Output visual diagram of pictureWrite a code golf problem in which Java winsGolf a Venn Diagram...
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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
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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:
In text form:
# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #
The winning code completes the following conditions in the least possible bytes.
code-golf
New contributor
$endgroup$
|
show 4 more comments
$begingroup$
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:
In text form:
# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #
The winning code completes the following conditions in the least possible bytes.
code-golf
New contributor
$endgroup$
$begingroup$
Nice challenge! For future challenges you may want to use The Sandbox
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– MilkyWay90
3 hours ago
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Also, will the frame height be given?
$endgroup$
– MilkyWay90
3 hours ago
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MilkyWay90, the frame is a constant width around the portrait so only one value is needed.
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– George Harris
3 hours ago
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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
|
show 4 more comments
$begingroup$
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:
In text form:
# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #
The winning code completes the following conditions in the least possible bytes.
code-golf
New contributor
$endgroup$
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:
In text form:
# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #
The winning code completes the following conditions in the least possible bytes.
code-golf
code-golf
New contributor
New contributor
edited 3 hours ago
Stephen
7,49223397
7,49223397
New contributor
asked 3 hours ago
George HarrisGeorge Harris
311
311
New contributor
New contributor
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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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
6 Answers
6
active
oldest
votes
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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)
)
$endgroup$
$begingroup$
Removing thee
assignment saves you two bytes, thef
assignment isn't saving you anything
$endgroup$
– Jo King
33 secs ago
add a comment |
$begingroup$
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))
$endgroup$
add a comment |
$begingroup$
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!
$endgroup$
add a comment |
$begingroup$
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 X
s.
.[$!..^*-$!;$!..^a+$!]='X'xx*;
# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
edited 14 mins ago
answered 1 hour ago
ArnauldArnauld
79k795328
79k795328
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 2 hours ago
NeilNeil
81.7k745178
81.7k745178
add a comment |
add a comment |
$begingroup$
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)
)
$endgroup$
$begingroup$
Removing thee
assignment saves you two bytes, thef
assignment isn't saving you anything
$endgroup$
– Jo King
33 secs ago
add a comment |
$begingroup$
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)
)
$endgroup$
$begingroup$
Removing thee
assignment saves you two bytes, thef
assignment isn't saving you anything
$endgroup$
– Jo King
33 secs ago
add a comment |
$begingroup$
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)
)
$endgroup$
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)
)
edited 2 hours ago
answered 2 hours ago
MilkyWay90MilkyWay90
523212
523212
$begingroup$
Removing thee
assignment saves you two bytes, thef
assignment isn't saving you anything
$endgroup$
– Jo King
33 secs ago
add a comment |
$begingroup$
Removing thee
assignment saves you two bytes, thef
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
add a comment |
$begingroup$
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))
$endgroup$
add a comment |
$begingroup$
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))
$endgroup$
add a comment |
$begingroup$
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))
$endgroup$
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))
answered 1 hour ago
zeveezevee
57029
57029
add a comment |
add a comment |
$begingroup$
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!
$endgroup$
add a comment |
$begingroup$
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!
$endgroup$
add a comment |
$begingroup$
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!
$endgroup$
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!
edited 50 mins ago
answered 1 hour ago
J42161217J42161217
13.3k21251
13.3k21251
add a comment |
add a comment |
$begingroup$
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 X
s.
.[$!..^*-$!;$!..^a+$!]='X'xx*;
# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #
$endgroup$
add a comment |
$begingroup$
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 X
s.
.[$!..^*-$!;$!..^a+$!]='X'xx*;
# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #
$endgroup$
add a comment |
$begingroup$
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 X
s.
.[$!..^*-$!;$!..^a+$!]='X'xx*;
# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #
$endgroup$
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 X
s.
.[$!..^*-$!;$!..^a+$!]='X'xx*;
# # # # # # # # #
# # # # # # # # #
# # + + + + + # #
# # + X X X + # #
# # + X X X + # #
# # + + + + + # #
# # # # # # # # #
# # # # # # # # #
answered 4 mins ago
Jo KingJo King
25k359128
25k359128
add a comment |
add a comment |
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.
George Harris is a new contributor. Be nice, and check out our Code of Conduct.
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
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
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MilkyWay90, the frame is a constant width around the portrait so only one value is needed.
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– George Harris
3 hours ago
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Thanks! Is the constant width always 2 (or is it the height of the frame)?
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– MilkyWay90
3 hours ago
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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. :)
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– George Harris
3 hours ago