Otared Kavian schrieb am 24.08.2023 um 01:04:

Hi Fabrice,

As Wolfgang points out, it is indeed possible to fill-in your table with Lua: maybe you were wondering how to fill the columns 2, 3 and 4. In this case you need to use the Lua function math.mod as in the following, which is a completed version of what Wolfgang sent:

I guess I completely missed the units digit part :)

%% begin filling-with-lua.tex \starttext

\startluacode context.startxtable{ align = "middle,lohi", bodyfont = "9pt", framecolor = "black" }     context.startxrow{ background = "color", backgroundcolor = "lightgray" }         context.startxcell{ nx = 4 }             context("Units digit of")   context.stopxcell()     context.stopxrow()     context.startxrow() context.startxcell{ width = "1cm" } context.im("a")         context.stopxcell()         context.startxcell{ width = "1cm" }           context.im("a^2") context.stopxcell()         context.startxcell{ width = "1cm" }             context.im("b")         context.stopxcell() context.startxcell{ width = "1cm" } context.im("2b^2")         context.stopxcell()     context.stopxrow()     for i = 0,9 do         context.startxrow()             for j = 1,4 do                 context.startxcell()                     if j == 1 then               context(i)                     elseif j == 2 then context(math.mod(i*i,10))  elseif j == 3 then context(i)  else context(math.mod(2*i*i,10))                     end context.stopxcell()             end         context.stopxrow()     end

I would drop the column check in this case.     for i = 0,9 do         context.startxrow()             context.startxcell()                 context(i)             context.stopxcell()             context.startxcell()                 context(math.mod(i*i,10))             context.stopxcell()             context.startxcell()                 context(i)             context.stopxcell()             context.startxcell()                 context(math.mod(2*i*i,10))             context.stopxcell()         context.stopxrow()     end Wolfgang

Otared Kavian schrieb am 24.08.2023 um 12:50:

...

On 24 Aug 2023, at 06:03, Wolfgang Schuster mailto:wolfgang.schuster.lists@gmail.com> wrote:

...

[…]

I would drop the column check in this case.

    for i = 0,9 do         context.startxrow()             context.startxcell()                 context(i)             context.stopxcell()             context.startxcell()                 context(math.mod(i*i,10))             context.stopxcell()             context.startxcell()                 context(i)             context.stopxcell()             context.startxcell()                 context(math.mod(2*i*i,10))             context.stopxcell()         context.stopxrow()     end

Wolfgang

Indeed this is much more elegant… It feels so good to be on this list and learn from such insights :-)

We don't even need the math.mod function because Lua added with version 5.1 a modulo operator, the loop to create the row can now be changed to     for i = 0,9 do         context.startxrow()             context.startxcell()                 context(i)             context.stopxcell()             context.startxcell()                 context(i * i % 10)             context.stopxcell()             context.startxcell()                 context(i)             context.stopxcell()             context.startxcell()                 context(2 * i * i % 10)             context.stopxcell()         context.stopxrow()     end Wolfgang

Hi Otared and Wolgang, Thanks for the answers. In fact, being familiar with Python and not at all with Lua, I didn't have the reflex to think of using a mathematical function which gives the remainder in the Euclidean division by 10, it's strange !! Fabrice Le jeu. 24 août 2023 à 04:06, Wolfgang Schuster < wolfgang.schuster.lists@gmail.com> a écrit :

Otared Kavian schrieb am 24.08.2023 um 01:04:

Hi Fabrice,

As Wolfgang points out, it is indeed possible to fill-in your table with Lua: maybe you were wondering how to fill the columns 2, 3 and 4. In this case you need to use the Lua function math.mod as in the following, which is a completed version of what Wolfgang sent:

I guess I completely missed the units digit part :)

%% begin filling-with-lua.tex \starttext

\startluacode context.startxtable{ align = "middle,lohi", bodyfont = "9pt", framecolor = "black" } context.startxrow{ background = "color", backgroundcolor = "lightgray" } context.startxcell{ nx = 4 } context("Units digit of") context.stopxcell() context.stopxrow() context.startxrow() context.startxcell{ width = "1cm" } context.im("a") context.stopxcell() context.startxcell{ width = "1cm" } context.im("a^2") context.stopxcell() context.startxcell{ width = "1cm" } context.im("b") context.stopxcell() context.startxcell{ width = "1cm" } context.im("2b^2") context.stopxcell() context.stopxrow() for i = 0,9 do context.startxrow() for j = 1,4 do context.startxcell() if j == 1 then context(i) elseif j == 2 then context(math.mod(i*i,10)) elseif j == 3 then context(i) else context(math.mod(2*i*i,10)) end context.stopxcell() end context.stopxrow() end

I would drop the column check in this case.

for i = 0,9 do context.startxrow() context.startxcell() context(i) context.stopxcell() context.startxcell() context(math.mod(i*i,10)) context.stopxcell() context.startxcell() context(i) context.stopxcell() context.startxcell() context(math.mod(2*i*i,10)) context.stopxcell() context.stopxrow() end

Wolfgang

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