Word Set Square
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.
[APL (Dyalog Unicode)], 23 byt …
5mo ago
Scala, 99 bytes ```scala s=> …
5mo ago
[V (vim)], 23 bytes æw| …
5mo ago
[Haskell], 110 bytes ```has …
5mo ago
Japt `-R`, 16 15 bytes …
~1mo ago
5 answers
APL (Dyalog Unicode), 23 bytes
Anonymous tacit prefix function. Reuires 0-based indexing (⎕IO←0)
(⊢⍪⍨¯1↓⊢,∘↑-∘⍳∘≢↑¨⊢)⊢,⌽
⌽ 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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V (vim), 23 bytes
æw|PòlÄxòjVLkîyllv$r $Ð
Hexdump:
00000000: e677 7c50 f26c c478 f26a 564c 6bee 796c .w|P.l.x.jVLk.yl
00000010: 6c76 2472 2024 d0 lv$r $.
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Scala, 99 bytes
s=>{val x=s+s.reverse
s(0)+"\n"+x.tail.init.zipWithIndex.map{case(c,i)=>c+" "*i+c+"\n"}.mkString+x}
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Haskell, 110 bytes
f a=s>>=(#)$a++reverse a
[email protected](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=' '
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Japt -R, 16 15 bytes
ê1
¬ËúEÄ hJDÃÆU
ê1\n¬ËúEÄ hJDÃÆU :Implicit input of string U
ê1 :Mirror
\n :Reassign to variable U
¬ :Split
Ë :Map each D at 0-based index E
ú : Left pad with spaces to length
EÄ : E+1
h : Replace the character at index ...
J : -1 (i.e., the last) with ...
D : D
à :End map
ÆU :Modify the last element and return U
:Implicit output joined with newlines
0 comment threads