Hi, In the codebelow,ifI commentthe two lines, it's okay, but if Iuncommentthose two lines, the numberoffaces of the cubesisnotwhitebutblack,andthere arepiecesoftextgreen.
\startsetups[table:initialize] \setupTABLE[height=2.5cm,align={middle,lohi}] \setupTABLE[column][2][[width=3cm,align={right,lohi},frame=off] \setupTABLE[column][3][[width=8cm,align={right,lohi},frame=off] \setupTABLE[1][1][[width=5cm,align={right,lohi},frame=off] \setupTABLE[1][2,3][bottomframe=off] \setupTABLE[1][3,4][topframe=off] \stopsetups \bTABLE[setups=table:initialize] \switchtobodyfont[11pt] \bTR \bTD \color[orange]{\tfa Suite d'instructions correspondant au programme de calcul :} \eTD \bTD \eTD \bTD \color[orange]{\tfa Valeur de la variable $X$ après l'exécution de chaque instruction :} \eTD \eTR \bTR \bTD $X$ prend la valeur $7$ \eTD \bTD \dontleavehmode \startMPcode input hvdm; l:=2.8mm; r:=0.6; alfa:=45; beta:=0; gamma:=0; defineDefaultArrow (l, r, alfa, beta, gamma); pickup pencircle scaled 0.4mm; arrowline (0cm,0cm)--(3cm,0cm) withcolor blue ;\stopMPcode \eTD \bTD On affecte à $X$ la valeur $7$ : la valeur de $X$ est \color[green]{$7$}. \eTD \eTR \bTR \bTD $X$ prend la valeur $2X$ \eTD \bTD \dontleavehmode \startMPcode input hvdm; l:=2.8mm; r:=0.6; alfa:=45; beta:=0; gamma:=0; defineDefaultArrow (l, r, alfa, beta, gamma); pickup pencircle scaled 0.4mm; arrowline (0cm,0cm)--(3cm,0cm) withcolor red ;\stopMPcode\eTD \bTD On affecte à $X$ la valeur $2X$.\\Comme la valeur de $X$ est \color[green]{$7$}, la nouvelle valeur de $X$ est : $2\times\color[green]{7}$.\\La valeur de $X$ est maintenant \color[green]{$14$}. \eTD \eTR \bTR \bTD $X$ prend la valeur $X+3$ \eTD \bTD \dontleavehmode \startMPcode input hvdm; l:=2.8mm; r:=0.6; alfa:=45; beta:=0; gamma:=0; defineDefaultArrow (l, r, alfa, beta, gamma); pickup pencircle scaled 0.4mm; arrowline (0cm,0cm)--(3cm,0cm) withcolor green ;\stopMPcode\eTD \eTR \bTD On affecte à $X$ la valeur $X+3$.\\Comme la valeur de $X$ est \color[green]{$14$}, la nouvelle valeur de $X$ est : $\color[green]{14}+3$.\\La valeur de $X$ est maintenant \color[green]{$17$}. \eTD \eTABLE{}
\startsetups[table:initialize] \setupTABLE[height=2cm,align={middle,lohi}] \setupTABLE[column][1][[width=4cm,align={right,lohi},frame=on, framecolor=H1prime] \setupTABLE[column][2][[width=5cm,align={middle,lohi},frame=off] \setupTABLE[column][3][[width=1cm,align={middle,lohi},frame=off] \setupTABLE[column][4][[width=9cm,align={right,lohi},frame=off] \setupTABLE[1][1][frame=off] \setupTABLE[1][2,3][bottomframe=off] \setupTABLE[1][3,4][topframe=off] \stopsetups \bTABLE[setups=table:initialize] \switchtobodyfont[10pt] \bTR \bTD \color[red]{ Suite d'instructions :} \eTD \bTD \eTD \bTD \eTD \bTD \color[orange]{ Valeur de la variable $A$ et valeur de la variable $B$ après l'exécution de chaque instruction :}\eTD \eTR \bTR \bTD $A$ prend la valeur $3$ \eTD \bTD \tikzset{% cube join/.style={ thick, -{Stealth}, }, cube face/.style={ minimum size=1cm, outer sep=0pt, draw=white, thick, line join=round, shading=ball, text=white, }, face color/.style={cube face/.append style={ball color=#1}}, pics/cube/.style args={#1 with #2}{ code={ \node [cube face, label={[name=-label]below:#2}] (-front) {#1}; \node [cube face] (-top) at (-front.north west) [anchor=south west, xslant=1, yscale=1/3] {}; \node [cube face] (-side) at (-front.south east) [anchor=south west, yslant=1, xscale=1/3] {}; }} } \starttikzpicture \pic [cube face/.append style={ball color=green}] at (-1, 1) {cube=3 with $A$};\pic [cube face/.append style={ball color=red}] at ( 1, 1) {cube={} with $B$};
\stoptikzpicture \eTD \bTD \dontleavehmode \startMPcode input hvdm; l:=2.8mm; r:=0.6; alfa:=45; beta:=0; gamma:=0; defineDefaultArrow (l, r, alfa, beta, gamma); pickup pencircle scaled 0.4mm; arrowline (1cm,0cm)--(0cm,0cm) ;\stopMPcode \eTD \bTD La valeur de $A$ est \color[green]{3} et $B$ n'a pas encore de valeur \eTD \eTR \bTR \bTD $B$ prend la valeur $A+1$ \eTD \bTD \tikzset{% cube join/.style={ thick, -{Stealth}, }, cube face/.style={ minimum size=1cm, outer sep=0pt, draw=white, thick, line join=round, shading=ball, text=white, font=\bfseries }, face color/.style={cube face/.append style={ball color=#1}}, pics/cube/.style args={#1 with #2}{ code={ \node [cube face, label={[name=-label]below:#2}] (-front) {#1}; \node [cube face] (-top) at (-front.north west) [anchor=south west, xslant=1, yscale=1/3] {}; \node [cube face] (-side) at (-front.south east) [anchor=south west, yslant=1, xscale=1/3] {}; }} } \starttikzpicture \pic [cube face/.append style={ball color=green}] at (-1, 1) {cube=3 with $A$}; \pic [cube face/.append style={ball color=red}] at ( 1, 1) {cube=4 with $B$}; \stoptikzpicture \eTD \bTD \dontleavehmode \startMPcode input hvdm; l:=2.8mm; r:=0.6; alfa:=45; beta:=0; gamma:=0; defineDefaultArrow (l, r, alfa, beta, gamma); pickup pencircle scaled 0.4mm; arrowline (1cm,0cm)--(0cm,0cm) ;\stopMPcode \eTD \bTD Comme la valeur de $A$ est \color[green]{$3$}, la valeur de $B$ est : $\color[geen]{3}+1$, soit \color[red]{$4$}.\\La valeur de $A$ ne change pas : elle reste égale à \color[green]{$3$}.\eTD \bTR \bTD $A$ prend la valeur $A+B$ \eTD \bTD \tikzset{% cube join/.style={ thick, -{Stealth}, }, cube face/.style={ minimum size=1cm, outer sep=0pt, draw=white, thick, line join=round, shading=ball, ball color=red, text=white, font=\bfseries }, face color/.style={cube face/.append style={ball color=#1}}, pics/cube/.style args={#1 with #2}{ code={ \node [cube face, label={[name=-label]below:#2}] (-front) {#1}; \node [cube face] (-top) at (-front.north west) [anchor=south west, xslant=1, yscale=1/3] {}; \node [cube face] (-side) at (-front.south east) [anchor=south west, yslant=1, xscale=1/3] {}; }} } \starttikzpicture \pic [cube face/.append style={ball color=green}] at (-1, 1) {cube=7 with $A$}; \pic [cube face/.append style={ball color=red}] at ( 1, 1) {cube=4 with $B$};
\stoptikzpicture \eTD \bTD \dontleavehmode \startMPcode input hvdm; l:=2.8mm; r:=0.6; alfa:=45; beta:=0; gamma:=0; defineDefaultArrow (l, r, alfa, beta, gamma); pickup pencircle scaled 0.4mm; arrowline (1cm,0cm)--(0cm,0cm) ;\stopMPcode \eTD \bTD Comme la valeur de $A$ est \color[green]{3} et que celle de $B$ est \color[red]{$4$}, la nouvelle valeur de $A$ est : $\color[green]{3}+\color[red]{4}$, c'est-à-dire \color[green]{$7$}.\\La valeur de $B$ ne change pas : elle reste égale à \color[red]{$4$}. \eTD \eTR \eTABLE{} \stoptext