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I got this challenge from Codingame and am curious about better solutions than mine:
Given a width via standard input draw a hollow square of '#' in given width and length.
Example:
5 results in
#####
# #
# #
# #
#####
I used python to solve this so i am particulary interested in other python code. But please feel free to post your solution in any language you want.
Pretty darn long for a "golfing" language, but I have forgotten how a lot of it works :P
i_2-_u'#*N+_'#' u*'#N+++u*u
Explanation:
i_2-_u # take input and triplicate, subtracting 2 (5 -> [3,3,5])
'#*N+_ # create the top and bottom rows
'#' u*'#N+++u* # create input - 2 copies of middle rows
u # rotate left 1 to get correct order, implicit output
Golfed:
void F(int n){Console.Write($"{new string('#',n)}\n");for(int i=2;i<n;i++)Console.Write($"#{new string(' ',n-2)}#\n");Console.Write(new string('#',n));}
Ungolfed:
void F(int n)
{
Console.Write($"{new string('#', n)}\n");
for (int i = 2; i < n; i++)
Console.Write($"#{new string(' ', n - 2)}#\n");
Console.Write(new string('#', n));
}
EDIT1: Loop range optimization.
Line split in two for readability:
#N::((var X (repeat "#" N))(print X)(each (seq 3 N) (scope #X::((print (+ "#"
(repeat " " (- N 2)) "#")))))(print X))
Sample usage:
% square.lithp
(
(import "lists")
(def s #N::((var X (repeat "#" N))(print X)(each (seq 3 N) (scope #X::((print (+ "#" (repeat " " (- N 2)) "#")))))(print X)))
(s 10)
)
Output:
$ ./run square.lithp
##########
# #
# #
# #
# #
# #
# #
# #
# #
##########
?sd0[1+[#]Pdld>k]sklkx0[1+[
]P0[#]P[1+[ ]Pdld2->x]sxlxxk[#]Pdld2->r]srld2<r[
]P0lkx
This is mostly straightforward. It's kinda late so I won't write up an explanation now, but I might edit one in later.
(Note: I'm new to dc so there are probably obvious improvements to be made here.)
def f(j):r=range(j);print'\n'.join(''.join(' #'[x*y<1or max(x,y)==j-1]for y in r)for x in r)
def s set
s n io.readnum()
s g goto
def w io.write "#"
s W io.writeln
if n=0
g 3
if n=1
g 4
label 0
if i=n
g 1
s i 1+i
w
g 0
label 1
s i 0
s m n+-2
W()
label a
if i=m
g b
s i 1+i
s j 0
w
label c
if j=m
g d
s j 1+j
io.write " "
g c
label d
W "#"
g a
label b
s i 0
label 2
if i=n
g 3
s i 1+i
w
g 2
label 4
w
label 3
This is so wrong... but anyway:
DECLARE @ INT=1,@n INT=10,@a VARCHAR(MAX),@b VARCHAR(MAX),@c CHAR='#',@s VARCHAR(MAX)=''WHILE @<=@n BEGIN SET @a=concat(@a,@c)SET @b=concat(@b,(SELECT CASE WHEN @=1THEN @c WHEN @=@n THEN @c ELSE' 'END))SET @=@+1 END SET @=1 WHILE @<=@n BEGIN SET @s=@s+(SELECT CASE WHEN @=1THEN @a WHEN @=@n THEN CHAR(13)+@a ELSE CHAR(13)+@b END)SET @=@+1 END PRINT @s
Ungolfed:
DECLARE @ INT=1,
@n INT=10,
@a VARCHAR(MAX),
@b VARCHAR(MAX),
@c CHAR='#',
@s VARCHAR(MAX)=''
WHILE @<=@n
BEGIN
SET @a=concat(@a,@c)
SET @b=concat(@b,(SELECT CASE WHEN @=1THEN @c WHEN @=@n THEN @c ELSE' 'END))
SET @=@+1;
END
SET @=1
WHILE @<=@n
BEGIN
SET @s=@s+
(SELECT
CASE
WHEN @=1THEN @a
WHEN @=@n THEN CHAR(13)+@a
ELSE CHAR(13)+@b
END
)
SET @=@+1
END
PRINT @s
x=input();h='#';print h*x+'\n'+(h+' '*(x-2)+h+'\n')*(x-2)+h*x
My first attempt at this problem, I will golf it down hopefully.
+5 bytes I forgot to put newlines between the lines :(
+4 bytes I forgot to add 2 to the last x :(
-7 bytes If I don't subtract 2 from the input at the beginning and subtract it where it would make sense to do so, I can actually save 7 bytes :)
-1 byte Thanks to @Cyoce for helping me save 1 byte by assigning '#' to a variable (which I called h because that's what he was thinking in the comments :P)
Still 1 byte longer than before... :(
'#' (I was imagining calling the variable h and my brain goofed
\$\endgroup\$
+++++[>++>+++++++<<-],[>>>+>+>+>+<<<<<<-]>>>>[-<<.>>]++++[>>>++++++++<<<-]>--<<[<<.>.>>>[>>.<<-]<<<.>->>>[<+<+>>-]<<[>>+<<-]>--<<]<<.>>>>>[<<<<.>>>>-]
Hopefully this can be golfed a bit more.
Explanation:
+++++[>++>+++++++<<-] Initializes to [0, 10, 35]. 10 is the charcode for a newline, 35 is the charcode for '#'
, Gets the input (as an ASCII char, which is converted to an integer)
[>>>+>+>+>+<<<<<<-] Copies the input to 4 different spaces
>>>>[-<<.>>] Prints the appropriate number of '#'
++++[>>>++++++++<<<-] Puts 32 in the next available space (32 is the charcode of a space)
>--<<-- Subtracts 2 from one of the input values and the value on which to loop
[
<<.>.>>> Print a newline and a '#'
[>>.<<-] Print n - 2 spaces
<<<.>->>> Print another '#'
[<+<+>>-]<<[>>+<<-]>--<< Copy n to two other spaces, then copy it back
]
<<.>>>>> Print a newline
[<<<<.>>>>-] Print the appropriate number of '#'
n=gets.to_i
puts ?#*n
if n>1
puts [?#+' '*(n-2)+?#]*(n-2),?#*n end
(format t"#~v@{#~}#
~@*~v@{~@*#~v@t#
~}~@*#~v{#~}#"(-(read)2)1)
For n=1 will give:
##
##
but I guess it's ok, since author's solution doesn't seem to work for n=1 (tested here but maybe it's because it's other Python version?)
Other than that it should work fine.
format ;printing function
~v@{#~} ;loop n times, where n equals current argument
;printing "#" - gives n "#"s
~@* ;go back to first argument - reuse (-(read)2)
~v@t ;add n spaces where n is current argument
'#Ž9¦Λ
'#Ž9¦Λ # full program
Λ # draw lines of length...
# implicit input...
Λ # made from...
'# # literal...
Λ # s with directions in [N, NE, E, SE, S, SW, W, NW] with indices in each character in...
Ž9¦ # 2460
# implicit output
Rj, 15 bytes₌Ẋ‹%ƛ1c\#*ð∨;∑²
I'm actually proud of this answer, because it took a bit of thinking and effort to obtain.
₌Ẋ‹%ƛ1c\#*ð∨;∑²
₌Ẋ‹ # Push cartesian_product(range(1, input + 1), range(1, input + 1)), input - 1
% # modulo those
ƛ ; # over each item X:
1c # does X contain 1?
\#* # repeat "#" that many times
ð∨ # and logical or with space
∑² # join into a single string and format as a grid
Tacit Function that takes number of lines as argument, returns hollow square
(' #'⌷⍨1∘+)¨(∘.∨⍨⍳∊1∘,)
(' #'⌷⍨1∘+)¨(∘.∨⍨⍳∊1∘,) 10
##########
# #
# #
# #
# #
# #
# #
# #
# #
##########
