Diamond creator +

Question

Challenge :

Given an integer n as input. Create a diamond that is 2x the given number n.

Input :

Input is integer n and 2 < n ≤ 3000.

Output :

Output will be a string and it will be in form of a diamond consisting of + with an addition line at the start showing n using +

Examples :

D(3) :

+++
  +
 +++
+++++
+++++
 +++
  +

D(5) :

+++++
    +
   +++
  +++++
 +++++++
+++++++++
+++++++++
 +++++++
  +++++
   +++
    +

D(6) : 

++++++
     +
    +++
   +++++
  +++++++
 +++++++++
+++++++++++
+++++++++++
 +++++++++
  +++++++
   +++++
    +++
     +

Winning Criteria :

This is code-golf so shortest code in bytes for each programming language wins.

Answer 1

brainfuck, 151 139 bytes

,[.[<]<+[>>]++++[-<++++++++>],]<[<]<<<++++++++++.>>[[>]>[-<+>]>[-<+>]>>[.>>]<<[<]<<.<<[..<<]<.>>-]>[[>]>[.>>]<<[<<]>.>>[..>>]<<,<[<]<<.>>>]

Try it online!

Takes input via unary, with +s as tally marks (allowed by the poster). Decided to rework this, as I thought the old one was a bit longer than it could be (though this one is too!).

Old Version (151 bytes):

>--[>+<++++++]<[->+>.<<]++++++++[-<+<++++>>]<++>>[<<.>>-[-<+<<.>>>]<[->+<]>>>+[-<.>>+<]>+[-<+>]<<<]>>[<<<<.>>[-<+<<.>>>]<[->+<]>+>>-[-<.>>+<]>-[-<+>]<]

Try it online!

Takes input as the starting cell. I couldn't think of a way to leverage the first half to help with the second, so there's a loop for each of them.

How It Works:

 >--[>+<++++++]  Create 43 ('+') two space to the left of n
 <[->+>.<<]      Print n '+'s while preserving n
 ++++++++[-<+<++++>>]<++  Create 32 (' ') and 10 ('\n')
                         Tape: 32 10 0 n 43 t
 >>
 [ Loop over the first half of the diamond
   <<.>>         Print a newline
   -[-<+<<.>>>]  Decrement n and print n spaces
   <[->+<]       Restore n
   >>>+[-<.>>+<] Increment t and print t '+'s
   >+[-<+>]<<<   Increment t again and restore it
]>>
[ Loop over the second half
  <<<<.>>        Print a newline
  [-<+<<.>>>]<   Print n spaces
  [->+<]>+       Restore and increment n
  >>-[-<.>>+<]   Decrement t and print t '+'s
  >-[-<+>]<      Decrement t again and restore it
]

And just for fun:

+++++++++
        >
       --[
      >+<++
     ++++]<[
    ->+>.<<]+
   +++++++[-<+
  <++++>>]<++>>
 [<<.>>-[-<+<<.>
>>]<[->+<]>>>+[-<
.>>+<]>+[-<+>]<<<
 ]>>[<<<<.>>[-<+
  <<.>>>]<[->+<
   ]>+>>-[-<.>
    >+<]>-[-<
     +>]<]++
      +++++
       +++
        +

Try it online!

Answer 2

Canvas, 9 bytes

+×Ｏ｛+×］±╪

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Explanation (some characters have been replaced to look monospace):

+×O{+×]±╪
+×         repeat "+" input times
  O        output that
   {  ]    map over 1..input
    +×       repeat "+" that many times
       ±   interpret the array as a 2D string, and reverse it
        ╪  quad-palindromize with 1 horizontal overlap and 0 vertical overlap

Answer 3

Python 3, 95 94 75 bytes

def f(n):a=[' '*(n+~i)+'+'*(i-~i)for i in range(n)];return['+'*n]+a+a[::-1]

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---

My first attempt at some golfing, any suggestions for improvement are welcome.

EDIT: saved 1 byte thanks to Kevin Cruijssen

EDIT: removed misunderstanding about byte count

EDIT: Saved many more bytes thanks to Jo King and user202729

Answer 4

05AB1E, 14 bytes

'+×sL·<'+×∊.c»

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Explanation

'+×              # push "+" repeated <input> times
   sL            # push range [1 ... input]
     ·<          # multiply each element by 2 and decrement (x*2-1)
       '+×       # replace each number with the corresponding number of "+"'s
          ∊      # mirror vertically
           .c    # center
             »   # join with the "+"-row created at the start

Also 14 bytes: L‚˜'+×ćs.∞∊.c»

Answer 5

Python 3, 79 78 bytes

def f(n):x=[('++'*i+'+').center(n*2)for i in range(n)];return[n*'+']+x+x[::-1]

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Thanks to this Tips for golfing Python answer for informing me about the .center function. Returns a list of strings.

Answer 6

R, 135 110 96 bytes

function(n){cat("+"<n,"
",sep="")
for(i in c(1:n,n:1))cat(" "<n-i,"+"<2*i-1,"
",sep="")}
"<"=rep

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@JayCe with the final cut.

The rep function is assigned to an existing infix operator, such as < or ^ so that rep("+", n) is equivalent to "<"("+", n) which can be written out using < as an infix operator as in "+" < n and shortened to "+"<n.

Answer 7

Charcoal, 15 bytes

Ｇ→→↙Ｎ+↓‖Ｍ↑×⊕ⅈ+‖

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

Ｇ→→↙Ｎ+

Print an inverted triangle of +s the height of the input and almost twice the width.

↓

Move the cursor down so it lands on the additional line after the reflection.

‖Ｍ↑

Make a mirror image of the triangle.

×⊕ⅈ+

Draw the additional line using the current column to avoid having to read the input again.

‖

Reflect the output so that the additional line points to the left.

Answer 8

Stax, 11 bytes

ñ1┌╙@↔g+┤a☻

Run and debug it

Answer 9

Python 3, 76 75 bytes

lambda n:['+'*n]+[' '*(n+~i)+'+'*(i-~i)for i in[*range(n),*range(n)[::-1]]]

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Answer 10

QB64, 82 79 bytes

INPUT n
?STRING$(n,43):FOR a=1TO 2*n:d=a-(a>n)*2*(n-a):?SPC(n-d);STRING$(2*d-1,43):NEXT

Answer 11

J, 29 22 bytes

-7 bytes thanks to Jonah!

' +'{~#&1,]>:[:+/~|@i:

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Original solution:

J, 29 bytes

'+'(,]\(}:@|."1,.])@,]\.)@$~]

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Explanation:

'+'$~] - generates the line at the start, which is a seed for the diamond:

   '+'$~]  3
+++

]\,]\. - finds the prefixes (]\) and suffixes (]\.) of the line, making "half" the diamond 

   '+'(]\,]\.)@$~] 3
+  
++ 
+++
+++
++ 
+  

}:@|."1,.] - makes the other "half" of the diamond by reversing each line (|."1)
and dropping its last '+' (}:) and stitches the first half to it (,.])

 '+'(]\(}:@|."1,.])@,]\.)@$~] 3
  +  
 +++ 
+++++
+++++
 +++ 
  +  

, - prepends the initial line to the diamond

'+'(,]\(}:@|."1,.])@,]\.)@$~] 3
+++  
  +  
 +++ 
+++++
+++++
 +++ 
  +  

Answer 12

Scala, 95 94 bytes

saved 1 byte(s) thanks to the comment.

---

Golfed version. Try it online!

def f(n:Int)={val a=(0 until n).map(i=>" "*(~i+n)+"+"*(2*i+1)).toSeq;Seq("+"*n)++a++a.reverse}

Ungolfed version.

object Main {
  def f(n: Int): Array[String] = {
    val a = (0 until n).map(i => " "*(n-1-i) + "+"*(2*i+1)).toArray
    Array("+"*n) ++ a ++ a.reverse
  }

  def main(args: Array[String]): Unit = {
    val result = f(10)
    println(result.mkString("\n"))
  }
}

Answer 13

JavaScript (Node.js), 106 105 bytes

• thanks to @Kevin Cruijssen for reducing by 1 byte

n=>[...Array(n*2+1)].map((_,i)=>" ".repeat(i?i>n?i+~n:n-i:0)+"+".repeat(i?i>n?4*n-2*i+1:i*2-1:n)).join`
`

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________________________________________________

Second approach

JavaScript (Node.js), 105 100 99 98 bytes

• thanks to @Kevin Cruijssen for reducing by 1 byte
• thanks to @ovs for reducing by 1 byte

n=>[X="+"[r="repeat"](n),...x=[...X].map((_,i)=>" "[r](n+~i)+"+"[r](i-~i)),...x.reverse()].join`
`

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Answer 14

Japt -R, 18 17 bytes

õÈ+Y î+Ãû ê1 iUî+

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Answer 15

PowerShell, 55 bytes

param($l)'+'*$l;1..$l+$l..1|%{" "*($l-$_)+'+'*($_*2-1)}

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Answer 16

Haskell, 85 82 bytes

^{Saved 3 bytes thanks to nimi!}

n!c=[1..n]>>c
f n|x<-[(n-i)!" "++(i*2-1)!"+"|i<-[1..n]]=unlines$n!"+":x++reverse x

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Answer 17

Python 3, 98 bytes

def d(s):print("+"*s);t=[("+"*i).center(2*s-1)for i in range(1,2*s,2)];print("\n".join(t+t[::-1]))

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Readable version:

def diamond(size):
    print(size * "+")
    top = [("+" * i).center(2*size - 1) for i in range(1, 2*size, 2)]
    print("\n".join(top))
    print("\n".join(reversed(top)))

Answer 18

05AB1E, 9 8 bytes

Dη€û.C∊»

Try it online! Takes input in unary with + as a tally mark; allowed as per OP's comment.

Dη€û.C∊»  # full program
       »  # join...
D         # implicit input...
       »  # and...
 η        # prefixes of...
D         # implicit input...
  €       # with each element...
   û      # concatenated with...
          # (implicit) current element in list...
   û      # reversed excluding the first character...
    .C    # with...
          # (implicit) each element...
    .C    # centered by spaces to length of longest element in this list (extra space on the left if unequal spaces)...
      ∊   # mirrored vertically...
       »  # by a newline
          # implicit output

Answer 19

Ruby, 71 61 bytes

->n{[?+*n]+(a=(1..n).map{|x|?\s*(n-x)+?+*(x+x-1)})+a.reverse}

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Answer 20

PHP, 103 bytes

for(;$i++<$argn;$s.="
".str_pad(str_pad("",$i*2-1,"+",2),$argn*2-1," ",2))echo"+";echo"$s
",strrev($s);

Run as pipe with `-nR´ or try it online.

Answer 21

PowerShell, 58 bytes

param($n)'+'*$n;1..$n+$n..1|%{" "*($n-$_)+"+"*$_+"+"*--$_}

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Simply a loop-up and -down, each iterating outputting the appropriate number of spaces and then the appropriate number of plus signs. Ho-hum.

Answer 22

F# (Mono), 123 bytes

let d n=
 let t n=String('+',n)
 let s n=t(n*2-1)
 [1..n]@[n.. -1..1]|>Seq.fold(fun a x->a+sprintf"\n%*s"(n+x-1)(s x))(t n)

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Answer 23

PHP 102 bytes

for($r=str_pad;$i++<$a;$s.="\n".$r($r("",$i*2-1,"+",2),$a*2-1," ",2))echo"+";echo"$s\n",strrev($s);

Ik know it can be much smaller than this ;) Greetz mangas

Answer 24

sed 4.2.2, 69

Score includes +1 for the -r option to sed.

h
y/1/+/
p
x
s/1\B/ /g
y/1/+/
p
:
s/ \+/+++/p
t
p
:a
s/\+\+/ /p
ta
d

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Answer 25

Ruby, 59 bytes

->i{[?+*i,b=(1..i).map{|c|' '*(i-c)+?+*(2*c-1)},b.reverse]}

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Answer 26

Yabasic, 102 bytes

An anonymous function that takes input as a unary number with + tally marks and outputs to the console.

Input""s$
n=Len(s$)
?s$
For i=-n To n
j=Abs(i)
If i For k=2To j?" ";Next:?Mid$(s$+s$,1,2*(n-j)+1)
Next

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Alternate Version, 117 bytes

An anonymous function answer that takes input as a decimal integer and outputs to the console.

Input""n
For i=1To n s$=s$+"+"Next
?s$
For i=-n To n
j=Abs(i)
If i For k=2To j?" ";Next:?Mid$(s$+s$,1,2*(n-j)+1)
Next

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Answer 27

JavaScript (Node.js), 183 bytes

a=x=>{g='\n';r=(m,n)=>String.prototype.repeat.call(m,n);k='+';l=r(k,x)+g;c=d='';for(i=0;i++<x;c+=r(' ',x-i)+r(k,i)+r(k,i-1)+g,d+=r(' ',i-1)+r(k,x+1-i)+r(k,x-i)+g);console.log(l+c+d);}

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Updated my answer thanks to @JoKing

Answer 28

APL (Dyalog Unicode), 25 bytes^SBCS

⍪∘⊖⍨c,⍨⌽1↓[2]c←↑,\⎕←⎕/'+'

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Explanation:

⍪∘⊖⍨c,⍨⌽1↓[2]c←↑,\⎕←⎕/'+'  ⍝ Full program
                       ⎕/'+'  ⍝ Get input from user as N, replicate '+' N times
                    ⎕←        ⍝ Print above string
                  ,\           ⍝ Find all prefixes of above string, e.g. '+','++','+++' etc.
                 ↑             ⍝ Mix the above into a matrix - right-pads with spaces as needed
               c←              ⍝ Assign above matrix to 'c' for 'corner'
          1↓[2]                ⍝ Drop the first column
        ⌽                     ⍝ Reverse the resulting matrix
     c,⍨                      ⍝ Append 'c' to above - this gives us the top half
⍪∘⊖⍨                         ⍝ Take the above, flip it about the horizontal axis,
                              ⍝ and append it to itself

Answer 29

Canvas (v8), 8 5 bytes

［］↔╪∔

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Takes input as unary using + as tally.

The other Canvas answer here uses v2.

Answer 30

Vyxal j, 6 bytes

¦ʁøĊmJ

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How?

¦ʁøĊmJ
¦      # Prefixes of implicit input
 ʁ     # Palindromize each
  øĊ   # Center
    m  # Mirror each
     J # Join the (implicit) input with this