $A$ is $e+(o-e)=o,$ which is equal to ${\rm
srank}(\lambda)$, as desired.

Similarly, we consider the case $e>o$. In this case, we aim to show that ${\rm srank}(\lambda)=e.$ By Theorem \[min bar\], if $\lambda_{2m}>0$, i.e., $\ell(\lambda)=2m$, then we have $${\rm srank}(\lambda)=\max(o,e+0)=e.$$ If $\lambda_{2m}=0$, i.e., $\ell(\lambda)=2m-1$, then we still have $${\rm srank}(\lambda)=\max(o,(e-1)+1)=e.$$

Let $$B=p_{\lambda_{i_1}+\lambda_{j_1}}\cdots
p_{\lambda_{i_o}+\lambda_{j_o}}p_{\lambda_{j_{o+1}}-1}p_{\lambda_{j_{o+2}}+1}\cdots
p_{\lambda_{j_{e-1}}-1}p_{\lambda_{j_e}+1}.$$ We proceed to prove that $B$ appears as a term of the lowest degree in the power sum expansion of $Q_{\lambda}$. Applying Lemma \[lemma1\] repeatedly, we deduce that if $p_{\lambda_{i_1}+\lambda_{j_1}}\cdots
p_{\lambda_{i_o}+\lambda_{j_o}}$ appears as a factor in the lowest degree terms of the product $Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$, then $$\label{match1}
\{\pi_1,\pi_2\}=\{i_1,j_1\},\ldots,\{\pi_{2o-1},\pi_{2o}\}=\{i_o,j_o\}.$$ On the other hand, iteration of Lemma \[lemma2\] reveals that if the power sum symmetric function $p_{\lambda_{j_{o+1}}-1}p_{\lambda_{j_{o+2}}+1}\cdots
p_{\lambda_{j_{e-1}}-1}p_{\lambda_{j_e}+1}$ appears as a term of the lowest degree in the power sum expansion of $Q_{(\lambda_{\pi_{2o+1}},\lambda_{\pi_{2o+2}})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$, then $$\label{match2}
\{\pi_{2o+1},\pi_{2o+2}\}=\{j_{o+1},j_{o+2}\},\ldots,\{\pi_{2m-1},\pi_{2m}\}=\{j_{e-1},j_e\}.$$ Therefore, if $B$ appears as a term of the lowest degree in the power sum expansion of $Q_{(\lambda_{\pi_1},\lambda_{\pi_2})}\cdots
Q_{(\lambda_{\pi_{2m-1}},\lambda_{\pi_{2m}})}$, then the matching $\pi$ is uniquely determined by and . Note that the degree of $B$ is $e$, which coincides with ${\rm
srank}(\lambda)$.

Since there is always a term of degree ${\rm srank}(\lambda)$ in the power sum expansion of $Q_\lambda$, the theorem follows.

Skew Schur’s $Q$-functions
==========================

In this section, we show that the srank ${\rm srank}(\lambda/\mu)$ is a lower bound of the lowest degree of the terms in the power sum expansion of the skew Schur’s $Q$-function $Q_{\lambda/\mu}$. Note that Clifford’s conjecture does not hold for skew shapes.

We first recall a definition of the skew Schur’s $Q$-function in terms of strip tableaux. The concept of strip tableaux were introuduced by Stembridge [@stemb1988] to describe the Morris rule for the evaluation of irreducible spin characters. Given a skew partition $\lambda/\mu$, the *$j$-th diagonal* of the skew shifted diagram $S(\lambda/\mu)$ is defined as the set of squares $(1,j), (2, j+1), (3, j+2), \ldots$ in $S(\lambda/\mu)$. A skew diagram $S(\lambda/\mu)$ is called a *strip* if it is rookwise connected and each diagonal contains at most one box. The *height* $h$ of a strip is defined to be the number of rows it occupies. A *double strip* is a skew diagram formed by the union of two strips which both start on the diagonal consisting of squares $(j,j)$. The *depth* of a double strip is defined to be $\alpha+\beta$ if it has $\alpha$ diagonals of length two and its diagonals of length one occupy $\beta$ rows. A *strip tableau* of shape $\lambda/\mu$ and type $\pi=(\pi_1,\ldots,\pi_k)$ is defined to be a sequence of shifted diagrams $$S(\mu)=S(\lambda^0)\subseteq S(\lambda^1)\subseteq \cdots \subseteq S(\lambda^k)=S(\lambda)$$ with $|\lambda^i/\lambda^{i-1}|=\pi_i$ ($1\leq i\leq k$) such that each skew shifted diagram $S(\lambda^i/\lambda^{i-1})$ is either a strip or a double strip.

The skew Schur’s $Q$-function can be defined as the weight generating function of strip tableaux in the following way. For a strip of height $h$ we assign the weight $(-1)^{h-1}$, and for a double strip of depth $d$ we assign the weight $2(-1)^{d-1}$. The weight of a strip tableau $T$, denoted $wt(T)$, is the product of the weights of strips and double strips of which $T$ is composed. Then the skew Schur’s $Q$-function $Q_{\lambda/\mu}$ is given by $$Q_{\lambda/\mu}=\sum_{\pi\in \mathcal{P}^o(|\lambda/\mu|)}\sum_{T}
2^{\ell(\pi)}wt(T)\frac{p_{\pi}}{z_{\pi}},$$ where $T$ ranges over all strip tableaux $T$ of shape $\lambda/\mu$ and type $\pi$, see [@stemb1988 Theorem 5.1].

J$\rm{\acute{o}}$zefiak and Pragacz [@jozpra1991] obtained the following Pfaffian formula for the skew Schur’s $Q$-function.

\[skewpf\] Let $\lambda, \mu$ be strict partitions with $m=\ell(\lambda)$, $n=\ell(\mu)$, $\mu\subset \lambda$, and let $M(\lambda,\mu)$ denote the skew-symmetric matrix $$\begin{pmatrix}
A & B\\ -B^t & 0
\end{pmatrix},$$ where $A=(Q_{(\lambda_i,\lambda_j)})$ and $B=(Q_{(\lambda_i-\mu_{n+1-j})})$.

Then

-   if $m+n$ is even, we have $Q_{\lambda/\mu}={\rm
    Pf}(M(\lambda,\mu))$;

-   if $m+n$ is odd, we have $Q_{\lambda/\mu}={\rm Pf}(M(\lambda,\mu^\prime))$, where $\mu^\prime=(\mu_1,\cdots,\mu_n,
    0)$.

A combinatorial proof of the above theorem was given by Stembridge [@stemb1990] in terms of lattice paths, and later, Hamel [@hamel1996] gave an interesting generalization by using the border strip decompositions of the shifted diagram.

Given a skew partition $\lambda/\mu$, Clifford [@cliff2003] constructed a bijection between skew bar tableaux of shape $\lambda/\mu$ and skew strip tableaux of the same shape, which preserves the type of the tableau. Using this bijection, it is straightforward to derive the following result.

The terms of the lowest degree in $Q_{\lambda/\mu}$ have degree at least ${\rm srank}(\lambda/\mu)$.

Different from the case of non-skew shapes, in general, the lowest degree terms in $Q_{\lambda/\mu}$ do not have the degree ${\rm
srank}(\lambda/\mu)$. For example, take the skew partition $(4,3)/3$. It is easy to see that ${\rm srank}((4,3)/3)=2$. While, using Theorem \[skewpf\] and Stembridge’s SF Package for Maple [@stem2], we obtain that $$Q_{(4,3)/3}={\rm Pf}
\begin{pmatrix}0 & Q_{(4,3)} & Q_{(4)} & Q_{(1)}\\[5pt]
Q_{(3,4)} & 0 & Q_{(3)} & Q_{(0)}\\[5pt]
-Q_{(4)} & -Q_{(3)}& 0  & 0\\[5pt]
-Q_{(1)} & -Q_{(0)}& 0  & 0
\end{pmatrix}=2p_1^4.$$ This shows that the lowest degree of $Q_{(4,3)/3}$ equals 4, which is strictly greater than ${\rm srank}((4,3)/3)$.

The srank of skew partitions {#sect4}
============================

In this section, we present an algorithm to determine the srank for the skew partition $\lambda/\mu$. In fact, the algorithm leads to a configuration of $0$’s. To obtain the srank of a skew partition, we need to minimize the number of bars by adjusting the positions of $0$’s. Given a configuration $\mathcal{C}$ of $0$’s in the shifted diagram $S(\lambda)$, let $$\kappa(\mathcal{C})=o_s+2e_s+\max(o_r,e_r+((e_r+o_r)\ \mathrm{mod}\  2)),$$ where $o_ r$ (resp. $e_r$) counts the number of nonempty rows in which there are an odd (resp. even) number of squares and no squares are filled with $0$, and $o_s$ (resp. $e_s$) records the number of rows in which at least one square is filled with $0$ but there are an odd (resp. nonzero even) number of blank squares.

If there exists at least one bar tableau of type $\lambda/\mu$ under some configuration $\mathcal{C}$, we say that $\mathcal{C}$ is *admissible*. For a fixed configuration $\mathcal{C}$, each row is one of the following eight possible types:

-   an even row bounded by an even number of $0$’s, denoted $(e,e)$,

-   an odd row bounded by an even number of $0$’s, denoted $(e,o)$,

-   an odd row bounded by an odd number of $0$’s, denoted $(o,e)$,

-   an even row bounded by an odd number of $0$’s, denoted $(o,o)$,

-   an even row without $0$’s, denoted $(\emptyset,e)$,

-   an odd row without $0$’s, denoted $(\emptyset,o)$,

-   an even row filled with $0$’s, denoted $(e,
    \emptyset)$,

-   an odd row filled with $0$’s, denoted $(o,
    \emptyset)$.

Given two rows with respective types $s$ and $s'$ for some configuration $\mathcal{C}$, if we can obtain a new configuration $\mathcal{C}'$ by exchanging the locations of $0$’s in these two rows such that their new types are $t$ and $t'$ respectively, then denote it by $\mathcal{C}'=\mathcal{C}\left( \left[\tiny{{s \atop
s'}}\right] \rightarrow \left[\tiny{{{t} \atop
{t'}}}\right]\right)$. Let $o_r,e_r,o_s,e_s$ be defined as above corresponding to configuration $\mathcal{C}$, and let $o_r',e_r',o_s',e_s'$ be those of $\mathcal{C}'$.

In the following we will show that how the quantity $\kappa(\mathcal{C})$ changes when exchanging the locations of $0$’s in $\mathcal{C}$.

\[varyzero1-1\] If $\mathcal{C}'=\mathcal{C}\left( \left[\tiny{{s \atop s'}}\right]
\rightarrow \left[\tiny{{s \atop s'}}\right]\right)$ or $\mathcal{C}'=\mathcal{C}\left( \left[\tiny{{s \atop s'}}\right]
\rightarrow \left[\tiny{{{s'} \atop s}}\right]\right)$, i.e., the types of the two involved rows are remained or exchanged, where $s,s'$ are any two possible types, then $\kappa({\mathcal{C}'})=
\kappa({\mathcal{C}})$.

\[varyzero1-6\] If $\mathcal{C}'=\mathcal{C}\left( \left[\tiny{{{(e,e)}
\atop\rule{0pt}{10pt} {(\emptyset,o)}}}\right] \rightarrow
\left[\tiny{{{(\emptyset,e)} \atop\rule{0pt}{10pt}
{(e,o)}}}\right]\right)$, then $\kappa({\mathcal{C}'})\leq
\kappa({\mathcal{C}})$.

In this case we have $$o_s^\prime=o_s+1, \quad e_s^\prime=e_s-1, \quad
o_r^\prime=o_r-1,\quad e_r^\prime=e_r+1.$$ Note that $o_r+e_r=\ell(\lambda)-\ell(\mu)$. Now there are two cases to consider.

**Case I.** The skew partition $\lambda/\mu$ satisfies that $\ell(\lambda)-\ell(\mu) \equiv 0\ (\mathrm{mod}\
2)$.

-   If $o_r\leq e_r$, then $o_r^\prime\leq e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C})&=o_s+2e_s+e_r,\\
    \kappa(\mathcal{C}')&=o_s+1+2(e_s-1)+e_r^\prime=o_s+2e_s+e_r=\kappa(\mathcal{C}).\end{aligned}$$

-   If $o_r\geq e_r+2$, then $o_r^\prime=o_r-1\geq
    e_r+1=e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C})&=o_s+2e_s+o_r,\\
    \kappa(\mathcal{C}')&=o_s+2e_s-1+o_r^\prime=o_s+2e_s+o_r-2<\kappa(\mathcal{C}).\end{aligned}$$

**Case II.** The skew partition $\lambda/\mu$ satisfies that $\ell(\lambda)-\ell(\mu) \equiv 1\ (\mathrm{mod}\
2)$.

-   If $o_r\leq e_r+1$, then $o_r^\prime<e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C})&=o_s+2e_s+e_r+1,\\
    \kappa(\mathcal{C}')&=o_s+2e_s-1+e_r^\prime+1=o_s+2e_s+e_r+1=\kappa(\mathcal{C}).\end{aligned}$$

-   If $o_r\geq e_r+3$, then $o_r^\prime=o_r-1\geq
    e_r+2>e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C})&=o_s+2e_s+o_r,\\
    \kappa(\mathcal{C}')&=o_s+2e_s-1+o_r^\prime=o_s+2e_s+o_r-2<\kappa(\mathcal{C}).\end{aligned}$$

Therefore, the inequality $\kappa({\mathcal{C}'})\leq
\kappa({\mathcal{C}})$ holds under the assumption.

\[varyzero2-6\] If $\mathcal{C}'=\mathcal{C}\left( \left[\tiny{{{(o,e)}
\atop\rule{0pt}{10pt} {(\emptyset,e)}}}\right] \rightarrow
\left[\tiny{{{(\emptyset,o)} \atop\rule{0pt}{10pt}
{(o,o)}}}\right]\right)$, then $\kappa({\mathcal{C}'})\leq
\kappa({\mathcal{C}})$.

In this case we have $$o_s^\prime=o_s+1, \quad e_s^\prime=e_s-1, \quad
o_r^\prime=o_r+1,\quad e_r^\prime=e_r-1.$$ Now there are two possibilities.

**Case I.** The skew partition $\lambda/\mu$ satisfies that $\ell(\lambda)-\ell(\mu) \equiv 0\ (\mathrm{mod}\
2)$.

-   If $o_r\leq e_r-2$, then $o_r^\prime\leq e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C})&=o_s+2e_s+e_r,\\
    \kappa(\mathcal{C}')&=o_s+1+2(e_s-1)+e_r^\prime=o_s+2e_s+e_r-2<\kappa(\mathcal{C}).\end{aligned}$$

-   If $o_r\geq e_r$, then $o_r^\prime=o_r+1>
    e_r-1=e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C})&=o_s+2e_s+o_r,\\
    \kappa(\mathcal{C}')&=o_s+2e_s-1+o_r^\prime=o_s+2e_s+o_r=\kappa(\mathcal{C}).\end{aligned}$$

**Case II.** The skew partition $\lambda/\mu$ satisfies that $\ell(\lambda)-\ell(\mu) \equiv 1\ (\mathrm{mod}\
2)$.

-   If $o_r\leq e_r-3$, then $o_r^\prime<e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C})&=o_s+2e_s+e_r+1,\\
    \kappa(\mathcal{C}')&=o_s+2e_s-1+e_r^\prime+1=o_s+2e_s+e_r-1<\kappa(\mathcal{C}).\end{aligned}$$

-   If $o_r\geq e_r-1$, then $o_r^\prime=o_r+1>
    e_r-1=e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C})&=o_s+2e_s+o_r,\\
    \kappa(\mathcal{C}')&=o_s+2e_s-1+o_r^\prime=o_s+2e_s+o_r=\kappa(\mathcal{C}).\end{aligned}$$

In both cases we have $\kappa({\mathcal{C}'})\leq
\kappa({\mathcal{C}})$, as required.

\[varyzero1-4\] If $\mathcal{C}'=\mathcal{C}\left( \left[\tiny{{{(e,e)}
\atop\rule{0pt}{10pt} {(o,e)}}}\right] \rightarrow
\left[\tiny{{{(o,o)} \atop\rule{0pt}{10pt}
{(e,o)}}}\right]\right)$, then $\kappa({\mathcal{C}'})<
\kappa({\mathcal{C}})$.

In this case, we have $$o_s^\prime=o_s+2, \quad e_s^\prime=e_s-2, \quad
o_r^\prime=o_r,\quad e_r^\prime=e_r.$$ Therefore, $$\kappa(\mathcal{C}')=o_s'+2e_s'+\max(o_r',e_r'+((e_r'+o_r')\
\mathrm{mod}\ 2))=\kappa(\mathcal{C})-2.$$ The desired inequality immediately follows.

\[varyzero3-5\] If $\mathcal{C}'=\mathcal{C}\left( \left[\tiny{{{(e,o)}
\atop\rule{0pt}{10pt} {(\emptyset,e)}}}\right] \rightarrow
\left[\tiny{{{(\emptyset,o)} \atop\rule{0pt}{10pt}
{(e,\emptyset)}}}\right]\right)$, then $\kappa({\mathcal{C}'})\leq
\kappa({\mathcal{C}})$.

Under this transformation we have $$o_s^\prime=o_s-1, \quad e_s^\prime=e_s, \quad
o_r^\prime=o_r+1,\quad e_r^\prime=e_r-1.$$ Since $o_r+e_r=\ell(\lambda)-\ell(\mu)$ is invariant, there are two cases.

**Case I.** The skew partition $\lambda/\mu$ satisfies that $\ell(\lambda)-\ell(\mu) \equiv 0\ (\mathrm{mod}\
2)$.

-   If $o_r\geq e_r$, then $o_r^\prime\geq e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C}')=o_s^\prime+2e_s^\prime+o_r^\prime
    =o_s-1+2e_s+o_r+1 =\kappa(\mathcal{C}).\end{aligned}$$

-   If $o_r\leq e_r-2$, then $o_r^\prime=o_r+1\leq
    e_r-1=e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C}')=o_s-1+2e_s+e_r^\prime=o_s+2e_s+e_r-2<\kappa(\mathcal{C}).\end{aligned}$$

**Case II.** The skew partition $\lambda/\mu$ satisfies that $\ell(\lambda)-\ell(\mu) \equiv 1\ (\mathrm{mod}\
2)$.

-   If $o_r\geq e_r+1$, then $o_r^\prime=o_r+1\geq
    e_r+2>e_r^\prime+1$ and $$\begin{aligned}
    \kappa(\mathcal{C}')=o_s^\prime+2e_s^\prime+o_r^\prime=o_s+2e_s+o_r=\kappa(\mathcal{C}).\end{aligned}$$

-   If $o_r\leq e_r-1$, then $o_r^\prime=o_r+1\leq
    e_r=e_r^\prime+1$ and $$\begin{aligned}
    \kappa(\mathcal{C}')=o_s^\prime+2e_s^\prime+e_r^\prime+1=o_s+2e_s+e_r-1<\kappa(\mathcal{C}).\end{aligned}$$

Hence the proof is complete.

\[varyzero5-8\] If $\mathcal{C}'=\mathcal{C}\left( \left[\tiny{{{(o,o)}
\atop\rule{0pt}{10pt} {(\emptyset,o)}}}\right] \rightarrow
\left[\tiny{{{(\emptyset,e)} \atop\rule{0pt}{10pt}
{(o,\emptyset)}}}\right]\right)$, then $\kappa({\mathcal{C}'})\leq
\kappa({\mathcal{C}})$.

In this case we have $$o_s^\prime=o_s-1, \quad e_s^\prime=e_s, \quad
o_r^\prime=o_r-1,\quad e_r^\prime=e_r+1.$$ There are two possibilities:

**Case I.** The skew partition $\lambda/\mu$ satisfies that $\ell(\lambda)-\ell(\mu)
\equiv 0\ (\mathrm{mod}\ 2)$.

-   If $o_r\geq e_r+2$, then $o_r^\prime=o_r-1\geq
    e_r+1=e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C}')=o_s^\prime+2e_s^\prime+o_r^\prime
    =o_s-1+2e_s+o_r-1<\kappa(\mathcal{C}).\end{aligned}$$

-   If $o_r\leq e_r$, then $o_r^\prime=o_r-1\leq
    e_r-1<e_r^\prime$ and $$\begin{aligned}
    \kappa(\mathcal{C}')=o_s-1+2e_s+e_r^\prime=o_s+2e_s+e_r=\kappa(\mathcal{C}).\end{aligned}$$

**Case II.** The skew partition $\lambda/\mu$ satisfies that $\ell(\lambda)-\ell(\mu) \equiv 1\ (\mathrm{mod}\
2)$.

-   If $o_r\geq e_r+3$, then $o_r^\prime=o_r-1\geq
    e_r+2=e_r^\prime+1$ and $$\begin{aligned}
    \kappa(\mathcal{C}')=o_s^\prime+2e_s^\prime+o_r^\prime=o_s+2e_s+o_r-2<
    \kappa(\mathcal{C}).\end{aligned}$$

-   If $o_r\leq e_r+1$, then $o_r^\prime=o_r-1\leq
    e_r<e_r^\prime+1$ and $$\begin{aligned}
    \kappa(\mathcal{C}')=o_s^\prime+2e_s^\prime+e_r^\prime+1=o_s+2e_s+e_r+1=\kappa(\mathcal{C}).\end{aligned}$$

Therefore, in both cases we have $\kappa({\mathcal{C}'})\leq
\kappa({\mathcal{C}})$.

\[varyzero2-3\] If $\mathcal{C}'=\mathcal{C}\left( \left[\tiny{{{(e,o)}
\atop\rule{0pt}{10pt} {(o,o)}}}\right] \rightarrow
\left[\tiny{{{(o,e)} \atop\rule{0pt}{10pt}
{(e,\emptyset)}}}\right]\right)$ or $\mathcal{C}'=\mathcal{C}\left(
\left[\tiny{{{(o,o)} \atop\rule{0pt}{10pt}  {(e,o)}}}\right]
\rightarrow \left[\tiny{{{(e,e)} \atop\rule{0pt}{10pt}
{(o,\emptyset)}}}\right]\right)$, then $\kappa({\mathcal{C}'})=
\kappa({\mathcal{C}})$.

In each case we have $$o_s^\prime=o_s-2, \quad e_s^\prime=e_s+1, \quad
o_r^\prime=o_r,\quad e_r^\prime=e_r.$$ Therefore $$\kappa(\mathcal{C}')=o_s'+2e_s'+\max(o_r',e_r'+((e_r'+o_r')\
\mathrm{mod}\ 2))=\kappa(\mathcal{C}),$$ as desired.

\[varyzero1-7\] If $\mathcal{C}'$ is one of the following possible cases: $$\begin{array}{ccc}
\mathcal{C}\left( \left[\tiny{{{(e,e)} \atop \rule{0pt}{10pt}
{(e,e)}}}\right] \rightarrow \left[\tiny{{{(e,e)} \atop
\rule{0pt}{10pt} {(e,\emptyset)}}}\right]\right), &
\mathcal{C}\left( \left[\tiny{{{(e,e)} \atop \rule{0pt}{10pt}
{(o,o)}}}\right] \rightarrow \left[\tiny{{{(o,o)}
\atop\rule{0pt}{10pt} {(e,\emptyset)}}}\right]\right), &
\mathcal{C}\left( \left[\tiny{{{(e,o)} \atop \rule{0pt}{10pt}
{(e,e)}}}\right] \rightarrow \left[\tiny{{{(e,o)} \atop
\rule{0pt}{10pt} {(e,\emptyset)}}}\right]\right),\\[10pt]
\mathcal{C}\left( \left[\tiny{{{(e,e)} \atop\rule{0pt}{10pt}
{(\emptyset,e)}}}\right] \rightarrow \left[\tiny{{{(\emptyset,e)}
\atop\rule{0pt}{10pt} {(e,\emptyset)}}}\right]\right), &
\mathcal{C}\left( \left[\tiny{{{(o,o)} \atop \rule{0pt}{10pt}
{(o,e)}}}\right] \rightarrow \left[\tiny{{{(o,o)} \atop
\rule{0pt}{10pt}  {(o,\emptyset)}}}\right]\right), &
\mathcal{C}\left( \left[\tiny{{ {(o,e)} \atop\rule{0pt}{10pt}
{(e,o)}}}\right] \rightarrow \left[\tiny{{{(e,o)}
\atop\rule{0pt}{10pt} {(o,\emptyset)}}}\right]\right),\\[10pt]
\mathcal{C}\left( \left[\tiny{{{(o,e)} \atop\rule{0pt}{10pt}
{(o,e)}}}\right] \rightarrow \left[\tiny{{{(o,e)}
\atop\rule{0pt}{10pt} {(o,\emptyset)}}}\right]\right), &
\mathcal{C}\left( \left[\tiny{{{(o,e)} \atop\rule{0pt}{10pt}
{(\emptyset,o)}}}\right] \rightarrow \left[\tiny{{{(\emptyset,o)}
\atop \rule{0pt}{10pt} {(o,\emptyset)}}}\right]\right),&
\end{array}$$ then $\kappa({\mathcal{C}'})< \kappa({\mathcal{C}})$.

In each case we have $$o_s^\prime=o_s, \quad e_s^\prime=e_s-1, \quad
o_r^\prime=o_r,\quad e_r^\prime=e_r.$$ Therefore $$\kappa(\mathcal{C}')=o_s'+2e_s'+\max(o_r',e_r'+((e_r'+o_r')\
\mathrm{mod}\ 2))<\kappa(\mathcal{C}),$$ as required.

Note that Lemmas \[varyzero1-1\]-\[varyzero1-7\] cover all possible transformations of exchanging the locations of $0$’s in two involved rows. Lemmas \[varyzero1-6\]-\[varyzero1-4\] imply that, to minimize the number of bars, we should put $0$’s in the skew shifted diagram such that there are as more as possible rows for which the first several squares are filled with $0$’s and then followed by an odd number of blank squares. Meanwhile, from Lemmas \[varyzero3-5\]-\[varyzero1-7\] we know that the number of rows fully filled with $0$’s should be as more as possible. Based on these observations, we have the following algorithm to determine the location of $0$’s for a given skew partition $\lambda/\mu$, where both $\lambda$ and $\mu$ are strict partitions. Using this algorithm we will obtain a shifted diagram with some squares filled with $0$’s such that the corresponding quantity $\kappa(\mathcal{C})$ is minimized. This property allows us to determine the srank of $\lambda/\mu$.

[**The Algorithm for Determining the Locations of $0$’s:**]{}

-   Let $\mathcal{C}_1=S(\lambda)$ be the initial configuration of $\lambda/\mu$ with blank square. Set $i=1$ and $J=\{1,\ldots,\ell(\lambda)\}$.

-   For $i\leq \ell(\mu)$, iterate the following procedure:

    -   If $\mu_i=\lambda_j$ for some $j\in J$, then we fill the $j$-th row of $\mathcal{C}_i$ with $0$.

    -   If $\mu_i\neq \lambda_j$ for any $j\in J$, then there are two possibilities.

        -   $\lambda_j-\mu_i$ is odd for some $j\in J$ and $\lambda_j>\mu_i$. Then we take the largest such $j$ and fill the leftmost $\mu_i$ squares with $0$ in the $j$-th row of $\mathcal{C}_i$.

        -   $\lambda_j-\mu_i$ is even for any $j\in J$ and $\lambda_j>\mu_i$. Then we take the largest such $j$ and fill the leftmost $\mu_i$ squares by $0$ in the $j$-th row of $\mathcal{C}_i$.

    Denote the new configuration by $\mathcal{C}_{i+1}$. Set $J=J\backslash \{j\}$.

-   Set $\mathcal{C}^{*}=\mathcal{C}_{i}$, and we get the desired configuration.

It should be emphasized that although the above algorithm does not necessarily generate a bar tableau, it is sufficient for the computation of the srank of a skew partition.

Using the arguments in the proofs of Lemmas \[varyzero1-1\]-\[varyzero1-7\], we can derive the following crucial property of the configuration $\mathcal{C}^*$. The proof is omitted since it is tedious and straightforward.

\[prop-min\] For any configuration ${\mathcal{C}}$ of $0$’s in the skew shifted diagram of $\lambda/\mu$, we have $\kappa({\mathcal{C}^*})\leq
\kappa({\mathcal{C}})$.

\[number of skew\] Given a skew partition $\lambda/\mu$, let $\mathcal{C}^*$ be the configuration of $0$’s obtained by applying the algorithm described above. Then $$\label{srank}
{\rm srank}(\lambda/\mu)=\kappa({\mathcal{C}^*}).$$

Suppose that for the configuration ${\mathcal{C}^*}$ there are $o_r^*$ rows of odd size with blank squares, and there are $o_s^*$ rows with at least one square filled with $0$ and an odd number of squares filled with positive integers. Likewise we let $e_r^*$ and $e_s^*$ denote the number of remaining rows. Therefore, $$\kappa(\mathcal{C}^*)=o_s^*+2e_s^*+\max(o_r^*,e_r^*+((e_r^*+o_r^*)\ \mathrm{mod}\  2)).$$ Since for each configuration $\mathcal{C}$ the number of bars in a minimal bar tableau is greater than or equal to $\kappa({\mathcal{C}})$, by Proposition \[prop-min\], it suffices to confirm the existence of a skew bar tableau, say $T$, with $\kappa({\mathcal{C}^*})$ bars.

Note that it is possible that the configuration ${\mathcal{C}^*}$ is not admissible. The key idea of our proof is to move $0$’s in the diagram such that the resulting configuration ${\mathcal{C}'}$ is admissible and $\kappa({\mathcal{C}'})=\kappa({\mathcal{C}^*})$. To achieve this goal, we will use the numbers $\{1,2,\ldots,\kappa({\mathcal{C}^*})\}$ to fill up the blank squares of $\mathcal{C}^*$ guided by the rule that the bars of Type $2$ or Type $3$ will occur before bars of Type $1$.

Let us consider the rows without $0$’s, and there are two possibilities: (A) $o_r^*\geq e_r^*$, (B) $o_r^*<e_r^*$.

In Case (A) we choose a row of even size and a row of odd size, and fill up these two rows with $\kappa({\mathcal{C}^*})$ to generate a bar of Type $3$. Then we continue to choose a row of even size and a row of odd size, and fill up these two rows with $\kappa({\mathcal{C}^*})-1$. Repeat this procedure until all even rows are filled up. Finally, we fill the remaining rows of odd size with $\kappa({\mathcal{C}^*})-e_r^*,
\kappa({\mathcal{C}^*})-e_r^*-1, \ldots,
\kappa({\mathcal{C}^*})-o_r^*+1$ to generate bars of Type $2$.

In Case (B) we choose the row with the $i$-th smallest even size and the row with the $i$-th smallest odd size and fill their squares with the number $\kappa({\mathcal{C}^*})-i+1$ for $i=1,\ldots,o_r^*$. In this way, we obtain $o_r^*$ bars of Type $3$. Now consider the remaining rows of even size without $0$’s. There are two subcases.

-   The remaining diagram, obtained by removing the previous $o_r^*$ bars of Type $3$, does not contain any row with only one square. Under this assumption, it is possible to fill the squares of a row of even size with the number $\kappa({\mathcal{C}^*})-o_r^*$ except the leftmost square. This operation will result in a bar of Type $1$. After removing this bar from the diagram, we may combine this leftmost square of the current row and another row of even size, if it exists, and to generate a bar of Type $3$. Repeating this procedure until there are no more rows of even size, we obtain a sequence of bars of Type $1$ and Type $3$. Evidently, there is a bar of Type $2$ with only one square. To summarize, we have $\max(o_r^*,e_r^*+((e_r^*+o_r^*)\
    \mathrm{mod}\ 2))$ bars.

-   The remaining diagram contains a row composed of the unique square filled with $0$. In this case, we will move this $0$ into the leftmost square of a row of even size, see Figure \[case2-2\]. Denote this new configuration by $\mathcal{C}^{\prime}$, and from Lemma \[varyzero5-8\] we see that $\kappa({\mathcal{C}^*})=\kappa({\mathcal{C}^{\prime}})$. If we start with ${\mathcal{C}'}$ instead of ${\mathcal{C}^*}$, by a similar construction, we get $\max(o_r',e_r'+((e_r'+o_r')\
    \mathrm{mod}\ 2))$ bars, occupying the rows without $0$’s in the diagram.

(300,100) (40,0)[(1,0)[20]{}]{}(40,20)[(1,0)[20]{}]{} (40,0)[(0,1)[20]{}]{}(60,0)[(0,1)[20]{}]{} (50,25)[$\vdots$]{}(20,40)[(1,0)[80]{}]{} (20,60)[(1,0)[80]{}]{} (20,40)(20,0)[5]{}[(0,1)[20]{}]{} (50,65)[$\vdots$]{} (0,80)(0,20)[2]{}[(1,0)[120]{}]{} (0,80)(20,0)[7]{}[(0,1)[20]{}]{}(48,6)[$0$]{} (130,50)[(1,0)[40]{}]{} (220,0)[(1,0)[20]{}]{}(220,20)[(1,0)[20]{}]{} (220,0)[(0,1)[20]{}]{}(240,0)[(0,1)[20]{}]{} (230,25)[$\vdots$]{}(200,40)[(1,0)[80]{}]{} (200,60)[(1,0)[80]{}]{} (200,40)(20,0)[5]{}[(0,1)[20]{}]{} (230,65)[$\vdots$]{} (180,80)(0,20)[2]{}[(1,0)[120]{}]{} (180,80)(20,0)[7]{}[(0,1)[20]{}]{}(188,86)[$0$]{}

Without loss of generality, we may assume that for the configuration ${\mathcal{C}^*}$ the rows without $0$’s in the diagram have been occupied by the bars with the first $\max(o_r^*,e_r^*+((e_r^*+o_r^*)\ \mathrm{mod}\ 2))$ positive integers in the decreasing order, namely, $(\kappa({\mathcal{C}^*}),
\ldots, 2, 1, 0)$. By removing these bars and reordering the remaining rows, we may get a shifted diagram with which we can continue the above procedure to construct a bar tableau.

At this point, it is necessary to show that it is possible to use $o_s^*+2e_s^*$ bars to fill this diagram. In doing so, we process the rows from bottom to top. If the bottom row has an odd number of blank squares, then we simply assign the symbol $o_s^*+2e_s^*$ to these squares to produce a bar of Type $1$. If the bottom row are completely filled with $0$’s, then we continue to deal with the row above the bottom row. Otherwise, we fill the rightmost square of the bottom row with $o_s^*+2e_s^*$ and the remaining squares with $o_s^*+2e_s^*-1$. Suppose that we have filled $i$ rows from the bottom and all the involved bars have been removed from the diagram. Then we consider the $(i+1)$-th row from the bottom. Let $t$ denote the largest number not greater than $o_s^*+2e_s^*$ which has not been used before. If all squares in the $(i+1)$-th row are filled with $0$’s, then we continue to deal with the $(i+2)$-th row. If the number of blank squares in the $(i+1)$-th row is odd, then we fill these squares with $t$. If the number of blank squares in the $(i+1)$-th row is even, then we are left with two cases:

-   The rows of the diagram obtained by removing the rightmost square of the $(i+1)$-th row have distinct lengths. In this case, we fill the rightmost square with $t$ and the remaining blank squares of the $(i+1)$-th row with $t-1$.

-   The removal of the rightmost square of the $(i+1)$-th row does not result in a bar tableau. Suppose that the $(i+1)$-th row has $m$ squares in total. It can only happen that the row underneath the $(i+1)$-th row has $m-1$ squares and all these squares are filled with $0$’s. By interchanging the location of $0$’s in these two rows, we get a new configuration $\mathcal{C}^{\prime}$, see Figure \[case2’\]. From Lemma \[varyzero2-3\] we deduce that $\kappa({\mathcal{C}^*})=\kappa({\mathcal{C}^{\prime}})$. So we can transform ${\mathcal{C}^*}$ to ${\mathcal{C}'}$ and continue to fill up the $(i+1)$-th row.

(340,40) (20,0)[(1,0)[120]{}]{} (0,20)(0,20)[2]{}[(1,0)[140]{}]{} (20,0)(20,0)[7]{}[(0,1)[40]{}]{} (0,20)[(0,1)[20]{}]{} (28,6)(20,0)[6]{}[$0$]{}(8,26)(20,0)[3]{}[$0$]{} (150,20)[(1,0)[40]{}]{} (220,0)[(1,0)[120]{}]{} (200,20)(0,20)[2]{}[(1,0)[140]{}]{} (220,0)(20,0)[7]{}[(0,1)[40]{}]{} (200,20)[(0,1)[20]{}]{} (228,6)(20,0)[3]{}[$0$]{}(208,26)(20,0)[6]{}[$0$]{}

Finally, we arrive at a shifted diagram whose rows are all filled up. Clearly, for those rows containing at least one $0$ there are $o_s^*+2e_s^*$ bars that are generated in the construction, and for those rows containing no $0$’s there are $\max(o_r^*,e_r^*+((e_r^*+o_r^*)\ \mathrm{mod}\ 2))$ bars that are generated. It has been shown that during the procedure of filling the diagram with nonnegative numbers if the configuration ${\mathcal{C}^*}$ is transformed to another configuration ${\mathcal{C}^{\prime}}$, then $\kappa({\mathcal{C}^\prime})$ remains equal to $\kappa({\mathcal{C}^*})$. Hence the above procedure leads to a skew bar tableau of shape $\lambda/\mu$ with $\kappa({\mathcal{C}^*})$ bars. This completes the proof.

[**Acknowledgments.**]{} This work was supported by the 973 Project, the PCSIRT Project of the Ministry of Education, the Ministry of Science and Technology, and the National Science Foundation of China.

[19]{}

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