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Challenges

Word Set Square

+7
−0

Challenge

Given a string, e.g. Hello, do the following:

Mirror it:

Hello -> HelloolleH

and create a right triangle using it as the sides:

H         
ee        
l l       
l  l      
o   o     
o    o    
l     l   
l      l  
e       e 
HelloolleH

Which looks sort of like a set square.

That's it!

Scoring

This is code-golf. Shortest answer in each language wins.

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11 answers

+6
−0

APL (Dyalog Unicode), 23 bytes

Anonymous tacit prefix function. Reuires 0-based indexing (⎕IO←0)

(⊢⍪⍨¯1↓⊢,∘↑-∘⍳∘≢↑¨⊢)⊢,⌽

Try it online!

 reverse the argument

⊢, prepend the argument

() apply the following tacit function to that:

 on the characters of the argument

 …↑¨ of each character, take this many characters:

  -∘⍳∘≢ the negative indices $0…\text{length}-1$

 …∘↑ mix list of strings into character matrix, then:

  ⊢, prepend the argument (vertically)

¯1↓ drop the last row

⊢⍪⍨ append the argument below

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+3
−0

V (vim), 23 bytes

æw|PòlÄxòjVLkîyllv$r $Ð

Try it online!

Hexdump:

00000000: e677 7c50 f26c c478 f26a 564c 6bee 796c  .w|P.l.x.jVLk.yl
00000010: 6c76 2472 2024 d0                        lv$r $.
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+3
−0

Japt -R, 16 15 14 bytes

pÔ¬Ëú°EDùEÃÆpÔ

Try it

pÔ¬Ëú°EDùEÃÆpÔ     :Implicit input of string
p                  :Append
 Ô                 :  Reverse
  ¬                :Split
   Ë               :Map each D at 0-based index E
    ú              :  Right pad
     °E            :    to length E, prefix incremented
       DùE         :    with D, left padded to length E
          Ã        :End map
           Æ       :Modify the last element, replacing it with
            p      :  Append to input
             Ô     :    Reverse
                   :Implicit output joined with newlines
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General comments (3 comments)
+3
−0

Haskell, 110 bytes

f a=s>>=(#)$a++reverse a
c@(a:b)!f=f c:b!f
_!_=[]
f#n=n!((n!).f)
s(p:q)(d:e)(b:c)|e==c||e==[]=b|q==c=d|0<1=' '

Try it online!

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+3
−0

Scala, 99 bytes

s=>{val x=s+s.reverse
s(0)+"\n"+x.tail.init.zipWithIndex.map{case(c,i)=>c+" "*i+c+"\n"}.mkString+x}

Try it in Scastie!

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+2
−0

JavaScript, 76 bytes

Outputs an array of lines.

s=>[...s+=[...s].reverse().join``].map((c,x)=>s[-~x]?c.padEnd(x)+(x?c:``):s)

Try it online!

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+2
−0

Vyxal J, 12 bytes

ḣmėƛ÷꘍ṘǏ;⁋?m

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ḣ            # Head extract (x[0], x[1:] )
 m           # Mirror the ToS (x[1:])
  ė          # Zip with 0...length
   ƛ    ;    # Map...
    ÷꘍       # Pad by spaces to correct amount
      ṘǏ     # Append the first character to the end
         ⁋   # Join by newlines
          ?m # Input mirrored
             # (J flag) Output stack joined by newlines
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+2
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JavaScript, 102 bytes

s=>[...x=s.split``].concat(x.reverse()).reduce((l,c,i,a)=>l+`
`+(a[i+1]?c+' '.repeat(i-1)+c:a.join``))

The first part creates the 'mirrored' string in an array. reduce starts with the first element (which conveniently handles the top vertex of the triangle) and calls the provided function for each later element, with the arguments being the previous value, the element, its index, and the array. a[i+1] is a way of checking for the last element: it is undefined (which is falsy) for the last element, and a character (truthy) for the rest.

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+2
−0

Ruby, 74 bytes

Full program:

s=gets+$_.reverse
puts [s[0]]+s[1..-2].chars.zip(0..).map{_1+' '*_2+_1}<<s

Note that this is not fit for interactive use because it looks weird when you input trailing newlines. In bash, try echo -n Hello | ruby wordsetsquare.rb

75 bytes, Proc

->n{n+=n.reverse;([n[0]]+n[1..-2].chars.zip(0..).map{_1+' '*_2+_1}<<n)*'
'}
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+1
−0

Ruby, 70 bytes

->s{s+=s.reverse;s[..-2].gsub(/./){t=$&+' '*$.+$/;t[-2]=$&;$.+=1;t}+s}

Try this online!

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+1
−0

Python 3, 196 195 145 102 91 89 87 79 bytes

x=input();x+=x[::-1];i=-1
for c in x[:-1]:print(c*(i>-1)+" "*i+c);i+=1
print(x)

Try it online!

Golfed 50 bytes thanks to @celtschk's advice. Golfed another 43 bytes thanks to @celtschk's advice. Golfed another 11 bytes thanks to @celtschk's advice. Golfed another 2 bytes thanks to @celtschk's advice. Golfed 2 bytes thanks to @Moshi's advice. Golfed another 8 bytes thanks to @Moshi's advice.

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4 comment threads

Sub 80 (1 comment)
bool(i) -> (i>0) (1 comment)
Further golfing opportunity (5 comments)
You can save quite a few bytes here. First, you can replace `print(" ",end='')` with `print(end=" ... (1 comment)

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