%\input amstex
%\input amsppt.sty
%\magnification1200

\NoRunningHeads
\TagsOnRight
\loadbold

\def\AndrAW{1}
\def\GaMiAB{2}
\def\HiGrAA{3}
\def\KratAK{4}
\def\KratBK{5}
\def\KupeAA{6}
\def\MacMAA{7}
\def\NoPSAA{8}
\def\PaStAA{9}
\def\ReWhAA{10}
\def\SagaAB{11}
\def\SagaAD{12}
\def\SagaAL{13}
\def\SchuAA{14}
\def\StanAA{15}
\def\StanAI{16}
\def\StWhAB{17}
\def\StemAG{18}
\def\VoWhAA{19}

\font\Bf=cmbx12
\font\Rm=cmr12
\def\LL{\leavevmode\setbox0=\hbox{L}\hbox to\wd0{\hss\char'40L}}
\def\al{\alpha}
\def\be{\beta}
\def\ga{\gamma}
\def\de{\delta}
\def\ep{\varepsilon}
\def\ze{\zeta}
\def\et{\eta}
\def\th{\theta}
\def\vt{\vartheta}
\def\io{\iota}
\def\ka{\kappa}
\def\la{\lambda}
\def\rh{\rho}
\def\si{\sigma}
\def\ta{\tau}
\def\ph{\varphi}
\def\ch{\chi}
\def\ps{\psi}
\def\om{\omega}
\def\Ga{\Gamma}
\def\De{\Delta}
\def\Th{\Theta}
\def\La{\Lambda}
\def\Si{\Sigma}
\def\Ph{\Phi}
\def\Ps{\Psi}
\def\Om{\Omega}
\def\row#1#2#3{#1_{#2},\ldots,#1_{#3}}
\def\rowup#1#2#3{#1^{#2},\ldots,#1^{#3}}
\def\x{\times}
\def\crf{}            %used for crossreferencing, Tex should ignore.
\def\rf{}             %used for refencing (section-numbers)
\def\rfnew{}          %used for new-section numbers
\def\P{{\Bbb P}}
\def\R{{\Bbb R}}
\def\X{{\Cal X}}
\def\C{{\Bbb C}}
\def\Mf{{\Cal Mf}}
\def\FM{{\Cal F\Cal M}}
\def\F{{\Cal F}}
\def\G{{\Cal G}}
\def\V{{\Cal V}}
\def\T{{\Cal T}}
\def\A{{\Cal A}}
\def\N{{\Bbb N}}
\def\Z{{\Bbb Z}}
\def\Q{{\Bbb Q}}
\def\ddt{\left.\tfrac \partial{\partial t}\right\vert_0}
\def\dd#1{\tfrac \partial{\partial #1}}
\def\today{\ifcase\month\or
 January\or February\or March\or April\or May\or June\or
 July\or August\or September\or October\or November\or December\fi
 \space\number\day, \number\year}
\def\nmb#1#2{#2} %zum Nummerieren
\def\dfrac#1#2{{\displaystyle{#1\over#2}}}
\def\tfrac#1#2{{\textstyle{#1\over#2}}}
\def\iprod#1#2{\langle#1,#2\rangle}
\def\pder#1#2{\frac{\partial #1}{\partial #2}}
\def\iint{\int\!\!\int}
\def\({\left(}
\def\){\right)}
\def\[{\left[}
\def\]{\right]}
\def\supp{\operatorname{supp}}
\def\Df{\operatorname{Df}}
\def\dom{\operatorname{dom}}
\def\Ker{\operatorname{Ker}}
\def\Tr{\operatorname{Tr}}
\def\Res{\operatorname{Res}}
\def\Aut{\operatorname{Aut}}
\def\kgV{\operatorname{kgV}}
\def\ggT{\operatorname{ggT}}
\def\diam{\operatorname{diam}}
\def\Im{\operatorname{Im}}
\def\Re{\operatorname{Re}}
\def\ord{\operatorname{ord}}
\def\rang{\operatorname{rang}}
\def\rng{\operatorname{rng}}
\def\grd{\operatorname{grd}}
\def\inv{\operatorname{inv}}
\def\maj{\operatorname{maj}}
\def\des{\operatorname{des}}
\def\varmaj{\operatorname{\overline{maj}}}
\def\vardes{\operatorname{\overline{des}}}
\def\pvarmaj{\operatorname{\overline{maj}'}}
\def\pmaj{\operatorname{maj'}}
\def\ln{\operatorname{ln}}
\def\der{\operatorname{der}}
\def\Hom{\operatorname{Hom}}
\def\tr{\operatorname{tr}}
\def\Span{\operatorname{Span}}
\def\grad{\operatorname{grad}}
\def\div{\operatorname{div}}
\def\rot{\operatorname{rot}}
\def\Sp{\operatorname{Sp}}
\def\sgn{\operatorname{sgn}}
\def\liml{\lim\limits}
\def\supl{\sup\limits}
\def\bigcupl{\bigcup\limits}
\def\bigcapl{\bigcap\limits}
\def\limsupl{\limsup\limits}
\def\liminfl{\liminf\limits}
\def\intl{\int\limits}
\def\suml{\sum\limits}
\def\maxl{\max\limits}
\def\minl{\min\limits}
\def\prodl{\prod\limits}
\def\tg{\operatorname{tan}}
\def\ctg{\operatorname{cot}}
\def\arctg{\operatorname{arctan}}
\def\arccot{\operatorname{arccot}}
\def\arcctg{\operatorname{arccot}}
\def\tgh{\operatorname{tanh}}
\def\ctgh{\operatorname{coth}}
\def\arcsinh{\operatorname{arcsinh}}
\def\arccosh{\operatorname{arccosh}}
\def\arctgh{\operatorname{arctanh}}
\def\arcctgh{\operatorname{arccoth}}
\def\3{\ss}
\catcode`\@=11
\def\dddot#1{\vbox{\ialign{##\crcr
      .\hskip-.5pt.\hskip-.5pt.\crcr\noalign{\kern1.5\p@\nointerlineskip}
      $\hfil\displaystyle{#1}\hfil$\crcr}}}

\newif\iftab@\tab@false
\newif\ifvtab@\vtab@false
\def\tab{\bgroup\tab@true\vtab@false\vst@bfalse\Strich@false%
   \def\\{\global\hline@@false%
     \ifhline@\global\hline@false\global\hline@@true\fi\cr}
   \edef\l@{\the\leftskip}\ialign\bgroup\hskip\l@##\hfil&&##\hfil\cr}
\def\endtab{\cr\egroup\egroup}
\def\vtab{\vtop\bgroup\vst@bfalse\vtab@true\tab@true\Strich@false%
   \bgroup\def\\{\cr}\ialign\bgroup&##\hfil\cr}
\def\endvtab{\cr\egroup\egroup\egroup}
\def\stab{\D@cke0.5pt\null 
 \bgroup\tab@true\vtab@false\vst@bfalse\Strich@true\Let@@\vspace@
 \normalbaselines\offinterlineskip
  \openup\spreadmlines@
 \edef\l@{\the\leftskip}\ialign
 \bgroup\hskip\l@##\hfil&&##\hfil\crcr}
\def\endstab{\crcr\egroup
 \egroup}
\newif\ifvst@b\vst@bfalse
\def\vstab{\D@cke0.5pt\null
 \vtop\bgroup\tab@true\vtab@false\vst@btrue\Strich@true\bgroup\Let@@\vspace@
 \normalbaselines\offinterlineskip
  \openup\spreadmlines@\bgroup}
\def\endvstab{\crcr\egroup\egroup
 \egroup\tab@false\Strich@false}

\newdimen\htstrut@
\htstrut@8.5\p@
\newdimen\htStrut@
\htStrut@12\p@
\newdimen\dpstrut@
\dpstrut@3.5\p@
\newdimen\dpStrut@
\dpStrut@3.5\p@
\def\openup{\afterassignment\@penup\dimen@=}
\def\@penup{\advance\lineskip\dimen@
  \advance\baselineskip\dimen@
  \advance\lineskiplimit\dimen@
  \divide\dimen@ by2
  \advance\htstrut@\dimen@
  \advance\htStrut@\dimen@
  \advance\dpstrut@\dimen@
  \advance\dpStrut@\dimen@}
\def\Let@@{\relax\iffalse{\fi%
    \def\\{\global\hline@@false%
     \ifhline@\global\hline@false\global\hline@@true\fi\cr}%
    \iffalse}\fi}
\def\matrix{\null\,\vcenter\bgroup
 \tab@false\vtab@false\vst@bfalse\Strich@false\Let@@\vspace@
 \normalbaselines\openup\spreadmlines@\ialign
 \bgroup\hfil$\m@th##$\hfil&&\quad\hfil$\m@th##$\hfil\crcr
 \Mathstrut@\crcr\noalign{\kern-\baselineskip}}
\def\endmatrix{\crcr\Mathstrut@\crcr\noalign{\kern-\baselineskip}\egroup
 \egroup\,}
\def\smatrix{\D@cke0.5pt\null\,
 \vcenter\bgroup\tab@false\vtab@false\vst@bfalse\Strich@true\Let@@\vspace@
 \normalbaselines\offinterlineskip
  \openup\spreadmlines@\ialign
 \bgroup\hfil$\m@th##$\hfil&&\quad\hfil$\m@th##$\hfil\crcr}
\def\endsmatrix{\crcr\egroup
 \egroup\,\Strich@false}
\newdimen\D@cke
\def\Dicke#1{\global\D@cke#1}
\newtoks\tabs@\tabs@{&}
\newif\ifStrich@\Strich@false
\newif\iff@rst

\def\Stricherr@{\iftab@\ifvtab@\errmessage{\noexpand\s not allowed
     here. Use \noexpand\vstab!}%
  \else\errmessage{\noexpand\s not allowed here. Use \noexpand\stab!}%
  \fi\else\errmessage{\noexpand\s not allowed
     here. Use \noexpand\smatrix!}\fi}
\def\format{\ifvst@b\else\crcr\fi\egroup\iffalse{\fi\ifnum`}=0 \fi\format@}
\def\format@#1\\{\def\preamble@{#1}%
 \def\Str@chfehlt##1{\ifx##1\s\Stricherr@\fi\ifx##1\\\let\Next\relax%
   \else\let\Next\Str@chfehlt\fi\Next}%
 \def\c{\hfil\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi\iftab@\else$\m@th\fi\the\hashtoks@\iftab@\else$\fi\hfil}%
 \def\r{\hfil\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi\iftab@\else$\m@th\fi\the\hashtoks@\iftab@\else$\fi}%
 \def\l{\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi\iftab@\else$\m@th\fi\the\hashtoks@\iftab@\else$\fi\hfil}%
 \def\s{\ifStrich@\ \the\tabs@\vrule width\D@cke\the\hashtoks@%
          \fi\the\tabs@\ }%
 \def\sa{\ifStrich@\vrule width\D@cke\the\hashtoks@%
            \the\tabs@\ %
            \fi}%
 \def\se{\ifStrich@\ \the\tabs@\vrule width\D@cke\the\hashtoks@\fi}%
 \def\cd{\hfil\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi$\dsize\m@th\the\hashtoks@$\hfil}%
 \def\rd{\hfil\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi$\dsize\m@th\the\hashtoks@$}%
 \def\ld{\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi$\dsize\m@th\the\hashtoks@$\hfil}%
 \ifStrich@\else\Str@chfehlt#1\\\fi%
 \setbox\z@\hbox{\xdef\Preamble@{\preamble@}}\ifnum`{=0 \fi\iffalse}\fi
 \ialign\bgroup\span\Preamble@\crcr}
\newif\ifhline@\hline@false
\newif\ifhline@@\hline@@false
\def\hlinefor#1{\multispan@{\strip@#1 }\leaders\hrule height\D@cke\hfill%
    \global\hline@true\ignorespaces}
\def\Item "#1"{\par\noindent\hangindent2\parindent%
  \hangafter1\setbox0\hbox{\rm#1\enspace}\ifdim\wd0>2\parindent%
  \box0\else\hbox to 2\parindent{\rm#1\hfil}\fi\ignorespaces}
\def\ITEM #1"#2"{\par\noindent\hangafter1\hangindent#1%
  \setbox0\hbox{\rm#2\enspace}\ifdim\wd0>#1%
  \box0\else\hbox to 0pt{\rm#2\hss}\hskip#1\fi\ignorespaces}
\def\item"#1"{\par\noindent\hang%
  \setbox0=\hbox{\rm#1\enspace}\ifdim\wd0>\the\parindent%
  \box0\else\hbox to \parindent{\rm#1\hfil}\enspace\fi\ignorespaces}
\let\plainitem@\item
\catcode`\@=13

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Input-Datei zum Erzeugen von Gitterpunktwegen.%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%entering a path: 
%  \Pfad(x-coordinate of starting point,y-coordinate of starting point),path
%  as 1-2-3-4-word\endPfad
%(1=step in x-direction, 2=step in y-direction, 3=upward diagonal step, 4=downward diagonal step)
%
%entering a dotted path: 
%  \SPfad(x-coordinate of starting point,y-coordinate of starting point),path
%  as 1-2-3-4-word\endSPfad
%(1=step in x-direction, 2=step in y-direction, 3=upward diagonal step, 4=downward diagonal step)
%
%coordinate-axes:
%  \Koordinatenachsen(length of positive x-axes, length of positive y-axes)(length of negative x-axes, length of negative y-axes)
%The length of negative axes are entered as negative numbers.
%
%lattice:
%  \Gitter(number of points in positive x-direction, number of points in positive y-direction)(number of points in negative x-direction, number of points in negative y-direction)
%
%diagonal lines:
%  \Diagonale(x-coordinate of SW-most point,y-coordinate of SW-most point)length of the projection on the x-axes
%
%antidiagonal lines:
%  \AntiDiagonale(x-coordinate of NW-most point,y-coordinate of NW-most point)length of the projection on the x-axes
%
%vectors:
%  \Vektor(x-coordinate of incline, y-coordinate of incline)length(x-coordinate of starting point, y-coordinate of starting point)
%
%labelling of points:
%  \Label[location?]{[label]}(x-coordinate,y-coordinate)
%where:
%  [location?]=\l,\lo,\lu,\r,\ro,\ru,\o,\u
%and l=left, r=right, u=bottom, o=top.
%In addition, if by \Einheit?cm the basic unit is changed, there exist
%\llo,\loo,\llu,\luu,\rro,\roo,\rru,\ruu.
%
%The basic unit can be changed by entering
%  \Einheit=?cm
%The default is \Einheit=0.5cm.
%
%The thickness of the paths can be changed by entering
%     \PfadDicke{?cm}
%The default is \PfadDicke=1pt.
%
%The following point sizes are available:
%\DuennPunkt, \NormalPunkt, \DickPunkt. Syntax:
%  \DickPunkt(x-coordinate,y-coordinate), etc.
%
%Besides, a circle is available by \Kreis. Syntax:
%  \Kreis(x-coordinate,y-coordinate)
%
\catcode`\@=11
\font\tenln    = line10
\font\tenlnw   = linew10

\newskip\Einheit \Einheit=0.5cm
\newcount\xcoord \newcount\ycoord
\newdimen\xdim \newdimen\ydim \newdimen\PfadD@cke \newdimen\Pfadd@cke

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%LaTeX counters, dimensions, variables for lines%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcount\@tempcnta
\newcount\@tempcntb

\newdimen\@tempdima
\newdimen\@tempdimb


\newdimen\@wholewidth
\newdimen\@halfwidth

\newcount\@xarg
\newcount\@yarg
\newcount\@yyarg
\newbox\@linechar
\newbox\@tempboxa
\newdimen\@linelen
\newdimen\@clnwd
\newdimen\@clnht

\newif\if@negarg

\def\@whilenoop#1{}
\def\@whiledim#1\do #2{\ifdim #1\relax#2\@iwhiledim{#1\relax#2}\fi}
\def\@iwhiledim#1{\ifdim #1\let\@nextwhile=\@iwhiledim
        \else\let\@nextwhile=\@whilenoop\fi\@nextwhile{#1}}

\def\@whileswnoop#1\fi{}
\def\@whilesw#1\fi#2{#1#2\@iwhilesw{#1#2}\fi\fi}
\def\@iwhilesw#1\fi{#1\let\@nextwhile=\@iwhilesw
         \else\let\@nextwhile=\@whileswnoop\fi\@nextwhile{#1}\fi}


\def\thinlines{\let\@linefnt\tenln \let\@circlefnt\tencirc
  \@wholewidth\fontdimen8\tenln \@halfwidth .5\@wholewidth}
\def\thicklines{\let\@linefnt\tenlnw \let\@circlefnt\tencircw
  \@wholewidth\fontdimen8\tenlnw \@halfwidth .5\@wholewidth}
\thinlines
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\PfadD@cke1pt \Pfadd@cke0.5pt
\def\PfadDicke#1{\PfadD@cke#1 \divide\PfadD@cke by2 \Pfadd@cke\PfadD@cke \multiply\PfadD@cke by2}
\long\def\LOOP#1\REPEAT{\def\BODY{#1}\ITERATE}
\def\ITERATE{\BODY \let\next\ITERATE \else\let\next\relax\fi \next}
\let\REPEAT=\fi
\def\Punkt{\hbox{\raise-2pt\hbox to0pt{\hss$\ssize\bullet$\hss}}}
\def\DuennPunkt(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-2.5pt\hbox to0pt{\hss$\bullet$\hss}\hss}}
\def\NormalPunkt(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-3pt\hbox to0pt{\hss\twelvepoint$\bullet$\hss}\hss}}
\def\DickPunkt(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-4pt\hbox to0pt{\hss\fourteenpoint$\bullet$\hss}\hss}}
\def\Kreis(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-4pt\hbox to0pt{\hss\fourteenpoint$\circ$\hss}\hss}}

%%%%%%%%%%%%%%%%%%%%%
%LaTeX line macros%
%%%%%%%%%%%%%%%%%%%%%
\def\Line@(#1,#2)#3{\@xarg #1\relax \@yarg #2\relax
\@linelen=#3\Einheit
\ifnum\@xarg =0 \@vline
  \else \ifnum\@yarg =0 \@hline \else \@sline\fi
\fi}

\def\@sline{\ifnum\@xarg< 0 \@negargtrue \@xarg -\@xarg \@yyarg -\@yarg
  \else \@negargfalse \@yyarg \@yarg \fi
\ifnum \@yyarg >0 \@tempcnta\@yyarg \else \@tempcnta -\@yyarg \fi
\ifnum\@tempcnta>6 \@badlinearg\@tempcnta0 \fi
\ifnum\@xarg>6 \@badlinearg\@xarg 1 \fi
\setbox\@linechar\hbox{\@linefnt\@getlinechar(\@xarg,\@yyarg)}%
\ifnum \@yarg >0 \let\@upordown\raise \@clnht\z@
   \else\let\@upordown\lower \@clnht \ht\@linechar\fi
\@clnwd=\wd\@linechar
\if@negarg \hskip -\wd\@linechar \def\@tempa{\hskip -2\wd\@linechar}\else
     \let\@tempa\relax \fi
\@whiledim \@clnwd <\@linelen \do
  {\@upordown\@clnht\copy\@linechar
   \@tempa
   \advance\@clnht \ht\@linechar
   \advance\@clnwd \wd\@linechar}%
\advance\@clnht -\ht\@linechar
\advance\@clnwd -\wd\@linechar
\@tempdima\@linelen\advance\@tempdima -\@clnwd
\@tempdimb\@tempdima\advance\@tempdimb -\wd\@linechar
\if@negarg \hskip -\@tempdimb \else \hskip \@tempdimb \fi
\multiply\@tempdima \@m
\@tempcnta \@tempdima \@tempdima \wd\@linechar \divide\@tempcnta \@tempdima
\@tempdima \ht\@linechar \multiply\@tempdima \@tempcnta
\divide\@tempdima \@m
\advance\@clnht \@tempdima
\ifdim \@linelen <\wd\@linechar
   \hskip \wd\@linechar
  \else\@upordown\@clnht\copy\@linechar\fi}

\def\@hline{\ifnum \@xarg <0 \hskip -\@linelen \fi
\vrule height\Pfadd@cke width \@linelen depth\Pfadd@cke
\ifnum \@xarg <0 \hskip -\@linelen \fi}

\def\@getlinechar(#1,#2){\@tempcnta#1\relax\multiply\@tempcnta 8
\advance\@tempcnta -9 \ifnum #2>0 \advance\@tempcnta #2\relax\else
\advance\@tempcnta -#2\relax\advance\@tempcnta 64 \fi
\char\@tempcnta}

\def\Vektor(#1,#2)#3(#4,#5){\unskip\leavevmode
  \xcoord#4\relax \ycoord#5\relax
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \Vector@(#1,#2){#3}\hss}}

\def\Vector@(#1,#2)#3{\@xarg #1\relax \@yarg #2\relax
\@tempcnta \ifnum\@xarg<0 -\@xarg\else\@xarg\fi
\ifnum\@tempcnta<5\relax
\@linelen=#3\Einheit
\ifnum\@xarg =0 \@vvector
  \else \ifnum\@yarg =0 \@hvector \else \@svector\fi
\fi
\else\@badlinearg\fi}

\def\@hvector{\@hline\hbox to 0pt{\@linefnt
\ifnum \@xarg <0 \@getlarrow(1,0)\hss\else
    \hss\@getrarrow(1,0)\fi}}

\def\@vvector{\ifnum \@yarg <0 \@downvector \else \@upvector \fi}

\def\@svector{\@sline
\@tempcnta\@yarg \ifnum\@tempcnta <0 \@tempcnta=-\@tempcnta\fi
\ifnum\@tempcnta <5
  \hskip -\wd\@linechar
  \@upordown\@clnht \hbox{\@linefnt  \if@negarg
  \@getlarrow(\@xarg,\@yyarg) \else \@getrarrow(\@xarg,\@yyarg) \fi}%
\else\@badlinearg\fi}

\def\@upline{\hbox to \z@{\hskip -.5\Pfadd@cke \vrule width \Pfadd@cke
   height \@linelen depth \z@\hss}}

\def\@downline{\hbox to \z@{\hskip -.5\Pfadd@cke \vrule width \Pfadd@cke
   height \z@ depth \@linelen \hss}}

\def\@upvector{\@upline\setbox\@tempboxa\hbox{\@linefnt\char'66}\raise
     \@linelen \hbox to\z@{\lower \ht\@tempboxa\box\@tempboxa\hss}}

\def\@downvector{\@downline\lower \@linelen
      \hbox to \z@{\@linefnt\char'77\hss}}

\def\@getlarrow(#1,#2){\ifnum #2 =\z@ \@tempcnta='33\else
\@tempcnta=#1\relax\multiply\@tempcnta \sixt@@n \advance\@tempcnta
-9 \@tempcntb=#2\relax\multiply\@tempcntb \tw@
\ifnum \@tempcntb >0 \advance\@tempcnta \@tempcntb\relax
\else\advance\@tempcnta -\@tempcntb\advance\@tempcnta 64
\fi\fi\char\@tempcnta}

\def\@getrarrow(#1,#2){\@tempcntb=#2\relax
\ifnum\@tempcntb < 0 \@tempcntb=-\@tempcntb\relax\fi
\ifcase \@tempcntb\relax \@tempcnta='55 \or
\ifnum #1<3 \@tempcnta=#1\relax\multiply\@tempcnta
24 \advance\@tempcnta -6 \else \ifnum #1=3 \@tempcnta=49
\else\@tempcnta=58 \fi\fi\or
\ifnum #1<3 \@tempcnta=#1\relax\multiply\@tempcnta
24 \advance\@tempcnta -3 \else \@tempcnta=51\fi\or
\@tempcnta=#1\relax\multiply\@tempcnta
\sixt@@n \advance\@tempcnta -\tw@ \else
\@tempcnta=#1\relax\multiply\@tempcnta
\sixt@@n \advance\@tempcnta 7 \fi\ifnum #2<0 \advance\@tempcnta 64 \fi
\char\@tempcnta}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\def\Diagonale(#1,#2)#3{\unskip\leavevmode
  \xcoord#1\relax \ycoord#2\relax
      \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit
         \Line@(1,1){#3}\hss}}
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\def\bP{{\bar P}}
\def\bT{{\bar T}}
\def\bS{{\bar S}}

\topmatter 
\title An involution principle-free bijective proof of Stanley's
hook-content formula 
\endtitle 
\author C.~Krattenthaler\footnote"$^\dagger$"{Supported in part by EC's Human
Capital and Mobility Program, grant CHRX-CT93-0400 and the\linebreak 
\hbox{Austrian Science Foundation FWF, grant P10191-PHY}}
\endauthor 
\affil 
Institut f\"ur Mathematik der Universit\"at Wien,\\
Strudlhofgasse 4, A-1090 Wien, Austria.\\
e-mail: KRATT\@Pap.Univie.Ac.At\\
WWW: \tt http://radon.mat.univie.ac.at/People/kratt
\endaffil
\address Institut f\"ur Mathematik der Universit\"at Wien,
Strudlhofgasse 4, A-1090 Wien, Austria.
\endaddress
%\email KRATT\@Pap.Univie.Ac.At\endemail
%\dedicatory \enddedicatory
%\date \enddate
%\thanks \endthanks
\subjclass Primary 05A15;
 Secondary 05A17, 05A30, 05E10, 05E15, 11P81
\endsubjclass
\keywords Plane partitions, tableaux, hook-content formula,
Hillman--Grassl algorithm, jeu de taquin, involution principle
\endkeywords
\abstract 
A bijective proof for Stanley's hook-content formula for the generating
function for column-strict reverse plane partitions of a given shape 
is given that does not involve the involution principle of Garsia and
Milne. It is based on the Hillman--Grassl algorithm and
Sch\"utzenberger's jeu de taquin.


\endabstract
\endtopmatter
\document

\subhead 1. Introduction\endsubhead
The purpose of this article is to give a bijective proof for 
Stanley's hook-content formula \cite{\StanAA, Theorem~15.3} for a
certain plane partition generating function. In order to 
be able to state the formula we have to recall some basic notions
from partition theory. A {\it partition\/} is a sequence
$\la=(\la_1,\la_2,\dots,\la_r)$ with $\la_1\ge \la_2\ge\dots\ge\la_r>0$,
for some $r$. The {\it Ferrers
diagram\/} of $\lambda$
is an array of cells with $r$ left-justified rows and $\lambda_i$
cells in row $i$. Figure~1.a shows the Ferrers
diagram corresponding to $(4,3,3,1)$.
The {\it conjugate of\/} $\lambda$ is the partition 
$(\lambda^\prime_1, \dots, \lambda^\prime_{\lambda_1})$ where
$\lambda_j'$ is the length of the $j$-th column in the Ferrers diagram of
$\lambda$. We label the cell in the $i$-th row and $j$-th column of
(the Ferrers diagram of) $\la$ by the pair $(i,j)$. Also, if we write
$\rh\in\la$ we mean `$\rh$ is a cell of $\la$'. The {\it hook length\/}
$h_\rh$ of a cell $\rh=(i,j)$ of $\la$ is $(\la_i-j)+(\la'_j-i)+1$, \
the number of cells in the
{\it hook\/} of $\rh$, which is the set of cells that are either in
the same row as $\rh$ and to the right of $\rh$, or in the same column
as $\rh$ and below $\rh$, $\rh$ included. The {\it content\/} $c_\rh$
of a cell $\rh=(i,j)$ of $\la$ is $j-i$.
\vskip10pt
\vbox{
$$
\gather
\smatrix \format \sa\c\quad \s\c\quad \s\c\quad \s\c\quad \se\\
\hlinefor9\\
&\vphantom{f}&& && && &\\
\hlinefor9\\
&\vphantom{f}&& && &\\
\hlinefor7\\
&\vphantom{f}&& && &\\
\hlinefor7\\
&\vphantom{f} &\\
\hlinefor3
\endsmatrix
\hskip1cm
\smatrix \format\sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&\hbox to10pt{\hss 1\hss}&&\hbox to10pt{\hss 3\hss}&&\hbox to10pt{\hss
4\hss}&&\hbox to10pt{\hss 4\hss}&\\
\hlinefor9\\
&5&&5&&7&\\
\hlinefor7\\
&5&&6&&7&\\
\hlinefor7\\
&8&\\
\hlinefor3
\endsmatrix
\hskip1cm
\smatrix \format\sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&\hbox to10pt{\hss 1\hss}&&\hbox to10pt{\hss 3\hss}&&\hbox to10pt{\hss
4\hss}&&\hbox to10pt{\hss 4\hss}&\\
\hlinefor9\\
&4&&6&&7&\\
\hlinefor7\\
&5&&7&&8&\\
\hlinefor7\\
&8&\\
\hlinefor3
\endsmatrix\\
\smallmatrix\format\l\\
 \text{\eightpoint a. Ferrers diagram\quad \quad 
b. reverse plane\quad \quad 
c. column-strict reverse}\\%
\text{\eightpoint \hphantom{a. Ferrers diagram\quad \quad 
b. }\hbox to0pt{partition\hss}%
\hphantom{reverse plane\quad \quad
c. }plane partition }
\endsmallmatrix
\endgather
$$
\centerline{\eightpoint Figure 1}
}
\vskip10pt

Given a partition $\la=(\la_1,\la_2,\dots,\la_r)$, a {\it reverse plane
partition of shape\/} $\la$ is a filling $P$ of the cells of $\la$
with integers such that the entries along rows and along columns are
weakly increasing. Figure~1.b displays a reverse plane
partition of shape $(4,3,3,1)$. A reverse plane partition is called {\it
column-strict\/} if in addition columns are {\it strictly\/}
increasing. Figure~1.c displays a column-strict reverse plane partition of
shape $(4,3,3,1)$. We write $P_\rh$ for the entry in cell $\rh$ of $P$. We
call the sum of all the entries of a reverse plane partition $P$ the
{\it norm\/} of $P$, and denote it by $n(P)$.

Now we are in the position to state Stanley's hook-content formula
\cite{\StanAA, Theorem~15.3}. 

\proclaim{Theorem 1 (Stanley)}Let $\la=(\la_1,\la_2,\dots,\la_r)$ be a
partition and $a$ be an integer $\ge r$. The generating function 
$\sum _{} ^{} q^{n(P)}$, where the sum is over all column-strict reverse plane
partitions $P$ of shape $\la$ with entries between $1$ and $a$, is given by
$$q^{\sum _{i=1} ^{r}i\la_i}\prod _{\rh\in\la} ^{}\frac
{1-q^{a+c_\rh}} {1-q^{h_\rh}}.\tag1.1$$ 
\endproclaim

Stanley proved this theorem by showing that the
generating function in question equals a determinant, and then
evaluated the determinant. However, such a proof does not explain why
the generating function equals such a nice product. In particular, it
does not give any clue why in (1.1) the hook lengths and contents appear.
The desire to have an explanation for these phenomenons provides the
motivation for the search for a bijective proof of this result.
A bijective proof for Stanley's Theorem~1 was given earlier by Remmel and
Whitney \cite{\ReWhAA}. Though being a significant advance, one cannot
claim that this proof is really enlightening or explains formula
(1.1) in a satisfying way. Aside from making use of the
involution principle of Garsia and Milne \cite{\GaMiAB} (which creates
bijections in an indirect way), it 
was based on bijections that mimicked recurrence relations, which is
certainly not the most direct route to attack the problem. Our proof
of Theorem~1 explains the appearance of hook lengths and contents
in a straight-forward way. It is based on the Hillman--Grassl algorithm
\cite{\HiGrAA} and on Sch\"utzenberger's \cite{\SchuAA} jeu de taquin. It does
not need the involution principle. However, the careful reader will notice
that we set up a bijection between two sets of objects that are
different  from those for which Remmel and Whitney set up their
bijection. In order to find a bijection between the sets that Remmel
and Whitney consider, also we have to use the involution principle.
However, the resulting bijection is considerably simpler than Remmel
and Whitney's.

We remark that a bijective proof of Theorem~1 for one-rowed shapes
(i.e\. the case 
of (ordinary) partitions with $\le \la_1$ parts, each part $\le a$)
can be found in Sagan's paper \cite{\SagaAB, proof of Theorem~8}.
However, this proof is different from our proof when restricted to
one-rowed shapes. Besides, this proof does not seem to generalize to arbitrary
shapes. In the same paper, which has bijective proofs of hook formulas
as its theme, also the problem of finding a bijective proof of
MacMahon's product formula
\cite{\MacMAA, sec.~429, proof in
sec.~494} for the generating function of plane partitions of
rectangular shapes with bounded entries is posed. Since MacMahon's
formula is actually contained in Theorem~1 (which is seen by an easy
transformation of plane partitions into column-strict reverse plane
partitions), our bijection also provides a solution to this problem.
See also section~4.


Our paper is organized as follows. In section~2 our bijective proof of
Theorem~1 is described. (A complete example for our bijection is carried out in the appendix of this paper). Then, in section~3, we discuss some
related bijections. In particular, it is there where we explain how
our bijection
described in section~2 can be used to provide a much simpler
(involution principle-based) bijection between the sets that Remmel and
Whitney use in their bijective proof \cite{\ReWhAA} of Theorem~1.
Finally, in section~4, we list some more plane partition formulas that
also desire to receive bijective proofs.

In conclusion of the Introduction, a few comments on the relation of
the present work to the recent beautiful bijective proof
\cite{\NoPSAA} of the
Frame--Robinson--Thrall hook formula for the number of standard Young
tableaux of a given shape by Novelli, Pak and Stoyanovskii
(announced in \cite{\PaStAA}) are in order. Clearly, the
Novelli--Pak--Stoyanovskii algorithm as well as our algorithm
described in Section~2 are based on jeu de taquin (a modified jeu de taquin in
our case). Still, I am not able to name an immediate, direct
relation. However, I discovered that it is possible to merge the
Novelli--Pak--Stoyanovskii idea of how to keep track of the hooks
with the modified jeu de taquin idea of this paper to obtain a new
bijective proof of the hook-content formula (1.1), which also avoids the
involution principle. (In fact, it sets up a bijection between the
sets Remmel and Whitney consider in their paper \cite{\ReWhAA}.) This
will be the subject of a forthcoming publication \cite{\KratBK}.

\subhead 2. Bijective proof of Stanley's hook-content
formula\endsubhead
First, we rewrite (1.1) in the form (here CRPP is short for `column-strict
reverse plane partition')
$$\Big(\underset \text{with }1\le \text{entries}\le a\to {\sum _{P
\text{ a CRPP of 
shape $\la$}}} ^{}q^{n(P)}\Big)\cdot\prod _{\rh\in\la} ^{}\frac {1}
{1-q^{a+c_\rh}}
= q^{\sum _{i=1} ^{r}i\la_i}\prod _{\rh\in\la} ^{}\frac {1}
{1-q^{h_\rh}}.
\tag2.1$$

Let us call an {\it arbitrary\/} filling of the cells of $\la$ with
{\it nonnegative\/} integers a {\it tabloid of shape\/} $\la$. Furthermore,
let us define the {\it hook weight\/} $w_h(T)$ of a tabloid $T$ of
shape $\la$ by $\sum _{\rh\in\la} ^{}T_\rh\cdot h_\rh$, and the
{\it content weight\/} $w_c(T)$ of a tabloid $T$ by $\sum _{\rh\in\la}
^{}T_\rh\cdot(a+c_\rh)$. Then the right-hand side of (2.1) is the
generating function $\sum _{} ^{}q^{n(P_0)}q^{w_h(T_R)}$, where the sum
is over all pairs $(P_0,T_R)$, with $P_0$ being the ``minimal"
column-strict reverse plane partition of shape $\la$ with entries
$\ge1$, i.e\. the 
column-strict reverse plane partition with all entries in row $i$
equal to $i$ for all $i$, and with $T_R$ varying over all tabloids of
shape $\la$. Similarly, the left-hand side of (2.1) is the generating
function $\sum _{} ^{}q^{n(P_L)}q^{w_c(T_L)}$, where the sum is over all
pairs $(P_L,T_L)$, with $P_L$ varying over all column-strict reverse
plane partitions of shape $\la$ with entries between $1$ and $a$ and
$T_L$ varying over all tabloids of shape $\la$. So the task is to set
up a bijection that maps a right-hand side pair $(P_0,T_R)$ to a
left-hand side pair $(P_L,T_L)$, such that
$n(P_0)+w_h(T_R)=n(P_L)+w_c(T_L)$.

One step in our bijection was already done much earlier. In their
celebrated paper \cite{\HiGrAA}, Hillman and Grassl constructed an
algorithmic bijection between tabloids $T_R$ of shape $\la$ and reverse
plane partitions $\bar P_R$ of shape $\la$ with nonnegative entries
such that $n(\bar P_R)=w_h(T_R)$. If we add such a reverse plane
partition $\bar P_R$ to $P_0$ cell-wise, then we obtain a
{\it column-strict\/} reverse plane partition $P_R$ of shape $\la$ with
entries $\ge 1$, and we have $n(P_R)=n(P_0)+n(\bar
P_R)=n(P_0)+w_h(T_R)$. Therefore the new task is to set up a bijection
that maps a column-strict reverse plane partition of shape $\la$ with
entries $\ge 1$ to a pair $(P_L,T_L)$, where $P_L$ is a column-strict
reverse plane partition of shape $\la$ with entries between $1$ and
$a$, and where $T_L$ is a tabloid of shape $\la$, such that
$$n(P_R)=n(P_L)+w_c(T_L).\tag2.2$$

We claim that the following algorithm performs this task.

\bigskip
\noindent
{\bf Algorithm C}. The input for the algorithm is a column-strict
reverse plane partition $P_R$ of shape $\la$ with entries $\ge1$.

(C0) Set $(P,T):=(P_R,\bold 0)$, where $\bold 0$ denotes the tabloid
of shape $\la$ with $0$ in each cell.

\smallskip
(C1) If $P$ does not contain any entry $>a$ then stop. The output of
the algorithm is $(P,T)$.

Otherwise, consider all {\it corner cells\/} of $\la$ (which are the
cells with no right and bottom neighbour cells). Choose all corner cells that
contain the maximal entry of $P$, and among all these pick the
left-most, cell $\om$ say. (Note that the maximal entry of $P$ {\it
must\/} appear in a corner cell of $P$ since $P$ is a column-strict
reverse plane partition. Hence, in our
situation it must be $>a$.) Replace the entry $P_\om$ in cell $\om$ by
$P_\om-(a+c_\om)$. Call this entry {\it special\/}. Continue with (C2).

\smallskip
(C2) If a column-strict reverse plane partition is obtained then
continue with (C3).

If not, i.e\. if the special entry, $s$ say, violates increase along
rows or strict increase along columns, then we have the following
situation,
$$\smatrix \format \sa\c\s\c\se\\
\omit&\omit&\hlinefor3\\
\omit&&&y&\\
\hlinefor5\\
&x&&s&\\
\hlinefor5
\endsmatrix\ ,\tag2.3
$$
where at least one of $x>s$ and $y\ge s$ holds. (One of $x$ or $y$ is
also allowed to 
be actually not there.) If $x>y$ then do the move
$$\smatrix \format \sa\c\s\c\se\\
\omit&\omit&\hlinefor3\\
\omit&&&y&\\
\hlinefor5\\
&s+1&&x&\\
\hlinefor5
\endsmatrix\ .\tag2.4
$$
If $x\le y$ then do the move
$$\smatrix \format \sa\c\s\c\se\\
\omit&\omit&\hlinefor3\\
\omit&&&s-1&\\
\hlinefor5\\
&x&&y&\\
\hlinefor5
\endsmatrix\ .\tag2.5
$$
The new special entry in (2.4) is $s+1$, the new special entry in
(2.5) is $s-1$. Repeat (C2). (Note that always after either type of
move the only possible violations of increase along rows or strict
increase along columns involve the new special entry and the entry to
the left or/and above.)

\smallskip
(C3) Let $P$ be the column-strict reverse plane partition just
obtained. If we ended up with the special entry in cell $\rh$ then add
$1$ to the entry in cell $\rh$ of $T$. The tabloid thus obtained is
the new $T$. Continue with (C1).\quad \quad \qed

\bigskip
\example{Example C}A complete example for Algorithm~C can be found in
the appendix. There we choose $a=4$ and map the column-strict reverse plane
partition of shape $(4,3,3,1)$ on the left of Figure~2 to the pair on
the right of Figure~2, consisting of a column-strict reverse plane partition
of shape $(4,3,3,1)$ with entries $\le4$ and a tabloid of shape
$(4,3,3,1)$, such that the
weight property (2.2) holds. In fact, the norm of the column-strict
reverse plane partition on the left of Figure~2 is $49$, while the norm of the
column-strict reverse plane partition on the right is $26$ and the
content weight of the tabloid is $23$.
\vskip10pt
\vbox{
$$
\matrix \format \c&\quad \c\quad &\c\\
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&3&&8&\\
\hlinefor9\\
&2&&3&&4&\\
\hlinefor7\\
&5&&7&&8&\\
\hlinefor7\\
&7&\\
\hlinefor3
\endsmatrix
&\longleftrightarrow&
\(
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&1&&3&\\
\hlinefor9\\
&2&&2&&2&\\
\hlinefor7\\
&3&&3&&4&\\
\hlinefor7\\
&4&\\
\hlinefor3
\endsmatrix
, 
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&0&&1&&0&\\
\hlinefor9\\
&1&&0&&0&\\
\hlinefor7\\
&3&&1&&0&\\
\hlinefor7\\
&1&\\
\hlinefor3
\endsmatrix
\)\\
n(.)=49& &n(.)=26\ ,\ w_c(.)=23
\endmatrix
$$
\centerline{\eightpoint Figure 2}
}
\vskip10pt
The appendix has to be read in the following way. First of all, ignore
all double circles, and all even rows in the right columns. What the
left columns show is the pair $(P,T)$ that is obtained after each loop
(C1)-(C2)-(C3). Together with the pair $(P,T)$ a filling of the shape
$(4,3,3,1)$ is displayed that 
shows all values $P_\rh+4+c_\rh$ for all cells $\rh$ with $T_\rh\ne0$.
This will be important for understanding Lemma~C but can be ignored for
the moment.
At each stage, the entry that is chosen by (C1) is circled. Then each
intermediate step during the loop (C1)-(C2)-(C3) is displayed in the
odd rows of the right columns. The special entry is always
underlined. When a column-strict reverse plane partition is reached,
the special entry is boxed. The entry in the
corresponding cell of the tabloid is subsequently increased by $1$ in
step (C3).\quad \quad \qed
\endexample

It should be noticed that, aside from adding/subtracting $1$ to/from
the special entry, what happens from (2.3) to (2.4), respectively
(2.5), is a {\it jeu de taquin forward move\/} (cf\. \cite{\SchuAA,
sec.~2; \SagaAL, pp.~120/169}).

It is obvious that this algorithm maps $P_R$ to a pair $(P_L,T_L)$,
where $T_L$ is a tabloid of shape $\la$ and $P_L$ is a column-strict
reverse plane partition of shape $\la$ with entries $\le a$. In fact,
the entries of $P_L$ have to be $\ge1$. This is seen as follows. The
only problem could arise with our special entry. However, when we arrive
at (C3), a special
entry $\le0$ can only occur in cell $(1,1)$, because otherwise step
(C2) was not finished. Each loop (C1)-(C2)-(C3) of the algorithm
starts with some entry $P_{\om}>a$ in a corner cell $\om$. It is replaced
by $P_{\om}-(a+c_{\om})$. Then it is (possibly) moved according to (2.4) and
(2.5). It is easy to check that at each stage
during performing the steps (C2), the special entry, if located in
cell $\rh$, will equal $P_\om-(a+c_\rh)$. This is a property so
important that it has to be recorded for later use,
$$\text{(special entry in $\rh$)}=P_\om-(a+c_\rh).\tag2.6$$
Suppose we reach cell $(1,1)$. 
When we arrive at $(1,1)$, by
(2.6) and since $c_{(1,1)}=0$, our special entry has become
$P_{\om}-a$. But this is $\ge1$ since $P_{\om}>a$.

So we have an algorithm that maps right-hand side objects $P_R$ of
(2.1) to left-hand side objects $(P_L,T_L)$ of (2.1). Besides, this
mapping satisfies the weight property (2.2). This is immediate from
(2.6). 

What remains is to establish that our algorithm is actually a
{\it bijection\/} between right-hand side and left-hand side objects. This
will be accomplished by constructing an algorithm, Algorithm~C* below,
that will turn out to be the inverse of Algorithm~C. We could exhibit
Algorithm~C* immediately. However, we prefer to provide motivation
for the
definition of Algorithm~C* first, in form of the following lemma. On the
other hand, readers who are not interested in the details can safely
skip the lemma and jump to the description of Algorithm~C* at this
point.

\proclaim{Lemma C}Let $(P,T)$ be obtained after some loop
{\rm (C1)-(C2)-(C3)} during Algorithm~C. Suppose that the loop terminated in
cell $\ga$ when reaching (C3). Then among all cells $\rh$ with
$T_\rh\ne0$, $\ga$ is a 
cell for which the value $P_\rh+a+c_\rh$ attains its minimum, and if
there are several cells $\rh$ with $T_\rh\ne0$ where the minimum is
attained, then $\ga$ is the right-most and top-most of those. Besides,
there holds
$$P_\ga+a+c_\ga\ge \max\{\text{\rm entries in }P\}.\tag2.7$$
\endproclaim
\demo{Proof}We prove the assertions by induction on the number of
loops (C1)-(C2)-(C3).

The assertions are certainly true for the pair $(\bP,\bT)$ obtained after
the very first loop (C1)-(C2)-(C3) from $(P_R,\bold 0)$. This is
because there is only one cell $\ze$ in $\bT$ with $\bT_\ze\ne0$, and,
what regards (2.7), because 
$$\bP_\ze+a+c_\ze=\max\{\text{entries in
}P_R\}\ge \max\{\text{entries in }\bP\},\tag2.8$$ 
the equality holding because
of (2.6), the inequality holding because of the following facts.
In the transition from $P_R$
to $\bP$ the multiset of entries remains the same, except for one
entry, $(P_R)_\om$, where $\om$ is the corner cell that is chosen when
applying (C1) to $P_R$. At the beginning of the loop (C1)-(C2)-(C3)
leading from $P_R$ to $\bP$, $(P_R)_\om$ is replaced by $(P_R)_\om-(a+c_\om)$,
which is less than $(P_R)_\om$ because of $a\ge
r>\max\{-c_\rh:\rh\in\la\}$ 
(recall that $a\ge r$ is one of the assumptions in Theorem~1.) Then
the special entry $(P_R)_\om-(a+c_\om)$ is (possibly) moved inwards
according to (2.4) and (2.5). At the end of the loop (C1)-(C2)-(C3) a
column-strict reverse plane partition is obtained, therefore the
special entry in the end has to be $\le\max\{\text{entries in }P_R\}$ in
any case.

\vskip10pt
\vbox{
$$
\underset
\text {from $\om$ to $\ze$, and the four regions determined by $\ga$.}
\to{
\underset
\text {The cells $\ga$, $\ze$, $\om$, the jeu de taquin path}
\to{
\Einheit.15cm
\PfadDicke{.5pt}
\Pfad(0,0),111111222211111111222222111111222211112222222222\endPfad
\Pfad(0,0),222222222222222222222222111111111111111111111111\endPfad
\PfadDicke{.75pt}
\Pfad(15,11),111\endPfad
\Pfad(15,11),2222\endPfad
\Pfad(11,15),1111\endPfad
\Pfad(11,15),222\endPfad
\SPfad(6,13),11111111111111\endSPfad
\SPfad(5,0),22222222222\endSPfad
\SPfad(0,13),1111\endSPfad
\SPfad(5,14),222222222\endSPfad
\PfadDicke{1pt}
\Pfad(18,10),1122\endPfad
\Pfad(18,10),2211\endPfad
\Pfad(4,12),1122\endPfad
\Pfad(4,12),2211\endPfad
\Pfad(10,18),1122\endPfad
\Pfad(10,18),2211\endPfad
\Label\lu{\ga\kern5pt}(4,12)
\Label\r{\om\kern5pt}(20,11)
\Label\ro{\kern5pt\ze}(12,20)
\Label\u{\kern2pt}(0,-1)
\hskip3.6cm
}}
$$
\centerline{\eightpoint Figure 3}
}
\vskip10pt

So, let us assume that the assertions are true
for $(P,T)$, obtained after some loop (C1)-(C2)-(C3). Let $\ga$ be the
cell where the last loop, which gave rise to $(P,T)$, terminated
at (C3). Let $\om$ be the cell where the next loop starts, i.e\. the
corner cell of $P$ chosen by (C1), and let $\ze$ be the cell where this
next loop terminates at (C3). See Figure~3. 
Let $(\bP, \bT)$ be the outcome after
this loop. Note that by construction the cells $\rh$ with $\bT_\rh\ne0$ are
$\ze$ and the cells with $T_\rh\ne0$. In particular, $T_\rh\ne0$
implies $\bT_\rh\ne0$.

\smallskip
First we prove (2.7) for $\bP$ and $\ze$. By (2.6),
the entry of $\bP$ in cell $\ze$ is 
$$\bP_\ze=P_\om-(a+c_\ze).\tag2.9$$
This immediately implies
$$\bP_\ze+a+c_\ze=P_\om=\max\{\text{entries in }P\}\ge
\max\{\text{entries in }\bP\}.\tag2.10$$
The last inequality follows from the arguments that proved
(2.8), just replace $P_R$ by $P$ in the paragraph after (2.8).
Obviously, (2.10) proves (2.7) for $\bP$ and $\ze$, as desired.

\smallskip
Next we show that $\bP_\rh+a+c_\rh$, evaluated at cells $\rh$ with
$\bT_\rh\ne0$, attains its minimal value at $\ze$. By induction
hypothesis, (2.7) holds for $P$ and $\ga$. Hence we have
$$P_\ga+a+c_\ga\ge\max\{\text{entries in }P\}=P_\om=\bP_\ze+a+c_\ze.
\tag2.11$$
Let $\rh$ be any cell with $\bT_\rh\ne0$. Recall that this means $\rh=\ze$ or
$T_\rh\ne0$. We want to show
$$\bP_\rh+a+c_\rh\ge \bP_\ze+a+c_\ze.\tag2.12$$
If $\rh=\ze$ then there is nothing to show. So let $\rh\ne \ze$. Then we
have $T_\rh\ne0$. By induction hypothesis for $\ga$, we have
$P_\ga+a+c_\ga\le P_\rh+a+c_\rh$ for all cells with $T_\rh\ne0$. So,
if we suppose $P_\rh=\bP_\rh$, then
we conclude, using (2.11),
$$\bP_\rh+a+c_\rh=P_\rh+a+c_\rh\ge P_\ga+a+c_\ga\ge \bP_\ze+a+c_\ze,$$
which verifies (2.12) in this case.
However, the only entries that are changed during the
performance of the loop (C1)-(C2)-(C3) are located in cells (weakly)
to the right and (weakly) below of cell $\ze$, see Figure~3. 
For these cells there holds
the following basic computation. For convenience, let $\ze=(i_1,j_1)$, and let
$\rh=(i_2,j_2)$ be in this region to the right and below of $\ze$, i.e\.
$i_1\le i_2$ and $j_1\le j_2$. Then, since $\bP$ is a column-strict
reverse plane partition, we have
$$\align \bP_\rh+a+c_\rh&=\bP_\rh+a+(j_2-i_2)\\
&\ge (\bP_\ze+i_2-i_1)+a+(j_2-i_2)=\bP_\ze+a+j_2-i_1\\
&\ge \bP_\ze+a+j_1-i_1=\bP_\ze+a+c_\ze.
\tag2.13
\endalign$$
Therefore the value $\bP_\rh+a+c_\rh$ for {\it any\/} cell $\rh$ in
the region to the right and below of $\ze$ is at least $\bP_\ze+a+c_\ze$,
so also for the cells with $\bT_\rh\ne0$, which verifies (2.12) in
this case, too. For later reference
we remark that the computation (2.13) also shows that the only cells in
this region for which we could have equality are in the same column as
$\ze$.

\smallskip
Finally we show that $\ze$ is the right-most and top-most among all
cells $\rh$ with $\bT_\rh\ne0$ where $\bP_\rh+a+c_\rh$ attains its minimal
value. Let $\rh$ be a cell with $\bT_\rh\ne0$ where $\bP_\rh+a+c_\rh$
attains its minimal value. In particular, we have
$$\bP_\rh+a+c_\rh=\bP_\ze+a+c_\ze.\tag2.14$$
First, suppose that $\rh$ is a cell with $\bP_\rh\ne P_\rh$.
Then $\rh$ has to be located in the region (weakly) to the right and
(weakly) below of $\ze$. As we noted after the computation (2.13), the
only cells $\rh$ in this region for which we could have equality in
(2.13) lie in the same column as $\ze$. $\ze$ is the top-most and
right-most of these, in agreement with our claim.

The more delicate case is when $\rh$ is a cell with $\bT_\rh\ne0$ and
$\bP_\rh=P_\rh$. Of course, nothing is to show for $\rh=\ze$, so we
may assume $\rh\ne\ze$. This implies $T_\rh\ne0$, and so
$\bP_\rh+a+c_\rh=P_\rh+a+c_\rh\ge P_\ga+a+c_\ga$, the inequality
holding by induction hypothesis for $\ga$. Combining this with (2.11)
and (2.14) we are forced to conclude
$$P_\rh+a+c_\rh=P_\ga+a+c_\ga=\bP_\ze+a+c_\ze.\tag2.15$$

We shall show that $\ze$ has to lie in the region (weakly) to the right
and (weakly) above of $\ga$ (as is indicated in Figure~3). 
Since $\ga$ was the right-most and
top-most of all the cells $\rh$ with $T_\rh\ne0$ where the minimal
value of $P_\rh+a+c_\rh$ is attained, this would establish that $\ze$
is the right-most and top-most of all the cells $\rh$ with
$\bT_\rh\ne0$ where the minimal value of $\bP_\rh+a+c_\rh$ is
attained, as desired.

We prove the claim of the previous paragraph by excluding the other
three quarter regions that are determined by the horizontal line and the
vertical line running through $\ga$, see Figure~3. 

First, suppose that $\ze$ lies in the region strictly to the right and
(weakly) below of $\ga$. Then $P_\ga=\bP_\ga$, since $\ga$ was not met
during the loop (C1)-(C2)-(C3) leading from $P$ to $\bP$. However,
then the basic computation (2.13) applies with 
$\ze$ replaced by $\ga$, and 
$\rh$ replaced by $\ze$. Since we assumed that $\ze$ is strictly
to the right of $\ga$, the remark below (2.13) tells that actually
strict inequality in (2.13), with the above replacements, holds, i.e\.
$\bP_\ze+a+c_\ze>\bP_\ga+a+c_\ga$. This contradicts (2.15) because of
$P_\ga=\bP_\ga$. Thus, this region is excluded.

Next we show that $\ze$ cannot lie in the region (weakly) to the left
and (weakly) below of $\ga$, $\ga$ excluded. This would follow
immediately from the claim that if two successive loops (C1)-(C2)-(C3)
start with the same size of entry in the corner cells chosen by (C1)
(which applies in our case since the loop that lead to $P$ started
with an entry $P_\ga+a+c_\ga$ in some corner cell, and the loop that
lead from $P$ to $\bP$ started with $\bP_\ze+a+c_\ze$, both
quantities being the same by (2.15)) then the second path of moves has
to stay to the right of the first path of moves.

To check this claim, once again note that both loops started with the
same size of entries in the cells chosen by (C1). By the rules in (C1),
this means that either the 
second loop started strictly to the right of the first, or we started
in the same cell, where in this case the first loop started with a
left move (2.4). (Note that if we start with an upward move (2.5)
then, by column-strictness, a smaller entry is moved into the corner
cell than has been there before.) Now, it is an easy-to-check property
of our modified forward jeu de taquin (C2) that if the second ``jeu de
taquin path" is to the right of the first ``jeu de taquin path" somewhere,
then it has to stay to the right from thereon. To make this precise,
suppose that during the first loop the special entry, $s_1$ say, went
up by (2.5), see the left half of Figure~4. (The arrows mark the
direction of move of the special entry.)
\vskip10pt
\vbox{
$$
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&y&&z&\\
\hlinefor5\\
&s_1&&*&\\
\hlinefor5
\endsmatrix
\overset \text{during}\to{\overset \text{first loop}\to\longrightarrow}
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&*&&z&\\
\hlinefor5\\
&\vbox to10pt{\hsize5.27pt \vss \noindent$\uparrow$\newline$y$\vphantom{f}}&&*&\\
\hlinefor5
\endsmatrix
\hskip2cm
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&*&&z&\\
\hlinefor5\\
&y&&s_2&\\
\hlinefor5
\endsmatrix
\overset \text{during}\to {\overset \text{second loop}\to\longrightarrow}
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&*&&*&\\
\hlinefor5\\
&y&&\vbox to10pt{\hsize5.1pt \vss \noindent$\uparrow$\newline$z$\vphantom{f}}&\\
\hlinefor5
\endsmatrix
$$
\centerline{\eightpoint Figure 4}
}
\vskip10pt
\noindent
Since rows are weakly increasing, we have $y\le z$. Suppose that
during the second loop we reach the cell neighbouring $y$ and $z$ with
a special entry $s_2$, see the right half of Figure~4. Then the
definition of the algorithm forces us to stop here or 
to move up in the next step
(C2). We already checked that the second ``jeu de taquin path" starts
to the right of the first, therefore it has to stay to the right
always. So also this region is excluded. 

Finally, we examine if $\ze$ could be located in the region strictly
to the left and (weakly) above of $\ga$. 
Once
more the computation (2.13), with $\rh$
replaced by $\ga$, applies and, together with the remark below (2.13)
(note that we assumed $\ga$ to be strictly to the right of $\ze$),
implies $\bP_\ga+a+c_\ga>\bP_\ze+a+c_\ze$, which contradicts (2.15)
because of the simple fact $P_\ga\ge \bP_\ga$.

This completes the proof of the Lemma.\quad \quad \qed
\enddemo

From Lemma~C it is pretty obvious what the inverse algorithm of
Algorithm~C could be.

\bigskip
\noindent
{\bf Algorithm C*}. The input for the algorithm is a pair $(P_L,T_L)$,
where $P_L$ is a column-strict
reverse plane partition of shape $\la$ with entries between $1$ and
$a$, and where $T_L$ is a tabloid of shape $\la$.

(C*0) Set $(P,T):=(P_L,T_L)$.

\smallskip
(C*1) If $T=\bold 0$ then stop. The output of
the algorithm is $P$.

Otherwise, consider all cells $\rh$ with $T_\rh\ne0$. Among these
choose the cells for which $P_\rh+a+c_\rh$ is minimal, and among all
these pick the right-most and top-most, cell $\ze$ say. (Observe that
among two cells
attaining the same value of $P_\rh+a+c_\rh$ one is always (weakly) to
the right and (weakly) above of the other, again because of the
computation (2.13), with $\bP$ replaced by $P$, and the remark below
(2.13). So the right-most and top-most of these does exist.) 
Replace the entry $P_\ze$ in cell $\ze$ by
$P_\ze+a+c_\ze$. Call this entry {\it special\/}. Continue with (C*2).

\smallskip
(C*2) If the special entry, $s$ say, is located in a corner cell of
$\la$ then continue with (C*3).

If not, then we have the following
situation,
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&s&&x&\\
\hlinefor5\\
&y&\\
\hlinefor3
\endsmatrix\ .\tag2.16
$$
(One of $x$ or $y$ is
also allowed to 
be actually not there.) If $x<y$ then do the move
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&x&&s&\\
\hlinefor5\\
&y&\\
\hlinefor3
\endsmatrix\ .\tag2.17
$$
If $x\ge y$ then do the move
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&y&&x&\\
\hlinefor5\\
&s&\\
\hlinefor3
\endsmatrix\ .\tag2.18
$$
The new special entry in either case is $s$.
Repeat (C*2). 

\smallskip
(C*3) Let $P$ be the column-strict reverse plane partition just
obtained. (The fact that indeed a column-strict reverse plane
partition is obtained will be proved in the subsequent
Lemma~C*.) Subtract $1$ from the entry in cell $\ze$ of $T$.
The tabloid thus obtained is
the new $T$. Continue with (C*1).\quad \quad \qed

\bigskip
\example{Example C*}A complete example for Algorithm~C* can be found in
the appendix. There we choose $a=4$ and map the pair on
the right of Figure~2, consisting of a column-strict reverse plane partition
of shape $(4,3,3,1)$ with entries $\le4$ and a tabloid of shape
$(4,3,3,1)$, to the column-strict reverse plane
partition of shape $(4,3,3,1)$ on the left of Figure~2, such that the
weight property (2.2) holds. It is simply the inverse of the example
for Algorithm~C given in Example~C. Therefore, here the appendix has to
be read in the reverse direction,
and in the following way. First of all, ignore
all single circles, and all odd rows in the right columns. What the
left columns show is the pair $(P,T)$ that is obtained after each loop
(C*1)-(C*2)-(C*3) together with a filling of the shape $(4,3,3,1)$ that
shows all values $P_\rh+4+c_\rh$ for all cells $\rh$ with $T_\rh\ne0$.
At each stage, the entry that is chosen by (C*1) is doubly circled. Then each
intermediate step during the loop (C*1)-(C*2)-(C*3) is displayed in the
even rows of the right columns. The special entry is always
doubly underlined. When a column-strict reverse plane partition is reached,
the special entry is doubly boxed. The entry in the
corresponding cell of the tabloid is subsequently decreased by $1$ in
step (C*3).\quad \quad \qed
\endexample

Again, it should be noticed that (2.17) and (2.18) are exactly {\it jeu de
taquin backward moves\/} (cf\. \cite{\SchuAA, sec.~2; \SagaAL,
pp.~120/169}),
which reverse the forward moves (2.4) 
and (2.5), respectively, except for the subtraction/addition of $1$
in (2.4) and (2.5).

In order to show that Algorithm~C* is always well-defined, we have to
confirm that when arriving at (C*3) we always obtained a column-strict
reverse plane partition. This is established in the following lemma.
Besides, this lemma contains the facts about Algorithm~C* that are
needed to prove that the Algorithms~C and C* are inverses of each other.

\proclaim{Lemma C*}Let $(P,T)$ be obtained after some loop
{\rm (C*1)-(C*2)-(C*3)} during Algorithm~C*. Then for all cells $\rh$ with
$T_\rh\ne0$ there holds
$$P_\rh+a+c_\rh\ge\max\{\text{\rm entries in }P\}.\tag2.19$$
Also, $P$ is a column-strict reverse plane partition. Besides, if
$\om$ is the corner cell that contained the special entry at the end
of the loop {\rm (C*1)-(C*2)-(C*3)} that lead to $P$, then $\om$ is the
left-most corner cell in $P$ that contains the maximal entry of $P$.
\endproclaim
\demo{Proof}We prove the assertions by induction on the number of
loops (C*1)-(C*2)-(C*3).

To begin with, we know that when we start with Algorithm~C* we have
a pair $(P,T)$, where $P$ is a column-strict reverse plane partition
with entries between $1$ and $a$. So for {\it any\/} cell $\rh=(i,j)$ we have
$$\align P_\rh+a+c_\rh&=P_{(i,j)}+a+c_{(i,j)}\\
&\ge i+a+(j-i)\\
&>a\\
&>\max\{\text{entries in }P\}.
\tag2.20
\endalign$$
So the assertion (2.19) and the assertion that $P$ is a column-strict reverse
plane partition hold at the very beginning. This will suffice for the
start of the induction.

As induction hypothesis let us assume that the assertions of the Lemma
are true for $(P,T)$ and all preceding pairs occuring in step (C*3)
during the process of the algorithm, except of course that the
assertion about the corner cell $\om$ does not hold for the initial
pair (because it does not make sense for the initial pair).

Let $\ze$ be the cell where the loop (C*1)-(C*2)-(C*3) starts from
$(P,T)$, i.e\. the cell that is chosen by applying (C*1) to $(P,T)$,
and let $\ka$ be the corner cell where the loop stops at (C*3). See
Figure~5.
Furthermore, let $(\bP,\bT)$ be the outcome after this loop. Then, by
definition of the algorithm we have
$$\bP_\ka=P_\ze+a+c_\ze.\tag2.21$$
Note that, also by definition of the algorithm,
the cells $\rh$ with $\bT_\rh\ne0$ are those with
$T_\rh\ne0$, except possibly for $\ze$. In particular, $\bT_\rh\ne0$
implies $T_\rh\ne0$.

\vskip10pt
\vbox{
$$
\underset
\text {from $\ga$ to $\om$ and from $\ze$ to $\ka$.}
\to{
\underset
\text {The cells $\ga$, $\om$, $\ze$, $\ka$, the jeu de taquin paths}
\to{
\Einheit.15cm
\PfadDicke{.5pt}
\Pfad(0,0),111111222211111111222222111111222211112222222222\endPfad
\Pfad(0,0),222222222222222222222222111111111111111111111111\endPfad
\PfadDicke{.75pt}
\Pfad(15,11),111\endPfad
\Pfad(15,11),2222\endPfad
\Pfad(11,15),1111\endPfad
\Pfad(11,15),222\endPfad
\Pfad(13,6),22\endPfad
\Pfad(10,8),111\endPfad
\Pfad(10,8),22222\endPfad
\Pfad(6,13),1111\endPfad
\PfadDicke{1pt}
\Pfad(18,10),1122\endPfad
\Pfad(18,10),2211\endPfad
\Pfad(12,4),1122\endPfad
\Pfad(12,4),2211\endPfad
\Pfad(4,12),1122\endPfad
\Pfad(4,12),2211\endPfad
\Pfad(10,18),1122\endPfad
\Pfad(10,18),2211\endPfad
\Label\lu{\ze\kern5pt}(4,12)
\Label\r{\om\kern5pt}(20,11)
\Label\ro{\kern5pt\ga}(12,20)
\Label\r{\kern5pt\ka}(10,3)
\Label\u{\kern2pt}(0,-1)
\hskip3.6cm
}}
$$
\centerline{\eightpoint Figure 5}
}
\vskip10pt

\smallskip
First we prove (2.19) for $\bP$. Let $\rh$ be any cell different from
$\ka$ with $\bT_\rh\ne0$. By definition of (C*2) we have $\bP_\rh\ge
P_\rh$. Besides, we already saw that $\bT_\rh\ne0$ implies
$T_\rh\ne0$. Therefore, by construction of $\ze$ in (C*1), we have
$$\bP_\rh+a+c_\rh\ge P_\rh+a+c_\rh\ge P_\ze+a+c_\ze.\tag2.22$$
Note that (2.22) also
holds for $\rh=\ka$ since by (2.21) we have
$\bP_\ka+a+c_\ka=(P_\ze+a+c_\ze)+a+c_\ka\ge P_\ze+a+c_\ze$, the
inequality being true because of $a\ge r> \max\{-c_\rh:\rh\in\la\}$
(recall that $a\ge r$ is one of the assumptions in Theorem~1.)
Also by construction of $\ze$, we have $T_\ze\ne0$, and hence by
induction hypothesis (2.19) that $P_\ze+a+c_\ze\ge\max\{\text{entries
in }P\}$. This implies that $P_\ze+a+c_\ze=\max\{\text{entries in
}\bP\}$ since the set of entries of $\bP$ is the same as the set of
entries in $P$ except for the special entry $P_\ze+a+c_\ze$ created in
(C*1) and finally located in cell $\ka$ in $\bP$. Hence, (2.22) proves
(2.19) with $P$ replaced by $\bP$, as desired. 

\smallskip
Now we prove that $\bP$ is a column-strict reverse plane partition. If
$P_\ze+a+c_\ze>\max\{\text{entries in }P\}$ then this assertion
certainly holds, since $\bP_\ka=P_\ze+a+c_\ze$ is the only new entry in
$\bP$. Note in particular, that by (2.20) we are in this case at the
very beginning.

By induction hypothesis, (2.19) holds for $\ze$, so the only other case
is 
$$P_\ze+a+c_\ze=\max\{\text{entries in }P\}.\tag2.23$$
Observe that the only difficulty arises when we reach corner cell
$\ka$ at the end of a loop (C*1)-(C*2)-(C*3) from above, and when in
addition $P_\ka=\bP_\ka$ holds. In this case column-strictness of $\bP$ would
be violated. We have to show that this case cannot occur.

Let $(P',T')$ be the pair preceding $(P,T)$, i.e\. $(P,T)$ is obtained
by applying one loop (C*1)-(C*2)-(C*3) to $(P',T')$. As we just noted,
$(P',T')$ exists, since if $(P,T)$ were the initial pair we would not
be in this case
because of (2.20). Furthermore, let $\ga$ be the cell where this loop
starts, and let $\om$ be the corner cell where it stops, see Figure~5. 
By definition
of the algorithm we have
$$P_\om=P'_\ga+a+c_\ga.\tag2.24$$ 
Now, by
induction hypothesis and the definition of the algorithm,
$$P'_\ga+a+c_\ga=\max\{\text{entries in }P\}.\tag2.25$$
Furthermore, there holds $P'_\ze\le P_\ze$, by the definition of (C*1)
if $\ze\ne\om$, but also if $\ze=\om$ because of (2.24) and the
induction hypothesis (2.19) for $P'$ and $\rh=\ga$.
Hence, again by definition of (C*1), $P'_\ga+a+c_\ga\le
P'_\ze+a+c_\ze\le P_\ze+a+c_\ze$.
Combining this with (2.23) and (2.25), we are forced to conclude
$$P'_\ga+a+c_\ga=P'_\ze+a+c_\ze\,(=P_\ze+a+c_\ze).\tag2.26$$
Therefore, again by definition of (C*1), $\ze$ lies (weakly) to the
left and (weakly) below of $\ga$, as is indicated in Figure~5.

It is an easy-to-check property of backward jeu de taquin (C*2) that
if the second ``jeu de
taquin path" is to the left of the first ``jeu de taquin path" somewhere,
then it has to stay to the left from thereon. To be precise,
suppose that during the first loop (C*1)-(C*2)-(C*3) the special
entry, $s_1$ say, went 
down by (2.18), see the left half of Figure~6. (Again, the arrows mark the
direction of move of the special entry.)
\vskip10pt
\vbox{
$$
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&*&&s_1&\\
\hlinefor5\\
&z&&y&\\
\hlinefor5
\endsmatrix
\overset \text{during}\to{\overset \text{first loop}\to\longrightarrow}
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&*&&y&\\
\hlinefor5\\
&z&&\vbox to10pt{\hsize5.27pt \vss \noindent$\downarrow$\newline$*$\vphantom{f}}&\\
\hlinefor5
\endsmatrix
\hskip2cm
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&s_2&&y&\\
\hlinefor5\\
&z&&*&\\
\hlinefor5
\endsmatrix
\overset \text{during}\to {\overset \text{second loop}\to\longrightarrow}
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&z&&y&\\
\hlinefor5\\
&\vbox to10pt{\hsize5.1pt \vss \noindent$\downarrow$\newline$*$\vphantom{f}}&&*&\\
\hlinefor5
\endsmatrix
$$
\centerline{\eightpoint Figure 6}
}
\vskip10pt
\noindent
Since rows are weakly increasing, we have $z\le y$. Suppose that
during the second loop we reach the cell neighbouring $y$ and $z$ with
a special entry $s_2$, see the right half of Figure~6. Then the
definition of the algorithm forces us to stop here or 
to move down in the next step (C*2). 

We already saw that the second ``jeu de taquin path" starts at $\ze$,
which is (weakly) to the left and (weakly) below of $\ga$, the
starting cell of the first ``jeu de taquin path". Therefore, if we
suppose that the
first path does not meet $\ze$, we are to the left of the first path
when we start the second path. Then, if at the end of the
second path we reach the same corner cell as the first path did, 
we have to reach it from the left. As noted above, this guarantees
that $\bP$ is a column-strict reverse plane partition. If we do not
reach the same corner cell, then we reach a corner cell $\ka$ to the
left. In this case our induction hypothesis for $\om$, the corner cell
that was reached at the end of the first path, says that $P_\ka$ is
strictly less than $P_\om=\max\{\text{entries in }P\}$. Therefore, by
using (2.21) and (2.23) we have $P_\ka<\max\{\text{entries in
}P\}=\bP_\ka$. Hence, 
$\bP$ will be a column-strict reverse plane partition, regardless from
which direction we reached cell $\ka$.

Now we have to consider the only remaining case that the first
path, starting at $\ga$, meets $\ze$. In this case, $\ze$ would have
to lie in the same column as $\ga$ (and below). First suppose that
$\ze$ is left by the first path by a downward move. In this case, by
(2.18), we would have $P_\ze>P'_\ze$, which contradicts (2.26).
So the first path has to leave $\ze$ by a right move. Hence, we are to
the left of the first path at the beginning of the second path. Thus,
the above considerations apply again.

\smallskip
Finally, we prove that $\ka$ is the left-most corner cell in $\bP$
that contains the maximal entry in $\bP$. This is trivially true if
$P_\ze+a+c_\ze>\max\{\text{entries in }P\}$, again by remembering
(2.21). Note that this inequality 
is in particular true at the very beginning of Algorithm~C*, because in
this case (2.20) holds even for all cells $\rh$, so also for $\ze$.
Because of the induction hypothesis (2.19), the only other case is
$P_\ze+a+c_\ze=\max\{\text{entries in }P\}$. Since we are {\it not\/}
at the very beginning, we are allowed to assume that this last
assertion of Lemma~C* holds for $P$ and $\om$. However, we already
considered the case $P_\ze+a+c_\ze=\max\{\text{entries in }P\}$
before (see (2.23)) and showed that the ``jeu de taquin 
path" leading from $\ze$ to $\ka$ has to stay to the left of the ``jeu
de taquin path" leading from $\ga$ to $\om$. Hence $\ka$ is (weakly)
to the left of $\om$. By induction hypothesis, $\om$ was the left-most
corner cell containing the maximal entry of $P$. So $\ka$, which by (2.21)
contains $P_\ze+a+c_\ze=\max\{\text{entries in }P\}$ in $\bP$, is the
left-most corner cell of $\bP$ containing $\max\{\text{entries in
}\bP\}$ ($=\max\{\text{entries in }P\}$), as desired.

This completes the proof of the Lemma.\quad \quad \qed
\enddemo

From Lemmas~1 and 2 it is abundantly clear that the Algorithms~C and
C* are inverses of each other. This finishes the bijective proof of
(2.1).

\subhead 3. Related bijections\endsubhead
In this section we discuss bijections related to the bijection in
section~2. It is mainly intended to serve the true purist among
combinatorialists. 

It is two items that we want to address in this section. First, one
might argue that we did not prove Theorem~1 directly, but the variant
(2.1). Well, as we show in the first part of this section, it is not
difficult to construct a bijective proof of Theorem~1 directly, i.e\.
of
$$\Big(\underset \text{with }1\le \text{entries}\le a\to {\sum _{P
\text{ a CRPP of 
shape $\la$}}} ^{}q^{n(P)}\Big)
= q^{\sum _{i=1} ^{r}i\la_i}\prod _{\rh\in\la} ^{}\frac {1}
{1-q^{h_\rh}}\cdot
\prod _{\rh\in\la} ^{}
(1-q^{a+c_\rh}),
\tag3.1$$
by using our Algorithms~C and C*. Secondly, Remmel and Whitney
\cite{\ReWhAA}, in the first bijective proof of Stanley's hook-content
formula Theorem~1, gave an involution principle-based bijective proof
for even another variant of Theorem~1, namely
$$\Big(\underset \text{with }1\le \text{entries}\le a\to {\sum _{P
\text{ a CRPP of 
shape $\la$}}} ^{}q^{n(P)}\Big)\cdot\prod _{\rh\in\la} ^{}[h_\rh]
=q^{\sum _{i=1} ^{r}i\la_i}\prod _{\rh\in\la} ^{}[a+c_\rh],\tag3.2
$$
where $[n]:=1+q+q^2+\dots+q^{n-1}$. While we do not know how to
construct a direct bijection (in the sense of avoiding the involution
principle) for (3.2), we are able to find a much simpler involution
principle-based bijective proof of (3.2) by using our Algorithms~C and
C*. This is subject of the second part of this section.

Before we start, we introduce some special tabloids to be used in the
course of the following bijections. Let $f$ be some function from the
set of all cells of a Ferrers diagram $\la$ into the integers, that
maps the cell $\rh$ to the value $f_\rh$, say. Then we call a tabloid
$T$ of shape $\la$ a {\it $(<f)$-tabloid\/} if $T_\rh<f_\rh$ for all
cells $\rh\in\la$, and we call $T$ a {\it $(0\text{--}f)$-tabloid\/}
if $T_\rh$ equals $0$ or $f_\rh$, for all cells $\rh\in\la$. The sign
$\sgn(T)$ of a $(0\text{--}f)$-tabloid is always defined to be
$(-1)^{\text{number of nonzero entries in $T$}}$.

{\smc Bijective proof of (3.1)}. As was already used in section~2, the
Hillman--Grassl correspondence \cite{\HiGrAA} bijectively shows that
$q^{\sum _{i=1} ^{r}i\la_i}\big/\prod _{\rh\in\la} ^{}(1-q^{h_\rh})$
is the generating function $\sum _{} ^{}q^{n(P_R)}$, where the sum is
over all column-strict reverse plane partitions of shape $\la$ with
entries $\ge1$. Thus the right-hand side of (3.1) is the generating
function\linebreak $\sum _{} ^{}\sgn(T_R)q^{n(P_R)}q^{n(T_R)}$, where
the sum 
is over all pairs $(P_R,T_R)$, with $P_R$ being a column-strict
reverse plane partition of shape $\la$ with entries $\ge1$ and with
$T_R$ being a $(0\text{--}(a+c))$-tabloid of shape $\la$ (viewing
$(a+c)$ as a function that maps a cell $\rh\in\la$ to $a+c_\rh$).
There is a straight-forward embedding of the set of column-strict
reverse plane partitions $P$ of shape $\la$ with entries between $1$
and $a$ (the ``left-hand side objects" in (3.1)) into the above pairs,
namely by mapping $P$ to $(P,\bold0)$. Therefore, in order to prove
(3.1) bijectively, we have to find a sign-reversing (with respect to
$\sgn(T_R)$) and weight-preserving (with respect to
$q^{n(P_R)+n(T_R)}$) involution on the set of all pairs $(P_R,T_R)$ as
above, where either $P_R$ contains some entry $>a$ or where $T_R$ contains
some nonzero entry.

Such an involution is simply described. Fix some total ordering of the
cells of $\la$. Consider such a pair $(P_R,T_R)$. Apply Algorithm~C to
$P_R$ to obtain $(\bP_R,\bT_R)$, where $\bP_R$ is a column-strict
reverse plane partition of shape $\la$ with entries between $1$ and
$a$ and where $\bT_R$ is some tabloid of shape $\la$. Pick the least
cell $\rh$ (in the chosen fixed total order) such that $\bT_R$ or
$T_R$ contain a nonzero entry in this cell. (Note that by assumption
there must be at least one such cell.) If the entry is nonzero in
$T_R$ then replace it by $0$, thus obtaining $T'_R$, and add $1$ to
the entry in cell $\rh$ of $\bT_R$, thus obtaining $\bT'_R$.
Otherwise, replace the $0$ in cell $\rh$ of $T_R$ by $a+c_\rh$, thus
otaining $T'_R$, and subtract $1$ from the entry in cell $\rh$ of
$\bT_R$, thus obtaining $\bT_R$. Apply Algorithm~C* to the pair
$(\bP_R,\bT'_R)$, thus obtaining $P'_R$. The image of $(P_R,T_R)$ under
our involution is defined to be $(P'_R,T'_R)$. The reader should not
have any difficulty to verify that this mapping is indeed an
involution and is sign-reversing and weight-preserving in the above
sense.\quad \quad \qed

{\smc Bijective proof of (3.2)}. First we have to recall the
involution principle of Garsia and Milne \cite{\GaMiAB} (see also
\cite{\StWhAB, sec.~4.6}). Let $X$ be a finite set with a signed
weight function $w$ defined on it. Furthermore, let $X_L$ and $X_R$ be
subsets of $X$, both of which containing elements with positive sign
only. Suppose that there is a sign-reversing and weight-preserving
involution $i_L$ on $X$ that fixes $X_L$ and a sign-reversing and
weight-preserving involution $i_R$ on $X$ that fixes $X_R$. Then there
must be a weight-preserving bijection between $X_L$ and $X_R$. And such
a bijection can be constructed explicitly by mapping $x\in X_L$ to
$(i_L\circ i_R)^n(x)$ where $n$ is the least integer such that
$(i_L\circ i_R)^n(x)$ is in $X_R$.

\medskip
Now we turn to our promised bijective proof of (3.2).
The right-hand side of (3.2) can be seen as the generating function
$\sum _{} ^{} q^{n(P_0)}q^{n(T_R)}$, where the sum is over all pairs
$(P_0,T_R)$, with $P_0$ being the ``minimal" column-strict plane
partition with entries $\ge1$ as explained in section~2, and with
$T_R$ being a $(<a+c)$-tabloid of shape $\la$. Call this set of pairs
$O_R$. Similarly, the left-hand side of (3.2) can be seen as the
generating function $\sum _{} ^{}q^{n(P_L)}q^{n(T_L)}$, where the sum
is over all pairs $(P_L,T_L)$, with $P_L$ being a column-strict
reverse plane partition of shape $\la$ with entries between $1$ and
$a$, and with $T_L$ being a $(<h)$-tabloid. Call this set of pairs
$O_L$.

We need to set up a bijection between these two sets of objects. We
want to use the involution principle. Therefore we have to say which
choices we take for the set $X$, the signed weight $w$, the subsets
$X_L$ and $X_R$, and the involutions $i_L$ and $i_R$. Of course, $O_L$
and $O_R$ should
correspond to $X_L$ and $X_R$, respectively, the latter being subsets of the
bigger set $X$, that has to be described next.

We define $X$ to be the set of all triples $(P,T^1,T^2)$ where $P$ is
a column-strict reverse plane partition os shape $\la$ with entries
$\ge 1$, $T^1$ is a $(0\text {--}\min\{h,a+c\})$-tabloid of shape $\la$, and $T^2$ is a
$(<\max\{h,a+c\})$-tabloid of shape $\la$. The signed weight $w$ 
on $X$ is defined by
$$w\big((P,T^1,T^2)\big)=\sgn(T^1)\,q^{n(P)+n(T^1)+n(T^2)}.\tag3.3$$

We define the set $X_L$ to be the subset of all triples $(P,T^1,T^2)$,
where $P$ is a column-strict reverse plane partition of shape $\la$
with entries between $1$ and $a$, where $T^1=\bold0$, and where $T^2$
is a $(<h)$-tabloid. Note in particular that the sign of
$w\big((P,T^1,T^2)\big)$ for all these 
triples is $\sgn(\bold0)=1$ which is positive. Obviously, $X_L$ is in
bijection with $O_L$. Besides, there holds
$w\big((P,\bold0,T^2)\big)=q^{n(P)+n(T^2)}$, which is exactly what we
need.

We define the set $X_R$ to be the subset of $X$ consisting of all
triples $(P_0,T^1,T^2)$, where $T^1=\bold0$ and where $T^2$ is a
$(<a+c)$-tabloid. Again, the sign of $w\big((P_0,T^1,T^2)\big)$ for
all these triples is positive. Obviously, $X_R$ is in bijection with
$O_R$, and $w\big((P_0,\bold0,T^2)\big)=q^{n(P_0)+n(T^2)}$, which is
exactly what we need.

For defining the involutions $i_L$ and $i_R$ we fix a total ordering
of the cells of $\la$, as we did before in the bijective proof of
(3.1). 

First we define the involution $i_L$ that fixes $X_L$. Let
$(P,T^1,T^2)$ be a triple in $X$ that is not in $X_L$, i.e\. $P$
contains an entry $>a$ or $T^1$ contains a nonzero entry. We
distinguish between two cases. For the first case we assume that there
is a cell $\rh$ in $\la$ with
$$h_\rh\le T^1_\rh+T^2_\rh<a+c_\rh.\tag3.4$$
Without loss of generality, let $\rh$ be the least such cell in our
fixed total order. Note in particular, that (3.4) implies
$h_\rh<a+c_\rh$, and therefore by definition of $T^1$ and $T^2$ we
have $T^1_\rh=0$ or $h_\rh$, and $0\le T^2_\rh<a+c_\rh$. If
$T^1_\rh=0$ then we replace $T^1_\rh$ by $h_\rh$, thus obtaining
$\bT^1$, and we replace $T^2_\rh$ by $T^2_\rh-h_\rh$, thus obtaining
$\bT^2$. If
$T^1_\rh=h_\rh$ then we replace $T^1_\rh$ by $0$, thus obtaining
$\bT^1$, and we replace $T^2_\rh$ by $T^2_\rh+h_\rh$, thus obtaining
$\bT^2$. It is easy to check that because of (3.4) in both cases
$\bT_1$ will be a $(0\text{--}\min\{h,a+c\})$-tabloid and $\bT^2$ will
be a $(<\max\{h,a+c\})$-tabloid. We define $i_L\big((P,T^1,T^2)\big)$
to be $(P,\bT^1,\bT^2)$. Obviously, there holds
$w\big((P,T^1,T^2)\big)=-w\big((P,\bT^1,\bT^2)\big)$, as required.

In the second case, i.e\. if we have a triple $(P,T^1,T^2)$ where
(3.4) is false for any cell $\rh$, then in the first step we transform
$(P,T^1,T^2)$ into a quadruple $(P',S^0,S^1,S^2)$, where $P'$ is a
column-strict reverse plane partition of shape $\la$ with entries
between $1$ and $a$, $S^0$ is a tabloid of shape $\la$, $S^1$ is a
$(<h)$-tabloid of shape $\la$, and $S^2$ is a
$(0\text{--}(a+c))$-tabloid of shape $\la$, such that
$$w\big((P,T^1,T^2)\big)=\sgn(S^2)\,q^{n(P')+w_c(S^0)+n(S^1)+n(S^2)}.\tag3.5$$
The pair $(P',S^0)$ is obtained by applying Algorithm~C to $P$. $S^1$
and $S^2$ are obtained by doing the following operation on $T^1$ and
$T^2$ for each cell $\rh$ in $\la$. If $h_\rh\ge a+c_\rh$ then
interchange $T^1_\rh$ and $T^2_\rh$. If $h_\rh<a+c_\rh$ and if
$T^1_\rh+T^2_\rh<h_\rh$ (which implies $T^1_\rh=0$) then again
interchange $T^1_\rh$ and $T^2_\rh$. If $h_\rh<a+c_\rh$ and if
$T^1_\rh+T^2_\rh\ge a+c_\rh$ (which implies $T^1_\rh=h_\rh$)
then replace $T^1_\rh$ by $T^1_\rh+T^2_\rh-(a+c_\rh)$ and $T^2_\rh$ by
$a+c_\rh$. By a simple case-by-case analysis the following facts can
be checked: The new tabloid $S^1$ is
a $(<h)$-tabloid and the new tabloid $S^2$ is a
$(0\text{--}(a+c))$-tabloid, 
in all cases. The relation (3.5) is satisfied. Also,
this transformation can be reversed. Finally,
$T^1\ne\bold0$ if and only if $S^2\ne\bold0$ under this
transformation.

In the second step, we choose the least cell $\rh$ in our fixed total
order such that $S^0_\rh$ or $S^2_\rh$ is nonzero. If $S^2_\rh$ is
nonzero then replace $S^2_\rh$ by $0$, thus obtaining $\bS^2$, and add
$1$ to $S^0_\rh$, thus obtaining $\bS^0$. If $S^2_\rh$ is
$0$ then replace $S^2_\rh$ by $a+c_\rh$, thus obtaining $\bS^2$, and subtract
$1$ from $S^0_\rh$, thus obtaining $\bS^0$. Finally, in the third step
we apply the inverse of the above transformation to
$(P',\bS^0,S^1,\bS^2)$, thus obtaining $(\bP,\bT^1,\bT^2)$. We define
$i_L\big((P,T^1,T^2)\big)$ to be $(\bP,\bT^1,\bT^2)$. By construction
we have $w\big((P,T^1,T^2)\big)=-w(\bP,\bT^1,\bT^2)$, as required.

We leave it to the reader to verify that $i_L$ is a sign-reversing and
weight-preserving (with respect to the signed weight (3.3)) involution
on $X\setminus X_L$.

The definition of $i_R$ proceeds in a similar spirit. Let
$(P,T^1,T^2)$ be a triple in $X$ with either $P$ being different from
$P_0$ or $T^1$ containing a nonzero entry.

Again, we distinguish between two cases. For the first case we assume
that there 
is a cell $\rh$ in $\la$ with
$$a+c_\rh\le T^1_\rh+T^2_\rh<h_\rh.\tag3.6$$
Without loss of generality, let $\rh$ be the least such cell in our
fixed total order. Note in particular, that (3.6) implies
$a+c_\rh<h_\rh$, and therefore by definition of $T^1$ and $T^2$ we
have $T^1_\rh=0$ or $a+c_\rh$, and $0\le T^2_\rh<h_\rh$. If
$T^1_\rh=0$ then we replace $T^1_\rh$ by $a+c_\rh$, thus obtaining
$\bT^1$, and we replace $T^2_\rh$ by $T^2_\rh-(a+c_\rh)$, thus obtaining
$\bT^2$. If
$T^1_\rh=a+c_\rh$ then we replace $T^1_\rh$ by $0$, thus obtaining
$\bT^1$, and we replace $T^2_\rh$ by $T^2_\rh+a+c_\rh$, thus obtaining
$\bT^2$. It is easy to check that because of (3.6) in both cases
$\bT_1$ will be a $(0\text{--}\min\{h,a+c\})$-tabloid and $\bT^2$ will
be a $(<\max\{h,a+c\})$-tabloid. We define $i_L\big((P,T^1,T^2)\big)$
to be $(P,\bT^1,\bT^2)$. Obviously, there holds
$w\big((P,T^1,T^2)\big)=-w\big((P,\bT^1,\bT^2)\big)$, as required.

In the second case, i.e\. if we have a triple $(P,T^1,T^2)$ where
(3.6) is false for any cell $\rh$, then in the first step we transform
$(P,T^1,T^2)$ into a quadruple $(P_0,S^0,S^1,S^2)$, 
where $S^0$ is a tabloid of shape $\la$, $S^1$ is a
$(<(a+c))$-tabloid of shape $\la$, and $S^2$ is a
$(0\text{--}h)$-tabloid of shape $\la$, such that
$$w\big((P,T^1,T^2)\big)=\sgn(S^2)\,q^{n(P_0)+w_h(S^0)+n(S^1)+n(S^2)}.\tag3.7$$
The pair $(P_0,S^0)$ is obtained from $P$ by using the Hillman--Grassl
algorithm as described in section~2. $S^1$
and $S^2$ are obtained by doing the following operation on $T^1$ and
$T^2$ for each cell $\rh$ in $\la$. If $a+c_\rh\ge h_\rh$ then
interchange $T^1_\rh$ and $T^2_\rh$. If $a+c_\rh<h_\rh$ and if
$T^1_\rh+T^2_\rh<a+c_\rh$ (which implies $T^1_\rh=0$) then again
interchange $T^1_\rh$ and $T^2_\rh$. If $a+c_\rh<h_\rh$ and if
$T^1_\rh+T^2_\rh\ge h_\rh$ (which implies $T^1_\rh=a+c_\rh$)
then replace $T^1_\rh$ by $T^1_\rh+T^2_\rh-h_\rh$ and $T^2_\rh$ by
$h_\rh$. Similarly as above, by a simple case-by-case analysis the
following facts can be checked: The new tabloids $S^1$ 
is a $(<(a+c))$-tabloid and the new tabloid $S^2$ is a
$(0\text{--}h)$-tabloid, 
in all cases. The relation (3.7) is satisfied. 
This transformation can be reversed. And,
$T^1\ne\bold0$ if and only if $S^2\ne\bold0$ under this
transformation.

In the second step, we choose the least cell $\rh$ in our fixed total
order such that $S^0_\rh$ or $S^2_\rh$ is nonzero. If $S^2_\rh$ is
nonzero then replace $S^2_\rh$ by $0$, thus obtaining $\bS^2$, and add
$1$ to $S^0_\rh$, thus obtaining $\bS^0$. If $S^2_\rh$ is
$0$ then replace $S^2_\rh$ by $h_\rh$, thus obtaining $\bS^2$, and subtract
$1$ from $S^0_\rh$, thus obtaining $\bS^0$. Finally, in the third step
we apply the inverse of the above transformation to
$(P',\bS^0,S^1,\bS^2)$, thus obtaining $(\bP,\bT^1,\bT^2)$. We define
$i_R\big((P,T^1,T^2)\big)$ to be $(\bP,\bT^1,\bT^2)$. By construction
we have $w\big((P,T^1,T^2)\big)=-w(\bP,\bT^1,\bT^2)$, as required.

Also here, we leave it to the reader to verify that $i_R$ is a
sign-reversing and 
weight-preserving (with respect to the signed weight (3.5)) involution
on $X\setminus X_R$.\quad \quad \qed

\subhead 4. Conclusion\endsubhead
There are many other product formulas for plane partition generating
functions, and only for a few of them combinatorial proofs are known.
However, many of the formulas for which no combinatorial proof is
known allow nice combinatorial formulations. Therefore there should
also be nice combinatorial proofs that explain these forms of the
formulas. Below we list a few candidates that desire to be proved
combinatorially.

(1) In \cite{\KratAK} it is proven that the number of plane partitions
of a given staircase shape with bounded entries is a product
involving the hook-lengths and some generalized contents of the
staircase shape. Is it possible to extend the ideas of section~2 to
this case? Of course, it cannot be that easy since there does not
exist a {\it weighted\/} version of this result.

(2) Counting plane partitions subject to various symmetries
acquired a lot of attention during the past 25 years. There
are 10 symmetry classes (see the survey in Stanley's paper
\cite{\StanAI}), for each of which there is a nice product formula. Now
all the formulas are proven (except for the $q$-analogue of Case~4;
the formulas that were only conjectured at the time of \cite{\StanAI}
are proved in \cite{\AndrAW, \KupeAA, \StemAG}). Case~1, MacMahon's
generating function \cite{\MacMAA, sec.~429, proof in
sec.~494} for plane partitions of rectangular shape with
bounded entries, is easily seen to be contained in Theorem~1.
Therefore this paper provides a bijective proof for Case~1. In all
other cases the existing proofs are to the more or lesser extent
non-combinatorial. Case~2, the generating function for symmetric plane
partitions of square shape with bounded entries, looks like a
promising candidate to be proved combinatorially. This is because
symmetric plane partitions can be seen as shifted plane partitions, by
forgetting about one half of the symmetric plane partition, where the
entries of 
the shifted plane partition
off the diagonal contribute twice their size to the norm. There exists
a shifted version of the Hillman--Grassl correspondence, due to
Sagan \cite{\SagaAB, sec.~3,4}, that would provide the analogue for
the first step in our bijection in section~2. So it remains to find
the shifted analogues of Algorithms~C and C*. Though there is also a
shifted version of jeu de taquin (cf\. \cite{\SagaAD}), it does not
seem to help here. And the fact that there is a nice product formula
for the generating function for symmetric plane partitions of square
shape only (and not for an arbitrary fixed symmetric shape) adds to
the evidence that a new idea is needed in this case.

On the other hand, in all the Cases~1--4 the formulas can be written
in a form such that the product is indexed by the boxes of the region
in which the plane partitions under consideration are contained, which
is a highly combinatorial description. So there must be bijective proofs
explaining these forms of the formulas.



\remark{Acknowledgement}This work was carried out while the author
visited the University of California at San Diego. He thanks the
University of California and in particular Adriano Garsia for making
this visit possible. Besides, he is indebted to Adriano Garsia for
drawing his attention to the problem of finding ``nice" combinatorial
proofs of hook formulas.
\endremark

\subhead Appendix\endsubhead
The appendix contains a complete example for Algorithms~C and C* for
$\la=(4,3,3,1)$ and $a=4$, setting up a mapping between the two sides
of Figure~2. See the specific descriptions given in Examples~C and C*
of how to read the following
tables.

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\newpage
$$
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&&
%9. Zeile, 2. Spalte
\overset\hbox{\kern2.5pt(C1)\kern2.5pt}\to\longrightarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
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\overset\hbox{\kern2.5pt(C3)\kern2.5pt}\to\longrightarrow
\\&&\\
%10. Zeile, 1. Spalte
\hskip46pt 
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&-&&-&&8&&-&\\
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&&
%10. Zeile, 2. Spalte
\overset\hbox{(C*3)}\to\longleftarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
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\overset\hbox{(2.18)}\to\longleftarrow
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&5&\\
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\endsmatrix
\overset\hbox{(C*1)}\to\longleftarrow
\\&&\\\hlinefor3\\&&\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%11. Zeile, 1. Spalte
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
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&&
%11. Zeile, 2. Spalte
\overset\hbox{\kern2.5pt(C1)\kern2.5pt}\to\longrightarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
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\overset\hbox{\kern2.5pt(2.5)\kern2.5pt}\to\longrightarrow
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\hlinefor9\\
&1&&1&&2&&3&\\
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&4&\\
\hlinefor3
\endsmatrix
\overset\hbox{\kern2.5pt(C3)\kern2.5pt}\to\longrightarrow
\\&&\\
%14. Zeile, 1. Spalte
\hskip46pt 
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&-&&-&&8&&-&\\
\hlinefor9\\
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\hlinefor3
\endsmatrix
&&
%14. Zeile, 2. Spalte
\overset\hbox{(C*3)}\to\longleftarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
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\endsmatrix
\overset\hbox{(2.18)}\to\longleftarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
\hlinefor9\\
&2&&3&&4&\\
\hlinefor7\\
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\hlinefor7\\
&4&\\
\hlinefor3
\endsmatrix
\overset\hbox{(C*1)}\to\longleftarrow
%\\&&\\\hlinefor3\\&&\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\endsmatrix
$$

\newpage
$$
\smatrix \format \l\s\l\\
%13. Zeile, 1. Spalte
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
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&2&&3&&4&\\
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\ 
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&&
%13. Zeile, 2. Spalte
\overset\hbox{\kern2.5pt(C1)\kern2.5pt}\to\longrightarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
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\overset\hbox{\kern2.5pt(2.5)\kern2.5pt}\to\longrightarrow
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\endsmatrix
\overset\hbox{\kern2.5pt(2.4)\kern2.5pt}\to\longrightarrow
\\&&\\
%14. Zeile, 1. Spalte
\hskip46pt 
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
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&&
%14. Zeile, 2. Spalte
\overset\hbox{(C*3)}\to\longleftarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
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\endsmatrix
\overset\hbox{(2.17)}\to\longleftarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
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&2&&3&&4&\\
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\overset\hbox{(2.18)}\to\longleftarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
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\endsmatrix
\overset\hbox{(2.17)}\to\longleftarrow
\\&&\\\hlinefor3\\&&\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%15. Zeile, 1. Spalte
&&
\overset\hbox{\kern2.5pt(2.4)\kern2.5pt}\to\longrightarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
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\endsmatrix
\overset\hbox{\kern2.5pt(C3)\kern2.5pt}\to\longrightarrow
\\&&\\
%16. Zeile, 1. Spalte
&&
\overset\hbox{(2.17)}\to\longleftarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
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&4&\\
\hlinefor3
\endsmatrix
\overset\hbox{(C*1)}\to\longleftarrow
\endsmatrix
$$

\newpage
$$
\smatrix \format \l\s\l\\
%17. Zeile, 1. Spalte
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
\hlinefor9\\
&\DKreisum2&&2&&4&\\
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\ 
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&&
%17. Zeile, 2. Spalte
\overset\hbox{\kern2.5pt(C1)\kern2.5pt}\to\longrightarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
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\overset\hbox{\kern2.5pt(2.4)\kern2.5pt}\to\longrightarrow
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%18. Zeile, 1. Spalte
\hskip46pt 
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
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&&
%18. Zeile, 2. Spalte
\overset\hbox{(C*3)}\to\longleftarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&2&&3&\\
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\endsmatrix
\overset\hbox{(2.17)}\to\longleftarrow
\\&&\\\hlinefor3\\&&\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%19. Zeile, 1. Spalte
&&
%19. Zeile, 2. Spalte
\overset\hbox{\kern2.5pt(2.4)\kern2.5pt}\to\longrightarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
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\overset\hbox{\kern2.5pt(C3)\kern2.5pt}\to\longrightarrow
\\&&\\
%20. Zeile, 1. Spalte
\hskip46pt 
&&
%20. Zeile, 2. Spalte
\overset\hbox{(2.17)}\to\longleftarrow
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
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\overset\hbox{(C*1)}\to\longleftarrow
\\&&\\\hlinefor3\\&&\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%21. Zeile, 1. Spalte
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&\DKreisum1&&1&&1&&3&\\
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&&
\\&&\\
%22. Zeile, 1. Spalte
\hskip46pt 
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&5&&-&&7&&-&\\
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&&
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$$

\newpage



\Refs

\ref\no \AndrAW\by G. E. Andrews \yr 1994 \paper Plane partitions V:
The t.s.s.c.p.p\. conjecture\jour J.~Combin\. Theory Ser\. A \vol 66
\pages 28--39\endref  

\ref\no \GaMiAB\by A. M. Garsia and S. C. Milne \yr 1981 \paper Method
for constructing bijections for classical partition identities\jour
Proc\. Nat\. Acad\. Sci\. U.S.A.\vol 78\pages 2026--2028\endref  

\ref\no \HiGrAA\by A. P. Hillman and R. M. Grassl \yr 1976 \paper
Reverse plane partitions and tableau hook numbers\jour J. Combin\.
Theory Ser.~A\vol 21\pages 216--221\endref  

\ref\no \KratAK\by C.    Krattenthaler \yr 1990 \paper A determinant
evaluation and some enumeration results for plane partitions\inbook
Number-Theoretic Analysis\eds E.~Hlawka, R.~F.~Tichy\publ Lect\. Notes
in Math. 1452, Springer-Verlag\publaddr Berlin, New York\pages
121--131\endref    

\ref\no \KratBK\by C.    Krattenthaler \yr \paper Another involution 
principle-free bijective proof of Stanley's hook-content formula\jour 
preprint\vol \pages \endref

\ref\no \KupeAA\by G.    Kuperberg \yr 1994 \paper Symmetries of plane
partitions and the permanent-determinant method\jour J. Combin\.
Theory Ser.~A\vol 68 \pages 115--151\endref

\ref\no \MacMAA\by P. A. MacMahon \yr 1960 \book Combinatory Analysis
\publ Cambridge University Press, 1916; reprinted by Chelsea, New York
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\ref\no \NoPSAA\by J. C. Novelli, I. M. Pak and A. V. Stoyanovsii 
\yr 1997 \paper A direct bijective proof of the hook-length formula\jour 
Discrete Math\. Theoret\. Computer Science\vol 1\pages paper~053-067\endref

\ref\no \PaStAA\by I. M. Pak and A. V. Stoyanovskii \yr 1992 \paper 
A bijective proof of the hook-length formula and its analogues\jour 
Funkt\. Anal\. Priloz\.\vol 26\rm , No.~3\pages 80--82\finalinfo English 
translation in Funct\. Anal\. Appl\. {\bf 26} (1992), 216--218\endref

\ref\no \ReWhAA\by J. B. Remmel and R. Whitney \yr 1983 \paper A
bijective proof of the hook formula for the number of column-strict
tableaux with bounded entries\jour Europ\. J. Combin\.\vol 4\pages
45--63\endref   

\ref\no \SagaAB\by B. E. Sagan \yr 1982 \paper Enumeration of
partitions with hooklengths\jour Europ\. J. Combin\.\vol 3\pages
85--94\endref   

\ref\no \SagaAD\by B. E. Sagan \yr 1987 \paper Shifted tableaux, Schur
$Q$-functions, and a conjecture of R. Stanley\jour J. Combin\. Theory
Ser.~A\vol 45\pages 62--103\endref  

\ref\no \SagaAL\by B. E. Sagan \yr 1991 \book The symmetric group\publ
Wadsworth \& Brooks/Cole\publaddr Pacific Grove, California\endref 

\ref\no \SchuAA\by M.-P. Sch\"utzenberger \yr 1977 \paper La
correspondance de Robinson\inbook Combinatoire et Repr\'esentation du
Groupe Sym\'etrique\publ Lecture Notes in Math\., vol.~579,
Springer--Verlag\publaddr Ber\-lin--Hei\-del\-berg--New York\pages
59--113\endref    

\ref\no \StanAA\by R. P. Stanley \yr 1971 \paper Theory and
applications of plane partitions: Part~2\jour Stud\. Appl\. Math\vol
50\pages 259--279 \endref   

\ref\no \StanAI\by R. P. Stanley \yr 1986 \paper Symmetries of plane
partitions\jour J. Combin. Theory A\vol 43\pages 103--113 \moreref
\jour Erratum \vol44 \yr 1987\page310 \endref  

\ref\no \StWhAB\by D.    Stanton and D. White \yr 1986 \book
Constructive Combinatorics\bookinfo Undergraduate Texts in Math\.\publ
Sprin\-ger--Ver\-lag\publaddr New York, Berlin Heidelberg, Tokyo\endref  

\ref\no \StemAG\by J. R. Stembridge \yr 1995 \paper The enumeration of
totally symmetric plane partitions\jour Adv\. in Math\.\vol 111\pages 
227--245\endref


\endRefs


\enddocument
