Add flow layout: cursor placement, left rail, declared band heights
Figures were positioned by hand-written offsets. Every gap was a magic number tuned against the content that happened to be there, so a label that grew two characters landed on the next tensor, and two stages started from two different x shared no rail. Both failures compile cleanly. Replace it with a cursor. Objects placed with an empty coordinate argument reserve their own width -- including a stack's offset sheets and a bracket's overhang -- and gaps are declared once (\stgutter, \strowgap, \stblockgap). The gap belongs to the object that follows it and the first object in a band gets none, so every band starts flush on a shared rail and gap=0pt states that two shards tile exactly. \stlink makes a connector's label a flow object, which is what removes the label-wider-than-its-arrow failure entirely. \stcol is a vertical sub-flow for a split along the contracted axis. \strow declares its height, so an object that does not fit -- or a column that does not add up to what it declared, and is therefore drawn off-center -- becomes a package warning, which build.sh fails on. Absolute placement is unchanged: passing a coordinate takes the original code path, and \sttrack folds a hand-placed node back into the cursor. All three golden examples and the new tests/flow.tex are converted and build clean.
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dela
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@@ -2,6 +2,7 @@
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% Shows: leading axes as stack depth, a transpose that physically swaps the
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% face, equal edge length on the contracted axis, and a Boolean mask drawn in
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% a different grammar from the scores it gates.
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% Layout is entirely by cursor: no absolute coordinate appears below.
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% ../scripts/build.sh mha-causal.tex
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\documentclass[border=10pt]{standalone}
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\usepackage[cjk]{supertensor}
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@@ -19,79 +20,66 @@
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\begin{document}
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\begin{tikzpicture}
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% ================================================================= formula ==
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\node (F) at (0,0) {\stformula{$\displaystyle
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\mathbf A^{(i)}=\mathrm{softmax}\!\left(
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\frac{\mathbf Q^{(i)}\mathbf K^{(i)\top}}{\sqrt{d_h}}+\mathbf M\right),\qquad
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\mathbf O^{(i)}=\mathbf A^{(i)}\mathbf V^{(i)}$}};
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% ============================================================ stage A row ===
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\node[st stage, below=7mm of F] (SA) {每头打分：沿 $d_h$ 收缩};
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\coordinate (a) at ($(SA)+(-3.9,-1.9)$);
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\ststack[role=q, bracket=true]{Q}{(a)}{T}{dh}{3}
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\node[st op, right=6mm of Q] (mA) {$\times$};
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% K^T: the face is physically swapped, not relabelled. Its height equals Q's
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% width -- that is the contracted axis d_h, drawn at one edge length.
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\ststack[role=k, bracket=true]{KT}{($(mA)+(1.9,0)$)}{dh}{T}{3}
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\node[st op, right=6mm of KT] (eA) {$=$};
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\ststack[role=s]{S}{($(eA)+(2.0,0)$)}{T}{T}{3}
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\node[inner sep=0pt, fit=(Q)(KT)(S)] (rowA) {};
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\stlane{rowA}
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\ststage{SA}{每头打分：沿 $d_h$ 收缩}
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\strow{rowA}{T}
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\ststack[role=q, bracket=true]{Q}{}{T}{dh}{3}
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\stglyph{mA}{$\times$}
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% K^T: the face is physically swapped, not relabelled. Its height equals Q's
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% width -- that is the contracted axis d_h, drawn at one edge length.
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\ststack[role=k, bracket=true]{KT}{}{dh}{T}{3}
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\stglyph{eA}{$=$}
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\ststack[role=s]{S}{}{T}{T}{3}
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\strowend
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\stcaption{Q}{$\mathbf Q^{(i)}$}{$h\times T\times d_h$}
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\stcaption{KT}{$\mathbf K^{(i)\top}$}{$h\times d_h\times T$}
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\stcaption{S}{$\mathbf S^{(i)}$}{$h\times T\times T$}
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\stnolane
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% ============================================================ stage B row ===
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\node[st stage, below=9mm of Q-shape.south west, anchor=north west] (SB)
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{因果掩码与加权求和};
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\coordinate (b) at ($(SB)+(0.6,-2.0)$);
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% The mask is a Boolean support, not a magnitude: one flat level, exact
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% triangle, no stack -- it is shared by every head.
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\stface[role=w, pattern=data,
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data={300000,330000,333000,333300,333330,333333}]{M}{(b)}{T}{T}
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\ststack[role=s, pattern=causal]{A}{($(M.east)+(3.75,0)$)}{T}{T}{3}
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\starrowlabel{M.east}{A.west}{softmax}
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\node[st op, right=6mm of A] (mB) {$\times$};
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\ststack[role=v, bracket=true]{V}{($(mB)+(1.4,0)$)}{T}{dh}{3}
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\node[st op, right=6mm of V] (eB) {$=$};
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\ststack[role=v]{O}{($(eB)+(1.4,0)$)}{T}{dh}{3}
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\node[inner sep=0pt, fit=(M)(A)(V)(O)] (rowB) {};
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\stlane{rowB}
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\ststage{SB}{因果掩码与加权求和}
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\strow{rowB}{T}
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% The mask is a Boolean support, not a magnitude: one flat level, exact
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% triangle, no stack -- it is shared by every head.
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\stface[role=w, pattern=data,
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data={300000,330000,333000,333300,333330,333333}]{M}{}{T}{T}
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\stlink{lB}{softmax}
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\ststack[role=s, pattern=causal]{A}{}{T}{T}{3}
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\stglyph{mB}{$\times$}
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\ststack[role=v, bracket=true]{V}{}{T}{dh}{3}
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\stglyph{eB}{$=$}
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\ststack[role=v]{O}{}{T}{dh}{3}
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\strowend
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\stcaption{M}{$\mathbf M$}{$T\times T$}
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\stcaption{A}{$\mathbf A^{(i)}$}{$h\times T\times T$}
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\stcaption{V}{$\mathbf V^{(i)}$}{$h\times T\times d_h$}
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\stcaption{O}{$\mathbf O^{(i)}$}{$h\times T\times d_h$}
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\stnolane
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% ============================================================ stage C row ===
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\node[st stage, below=9mm of M-shape.south west, anchor=north west] (SC)
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{沿 $d_h$ 拼接后投影};
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\coordinate (c) at ($(SC)+(1.2,-2.0)$);
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% Concatenation reverses the split: three h-shards of width d_h tile a face of
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% width d exactly.
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\stface[role=v]{C1}{(c)}{T}{dh}
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\stface[role=v]{C2}{($(C1.east)+(1.5*\stunit,0)$)}{T}{dh}
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\stface[role=v]{C3}{($(C2.east)+(1.5*\stunit,0)$)}{T}{dh}
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\node[st op, right=6mm of C3] (mC) {$\times$};
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\stface[role=w]{WO}{($(mC)+(2.5,0)$)}{d}{d}
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\node[st op, right=6mm of WO] (eC) {$=$};
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\stface[role=v, bracket=true]{Y}{($(eC)+(2.5,0)$)}{T}{d}
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\node[inner sep=0pt, fit=(C1)(WO)(Y)] (rowC) {};
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\stlane{rowC}
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\ststage{SC}{沿 $d_h$ 拼接后投影}
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\strow{rowC}{d} % the d x d projection is the tallest object here
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% Concatenation reverses the split: gap=0pt makes the three h-shards of width
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% d_h tile a face of width d exactly, with no eyeballed offset.
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\stface[role=v]{C1}{}{T}{dh}
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\stface[role=v, gap=0pt]{C2}{}{T}{dh}
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\stface[role=v, gap=0pt]{C3}{}{T}{dh}
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\stglyph{mC}{$\times$}
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\stface[role=w]{WO}{}{d}{d}
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\stglyph{eC}{$=$}
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\stface[role=v, bracket=true]{Y}{}{T}{d}
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\strowend
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\stcaption{C2}{$[\,\mathbf O^{(1)}\mid\mathbf O^{(2)}\mid\mathbf O^{(3)}\,]$}{$T\times d$}
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\stcaption{WO}{$\mathbf W_O$}{$d\times d$}
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\stcaption{Y}{$\mathbf Y$}{$T\times d$}
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\stnolane
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% ================================================================= formula ==
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% Placed last so it is centered on the figure that was actually drawn.
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\sttopformula{F}{$\displaystyle
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\mathbf A^{(i)}=\mathrm{softmax}\!\left(
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\frac{\mathbf Q^{(i)}\mathbf K^{(i)\top}}{\sqrt{d_h}}+\mathbf M\right),\qquad
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\mathbf O^{(i)}=\mathbf A^{(i)}\mathbf V^{(i)}$}
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% ============================================================== meaning box ==
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\node[inner sep=0pt, fit=(F)(rowA)(rowB)(rowC)(Y-shape)(C2-shape)] (all) {};
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\stbbox{all}
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\stmeaningbox{mb}{16.8cm}{all}
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{$T$ 序列长度，$d_h$ 单头宽度，$h$ 头数（图中 $h=3$，即堆叠的三张面），
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$d=h\,d_h$；批轴 $B$ 省略}
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