| Bytes | Lang | Time | Link |
|---|---|---|---|
| 025 | Vyxal 3 Mj | 241114T151325Z | pacman25 |
| 013 | Vyxal | 241114T232311Z | emanresu |
| 033 | Japt h | 241114T170349Z | Shaggy |
| 023 | Uiua SBCS | 240405T205049Z | chunes |
| 102 | Excel | 240405T233000Z | z.. |
| 139 | JavaScript Node.js | 240405T115400Z | l4m2 |
| 550 | Scratch | 240403T173023Z | Patric |
| 184 | Python 2 | 160901T174930Z | Neorej |
| 051 | Perl | 160901T095857Z | Ton Hosp |
| 050 | Pyth | 160830T140538Z | Leaky Nu |
| 153 | Python 3 | 160826T223327Z | RootTwo |
| 139 | PowerShell v3+ | 160825T142423Z | AdmBorkB |
| 155 | R | 160825T143140Z | Rudier |
| 087 | Retina | 160826T085746Z | Martin E |
| 017 | Jelly | 160825T163252Z | Dennis |
| 062 | Perl 6 | 160825T210601Z | smls |
| 070 | Mathematica | 160825T115129Z | LegionMa |
| 040 | 05AB1E | 160825T134055Z | Emigna |
| 168 | JavaScript ES6 | 160825T161741Z | Neil |
| 317 | Scala | 160825T160708Z | AmazingD |
| 212 | Python | 160825T150659Z | Jonathan |
| 146 | VBA Excel | 160825T131242Z | Anastasi |
| 251 | C# | 160825T123027Z | TheLetha |
Vyxal 3 -Mj, 28 25 bytes
ᵛÞỊ∦ÞỊϩfSᵛэÞhᴴv/Rꜝ\Ẋ1ÞẠ“
neat
-3 by using 1 and 0 instead of . and *
ƛᵛ•=ÞỊ}∦ÞỊϩfSᵛэÞhᴴv/Rꜝ\Ẋ•ÞẠṠ
ƛᵛ•=ÞỊ} # positions of asterisks in each line
∦ # parallel apply wrap
ÞỊ # line has an asterisk
ϩfS # flatten and sort
# this gives a list with the lines and columns that contain asterisks
ᵛэ # to each list
Þhᴴv # get min/max and decrement the head
/Rꜝ # rangeify and increment
\Ẋ # cartesian product
•ÞẠ # put an asterisk in the coordinates
Ṡ # vectorised sum to format properly
💎
Created with the help of Luminespire.
Vyxal, 13 bytes
⌊:ÞTƒṡvfΠÞǔ⋎⁋
Try it Online! Kinda clunky but it works, takes a char matrix of 1s/0s as input and outputs a grid of 1s/0s. This could be 1 bytes if taking a boolean matrix was allowed :\
Because of how it's written, this (with fluid I/O) actually works for multidimensional arrays of any size. That said, due to how it's written it'll perform badly on sufficiently large arrays - add a vU after the vf to fix this.
⌊ # Parse input
ÞT # Get the multidimensional indices of 1s in the input
ƒṡ # Reduce these coordinates by inclusive ranges,
# creating a ragged array with every number between the minimum and maximum (trick stolen from Dennis's Jelly answer)
vf # Flatten this, getting lists with every number in the range of each coordinate
Π # Get all pairs of coordinates
Þǔ # and create an array with 1s at these multidimensional indices
: ⋎ # and bitwise OR with the input
⁋ # Format as a grid
Japt -h, 33 bytes
Started out at well over 40 bytes, so I'll have to be happy with this for now.
Takes input as an array of lines, using 0 for . and 1 for #.
4Æ=ËhV=UËbÍÃfÌrm VnUËaÍÃrÔ çDÍÎÃz
Uiua SBCS, 28 23 bytes
⍜⊙↻⍜↙∵⋅@#-,+1⊃/↥/↧⊚=@#.
⍜⊙↻⍜↙∵⋅@#-,+1⊃/↥/↧⊚=@#.
. # duplicate
=@# # mask of hash characters
⊚ # coordinates
⊃/↥/↧ # maxes and mins of both columns
+1 # increment the maxes
-, # subtract mins from maxes non-destructively
⍜⊙↻ # start in upper left of bounding box...
⍜↙ # ...taking a rectangle of the specified size...
∵ # for each character in this rectangle...
⋅@# # pop it and push a hash character
Excel, 102 bytes
=LET(a,A1:J5,x,ROW(a),y,COLUMN(a),r,IF(a,x),c,IF(a,y),(x>=MIN(r))*(x<=MAX(r))*(y>=MIN(c))*(y<=MAX(c)))
JavaScript (Node.js), 139 bytes
s=>(L=R=U=D=a=>a.replace(/[^]/g,c=>c<1?X=++Y&&c:c<'$'?(L=L<X?L:X,R=R>++X?R:X,U=U<Y?U:Y,D=D>Y?D:Y,c):X<R&++X>L&Y>=U&Y<=D?'#':c,X=Y=0))(L(s))
Run the unnamed function twice,
Scratch, 550 bytes
(Hardly a golf, rather an exercise in Scratch to operate on strings and matrices)
Instructions:
Create a sprite with two costumes. Replace Costume 1 with a dot . and Costume 2 with a hash #.
The program asks for two inputs: the "flat" matrix s as a string and k - number of columns. This allows to reconstruct a 2D matrix in Scratch (since there are no arrays in Scratch).
Variables m and p were initialized to 1000 assuming that the matrice would not be larger than 999x999
erase all
ask s? and wait
set s to answer
ask s? and wait
set k to answer
set j to 1
set m to 1000
set n to 0
set p to 1000
set q to 0
repeat length of s
if letter j of s = # then
set d to (j-1) mod k+1
set c to (j-d)/k+1
if c>q then
set q to c
if c<p then
set p to c
if d>n then
set n to d
if d<m then
set m to d
change j by 1
set j to 1
show
repeat length of s
set d to (j-1)mod k+1
set c to (j-d)/k+1
if c>(p+1) and d>(m-1) and c<(q-1) and d<(n+1) then
switch costume to 2
else
switch costume to 1
go to x: d*20 y: c*-20
stamp
change j by 1
hide
Test Cases
(Left side - input, right side - output)
Python 2, 184 bytes
def c(i):
m=n=();e,z=enumerate,'for j,r in e(i):\n for k,c in e(r):%s'
exec z%'\n if"#"==c:m+=j,;n+=k,'
exec z%'\n if min(m)<=j<=max(m)<[]>min(n)<=k<=max(n):i[j][k]="#"'
return i
Input and output are a list of strings.
Try it on Ideone (fork of Jonathan Allan's test page)
Perl, 51 bytes
Includes +2 for -0p
Give input on STDIN, off character is A, on character is a, e.g.:
bounding.pl
AAAAAAAAAA
AAaAAAAAAA
AAAAaAAaAA
AAAaAAAAAA
AAAAAAAAAA
^D
bounding.pl:
#!/usr/bin/perl -0p
s%(?=\D*a).+%$a|=$&%eg;s%.*a.*%$a%g;s/a.*a/\L$&/g
Same length:
#!/usr/bin/perl -0p
s%.+%${a./a/g}|=$&%eg;s%.*a.*%$a1%g;s/a.*a/\L$&/g
Python 3, 153 bytes
r=lambda w:list(zip(*w[::-1]))
f=lambda w,n=4:list(map(''.join,n and(('#'in w[0])and r(r(r(f(r(w),n-1))))or[w[0]]+foo(w[1:],n))or['#'*len(w[0])]*len(w)))
Input and output are a list of strings.
ungolfed
r=lambda w:list(zip(*w[::-1])) # rotate grid cw 90 degrees
def f(w,n=4):
if n:
if '#' in w[0]:
u = r(r(r(f(r(w), n-1))))
else:
u = [w[0]] + foo(w[1:], n)
else:
u = ['#'*len(w[0])]*len(w)
return list(map(''.join,u))
theory of operation
The main idea is to remove rows and columns around the outside of the array if they don't have a '#'. Whatever is left should be filled in with '#'s.
It is implemented using a recursive function.
Case 1: row 0 doesn't contain a '#'. Result is row 0 + recursive call on remaining rows.
Case 2: row 0 does contain a '#'. No more rows can be removed. Rotate array cw so that column 0 is now row 0. Then recursively process the rotated array. The result is rotated ccw.
Base case: The array has been rotated 4 times, meaning that all outer rows/columns have been removed if possible. Whatever remains should be filled in with '#'s
PowerShell v3+, 215 162 148 144 139 bytes
param($n)$n|%{(((-join(0..($n[0].length-1)|%{$i=$_;+('1'-in(0..($n.length-1)|%{$n[$_][$i]}))}))-replace'(?<=1.*?).(?=.*?1)',1),$_)[0-ge$_]}
Takes input as an array of strings $n, with 0 instead of . and 1 instead of #. Then, we loop through $n, each iteration testing whether the current string is smaller than 0 (i.e., there's a 1 in it), and if so, output a string. Uses a pseudo-ternary in place of an if/else operation.
The string is constructed from loops through the width of the input string. Each iteration, we tack on a 0 or a 1 depending upon if 1 is found somewhere in the corresponding vertical column. For the last test case, for example, this will result in a string like 0011001001. Requires v3+ for the -in operator. That string is paired with a fancy-dancy regex replace to replace any "inner" 0s with 1s. Much thanks to Business Cat in chat for the assist on that. Our string would be 0011111111 at this point.
Else, output the current (all-zeros) string $_.
The resulting strings are left on the pipeline, and output is implicit. The default Write-Output for an array of strings is with a newline between each element, so that's visually what happens.
Examples
PS C:\Tools\Scripts\golfing> .\highlight-the-bounding-box-cartesian.ps1 '0000000001','0010000000','0000100100','0001000000'
0011111111
0011111111
0011111111
0011111111
PS C:\Tools\Scripts\golfing> .\highlight-the-bounding-box-cartesian.ps1 '0000000000','0000000000','0000100000','0001000000'
0000000000
0000000000
0001100000
0001100000
R, 158 155 bytes
This program takes in input points . and hashtags #, line by line.
v=c();f=which((d=matrix(strsplit(paste0(a<-scan(,""),collapse=""),"")[[1]],nr=sum(a<0),b=T))=="#",a=T);d[min(f[,1]):max(f[,1]),min(f[,2]):max(f[,2])]="#";d
Ungolfed :
a<-scan(,"") #Input
v=c() #Empty vector
f=which((d=(matrix(strsplit(paste0(a,collapse=""),"")[[1]],nr=length(a),b=T)))=="#",a=T) #Main work is here !
d[min(f[,1]):max(f[,1]),min(f[,2]):max(f[,2])]="#" #Creates
#the new figure
d #Displays it
Here are the details of the third line :
paste0(a,collapse="")
#Collapses the input into a single string
strsplit(paste0(a,collapse=""),"")[[1]]
#Split this string character-wise
matrix(strsplit(paste0(a,collapse=""),"")[[1]],nr=sum(a<0),b=T)
#Creates and fills (by row) a matrix with number of row the number of line of the input
which((d=(matrix(strsplit(paste0(a,collapse=""),"")[[1]],nr=l,b=T)))=="#",a=T)
#Gives the index of the matrix's elements that are "#"
Retina, 87 bytes
Byte count assumes ISO 8859-1 encoding.
Tm`A` `^\GA+¶|(¶A+)+\Z|^(A+?)(?<=(?=\D*^\2Z)\A\D*)|(A+)$(?=\D*\Z(?<!(?<!\3)$\D*))
T`p`L
Uses A for . and Z for #.
Jelly, 21 19 18 17 bytes
|/Tr/FṬ|
ỴµZÇZ&ÇY
This is a full program. Input and output are strings of 0's and 1's, delimited by linefeeds.
Try it online! or verify all test cases.
How it works
ỴµZÇZ&ÇY Main link. Argument: s (string)
Ỵ Split s at linefeeds into the array A.
µ Begin a new, monadic chain. Argument: A
Z Zip/transpose A.
Ç Apply the helper link to the transpose.
Z Zip/transpose to restore the original order.
Ç Apply the helper link to A.
& Take the bitwise AND of both results.
Y Join, separating by linefeeds.
|/Tr/FṬ| Helper link. Argument: A (array of strings)
|/ Reduce A columnwise by bitwise OR. This casts to integer.
T Truth; yield the indices of 1's.
r/ Reduce by range. This yields an exponentially growing, nested, ragged
array that contains all integers between the lowest and highest index
in the previous result, at least once but possibly multiple times.
F Flatten the result.
Ṭ Untruth; yield an array with 1's at the specified indices.
Multiple occurrences of the same index are ignored.
| Take the bitwise OR of the result and each row of A.
Perl 6, 62 bytes
{.[.grep(/a/,:k).minmax;$_».grep('a',:k).flat.minmax]='a'xx*}
An anonymous routine that can be passed an array of arrays of characters (representing the matrix) as argument, and modifies it in-place so that the calling scope has the modified array afterwards.
Uses a instead of # as the "on" character. The "off" character can be anything, it doesn't care.
Mathematica, 91 70 bytes
21 bytes saved due to @MartinEnder.
ReplacePart["."+0#,Tuples[Range@@@MinMax/@(#~Position~"#")]]->"#"]&
Anonymous function. Takes a character matrix as input an returns a character matrix as output. The Unicode character is U+F3C7 for \[Transpose].
05AB1E, 70 68 69 61 58 60 40 bytes
€S`¹gG~}Dg©L*0KŸ<U¹v¼y1åi®FXNå}ë0®×}J}¾ä
Explanation
€S` # split each string in input to a charlist and place separately on stack
¹gG~} # OR the char arrays to produce a single list with 1's in the columns that have 1's and 0 in the rest
Dg L* # multiply by indices (1-indexed)
© # store row length in register
0K # remove 0's (the indices which should not have 1's
Ÿ<U # store a list of the indices that should have 1's in X
¹v } # for each string in input
¼ # increase counter
y1åi ë } # if the row contains at least one 1
®FXNå} # push 1 for indices which should have 1 and else 0
0®× # else push a row of 0's
J # join into a string
¾ä # split the string in rows
JavaScript (ES6), 168 bytes
s=>/^#/gm.test(s)?/#$/gm.test(s)?s.replace(/^.*#[^]*#.*$/m,s=>s.replace(/./g,'#'))?f(s.replace(/.$/gm,'')).replace(/$/gm,'.'):f(s.replace(/^./gm,'')).replace(/^/gm,'.')
Takes input as a multiline string. Works by recursively stripping leading and trailing .s from all lines until at least one line begins and one ends with a #, then selects as many lines as possible but starting and finishing on lines containing # and changes all the .s to #. Probably readily golfable.
Scala, 317 characters
val a=input.split("\n");val e=a.map{s=>(s.indexOf("#"),s.lastIndexOf("#"))}.zipWithIndex.filter(_._1._1!= -1);val b=(e.map{s=>s._1._1}.min,e.map{s=>s._1._2}.max,e.head._2,e.last._2);print((0 to a.length-1).map{y=>(0 to a(y).length-1).map{x=>if(x>=b._1&&x<=b._2&&y>=b._3&&y<=b._4)"#" else "."}.mkString+"\n"}.mkString)
More readable version, could probably have golfed it more:
val a=input.split("\n")
val e=a.map{s=>
(s.indexOf("#"),s.lastIndexOf("#"))
}.zipWithIndex // Need the indexes for the Y values
.filter(_._1._1!= -1) // Ugly because of tupleception: (actual tuple, index)
val b=(
e.map{s=>s._1._1}.min,
e.map{s=>s._1._2}.max,
e.head._2,
e.last._2)
print(
(0 to a.length-1).map{y=>
(0 to a(y).length-1).map{x=>
if(x>=b._1&&x<=b._2&&y>=b._3&&y<=b._4)"#"
else "."
}.mkString+"\n"
}.mkString
)
Python, 219 212 bytes
def b(a):j=len(a[0]);g=range;z=g(len(a));h=[i for i in z if'#'in a[i]];w=[i for i,c in[(i,[r[i]for r in a])for i in g(j)]if'#'in c];return[[any((r<h[0],h[-1]<r,c<w[0],w[-1]<c))and'.'or'#'for c in g(j)]for r in z]
(Although I think another method may well be shorter)
Takes and returns a list of list of chars.
Test it on ideoone
VBA Excel, 150 bytes 146 bytes
Instruction:
Create a workbook with two blank worksheets: Sheet1 and Sheet2. Set the input in Sheet1 and then put the following code in the Sheet1 code module
Sub A:For Each C In UsedRange:If C.Value="#"Then Sheet2.Range(C.Address)="#"
Next:For Each C In Sheet2.UsedRange:Range(C.Address)="#":Next:End Sub
Ungolfed the code:
Sub A()
For Each C In UsedRange
If C.Value = "#" Then Sheet2.Range(C.Address) = "#"
Next
For Each C In Sheet2.UsedRange
Range(C.Address) = "#"
Next
End Sub
Explanation:
- Loop through every cell in the used range Sheet1
- Set the conditional statement to copy every cell contains character hashtag (#) in the used range Sheet1 and paste it to the cell in Sheet2 with the same address as Sheet1.
- Loop through once again every cell in the used range Sheet2 to copy every cell address in it and then use it to assign character hashtag (#) to the cell in Sheet1 with the same address as the used range Sheet2.
Example I/O:
Caveat: Make sure every cell in Sheet2 always blank every time you run the program.
C#, 262 251 bytes
s=>{int l,t,r,b,i,j,k;l=t=r=b=i=-1;for(;++i<s.Length;){j=s[i].IndexOf('#');if(j>-1){k=s[i].LastIndexOf('#');l=l==-1|j<l?j:l;t=t==-1?i:t;r=k>r?k:r;b=i;}}for(i=t;i<=b;++i)for(j=l;j<=r;){var c=s[i].ToCharArray();c[j++]='#';s[i]=new string(c);}return s;};
Will golf it further when I have more time.
It compiles into a Func<string[], string[]>.
Formatted version:
s =>
{
int l, t, r, b, i, j, k;
l = t = r = b = i = -1;
for (; ++i < s.Length;)
{
j = s[i].IndexOf('#');
if (j > -1)
{
k = s[i].LastIndexOf('#');
l = l == -1 | j < l ? j : l;
t = t == -1 ? i : t;
r = k > r ? k : r;
b = i;
}
}
for (i = t; i <= b; ++i)
for (j = l; j <= r;)
{
var c = s[i].ToCharArray();
c[j++] = '#';
s[i] = new string(c);
}
return s;
};