Lay out the Carpet

40

5

Inspired by this SO question.

Challenge:

Input:

  • A string \$s\$
  • A character \$c\$

Output:

Create a diamond-square ASCII art of the string in all four directions, with the first character of the string in the center and going outwards. Which is inside a square ASCII-art carpet, with the character as filler. This may sound pretty vague, so here an example:

Input: \$s\$ = string, \$c\$ = .
Output:

..........g..........
........g.n.g........
......g.n.i.n.g......
....g.n.i.r.i.n.g....
..g.n.i.r.t.r.i.n.g..
g.n.i.r.t.s.t.r.i.n.g
..g.n.i.r.t.r.i.n.g..
....g.n.i.r.i.n.g....
......g.n.i.n.g......
........g.n.g........
..........g..........

Challenge rules:

  • Input-string may also be a list of characters
  • Output may also be a list of string-lines or matrix of characters
  • Input-string and character are guaranteed to be non-empty
  • The string is guaranteed to not contain the character
  • Both string and character will only be printable ASCII (unicode range [32,126], space ' ' to and including tilde '~')

General rules:

  • This is , so shortest answer in bytes wins.
    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
  • Default Loopholes are forbidden.
  • If possible, please add a link with a test for your code (i.e. TIO).
  • Also, adding an explanation for your answer is highly recommended.

Test cases:

Input: \$s\$ = 11111, \$c=\$ = 0
Output:

00000000100000000
00000010101000000
00001010101010000
00101010101010100
10101010101010101
00101010101010100
00001010101010000
00000010101000000
00000000100000000

Input: \$s\$ = 12345ABCDEF, \$c\$ = #
Output:

####################F####################
##################F#E#F##################
################F#E#D#E#F################
##############F#E#D#C#D#E#F##############
############F#E#D#C#B#C#D#E#F############
##########F#E#D#C#B#A#B#C#D#E#F##########
########F#E#D#C#B#A#5#A#B#C#D#E#F########
######F#E#D#C#B#A#5#4#5#A#B#C#D#E#F######
####F#E#D#C#B#A#5#4#3#4#5#A#B#C#D#E#F####
##F#E#D#C#B#A#5#4#3#2#3#4#5#A#B#C#D#E#F##
F#E#D#C#B#A#5#4#3#2#1#2#3#4#5#A#B#C#D#E#F
##F#E#D#C#B#A#5#4#3#2#3#4#5#A#B#C#D#E#F##
####F#E#D#C#B#A#5#4#3#4#5#A#B#C#D#E#F####
######F#E#D#C#B#A#5#4#5#A#B#C#D#E#F######
########F#E#D#C#B#A#5#A#B#C#D#E#F########
##########F#E#D#C#B#A#B#C#D#E#F##########
############F#E#D#C#B#C#D#E#F############
##############F#E#D#C#D#E#F##############
################F#E#D#E#F################
##################F#E#F##################
####################F####################

Input: \$s\$ = @+-|-o-|-O, \$c\$ = :
Output:

::::::::::::::::::O::::::::::::::::::
::::::::::::::::O:-:O::::::::::::::::
::::::::::::::O:-:|:-:O::::::::::::::
::::::::::::O:-:|:-:|:-:O::::::::::::
::::::::::O:-:|:-:o:-:|:-:O::::::::::
::::::::O:-:|:-:o:-:o:-:|:-:O::::::::
::::::O:-:|:-:o:-:|:-:o:-:|:-:O::::::
::::O:-:|:-:o:-:|:-:|:-:o:-:|:-:O::::
::O:-:|:-:o:-:|:-:+:-:|:-:o:-:|:-:O::
O:-:|:-:o:-:|:-:+:@:+:-:|:-:o:-:|:-:O
::O:-:|:-:o:-:|:-:+:-:|:-:o:-:|:-:O::
::::O:-:|:-:o:-:|:-:|:-:o:-:|:-:O::::
::::::O:-:|:-:o:-:|:-:o:-:|:-:O::::::
::::::::O:-:|:-:o:-:o:-:|:-:O::::::::
::::::::::O:-:|:-:o:-:|:-:O::::::::::
::::::::::::O:-:|:-:|:-:O::::::::::::
::::::::::::::O:-:|:-:O::::::::::::::
::::::::::::::::O:-:O::::::::::::::::
::::::::::::::::::O::::::::::::::::::

Input: \$s\$ = AB, \$c\$ = c
Output:

ccBcc
BcAcB
ccBcc

Input: \$s\$ = ~, \$c\$ = X
Output:

~

Input: \$s\$ = /\^/\, \$c\$ = X
Output:

XXXXXXXX\XXXXXXXX
XXXXXX\X/X\XXXXXX
XXXX\X/X^X/X\XXXX
XX\X/X^X\X^X/X\XX
\X/X^X\X/X\X^X/X\
XX\X/X^X\X^X/X\XX
XXXX\X/X^X/X\XXXX
XXXXXX\X/X\XXXXXX
XXXXXXXX\XXXXXXXX

Kevin Cruijssen

Posted 2019-03-26T08:51:11.407

Reputation: 67 575

Can the string contain spaces? – Emigna – 2019-03-26T09:16:18.193

1@Emigna Yes, all printable ASCII (unicode range [32,126]) are valid input-characters. – Kevin Cruijssen – 2019-03-26T09:21:01.563

1This becomes wonderful to debug if you use characters that visually look like a single character, e.g. ()()(). – Filip Haglund – 2019-03-26T10:09:12.507

What should happen if $s$ is empty? – Solomon Ucko – 2019-03-26T19:09:08.397

@SolomonUcko From the rules section: "Input-string and character are guaranteed to be non-empty" :) – Kevin Cruijssen – 2019-03-26T19:27:36.640

@KevinCruijssen Oops, missed that! Note that the character is non-empty by definition, but there's nothing wrong with a bit of clarification (what you already have). – Solomon Ucko – 2019-03-26T19:29:12.103

Answers

3

Canvas, 8 bytes

Ｒ±［］／┼┼＊

Try it here!

7 bytes but mirrors a bunch of chars.

dzaima

Posted 2019-03-26T08:51:11.407

Reputation: 19 048

6

R, 118 95 92 bytes

function(a,d,n=length(a),I=c(n:1,1:n)[-n])for(i in I-1)write(c(a,d)[pmin(I+i,n+1)],1,n*2,,d)

Try it online!

Thanks to:

  • Giuseppe for fixing an error and a golf
  • Aaron Hayman for 22 bytes worth of golf

Kirill L.

Posted 2019-03-26T08:51:11.407

Reputation: 6 693

I guess I need to drop my aversion to for loops in R, at least for golfing. – Aaron Hayman – 2019-03-26T15:29:36.850

@Giuseppe, thanks, I shouldn't be so lazy about inclusion of extra test cases! – Kirill L. – 2019-03-26T15:46:39.267

1

This for 98 looks closer to your solution than mine Try it online!

– Aaron Hayman – 2019-03-28T13:10:52.437

@Aaron, Wow, wouldn't even think that this approach could be pushed down so much! Golfed a few more bytes on top of that. – Kirill L. – 2019-03-28T15:42:41.593

1

can take off another two with a bit of rearrangement: Try it online!

– Aaron Hayman – 2019-03-28T16:02:48.997

Alternately, if you prefer outer to for loops, you can get 93 bytes this way

– Giuseppe – 2019-03-28T16:29:45.673

1

@AaronHayman or else this 92 byter combining the pmin logic with the rearrangement :-)

– Giuseppe – 2019-03-28T16:42:35.097

5

J, 59 56 bytes

,{~[:((0-2*#)}.\[:,0,:"0({:>:t)*t=:]+/<:)[:(|.@}.,])#\@]

Try it online!

Too long solution for J ... (entirely my fault)

Galen Ivanov

Posted 2019-03-26T08:51:11.407

Reputation: 13 815

2

Here’s 39

– FrownyFrog – 2019-03-27T14:09:27.673

@FrownyFrog Thanks, I think you should post it as your own solution – Galen Ivanov – 2019-03-27T16:23:57.883

5

R, an ugly 118 bytes version

By letting the input be a vector of single characters, and outputting a matrix instead of printing nice ascii art.

function(s,C,l=length(s),L=4*l-3,k=2*l-1,y=abs(rep(1:k,L)-l)+abs(rep(1:L,e=k)-k)/2+1)matrix(ifelse(y%%1|y>l,C,s[y]),k)

Try it online!

R, 161 157 bytes

saved 4 bytes by using ifelse instead of conditionally modifying y 

function(S,C,l=nchar(S),L=4*l-3,k=2*l-1,y=abs(rep(1:L,k)-k)/2+abs(rep(1:k,e=L)-l)+1)cat(rbind(matrix(ifelse(y%%1|y>l,C,el(strsplit(S,''))[y]),L),'
'),sep='')

Try it online!

ungolfed and commented

function(S,C){
    s=el(strsplit(S,''))
    l=nchar(S)
    L=4*l-3
    k=2*l-1
    y=abs(rep(1:L,k)-k)/2+abs(rep(1:k,e=L)-l)+1 # distance from centre
    y[!!y%%1]=l+1  # set non integers to one more than length of string
    y[y>l]=l+1     # set number beyond length of string to one more than length of string
    M = rbind(matrix(c(s,C)[y],L),'\n') # build matrix and add line returns
    cat(M,sep='') # print the matrix as a string
}

hmmm, seems like the longest answer so far!

Aaron Hayman

Posted 2019-03-26T08:51:11.407

Reputation: 481

Oh dear, should have looked at the question better – Aaron Hayman – 2019-03-26T13:18:33.443

1@KevinCruijssen I have fixed it now – Aaron Hayman – 2019-03-26T13:47:51.780

1

for +15 bytes, you can make your ugly answer ascii-art: Try it online!

– Giuseppe – 2019-03-26T16:58:10.283

4

Python 2, 97 96 90 84 bytes

def f(s,c):r=range(len(s));return[c.join(c*i+s[:i:-1]+s[i:]+c*i)for i in r[:0:-1]+r]

Try it online!

TFeld

Posted 2019-03-26T08:51:11.407

Reputation: 19 246

4

05AB1E, 15 11 bytes

.sûsζøsýí€û

Try it online! or as a Test Suite

Explanation

.s            # push suffixes of input
  û           # palendromize this list
   sζ         # transpose using the second input as filler
     ø        # transpose back
      sý      # merge each on the second input
        í     # reverse each row
         €û   # palendromize each row

Emigna

Posted 2019-03-26T08:51:11.407

Reputation: 50 798

1@KevinCruijssen: Yeah. Fortunately it didn't cost any bytes to fix it. I still feel as if there should be a better way of doing this though, so I'll keep looking. – Emigna – 2019-03-26T09:49:26.770

"Output may also be a list of string-lines or matrix of characters", so you can move the » to the footer. :) – Kevin Cruijssen – 2019-03-26T10:37:24.150

@KevinCruijssen Ah right. I glanced over that part. Thanks :) – Emigna – 2019-03-26T12:01:41.640

4

K (ngn/k), 38 bytes

{1_',/'y,''2{+x,1_|x}/(#x)'1_,/+y,'|x}

Try it online!

{ } function with arguments x (the string s) and y (the character c)

|x reverse x

y,' prepend y to each

+ transpose

,/ concat

1_ drop first char

at this point we have a string of length(x) instances of y followed by the characters from x

#x length of x

(#x)' sliding window of that many consecutive chars

2{ }/ do twice

+x,1_|x join x with the reversed x without its first element, and transpose

y,'' prepend y to each each

,/' concat each

1_' drop one from each

ngn

Posted 2019-03-26T08:51:11.407

Reputation: 11 449

4

J, 35 34 33 bytes

,{~1j1(}:@#"1{:(<*-)1-|+/|)i:@-&#

Try it online!

Right to left:
-&# Length of \$c\$ minus the length of \$s\$
i: Range from negative n to n
1-|+/| Absolute value, outer sum with absolute value, subtract from 1
{: (<*-) Compare the matrix with the end of the range (the -&#), 0 for less than or equal, 1 otherwise. Also subtract the matrix from the end of range. Multiply together. The double subtraction saves a byte and gives something like this

 0  0 _1  0  0
 0 _1 _2 _1  0
_1 _2 _3 _2 _1
 0 _1 _2 _1  0
 0  0 _1  0  0

Negative indices start from -1 like in python. The only thing left is to insert the columns of zeroes.

1j1( #"1 for each row, repeat each element 1 time and pad with one 0
  }:@ then drop the last one (zero)
,{~concat \$c\$ and \$s\$ and index into that

Many thanks to Galen Ivanov for the algorithm.

FrownyFrog

Posted 2019-03-26T08:51:11.407

Reputation: 3 112

Would you mind adding an explanation? I'm not that familiar with J. – Kevin Cruijssen – 2019-03-29T14:02:29.723

3

Japt, 15 bytes

Returns an array of lines

Ôå+ ®¬qV êÃûV ê

Try it

Ôå+ ®¬qV êÃûV ê     :Implicit input of strings U=s & V=c
Ô                   :Reverse U
 å+                 :Prefixes
    ®               :Map
     ¬              :  Split
      qV            :  Join with V
         ê          :  Palindromise
          Ã         :End map
           ûV       :Centre pad each string with V, to the length of the longest
              ê     :Palindromise

Shaggy

Posted 2019-03-26T08:51:11.407

Reputation: 24 623

3

Charcoal, 15 bytes

ＵＢηＥθ✂θκ‖Ｏ↑←ＵＥ¹

Try it online! Link is to verbose version of code. Originally submitted as a comment on the now deleted sandbox post. Explanation:

ＵＢη

Set the background to the second input c.

Ｅθ✂θκ

Map over the first input s to generate all suffixes and implicitly print them on separate lines.

‖Ｏ↑←

Reflect horizontally and vertically.

ＵＥ¹

Add extra space horizontally.

Neil

Posted 2019-03-26T08:51:11.407

Reputation: 95 035

3

Ruby, 95 84 75 bytes

->a,c{(1...2*z=a.size).map{|i|s=a[j=(z-i).abs,z]*c+c*2*j;s.reverse.chop+s}}

Try it online!

Takes input string as an array of chars. Returns an array of strings.

Kirill L.

Posted 2019-03-26T08:51:11.407

Reputation: 6 693

2

Wolfram Language (Mathematica), 121 bytes

Join[q=Table[(v=Table[#2,2(l-k)])<>Riffle[Join[(h=Reverse)[n=(g=Take)[#,-k]],g[n,-k+1]],#2]<>v,{k,l=Length@#}],Rest@h@q]&

Try it online!

J42161217

Posted 2019-03-26T08:51:11.407

Reputation: 15 931

2

Japt, 16 bytes

Note: I'll golf it :)

Ôå+ ®¬qVÃùV mê ê

Try it online!

Luis felipe De jesus Munoz

Posted 2019-03-26T08:51:11.407

Reputation: 9 639

It's fairly similar to the other Japt answer, but in a different order, isn't it? I don't know Japt, but I see similar characters in both answers. ;) – Kevin Cruijssen – 2019-03-26T11:56:32.950

@KevinCruijssen Yeah, both are almost the same, for now – Luis felipe De jesus Munoz – 2019-03-26T12:09:46.633

@KevinCruijssen, Luis developed this independently from my solution. – Shaggy – 2019-03-26T14:31:32.330

2

PowerShell, 120 bytes

param($s,$c)($s,(($l=$s.length-1)..0+1..$l|%{($x=$c*(2*$_))+($s[($_,($l..$_+($_+1)..$l))[$_-ne$l]]-join$c)+$x}))[$l-gt0]

Try it online!

Some days, having index ranges instead of slices really hurts. Today is one of those days. Due to conjoined ranges messing up when dealing with single elements (e.g. returning 0..0+1..0), special-casing is used to avoid it altogether (at the cost of many bytes).

Veskah

Posted 2019-03-26T08:51:11.407

Reputation: 3 580

2

Japt, 15 bytes

Ôå+ ê ®ê ¬qV
ûV

Try it

Oliver

Posted 2019-03-26T08:51:11.407

Reputation: 7 160

2

PowerShell, 82 83 bytes

+2 bytes thanks Veskah: the single character case bug fixed

-1 byte: The rule Input-string may also be a list of characters used

$c,$s=$args
($s|%{(-join$s|% s*g $i)+$c*$i++})[($r=$i..0+1..$i)]|%{"$_"[$r]-join$c}

Try it online!

Less golfed:

$c,$s=$args
$southEast = $s|%{
    (-join$s|% substring $i) + $c*$i++
}
$range=$i..0+1..$i
$southEast[$range]|%{
    "$_"[$range]-join$c
}

mazzy

Posted 2019-03-26T08:51:11.407

Reputation: 4 832

1Looks like this breaks for the single character case. There's just an empty line in the TIO link for ~ – Veskah – 2019-03-26T17:21:14.177

Indeed. Thanks! – mazzy – 2019-03-26T19:54:14.623

2

Jelly, 11 bytes

jÐƤṚzṚŒḄZŒḄ

Try it online!

Left argument: \$s\$.
Right argument: \$c\$ (as a single character, not as a string).
Output: List of Jelly strings (appears as a list of lists of 1-char Python strings, replace ŒṘ with Y to see the \n-joined output).

Erik the Outgolfer

Posted 2019-03-26T08:51:11.407

Reputation: 38 134

2

Pip, 24 20 bytes

QPRV:_JbMa@>RV,#aZDb

Use the -l flag to get human-readable output. Try it online!

Explanation

QPRV:_JbMa@>RV,#aZDb
                      a,b are cmdline args (implicit)
                a     1st cmdline arg (the string)
               #      Length
              ,       Range
            RV        Reverse
         a@>          Take slices of a starting at those indices
                 ZDb  Zip the list of slices together, filling out missing values in
                      the matrix with b (the character)
        M             To each row, map this function:
     _Jb               Join on b
  RV:                 Reverse (making top row the bottom and vice versa)
QP                    Quad-palindromize: reflect downward and rightward, with overlap

For example, with inputs of abcd and .:

RV,#a
 [3 2 1 0]
a@>
 ["d" "cd" "bcd" "abcd"]
ZDb
 [['d 'c 'b 'a] ['. 'd 'c 'b] ['. '. 'd 'c] ['. '. '. 'd]]
_JbM
 ["d.c.b.a" "..d.c.b" "....d.c" "......d"]
RV:
 ["......d" "....d.c" "..d.c.b" "d.c.b.a"]
QP
 ["......d......" "....d.c.d...." "..d.c.b.c.d.." "d.c.b.a.b.c.d" "..d.c.b.c.d.." "....d.c.d...." "......d......"]

DLosc

Posted 2019-03-26T08:51:11.407

Reputation: 21 213

2

Perl 6, 79 bytes

->\c{{map {join c,g $_ X||c},g .[^*X+(^$_,)]}o*.comb}
my&g={.[$_-1...0...$_-1]}

Try it online!

Anonymous codeblock that takes input curried (like f(char)(string)) and returns a list of lines. I think a different approach would be shorter.

Explanation:

my&g={.[$_-1...0...$_-1]}  # Helper function to palindromise a list
->\c{                                                }  # Code block that takes a char
     {                                       }o*.comb   # And returns a function
                                .[^*X+(^$_,)]  # Get all prefixes with end padding
                                               # e.g. "str" => [["r",Nil,Nil]
                                                                ["t","r",Nil]
                                                                ["s","t","r"]]
                              g   # Palindromise the lsit
       map {                },    # Map each element to
                     $_ X||c      # Replace all Nils with the character
                   g              # Palindromise it
            join c,               # And join by the character

Jo King

Posted 2019-03-26T08:51:11.407

Reputation: 38 234

2

Attache, 57 bytes

${q:=#x-1Bounce!Bounce@Join&y@PadLeft&y&#x=>x[q::(q::0)]}

Try it online! Output is a list of lines.

Explanation

?? parameters: x, y
${
    ?? q is length of x - 1
    q:=#x-1
    ?? Reflect, collapsing middle:
    Bounce!
        ?? Function:
            ?? Reflect,
            Bounce@
            ?? Joined by y,
            Join&y@
            ?? padded to the length of x with y
            PadLeft&y&#x
        ?? Mapped over
        =>
            ?? The elements of x at
            x[
                ?? decreasing range from q to
                q::(
                    ?? each element in the range from q to 0
                    q::0
                )
            ]
}

Conor O'Brien

Posted 2019-03-26T08:51:11.407

Reputation: 36 228

2

APL (Dyalog Classic), 32 31 bytes

(⍉1↓¯1↓⊖⍪1↓⊢)⍣2∘↑(≢⊣),/,¨,2⍴¨⊢¨

Try it online!

ngn

Posted 2019-03-26T08:51:11.407

Reputation: 11 449

1

Perl 5 with -lF, -M5.010, 71 bytes

$"=<>;$A=abs,$_="@F[$A..$#F]".$"x($A*2),/./,say reverse.$' for-$#F..$#F

Try it online!

Dom Hastings

Posted 2019-03-26T08:51:11.407

Reputation: 16 415

1

TSQL query, 191 bytes

In MS-SQL Server Management Studio press Ctrl-T before running this query, this will change the output to text.

This script is building up the output from left to right in one long "string", calculating the value to put in each position. The output is limited to 4096 characters.

Golfed:

SELECT
string_agg(iif(h>k/2,@y,substring(@,h+1,1))+iif(-~n%k=0,'
',@y),'')FROM(SELECT
abs(k/2-n%k)+abs(k/2-n/k)h,*FROM(SELECT
number n,len(@)*2-1k,*FROM spt_values)c)d
WHERE n<k*k and'P'=type

Ungolfed:

USE master
DECLARE 
@y char='.',
@ varchar(20) = 'abcd'

SELECT
  string_agg(iif(h>k/2,@y,substring(@,h+1,1))+iif(-~n%k=0,'
',@y),'')
FROM
(
  SELECT
    abs(k/2-n%k)+abs(k/2-n/k)h,*
  FROM
  (
    SELECT
      number n,
      len(@)*2-1k,*
    FROM spt_values
  )c
)d
WHERE n<k*k and'P'=type

I had to make some changes to format the output in the online version.

Try it online

t-clausen.dk

Posted 2019-03-26T08:51:11.407

Reputation: 2 874

1

C# (Visual C# Interactive Compiler), 249 bytes

s=>c=>{var r=s.Select((x,_)=>{int k=s.Length;var m=s.Substring(_,k-_).Aggregate("",(a,b)=>a+c+b);return new string(m.Skip(2).Reverse().Concat(m.Skip(1)).ToArray()).PadLeft(2*k-3+m.Length,c).PadRight(4*k-3,c);});return r.Skip(1).Reverse().Concat(r);}

Try it online!

This must be improvable...

Expired Data

Posted 2019-03-26T08:51:11.407

Reputation: 3 129

1213 – dana – 2019-03-31T12:10:07.793

1

JavaScript (Node.js), 143 bytes

(s,c)=>{q=Math.abs;m=(l=s.length*4-3)-1;for(i=j=0;j<l/2;(i=(++i)%l)?0:++j){p=s[q(i-m/2)/2+q(j-m/4)];process.stdout.write((i?"":"\n")+(p?
p:c))}}

Try it online!

A bit more thinkering would lead to calculating in terms of a one-dimensional array, and less bytes.

Igor Sowinski

Posted 2019-03-26T08:51:11.407

Reputation: 29

1

Kotlin, 250 bytes

Note: Kotlin tio currently fails to return a new class so this code gets an null pointer exception. This also occurs for codes I previously posted that worked at that time. I assume it will eventually get fixed, but couldn't find a support contact to report the issue to. It can also be run here.

{s:String,c:Char->val h=s.length*2-1
val w=h*2-1
val m=Array(h){Array(w){c}}
for(i in s.indices)for(r in 0..h-1){val o=(i-Math.abs(h/2-r))*2
if(o>=0){m[r][w/2+o]=s[i].toChar()
m[r][w/2-o]=s[i].toChar()}}
m.map{it.joinToString("")}.joinToString("\n")}

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JohnWells

Posted 2019-03-26T08:51:11.407

Reputation: 611

You can contact @Dennis in The Ninteenth Byte chat. He's the moderator for TIO. Also, I allow returning a list of strings instead of actually printing, so I think (not sure) you can remove the .joinToString("\n") from the byte-count (and do that in the footer outside the function).

– Kevin Cruijssen – 2019-03-29T22:59:57.633

1

JavaScript (Node.js), 101 bytes

s=>c=>[l=s.length-1,...Array(l*2)].map((x,i,a,m=Math.abs)=>a.map((y,j)=>s[m(l-i)+m(l-j)]||c).join(c))

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dana

Posted 2019-03-26T08:51:11.407

Reputation: 2 541

1close :( – ASCII-only – 2019-04-10T06:42:40.083

1

Java (JDK), 213 199 198 bytes

a->b->{int i=0,l=a.length()-1;String s=a,r[]=new String[l-~l],p;for(;i<=l;s=s.substring(1))r[l+i]=r[l-i]=new StringBuffer(p=b.join(b,s.split(""))+b.repeat(2*i++)).reverse()+p.substring(1);return r;}

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-14 bytes thanks to @KevinCruijssen
-1 byte thanks to @ceilingcat

Ungolfed

a->
    b-> {
        int i = 0, l = a.length() - 1;
        String s = a, r[]=new String[a.length()*2-1],p;
        for (; i<=l; s=s.substring(1))
            r[l+i]
              = r[l-i++]
              =   new StringBuffer(
                                   p =   String.join(b,s.split(""))
                                       + b.repeat(2*i)
                                  ).reverse()
                + p.substring(1);
        return r;
    }

Sara J

Posted 2019-03-26T08:51:11.407

Reputation: 2 576

1

Nice answer, you can golf it by 14 bytes below 200 however. :)

– Kevin Cruijssen – 2019-04-03T06:37:17.540

@KevinCruijssen Well spotted, thanks! – Sara J – 2019-04-03T10:09:25.840

@ceilingcat Nice thinking, thanks! – Sara J – 2019-09-20T01:03:39.323

1

Gaia, 19 bytes

:ṫl¤┅v;@&Z¦<¦v¦ṫ¦ṫṣ

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Explanation to follow.

Giuseppe

Posted 2019-03-26T08:51:11.407

Reputation: 21 077

1

Wolfram Language (Mathematica), 68 bytes

Table[#[[1+Abs@y+Abs@x/2]]/._@__:>#2,{y,l=1-Tr[1^#],-l},{x,2l,-2l}]&

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Takes a list of characters (along with the filler character) as input, and outputs a matrix of characters.

Table[                              ,{y,l=1-Tr[1^#],-l},{x,2l,-2l}]&    (* make a (2*len-1 x 4*len-3) table centered at (0,0) *)
      #[[               ]]                                              (*  where the value at each point is the string indexed at *)
         1+Abs@y+Abs@x/2                                                (*   (1 + |y-coordinate| + |x-coordinate|/2) *)
                          /._@__:>#2                                    (*  if that indexing fails, use the filler character instead *)

To take the index of the list of characters, we use list[[index]], which is internally expanded to Part[list, index]. If the index is valid, that expression evaluates to the value at that index. If not - if the index is not an integer or is out of range - the expression remains unevaluated.
The simplest (shortest) pattern which matches Part[...] but not a single character is _@__, which matches any expression with one or more arguments.

attinat

Posted 2019-03-26T08:51:11.407

Reputation: 3 495

Asked: 2019-03-26T08:51:11.407

Viewed: 2 778 times

Active: 2019-09-20T06:36:45.043