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Output visual diagram of picture


Write a code golf problem in which Java winsGolf a Venn Diagram generatorBooks on a ShelfDetermine the Dimensions of a Rotated RectangleDraw a Houndstooth PatternDraw and label an ASCII hexagonal gridGolf me an ASCII AlphabetOutput a pretty boxASCII-Art Venn DiagramTatamibari solver













6












$begingroup$


Write a program that inputs the dimensions of a painting, the matting width, and the frame width for a framed portrait. The program should output a diagram using the symbol ‘X ’ for the painting, ‘+’ for the matting, and ‘# ’ for the framing. The symbols must be space-separated.



INPUT: 3 2 1 2
(Width, Height, Matte Width, Frame Width)



OUTPUT:



First 3 and 2 are painting width and height. 1 is the matte width around it. 2 is the frame width around the whole thing.



In text form:



# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #


The winning code completes the following conditions in the least possible bytes.










share|improve this question









New contributor




George Harris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$












  • $begingroup$
    Nice challenge! For future challenges you may want to use The Sandbox
    $endgroup$
    – MilkyWay90
    3 hours ago










  • $begingroup$
    Also, will the frame height be given?
    $endgroup$
    – MilkyWay90
    3 hours ago










  • $begingroup$
    MilkyWay90, the frame is a constant width around the portrait so only one value is needed.
    $endgroup$
    – George Harris
    3 hours ago










  • $begingroup$
    Thanks! Is the constant width always 2 (or is it the height of the frame)?
    $endgroup$
    – MilkyWay90
    3 hours ago












  • $begingroup$
    Well, the program should be able to handle any case, no? Typically it should be assumed any of the numbers are subject to change. Just given the four inputs, you must produce the visual output. :)
    $endgroup$
    – George Harris
    3 hours ago
















6












$begingroup$


Write a program that inputs the dimensions of a painting, the matting width, and the frame width for a framed portrait. The program should output a diagram using the symbol ‘X ’ for the painting, ‘+’ for the matting, and ‘# ’ for the framing. The symbols must be space-separated.



INPUT: 3 2 1 2
(Width, Height, Matte Width, Frame Width)



OUTPUT:



First 3 and 2 are painting width and height. 1 is the matte width around it. 2 is the frame width around the whole thing.



In text form:



# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #


The winning code completes the following conditions in the least possible bytes.










share|improve this question









New contributor




George Harris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Nice challenge! For future challenges you may want to use The Sandbox
    $endgroup$
    – MilkyWay90
    3 hours ago










  • $begingroup$
    Also, will the frame height be given?
    $endgroup$
    – MilkyWay90
    3 hours ago










  • $begingroup$
    MilkyWay90, the frame is a constant width around the portrait so only one value is needed.
    $endgroup$
    – George Harris
    3 hours ago










  • $begingroup$
    Thanks! Is the constant width always 2 (or is it the height of the frame)?
    $endgroup$
    – MilkyWay90
    3 hours ago












  • $begingroup$
    Well, the program should be able to handle any case, no? Typically it should be assumed any of the numbers are subject to change. Just given the four inputs, you must produce the visual output. :)
    $endgroup$
    – George Harris
    3 hours ago














6












6








6





$begingroup$


Write a program that inputs the dimensions of a painting, the matting width, and the frame width for a framed portrait. The program should output a diagram using the symbol ‘X ’ for the painting, ‘+’ for the matting, and ‘# ’ for the framing. The symbols must be space-separated.



INPUT: 3 2 1 2
(Width, Height, Matte Width, Frame Width)



OUTPUT:



First 3 and 2 are painting width and height. 1 is the matte width around it. 2 is the frame width around the whole thing.



In text form:



# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #


The winning code completes the following conditions in the least possible bytes.










share|improve this question









New contributor




George Harris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Write a program that inputs the dimensions of a painting, the matting width, and the frame width for a framed portrait. The program should output a diagram using the symbol ‘X ’ for the painting, ‘+’ for the matting, and ‘# ’ for the framing. The symbols must be space-separated.



INPUT: 3 2 1 2
(Width, Height, Matte Width, Frame Width)



OUTPUT:



First 3 and 2 are painting width and height. 1 is the matte width around it. 2 is the frame width around the whole thing.



In text form:



# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #


The winning code completes the following conditions in the least possible bytes.







code-golf






share|improve this question









New contributor




George Harris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




George Harris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 3 hours ago









Stephen

7,49223397




7,49223397






New contributor




George Harris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









George HarrisGeorge Harris

311




311




New contributor




George Harris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





George Harris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






George Harris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Nice challenge! For future challenges you may want to use The Sandbox
    $endgroup$
    – MilkyWay90
    3 hours ago










  • $begingroup$
    Also, will the frame height be given?
    $endgroup$
    – MilkyWay90
    3 hours ago










  • $begingroup$
    MilkyWay90, the frame is a constant width around the portrait so only one value is needed.
    $endgroup$
    – George Harris
    3 hours ago










  • $begingroup$
    Thanks! Is the constant width always 2 (or is it the height of the frame)?
    $endgroup$
    – MilkyWay90
    3 hours ago












  • $begingroup$
    Well, the program should be able to handle any case, no? Typically it should be assumed any of the numbers are subject to change. Just given the four inputs, you must produce the visual output. :)
    $endgroup$
    – George Harris
    3 hours ago


















  • $begingroup$
    Nice challenge! For future challenges you may want to use The Sandbox
    $endgroup$
    – MilkyWay90
    3 hours ago










  • $begingroup$
    Also, will the frame height be given?
    $endgroup$
    – MilkyWay90
    3 hours ago










  • $begingroup$
    MilkyWay90, the frame is a constant width around the portrait so only one value is needed.
    $endgroup$
    – George Harris
    3 hours ago










  • $begingroup$
    Thanks! Is the constant width always 2 (or is it the height of the frame)?
    $endgroup$
    – MilkyWay90
    3 hours ago












  • $begingroup$
    Well, the program should be able to handle any case, no? Typically it should be assumed any of the numbers are subject to change. Just given the four inputs, you must produce the visual output. :)
    $endgroup$
    – George Harris
    3 hours ago
















$begingroup$
Nice challenge! For future challenges you may want to use The Sandbox
$endgroup$
– MilkyWay90
3 hours ago




$begingroup$
Nice challenge! For future challenges you may want to use The Sandbox
$endgroup$
– MilkyWay90
3 hours ago












$begingroup$
Also, will the frame height be given?
$endgroup$
– MilkyWay90
3 hours ago




$begingroup$
Also, will the frame height be given?
$endgroup$
– MilkyWay90
3 hours ago












$begingroup$
MilkyWay90, the frame is a constant width around the portrait so only one value is needed.
$endgroup$
– George Harris
3 hours ago




$begingroup$
MilkyWay90, the frame is a constant width around the portrait so only one value is needed.
$endgroup$
– George Harris
3 hours ago












$begingroup$
Thanks! Is the constant width always 2 (or is it the height of the frame)?
$endgroup$
– MilkyWay90
3 hours ago






$begingroup$
Thanks! Is the constant width always 2 (or is it the height of the frame)?
$endgroup$
– MilkyWay90
3 hours ago














$begingroup$
Well, the program should be able to handle any case, no? Typically it should be assumed any of the numbers are subject to change. Just given the four inputs, you must produce the visual output. :)
$endgroup$
– George Harris
3 hours ago




$begingroup$
Well, the program should be able to handle any case, no? Typically it should be assumed any of the numbers are subject to change. Just given the four inputs, you must produce the visual output. :)
$endgroup$
– George Harris
3 hours ago










6 Answers
6






active

oldest

votes


















1












$begingroup$

JavaScript (ES6),  118  113 bytes





(w,h,M,F)=>(g=(c,n)=>'01210'.replace(/./g,i=>c(+i).repeat([F,M,n][i])))(y=>g(x=>'###+X#++'[y+x*5&7]+' ',w)+`
`,h)


Try it online!



Commented



(w, h, M, F) => (       // given the 4 input variables
g = ( // g = helper function taking:
c, // c = callback function returning a string to repeat
n // n = number of times the painting part must be repeated
) => //
'01210' // string describing the picture structure, with:
.replace( // 0 = frame, 1 = matte, 2 = painting
/./g, // for each character in the above string:
i => // i = identifier of the current area
c(+i) // invoke the callback function
.repeat // and repeat it ...
([F, M, n][i]) // ... either F, M or n times
) // end of replace()
)( // outer call to g:
y => // callback function taking y:
g( // inner call to g:
x => // callback function taking x:
'###+X#++' // figure out which character to use
[y + x * 5 & 7] // by applying a small hash function to (x, y)
+ ' ', // append a space
w // repeat the painting part w times
) // end of inner call
+ 'n', // append a line feed
h // repeat the painting part h times
) // end of outer call





share|improve this answer











$endgroup$





















    0












    $begingroup$


    Charcoal, 48 bytes



    NθNηNζNεUO⁺θ⊗⁺ζε⁺η⊗⁺ζε#Mε↘UO⁺θ⊗ζ⁺η⊗ζ+Mζ↘UOθηXUE¹


    Try it online! Link is to verbose version of code. Explanation:



    NθNηNζNε


    Input the four values.



    UO⁺θ⊗⁺ζε⁺η⊗⁺ζε#


    Draw the framing.



    Mε↘UO⁺θ⊗ζ⁺η⊗ζ+


    Move to and draw the matting.



    Mζ↘UOθηX


    Move to and draw the painting.



    UE¹


    Double-space the output horizontally.



    Alternative solution, also 48 bytes:



    NθNηNζNεUO⁺θ⁺ζε⁺η⁺ζε#UO⁺θζ⁺ηζ+UOθηX‖OO←θ‖OO↑ηUE¹


    Try it online! Link is to verbose version of code. Explanation:



    NθNηNζNε


    Input the four values.



    UO⁺θ⁺ζε⁺η⁺ζε#


    Draw the framing, but not to the left or above the painting.



    UO⁺θζ⁺ηζ+


    Draw the matting, but not to the left or above the painting.



    UOθηX


    Draw the painting.



    ‖OO←θ‖OO↑ηUE¹


    Reflect and double-space the output horizontally.






    share|improve this answer









    $endgroup$





















      0












      $begingroup$


      Python 3.8 (pre-release), 116 115 bytes





      lambda a,b,c,d,e='#',f='+':"n".join((g:=[e*(a+2*c+2*d)]*d+[(h:=e*d)+f*(a+c*2)+h]*c)+[h+f*c+'X'*a+f*c+h]*b+g[::-1])


      Try it online!



      First attempt at golfing, will be improved soon.
      a is width, b is height, c is matte width, and d is frame width.



      -1 bytes using the := operator to define h as e * d



      EXPLANATION:



      lambda a,b,c,d,e='#',f='+':          Define a lambda which takes in arguments a, b, c, and d (The width of the painting, the height of the painting, the padding of the matte, and the padding of the frame width, respectively). It also defines variables e and f as '#' and '+', respectively.
      "n".join( Turn the list into a string, where each element is separated by newlines
      (g:= Define g as (while still evaling the lists)...
      [e*(a+2*c+2*d)]*d+ Form the top rows (the ones filled with hashtags)
      [(h:=e*d)+f*(a+c*2)+h]*c Form the middle-top rows (uses := to golf this section)
      )+
      [h+f*c+'X'*a+f*c+h]*b+ Form the middle row
      g[::-1] Uses g to golf the code (forms the entire middle-bottom-to-bottom)
      )





      share|improve this answer











      $endgroup$













      • $begingroup$
        Removing the e assignment saves you two bytes, the f assignment isn't saving you anything
        $endgroup$
        – Jo King
        33 secs ago



















      0












      $begingroup$

      Javascript, 158 bytes



      (w,h,m,f)=>(q="repeat",(z=("#"[q](w+2*(m+f)))+`
      `)[q](f))+(x=((e="#"[q](f))+(r="+"[q](m))+(t="+"[q](w))+r+e+`
      `)[q](m))+(e+r+"X"[q](w)+r+e+`
      `)[q](h)+x+z)


      Can probably be trimmed down a little bit






      f=

      (w,h,m,f)=>(q="repeat",(z=("# "[q](w+2*(m+f))+`
      `)[q](f))+(x=((e="# "[q](f))+(r="+ "[q](m))+(t="+ "[q](w))+r+e+`
      `)[q](m))+(e+r+"X "[q](w)+r+e+`
      `)[q](h)+x+z)

      console.log(f(3,2,1,2))








      share|improve this answer









      $endgroup$





















        0












        $begingroup$


        Wolfram Language (Mathematica), 152 bytes



        (t=(p=Table)["# ",(x=2#4+2#3)+#2,x+#];p[t[[i,j]]="+ ",{j,z=#4+1,#4+2#3+#},{i,z,x-#4+#2}];p[t[[i,j]]="X ",{j,#3+z,#3+#4+#},{i,#3+z,#3+#4+#2}];""<>#&/@t)&


        Try it online!






        share|improve this answer











        $endgroup$





















          0












          $begingroup$


          Perl 6, 115 bytes





          ->a,b,c,d{$_=['#'xx$!*2+a]xx($!=c+d)*2+b;.[d..^*-d;d..^a+$!+c]='+'xx*;.[$!..^*-$!;$!..^a+$!]='X'xx*;.join("
          ")}


          Try it online!



          Roughly golfed anonymous codeblock utilising Perl 6's multi-dimensional list assignment. For example, @a[1;2] = 'X'; will assign 'X' to the element with index 2 from the list with index 1, and @a[1,2,3;3,4,5]='X'xx 9; will replace all the elements with indexes 3,4,5 of the lists with indexes 1,2,3 with 'X'.



          Explanation:



          First, we initialise the list as a a+2*(c+d) by b+2*(c+d) rectangle of #s.



          $_=['#'xx$!*2+a]xx($!=c+d)*2+a;
          State:
          # # # # # # # # #
          # # # # # # # # #
          # # # # # # # # #
          # # # # # # # # #
          # # # # # # # # #
          # # # # # # # # #
          # # # # # # # # #
          # # # # # # # # #


          Then we assign the inner rectangle of +s



          .[d..^*-d;d..^a+$!+c]='+'xx*;
          State:
          # # # # # # # # #
          # # # # # # # # #
          # # + + + + + # #
          # # + + + + + # #
          # # + + + + + # #
          # # + + + + + # #
          # # # # # # # # #
          # # # # # # # # #


          Finally, the innermost rectangle of Xs.



          .[$!..^*-$!;$!..^a+$!]='X'xx*;
          # # # # # # # # #
          # # # # # # # # #
          # # + + + + + # #
          # # + X X X + # #
          # # + X X X + # #
          # # + + + + + # #
          # # # # # # # # #
          # # # # # # # # #




          share









          $endgroup$













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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

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            1












            $begingroup$

            JavaScript (ES6),  118  113 bytes





            (w,h,M,F)=>(g=(c,n)=>'01210'.replace(/./g,i=>c(+i).repeat([F,M,n][i])))(y=>g(x=>'###+X#++'[y+x*5&7]+' ',w)+`
            `,h)


            Try it online!



            Commented



            (w, h, M, F) => (       // given the 4 input variables
            g = ( // g = helper function taking:
            c, // c = callback function returning a string to repeat
            n // n = number of times the painting part must be repeated
            ) => //
            '01210' // string describing the picture structure, with:
            .replace( // 0 = frame, 1 = matte, 2 = painting
            /./g, // for each character in the above string:
            i => // i = identifier of the current area
            c(+i) // invoke the callback function
            .repeat // and repeat it ...
            ([F, M, n][i]) // ... either F, M or n times
            ) // end of replace()
            )( // outer call to g:
            y => // callback function taking y:
            g( // inner call to g:
            x => // callback function taking x:
            '###+X#++' // figure out which character to use
            [y + x * 5 & 7] // by applying a small hash function to (x, y)
            + ' ', // append a space
            w // repeat the painting part w times
            ) // end of inner call
            + 'n', // append a line feed
            h // repeat the painting part h times
            ) // end of outer call





            share|improve this answer











            $endgroup$


















              1












              $begingroup$

              JavaScript (ES6),  118  113 bytes





              (w,h,M,F)=>(g=(c,n)=>'01210'.replace(/./g,i=>c(+i).repeat([F,M,n][i])))(y=>g(x=>'###+X#++'[y+x*5&7]+' ',w)+`
              `,h)


              Try it online!



              Commented



              (w, h, M, F) => (       // given the 4 input variables
              g = ( // g = helper function taking:
              c, // c = callback function returning a string to repeat
              n // n = number of times the painting part must be repeated
              ) => //
              '01210' // string describing the picture structure, with:
              .replace( // 0 = frame, 1 = matte, 2 = painting
              /./g, // for each character in the above string:
              i => // i = identifier of the current area
              c(+i) // invoke the callback function
              .repeat // and repeat it ...
              ([F, M, n][i]) // ... either F, M or n times
              ) // end of replace()
              )( // outer call to g:
              y => // callback function taking y:
              g( // inner call to g:
              x => // callback function taking x:
              '###+X#++' // figure out which character to use
              [y + x * 5 & 7] // by applying a small hash function to (x, y)
              + ' ', // append a space
              w // repeat the painting part w times
              ) // end of inner call
              + 'n', // append a line feed
              h // repeat the painting part h times
              ) // end of outer call





              share|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                JavaScript (ES6),  118  113 bytes





                (w,h,M,F)=>(g=(c,n)=>'01210'.replace(/./g,i=>c(+i).repeat([F,M,n][i])))(y=>g(x=>'###+X#++'[y+x*5&7]+' ',w)+`
                `,h)


                Try it online!



                Commented



                (w, h, M, F) => (       // given the 4 input variables
                g = ( // g = helper function taking:
                c, // c = callback function returning a string to repeat
                n // n = number of times the painting part must be repeated
                ) => //
                '01210' // string describing the picture structure, with:
                .replace( // 0 = frame, 1 = matte, 2 = painting
                /./g, // for each character in the above string:
                i => // i = identifier of the current area
                c(+i) // invoke the callback function
                .repeat // and repeat it ...
                ([F, M, n][i]) // ... either F, M or n times
                ) // end of replace()
                )( // outer call to g:
                y => // callback function taking y:
                g( // inner call to g:
                x => // callback function taking x:
                '###+X#++' // figure out which character to use
                [y + x * 5 & 7] // by applying a small hash function to (x, y)
                + ' ', // append a space
                w // repeat the painting part w times
                ) // end of inner call
                + 'n', // append a line feed
                h // repeat the painting part h times
                ) // end of outer call





                share|improve this answer











                $endgroup$



                JavaScript (ES6),  118  113 bytes





                (w,h,M,F)=>(g=(c,n)=>'01210'.replace(/./g,i=>c(+i).repeat([F,M,n][i])))(y=>g(x=>'###+X#++'[y+x*5&7]+' ',w)+`
                `,h)


                Try it online!



                Commented



                (w, h, M, F) => (       // given the 4 input variables
                g = ( // g = helper function taking:
                c, // c = callback function returning a string to repeat
                n // n = number of times the painting part must be repeated
                ) => //
                '01210' // string describing the picture structure, with:
                .replace( // 0 = frame, 1 = matte, 2 = painting
                /./g, // for each character in the above string:
                i => // i = identifier of the current area
                c(+i) // invoke the callback function
                .repeat // and repeat it ...
                ([F, M, n][i]) // ... either F, M or n times
                ) // end of replace()
                )( // outer call to g:
                y => // callback function taking y:
                g( // inner call to g:
                x => // callback function taking x:
                '###+X#++' // figure out which character to use
                [y + x * 5 & 7] // by applying a small hash function to (x, y)
                + ' ', // append a space
                w // repeat the painting part w times
                ) // end of inner call
                + 'n', // append a line feed
                h // repeat the painting part h times
                ) // end of outer call






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 14 mins ago

























                answered 1 hour ago









                ArnauldArnauld

                79k795328




                79k795328























                    0












                    $begingroup$


                    Charcoal, 48 bytes



                    NθNηNζNεUO⁺θ⊗⁺ζε⁺η⊗⁺ζε#Mε↘UO⁺θ⊗ζ⁺η⊗ζ+Mζ↘UOθηXUE¹


                    Try it online! Link is to verbose version of code. Explanation:



                    NθNηNζNε


                    Input the four values.



                    UO⁺θ⊗⁺ζε⁺η⊗⁺ζε#


                    Draw the framing.



                    Mε↘UO⁺θ⊗ζ⁺η⊗ζ+


                    Move to and draw the matting.



                    Mζ↘UOθηX


                    Move to and draw the painting.



                    UE¹


                    Double-space the output horizontally.



                    Alternative solution, also 48 bytes:



                    NθNηNζNεUO⁺θ⁺ζε⁺η⁺ζε#UO⁺θζ⁺ηζ+UOθηX‖OO←θ‖OO↑ηUE¹


                    Try it online! Link is to verbose version of code. Explanation:



                    NθNηNζNε


                    Input the four values.



                    UO⁺θ⁺ζε⁺η⁺ζε#


                    Draw the framing, but not to the left or above the painting.



                    UO⁺θζ⁺ηζ+


                    Draw the matting, but not to the left or above the painting.



                    UOθηX


                    Draw the painting.



                    ‖OO←θ‖OO↑ηUE¹


                    Reflect and double-space the output horizontally.






                    share|improve this answer









                    $endgroup$


















                      0












                      $begingroup$


                      Charcoal, 48 bytes



                      NθNηNζNεUO⁺θ⊗⁺ζε⁺η⊗⁺ζε#Mε↘UO⁺θ⊗ζ⁺η⊗ζ+Mζ↘UOθηXUE¹


                      Try it online! Link is to verbose version of code. Explanation:



                      NθNηNζNε


                      Input the four values.



                      UO⁺θ⊗⁺ζε⁺η⊗⁺ζε#


                      Draw the framing.



                      Mε↘UO⁺θ⊗ζ⁺η⊗ζ+


                      Move to and draw the matting.



                      Mζ↘UOθηX


                      Move to and draw the painting.



                      UE¹


                      Double-space the output horizontally.



                      Alternative solution, also 48 bytes:



                      NθNηNζNεUO⁺θ⁺ζε⁺η⁺ζε#UO⁺θζ⁺ηζ+UOθηX‖OO←θ‖OO↑ηUE¹


                      Try it online! Link is to verbose version of code. Explanation:



                      NθNηNζNε


                      Input the four values.



                      UO⁺θ⁺ζε⁺η⁺ζε#


                      Draw the framing, but not to the left or above the painting.



                      UO⁺θζ⁺ηζ+


                      Draw the matting, but not to the left or above the painting.



                      UOθηX


                      Draw the painting.



                      ‖OO←θ‖OO↑ηUE¹


                      Reflect and double-space the output horizontally.






                      share|improve this answer









                      $endgroup$
















                        0












                        0








                        0





                        $begingroup$


                        Charcoal, 48 bytes



                        NθNηNζNεUO⁺θ⊗⁺ζε⁺η⊗⁺ζε#Mε↘UO⁺θ⊗ζ⁺η⊗ζ+Mζ↘UOθηXUE¹


                        Try it online! Link is to verbose version of code. Explanation:



                        NθNηNζNε


                        Input the four values.



                        UO⁺θ⊗⁺ζε⁺η⊗⁺ζε#


                        Draw the framing.



                        Mε↘UO⁺θ⊗ζ⁺η⊗ζ+


                        Move to and draw the matting.



                        Mζ↘UOθηX


                        Move to and draw the painting.



                        UE¹


                        Double-space the output horizontally.



                        Alternative solution, also 48 bytes:



                        NθNηNζNεUO⁺θ⁺ζε⁺η⁺ζε#UO⁺θζ⁺ηζ+UOθηX‖OO←θ‖OO↑ηUE¹


                        Try it online! Link is to verbose version of code. Explanation:



                        NθNηNζNε


                        Input the four values.



                        UO⁺θ⁺ζε⁺η⁺ζε#


                        Draw the framing, but not to the left or above the painting.



                        UO⁺θζ⁺ηζ+


                        Draw the matting, but not to the left or above the painting.



                        UOθηX


                        Draw the painting.



                        ‖OO←θ‖OO↑ηUE¹


                        Reflect and double-space the output horizontally.






                        share|improve this answer









                        $endgroup$




                        Charcoal, 48 bytes



                        NθNηNζNεUO⁺θ⊗⁺ζε⁺η⊗⁺ζε#Mε↘UO⁺θ⊗ζ⁺η⊗ζ+Mζ↘UOθηXUE¹


                        Try it online! Link is to verbose version of code. Explanation:



                        NθNηNζNε


                        Input the four values.



                        UO⁺θ⊗⁺ζε⁺η⊗⁺ζε#


                        Draw the framing.



                        Mε↘UO⁺θ⊗ζ⁺η⊗ζ+


                        Move to and draw the matting.



                        Mζ↘UOθηX


                        Move to and draw the painting.



                        UE¹


                        Double-space the output horizontally.



                        Alternative solution, also 48 bytes:



                        NθNηNζNεUO⁺θ⁺ζε⁺η⁺ζε#UO⁺θζ⁺ηζ+UOθηX‖OO←θ‖OO↑ηUE¹


                        Try it online! Link is to verbose version of code. Explanation:



                        NθNηNζNε


                        Input the four values.



                        UO⁺θ⁺ζε⁺η⁺ζε#


                        Draw the framing, but not to the left or above the painting.



                        UO⁺θζ⁺ηζ+


                        Draw the matting, but not to the left or above the painting.



                        UOθηX


                        Draw the painting.



                        ‖OO←θ‖OO↑ηUE¹


                        Reflect and double-space the output horizontally.







                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered 2 hours ago









                        NeilNeil

                        81.7k745178




                        81.7k745178























                            0












                            $begingroup$


                            Python 3.8 (pre-release), 116 115 bytes





                            lambda a,b,c,d,e='#',f='+':"n".join((g:=[e*(a+2*c+2*d)]*d+[(h:=e*d)+f*(a+c*2)+h]*c)+[h+f*c+'X'*a+f*c+h]*b+g[::-1])


                            Try it online!



                            First attempt at golfing, will be improved soon.
                            a is width, b is height, c is matte width, and d is frame width.



                            -1 bytes using the := operator to define h as e * d



                            EXPLANATION:



                            lambda a,b,c,d,e='#',f='+':          Define a lambda which takes in arguments a, b, c, and d (The width of the painting, the height of the painting, the padding of the matte, and the padding of the frame width, respectively). It also defines variables e and f as '#' and '+', respectively.
                            "n".join( Turn the list into a string, where each element is separated by newlines
                            (g:= Define g as (while still evaling the lists)...
                            [e*(a+2*c+2*d)]*d+ Form the top rows (the ones filled with hashtags)
                            [(h:=e*d)+f*(a+c*2)+h]*c Form the middle-top rows (uses := to golf this section)
                            )+
                            [h+f*c+'X'*a+f*c+h]*b+ Form the middle row
                            g[::-1] Uses g to golf the code (forms the entire middle-bottom-to-bottom)
                            )





                            share|improve this answer











                            $endgroup$













                            • $begingroup$
                              Removing the e assignment saves you two bytes, the f assignment isn't saving you anything
                              $endgroup$
                              – Jo King
                              33 secs ago
















                            0












                            $begingroup$


                            Python 3.8 (pre-release), 116 115 bytes





                            lambda a,b,c,d,e='#',f='+':"n".join((g:=[e*(a+2*c+2*d)]*d+[(h:=e*d)+f*(a+c*2)+h]*c)+[h+f*c+'X'*a+f*c+h]*b+g[::-1])


                            Try it online!



                            First attempt at golfing, will be improved soon.
                            a is width, b is height, c is matte width, and d is frame width.



                            -1 bytes using the := operator to define h as e * d



                            EXPLANATION:



                            lambda a,b,c,d,e='#',f='+':          Define a lambda which takes in arguments a, b, c, and d (The width of the painting, the height of the painting, the padding of the matte, and the padding of the frame width, respectively). It also defines variables e and f as '#' and '+', respectively.
                            "n".join( Turn the list into a string, where each element is separated by newlines
                            (g:= Define g as (while still evaling the lists)...
                            [e*(a+2*c+2*d)]*d+ Form the top rows (the ones filled with hashtags)
                            [(h:=e*d)+f*(a+c*2)+h]*c Form the middle-top rows (uses := to golf this section)
                            )+
                            [h+f*c+'X'*a+f*c+h]*b+ Form the middle row
                            g[::-1] Uses g to golf the code (forms the entire middle-bottom-to-bottom)
                            )





                            share|improve this answer











                            $endgroup$













                            • $begingroup$
                              Removing the e assignment saves you two bytes, the f assignment isn't saving you anything
                              $endgroup$
                              – Jo King
                              33 secs ago














                            0












                            0








                            0





                            $begingroup$


                            Python 3.8 (pre-release), 116 115 bytes





                            lambda a,b,c,d,e='#',f='+':"n".join((g:=[e*(a+2*c+2*d)]*d+[(h:=e*d)+f*(a+c*2)+h]*c)+[h+f*c+'X'*a+f*c+h]*b+g[::-1])


                            Try it online!



                            First attempt at golfing, will be improved soon.
                            a is width, b is height, c is matte width, and d is frame width.



                            -1 bytes using the := operator to define h as e * d



                            EXPLANATION:



                            lambda a,b,c,d,e='#',f='+':          Define a lambda which takes in arguments a, b, c, and d (The width of the painting, the height of the painting, the padding of the matte, and the padding of the frame width, respectively). It also defines variables e and f as '#' and '+', respectively.
                            "n".join( Turn the list into a string, where each element is separated by newlines
                            (g:= Define g as (while still evaling the lists)...
                            [e*(a+2*c+2*d)]*d+ Form the top rows (the ones filled with hashtags)
                            [(h:=e*d)+f*(a+c*2)+h]*c Form the middle-top rows (uses := to golf this section)
                            )+
                            [h+f*c+'X'*a+f*c+h]*b+ Form the middle row
                            g[::-1] Uses g to golf the code (forms the entire middle-bottom-to-bottom)
                            )





                            share|improve this answer











                            $endgroup$




                            Python 3.8 (pre-release), 116 115 bytes





                            lambda a,b,c,d,e='#',f='+':"n".join((g:=[e*(a+2*c+2*d)]*d+[(h:=e*d)+f*(a+c*2)+h]*c)+[h+f*c+'X'*a+f*c+h]*b+g[::-1])


                            Try it online!



                            First attempt at golfing, will be improved soon.
                            a is width, b is height, c is matte width, and d is frame width.



                            -1 bytes using the := operator to define h as e * d



                            EXPLANATION:



                            lambda a,b,c,d,e='#',f='+':          Define a lambda which takes in arguments a, b, c, and d (The width of the painting, the height of the painting, the padding of the matte, and the padding of the frame width, respectively). It also defines variables e and f as '#' and '+', respectively.
                            "n".join( Turn the list into a string, where each element is separated by newlines
                            (g:= Define g as (while still evaling the lists)...
                            [e*(a+2*c+2*d)]*d+ Form the top rows (the ones filled with hashtags)
                            [(h:=e*d)+f*(a+c*2)+h]*c Form the middle-top rows (uses := to golf this section)
                            )+
                            [h+f*c+'X'*a+f*c+h]*b+ Form the middle row
                            g[::-1] Uses g to golf the code (forms the entire middle-bottom-to-bottom)
                            )






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 2 hours ago

























                            answered 2 hours ago









                            MilkyWay90MilkyWay90

                            523212




                            523212












                            • $begingroup$
                              Removing the e assignment saves you two bytes, the f assignment isn't saving you anything
                              $endgroup$
                              – Jo King
                              33 secs ago


















                            • $begingroup$
                              Removing the e assignment saves you two bytes, the f assignment isn't saving you anything
                              $endgroup$
                              – Jo King
                              33 secs ago
















                            $begingroup$
                            Removing the e assignment saves you two bytes, the f assignment isn't saving you anything
                            $endgroup$
                            – Jo King
                            33 secs ago




                            $begingroup$
                            Removing the e assignment saves you two bytes, the f assignment isn't saving you anything
                            $endgroup$
                            – Jo King
                            33 secs ago











                            0












                            $begingroup$

                            Javascript, 158 bytes



                            (w,h,m,f)=>(q="repeat",(z=("#"[q](w+2*(m+f)))+`
                            `)[q](f))+(x=((e="#"[q](f))+(r="+"[q](m))+(t="+"[q](w))+r+e+`
                            `)[q](m))+(e+r+"X"[q](w)+r+e+`
                            `)[q](h)+x+z)


                            Can probably be trimmed down a little bit






                            f=

                            (w,h,m,f)=>(q="repeat",(z=("# "[q](w+2*(m+f))+`
                            `)[q](f))+(x=((e="# "[q](f))+(r="+ "[q](m))+(t="+ "[q](w))+r+e+`
                            `)[q](m))+(e+r+"X "[q](w)+r+e+`
                            `)[q](h)+x+z)

                            console.log(f(3,2,1,2))








                            share|improve this answer









                            $endgroup$


















                              0












                              $begingroup$

                              Javascript, 158 bytes



                              (w,h,m,f)=>(q="repeat",(z=("#"[q](w+2*(m+f)))+`
                              `)[q](f))+(x=((e="#"[q](f))+(r="+"[q](m))+(t="+"[q](w))+r+e+`
                              `)[q](m))+(e+r+"X"[q](w)+r+e+`
                              `)[q](h)+x+z)


                              Can probably be trimmed down a little bit






                              f=

                              (w,h,m,f)=>(q="repeat",(z=("# "[q](w+2*(m+f))+`
                              `)[q](f))+(x=((e="# "[q](f))+(r="+ "[q](m))+(t="+ "[q](w))+r+e+`
                              `)[q](m))+(e+r+"X "[q](w)+r+e+`
                              `)[q](h)+x+z)

                              console.log(f(3,2,1,2))








                              share|improve this answer









                              $endgroup$
















                                0












                                0








                                0





                                $begingroup$

                                Javascript, 158 bytes



                                (w,h,m,f)=>(q="repeat",(z=("#"[q](w+2*(m+f)))+`
                                `)[q](f))+(x=((e="#"[q](f))+(r="+"[q](m))+(t="+"[q](w))+r+e+`
                                `)[q](m))+(e+r+"X"[q](w)+r+e+`
                                `)[q](h)+x+z)


                                Can probably be trimmed down a little bit






                                f=

                                (w,h,m,f)=>(q="repeat",(z=("# "[q](w+2*(m+f))+`
                                `)[q](f))+(x=((e="# "[q](f))+(r="+ "[q](m))+(t="+ "[q](w))+r+e+`
                                `)[q](m))+(e+r+"X "[q](w)+r+e+`
                                `)[q](h)+x+z)

                                console.log(f(3,2,1,2))








                                share|improve this answer









                                $endgroup$



                                Javascript, 158 bytes



                                (w,h,m,f)=>(q="repeat",(z=("#"[q](w+2*(m+f)))+`
                                `)[q](f))+(x=((e="#"[q](f))+(r="+"[q](m))+(t="+"[q](w))+r+e+`
                                `)[q](m))+(e+r+"X"[q](w)+r+e+`
                                `)[q](h)+x+z)


                                Can probably be trimmed down a little bit






                                f=

                                (w,h,m,f)=>(q="repeat",(z=("# "[q](w+2*(m+f))+`
                                `)[q](f))+(x=((e="# "[q](f))+(r="+ "[q](m))+(t="+ "[q](w))+r+e+`
                                `)[q](m))+(e+r+"X "[q](w)+r+e+`
                                `)[q](h)+x+z)

                                console.log(f(3,2,1,2))








                                f=

                                (w,h,m,f)=>(q="repeat",(z=("# "[q](w+2*(m+f))+`
                                `)[q](f))+(x=((e="# "[q](f))+(r="+ "[q](m))+(t="+ "[q](w))+r+e+`
                                `)[q](m))+(e+r+"X "[q](w)+r+e+`
                                `)[q](h)+x+z)

                                console.log(f(3,2,1,2))





                                f=

                                (w,h,m,f)=>(q="repeat",(z=("# "[q](w+2*(m+f))+`
                                `)[q](f))+(x=((e="# "[q](f))+(r="+ "[q](m))+(t="+ "[q](w))+r+e+`
                                `)[q](m))+(e+r+"X "[q](w)+r+e+`
                                `)[q](h)+x+z)

                                console.log(f(3,2,1,2))






                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered 1 hour ago









                                zeveezevee

                                57029




                                57029























                                    0












                                    $begingroup$


                                    Wolfram Language (Mathematica), 152 bytes



                                    (t=(p=Table)["# ",(x=2#4+2#3)+#2,x+#];p[t[[i,j]]="+ ",{j,z=#4+1,#4+2#3+#},{i,z,x-#4+#2}];p[t[[i,j]]="X ",{j,#3+z,#3+#4+#},{i,#3+z,#3+#4+#2}];""<>#&/@t)&


                                    Try it online!






                                    share|improve this answer











                                    $endgroup$


















                                      0












                                      $begingroup$


                                      Wolfram Language (Mathematica), 152 bytes



                                      (t=(p=Table)["# ",(x=2#4+2#3)+#2,x+#];p[t[[i,j]]="+ ",{j,z=#4+1,#4+2#3+#},{i,z,x-#4+#2}];p[t[[i,j]]="X ",{j,#3+z,#3+#4+#},{i,#3+z,#3+#4+#2}];""<>#&/@t)&


                                      Try it online!






                                      share|improve this answer











                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$


                                        Wolfram Language (Mathematica), 152 bytes



                                        (t=(p=Table)["# ",(x=2#4+2#3)+#2,x+#];p[t[[i,j]]="+ ",{j,z=#4+1,#4+2#3+#},{i,z,x-#4+#2}];p[t[[i,j]]="X ",{j,#3+z,#3+#4+#},{i,#3+z,#3+#4+#2}];""<>#&/@t)&


                                        Try it online!






                                        share|improve this answer











                                        $endgroup$




                                        Wolfram Language (Mathematica), 152 bytes



                                        (t=(p=Table)["# ",(x=2#4+2#3)+#2,x+#];p[t[[i,j]]="+ ",{j,z=#4+1,#4+2#3+#},{i,z,x-#4+#2}];p[t[[i,j]]="X ",{j,#3+z,#3+#4+#},{i,#3+z,#3+#4+#2}];""<>#&/@t)&


                                        Try it online!







                                        share|improve this answer














                                        share|improve this answer



                                        share|improve this answer








                                        edited 50 mins ago

























                                        answered 1 hour ago









                                        J42161217J42161217

                                        13.3k21251




                                        13.3k21251























                                            0












                                            $begingroup$


                                            Perl 6, 115 bytes





                                            ->a,b,c,d{$_=['#'xx$!*2+a]xx($!=c+d)*2+b;.[d..^*-d;d..^a+$!+c]='+'xx*;.[$!..^*-$!;$!..^a+$!]='X'xx*;.join("
                                            ")}


                                            Try it online!



                                            Roughly golfed anonymous codeblock utilising Perl 6's multi-dimensional list assignment. For example, @a[1;2] = 'X'; will assign 'X' to the element with index 2 from the list with index 1, and @a[1,2,3;3,4,5]='X'xx 9; will replace all the elements with indexes 3,4,5 of the lists with indexes 1,2,3 with 'X'.



                                            Explanation:



                                            First, we initialise the list as a a+2*(c+d) by b+2*(c+d) rectangle of #s.



                                            $_=['#'xx$!*2+a]xx($!=c+d)*2+a;
                                            State:
                                            # # # # # # # # #
                                            # # # # # # # # #
                                            # # # # # # # # #
                                            # # # # # # # # #
                                            # # # # # # # # #
                                            # # # # # # # # #
                                            # # # # # # # # #
                                            # # # # # # # # #


                                            Then we assign the inner rectangle of +s



                                            .[d..^*-d;d..^a+$!+c]='+'xx*;
                                            State:
                                            # # # # # # # # #
                                            # # # # # # # # #
                                            # # + + + + + # #
                                            # # + + + + + # #
                                            # # + + + + + # #
                                            # # + + + + + # #
                                            # # # # # # # # #
                                            # # # # # # # # #


                                            Finally, the innermost rectangle of Xs.



                                            .[$!..^*-$!;$!..^a+$!]='X'xx*;
                                            # # # # # # # # #
                                            # # # # # # # # #
                                            # # + + + + + # #
                                            # # + X X X + # #
                                            # # + X X X + # #
                                            # # + + + + + # #
                                            # # # # # # # # #
                                            # # # # # # # # #




                                            share









                                            $endgroup$


















                                              0












                                              $begingroup$


                                              Perl 6, 115 bytes





                                              ->a,b,c,d{$_=['#'xx$!*2+a]xx($!=c+d)*2+b;.[d..^*-d;d..^a+$!+c]='+'xx*;.[$!..^*-$!;$!..^a+$!]='X'xx*;.join("
                                              ")}


                                              Try it online!



                                              Roughly golfed anonymous codeblock utilising Perl 6's multi-dimensional list assignment. For example, @a[1;2] = 'X'; will assign 'X' to the element with index 2 from the list with index 1, and @a[1,2,3;3,4,5]='X'xx 9; will replace all the elements with indexes 3,4,5 of the lists with indexes 1,2,3 with 'X'.



                                              Explanation:



                                              First, we initialise the list as a a+2*(c+d) by b+2*(c+d) rectangle of #s.



                                              $_=['#'xx$!*2+a]xx($!=c+d)*2+a;
                                              State:
                                              # # # # # # # # #
                                              # # # # # # # # #
                                              # # # # # # # # #
                                              # # # # # # # # #
                                              # # # # # # # # #
                                              # # # # # # # # #
                                              # # # # # # # # #
                                              # # # # # # # # #


                                              Then we assign the inner rectangle of +s



                                              .[d..^*-d;d..^a+$!+c]='+'xx*;
                                              State:
                                              # # # # # # # # #
                                              # # # # # # # # #
                                              # # + + + + + # #
                                              # # + + + + + # #
                                              # # + + + + + # #
                                              # # + + + + + # #
                                              # # # # # # # # #
                                              # # # # # # # # #


                                              Finally, the innermost rectangle of Xs.



                                              .[$!..^*-$!;$!..^a+$!]='X'xx*;
                                              # # # # # # # # #
                                              # # # # # # # # #
                                              # # + + + + + # #
                                              # # + X X X + # #
                                              # # + X X X + # #
                                              # # + + + + + # #
                                              # # # # # # # # #
                                              # # # # # # # # #




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                                              $endgroup$
















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                                                0








                                                0





                                                $begingroup$


                                                Perl 6, 115 bytes





                                                ->a,b,c,d{$_=['#'xx$!*2+a]xx($!=c+d)*2+b;.[d..^*-d;d..^a+$!+c]='+'xx*;.[$!..^*-$!;$!..^a+$!]='X'xx*;.join("
                                                ")}


                                                Try it online!



                                                Roughly golfed anonymous codeblock utilising Perl 6's multi-dimensional list assignment. For example, @a[1;2] = 'X'; will assign 'X' to the element with index 2 from the list with index 1, and @a[1,2,3;3,4,5]='X'xx 9; will replace all the elements with indexes 3,4,5 of the lists with indexes 1,2,3 with 'X'.



                                                Explanation:



                                                First, we initialise the list as a a+2*(c+d) by b+2*(c+d) rectangle of #s.



                                                $_=['#'xx$!*2+a]xx($!=c+d)*2+a;
                                                State:
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #


                                                Then we assign the inner rectangle of +s



                                                .[d..^*-d;d..^a+$!+c]='+'xx*;
                                                State:
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # + + + + + # #
                                                # # + + + + + # #
                                                # # + + + + + # #
                                                # # + + + + + # #
                                                # # # # # # # # #
                                                # # # # # # # # #


                                                Finally, the innermost rectangle of Xs.



                                                .[$!..^*-$!;$!..^a+$!]='X'xx*;
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # + + + + + # #
                                                # # + X X X + # #
                                                # # + X X X + # #
                                                # # + + + + + # #
                                                # # # # # # # # #
                                                # # # # # # # # #




                                                share









                                                $endgroup$




                                                Perl 6, 115 bytes





                                                ->a,b,c,d{$_=['#'xx$!*2+a]xx($!=c+d)*2+b;.[d..^*-d;d..^a+$!+c]='+'xx*;.[$!..^*-$!;$!..^a+$!]='X'xx*;.join("
                                                ")}


                                                Try it online!



                                                Roughly golfed anonymous codeblock utilising Perl 6's multi-dimensional list assignment. For example, @a[1;2] = 'X'; will assign 'X' to the element with index 2 from the list with index 1, and @a[1,2,3;3,4,5]='X'xx 9; will replace all the elements with indexes 3,4,5 of the lists with indexes 1,2,3 with 'X'.



                                                Explanation:



                                                First, we initialise the list as a a+2*(c+d) by b+2*(c+d) rectangle of #s.



                                                $_=['#'xx$!*2+a]xx($!=c+d)*2+a;
                                                State:
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # # # # # # # #


                                                Then we assign the inner rectangle of +s



                                                .[d..^*-d;d..^a+$!+c]='+'xx*;
                                                State:
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # + + + + + # #
                                                # # + + + + + # #
                                                # # + + + + + # #
                                                # # + + + + + # #
                                                # # # # # # # # #
                                                # # # # # # # # #


                                                Finally, the innermost rectangle of Xs.



                                                .[$!..^*-$!;$!..^a+$!]='X'xx*;
                                                # # # # # # # # #
                                                # # # # # # # # #
                                                # # + + + + + # #
                                                # # + X X X + # #
                                                # # + X X X + # #
                                                # # + + + + + # #
                                                # # # # # # # # #
                                                # # # # # # # # #





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                                                answered 4 mins ago









                                                Jo KingJo King

                                                25k359128




                                                25k359128






















                                                    George Harris is a new contributor. Be nice, and check out our Code of Conduct.










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                                                    George Harris is a new contributor. Be nice, and check out our Code of Conduct.













                                                    George Harris is a new contributor. Be nice, and check out our Code of Conduct.












                                                    George Harris is a new contributor. Be nice, and check out our Code of Conduct.
















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