34
\$\begingroup\$

Given any string, print it in the form of a triangle where the text runs up and down along each diagonal. For example, an input of "Hello World" should output:

                    d
                  l  
                r   d
              o   l  
            W   r   d
              o   l  
        o   W   r   d
      l       o   l  
    l   o   W   r   d
  e   l       o   l  
H   l   o   W   r   d
  e   l       o   l  
    l   o   W   r   d
      l       o   l  
        o   W   r   d
              o   l  
            W   r   d
              o   l  
                r   d
                  l  
                    d

The space between each character in a row must be at least 1, in order to maintain the proper format.

\$\endgroup\$
  • 1
    \$\begingroup\$ May we assume the string isn't empty? \$\endgroup\$ – Mr. Xcoder Aug 14 '17 at 13:09
  • \$\begingroup\$ @Mr.Xcoder Yes you may \$\endgroup\$ – machiavelli Aug 14 '17 at 13:15
  • 1
    \$\begingroup\$ The space between each character in a row must be at least 1: Does this mean that H l o W r d is a valid center row? Asking because in your example each row has 3 spaces between each character. \$\endgroup\$ – Emigna Aug 14 '17 at 13:33
  • 1
    \$\begingroup\$ @Emigna I misunderstood you question, my apologies. Yes, your example would be valid. \$\endgroup\$ – machiavelli Aug 14 '17 at 13:37
  • 1
    \$\begingroup\$ Leading or trailing spaces allowed? \$\endgroup\$ – Luis Mendo Aug 14 '17 at 13:45

22 Answers 22

19
\$\begingroup\$

Charcoal, 10 7 bytes

↗ELθ✂θιUE¹

Try it online! Try it online! Links are to verbose version of code. Explanation:

    ↗       Print up and to the right
     ELθ✂θι All suffixes of the input, as a list down and to the right
    UE¹     Insert blank columns

First time I got to use the UE command.

\$\endgroup\$
  • \$\begingroup\$ I knew Charcoal would be one of the first answers here.. Was almost tempting to start myself, but I'm way to inexperienced with it to finish a solution in time, and it would get out-golfed anyway.. ;) \$\endgroup\$ – Kevin Cruijssen Aug 14 '17 at 13:11
  • 6
    \$\begingroup\$ @Emigna ...but this was my big chance to use UE... \$\endgroup\$ – Neil Aug 14 '17 at 14:25
  • 6
    \$\begingroup\$ @EriktheOutgolfer ...but this was my big chance to use UE.. \$\endgroup\$ – Neil Aug 14 '17 at 14:50
  • 1
    \$\begingroup\$ @Neil It's -3 bytes! -3 for a nice sacrifice! Who doesn't want some nice -3? \$\endgroup\$ – Erik the Outgolfer Aug 14 '17 at 14:53
  • 4
    \$\begingroup\$ @EriktheOutgolfer What you're supposed to say is, "You can't let 05AB1E beat you, can you?" \$\endgroup\$ – Neil Aug 14 '17 at 15:21
12
\$\begingroup\$

05AB1E, 10 8 7 bytes

Thanks to Emigna for saving 2 bytes!

ðâƶ.cðζ

Uses the 05AB1E encoding. Try it online!

\$\endgroup\$
  • \$\begingroup\$ As per this line The space between each character in a row must be at least 1 you can remove ¶«. (also verified the validity with OP ) \$\endgroup\$ – Emigna Aug 14 '17 at 13:39
  • \$\begingroup\$ @Emigna Thanks! :) \$\endgroup\$ – Adnan Aug 14 '17 at 13:46
  • \$\begingroup\$ Might want to update the Tio link :) \$\endgroup\$ – Mr. Xcoder Aug 14 '17 at 13:47
  • \$\begingroup\$ @Mr.Xcoder ninja'd \$\endgroup\$ – Adnan Aug 14 '17 at 13:47
  • 1
    \$\begingroup\$ Clever use of â in ðâ instead of Sð«! \$\endgroup\$ – Erik the Outgolfer Aug 14 '17 at 14:45
9
\$\begingroup\$

Python 2, 75 bytes

s=input()
k=l=len(s)
while k>1-l:k-=1;m=abs(k);print' '*m+' '.join(s[m::2])

Try it online!

Ruud saved 3 bytes.

\$\endgroup\$
  • 3
    \$\begingroup\$ This has been allowed by the OP: 78 bytes. The space on each row must be at least 1. \$\endgroup\$ – Mr. Xcoder Aug 14 '17 at 13:49
  • 1
    \$\begingroup\$ 75 bytes \$\endgroup\$ – Arfie Aug 16 '17 at 12:15
8
\$\begingroup\$

C, 86 78 73 70 chars

for(int i=1,j=1-n;i=putchar(j*j<i*i&i-j?s[i-1]?:13:32)^13?i+1:++j<n;);

Try it online!

Explanation

Naive implementation: two cycles, fill from top to bottom, left to right (99 bytes):

for(int j=1;j<n*2;j++){for(int i=0;i<n;i++)printf("%c ",(i+j)%2&&i+1>=abs(j-n)?s[i]:' ');puts("");}

Here, puts() just prints \n to the output. Let’s combine variable declarations and combine j++ with something (94 bytes):

for(int i,j=0;++j<n*2;){for(i=0;i<n;i++)printf("%c ",(i+j)%2&&i>=abs(j-n)?s[i]:' ');puts("");}

Good. Variable j has a range 0...2n; let it be within -n...n, this makes the math simpler. Notice that boolean expression at the right of && always has the value 0 or 1. This means we can replace && with &. 91 byte:

for(int i,j=-n;++j<n;){for(i=0;i<n;i++)printf("%c ",~(i+j)%2&i>=abs(j)?s[i]:' ');puts("");}

And now we realized we printing an extra space. And yeah, we don’t need printf() to print just a single symbol. 86 bytes:

for(int i,j=-n;++j<n;){for(i=0;i<n;i++)putchar(~(i+j)%2&i>=abs(j)?s[i]:' ');puts("");}

Even better. Notice that condition i * i>=j * j is same as i>=abs(j), but shorter. Let’s move puts() into for loop increment expression. And guess what? Actually, we don’t need the braces around i+j. 78 bytes:

for(int i,j=-n;++j<n;puts(""))for(i=0;i<n;i++)putchar(i*i>=j*j&~i+j?s[i]:' '); 

Did you know that putchar() returns the character it has printed? Let’s use XOR to test numbers for equivalence. Let’s replace space with its ASCII code, 32. Remember that end-of-line character code is 13. And finally: did you know that GCC/Clang do support https://en.wikipedia.org/wiki/Elvis_operator ? 73 bytes:

for(int i,j=-n;++j<n;)for(i=0;putchar(i*i>=j*j&~i+j?s[i]?:13:32)^13;i++);

Finally, guess what? We don’t need two for loops. We can replace ugly ~i+j with just i-j. 70 bytes:

for(int i=1,j=1-n;i=putchar(j*j<i*i&i-j?s[i-1]?:13:32)^13?i+1:++j<n;);

Future work: change loop direction? This might save some bytes, if done properly.

\$\endgroup\$
5
\$\begingroup\$

SOGL V0.12, 13 10 9 bytes

ēI*@∑}¹╚H

This uses a feature that I just added, but was documented a while ago.

Try it here!
In that link , is added because this expects the input on the stack and { added because otherwise , would get executed every time in the loop

implicitly start loop over POP
ē            increase the variable E, on first push which will be 0
 I           increase by 1
  *          multiply the current character that many times
   @∑        join with spaces
     }     end loop
      ¹    wrap everything on stack in an array
       ╚   center vertically
        H  rotate counter-clockwise
\$\endgroup\$
4
\$\begingroup\$

Haskell, 73 bytes

f s|l<-length s-1=[zipWith min s$(' '<$[1..abs x])++cycle"~ "|x<-[-l..l]]

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Gaia, 16 bytes

$:ċ⟪×$§×⟫†€|$¦tụ

Try it online!

Explanation

$                 Split into list of chars
 :ċ               Push [1 .. len(input)]
   ⟪×$§×⟫†        Apply this block to corresponding elements of the two lists:
    ×              Repetition
     $             Split into chars
      §×           Join with spaces
          €|      Centre align the rows
            $¦    Split each line into chars
              t   Transpose
               ụ  Join each row with spaces, then join the rows together with newlines
\$\endgroup\$
3
\$\begingroup\$

Python 2, 86 83 bytes

-3 thanks to officialaimm

lambda s,j=' '.join:map(j,zip(*(j(c*-~i).center(len(s)*2)for i,c in enumerate(s))))

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Jelly, 15 bytes

;€⁶ṁ"JUz⁶ZUŒBZG

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Octave, 59 63 58 57 bytes

@(s,a=@strjust)a([kron(+a(hankel(s)),[1 0;0 0]) '']',99)'

Try it online!

\$\endgroup\$
  • 1
    \$\begingroup\$ 99: good idea :-) \$\endgroup\$ – Luis Mendo Aug 14 '17 at 16:01
3
\$\begingroup\$

Java, 292 bytes (sorry)

public class D{
public static void main(String[]r){
String s=r[0];int L=s.length(),n=L*2-1,x=L-1,d=-1,i,j;boolean a=false,o=L%2==1;
for(i=0;i<n;i++){
for(j=0;j<L;j++)System.out.print(j<x||a&&j%2==(o?0:1)||!a&&j%2==(o?1:0)?' ':s.charAt(j));
System.out.println();
x+=d;if(x<0){x=0;d=1;}a=!a;}}}
\$\endgroup\$
  • 1
    \$\begingroup\$ You can remove the newlines, otherwise, this looks pretty golfed! \$\endgroup\$ – Zacharý Aug 14 '17 at 22:31
  • 1
    \$\begingroup\$ You can golf more: 1. boolean a=1<0,o=L%2>0;. 2. If you don't need i, use this loop: for(i=0;i++<n;). 3. You can get rid of o: j%2<L%2 then j%2>L%2. 4. Using d as flip takes so many characters: just do j<(x<0?-x:x). 5. You have way more variables than needed. 6. You don't need a full program: a lambda or method is enough. -- If you want a golfed Java example, check my answer. \$\endgroup\$ – Olivier Grégoire Aug 14 '17 at 22:51
3
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Haskell, 81 bytes

f s|l<-length s=[[last$' ':[s!!i|i>=n,mod(n+i)2<1]|i<-[0..l-1]]|n<-abs<$>[-l..l]]

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Java (OpenJDK 8), 116 bytes

s->{for(int l=s.length(),i=-l;++i<l;)System.out.printf("%"+l+"s%n",s.substring(i<0?-i:i).replaceAll("(.).","$1 "));}

Try it online!

Explanation

s->{                                // Consumer<String> lambda
 for(int l=s.length(),i=-l;++i<l;)  // For each length between l and 1 and back to l, 
  System.out.printf("%"+l+"s%n",    // Print with align to right
    s.substring(i<0?-i:i)           // skip the first |i| characters
     .replaceAll("(.).","$1 ")      // replace every even-positioned character with a space.
   );
}
\$\endgroup\$
3
\$\begingroup\$

C++, 135 bytes

Okay, here's my shot with C++:

auto f=[&](auto f,int y)->void{
  for(int i{};i<n;i++) putchar(y<0?f(f,y+1?i+1:n-1-i),'\n':i<y||i+y&1?' ':s[i]);
}; f(f,-1); f(f,-2);

Try It Online (ideone)!

\$\endgroup\$
3
\$\begingroup\$

Haskell, 140 137 bytes

(m#n)s=(\(i,x)->' ':(last$"  ":[x:" "|rem i 2==m&&i>n]))=<<zip[0..]s
g s=((++)=<<reverse.tail)$id=<<[[(0#n)s,(1#n)s]|n<-[-1,1..length s]]

Try it online!

Saved 3 bytes thanls to Challenger5

I don't think that's optimal...

f produces one of the lines (m = 0 or 1 is the modulo of the line number, n is the number of the line)

g intercalates "odd" and "even" lines, and add to the result a mirror of itself.

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  • \$\begingroup\$ You can save bytes by defining f as an infix function (as in (m#n)s=...) rather than a prefix function. \$\endgroup\$ – Esolanging Fruit Aug 16 '17 at 7:01
2
\$\begingroup\$

Pyth, 25 bytes

j+_Km++*d;@Qdtj;%2>QdUQtK

Try it here.

\$\endgroup\$
2
\$\begingroup\$

Jelly, 18 bytes

JrLm€2Ṭ;€0a⁸o⁶ṚŒḄG

Try it online!

\$\endgroup\$
2
\$\begingroup\$

Mathematica 105 Bytes

(c=Characters@#;l=Length@c;StringRiffle@Table[If[Abs[j-l]<i&&EvenQ[j+i],c[[i]]," "],{j,1,2l+1},{i,1,l}])&

Maybe I could shave off another byte or two, but the character count overhead of dealing with strings in Mathematica makes simple challenges like this uncompetitive.

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2
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J, 54 bytes

[:|:|.@i.@#(>@],~' '#~[)"_1[:(,' '&,)/&.>>:@i.@#<@#"0]

Try it online! (note that the output on TIO has a newline and three spaces, but that isn't from the function call -- it's probably just what the J interpreter does automatically).

I think the general idea for solving this is right, but there are small things that I'm probably doing sub-optimally that are adding to the bytecount.

Previous variants

55 bytes

<:@+:@#{.[:|:|.@i.@#(],~' '#~[)"_1>:@i.@#(,' '&,)/@#"0]

56 bytes

<:@+:@#{.[:|:|.@i.@#(],~' '#~[)"_1#{.[:(,' '&,)//.[:]\.]

Explanation

This will be split into a few functions. Also, I wasn't as thorough with latter parts of the explanation, so let me know if you want a better explanation for a certain part and I can edit that in.

dup   =. >:@i.@# <@#"0 ]
space =. (,' '&,)/&.>
pad   =. |.@i.@# (>@],~' '#~[)"_1 ]
trans =. |:
  • dup duplicates each character as many times as its index (plus one) in the string
  • space inserts spaces between each character
  • pad pads the characters with the right amount of spaces
  • trans transposes the resulting matrix

Sample call:

   trans pad space dup 'abc'
  c
 b 
a c
 b 
  c

Dup

>:@i.@# <@#"0 ]
>:@i.@#         Indices of each character plus one
      #          Length of the string
   i.            Range [0,length)
>:               Add one
        <@#"0 ] Duplicate each character as many times as it index (plus one)
           "0   For each
          #   ]  Copy the character
>:@i.@#           as many times as it index
        <        Box the result

The results are boxed to prevent J from padding the ends with spaces (since they're of uneven length).

Sample call:

   dup 'abc'
┌─┬──┬───┐
│a│bb│ccc│
└─┴──┴───┘

Space

(,' '&,)/&.>
         &.>  For each boxed element
(,' '&,)/      Insert spaces between each

Sample call:

   space dup 'abc'
┌─┬───┬─────┐
│a│b b│c c c│
└─┴───┴─────┘

Pad

|.@i.@# (>@],~' '#~[)"_1 ]
        (>@],~' '#~[)      Pad the right arg with spaces given by the left arg
|.@i.@#                    Indices in reverse order
   i. #                     Range [0,length)
|.                          Reverse

Basically, pad the first element with length - 1 spaces, the second with length - 2, etc. It also removes the boxing.

Sample call:

   pad space dup 'abc'
  a  
 b b 
c c c

Transpose

This is just the built-in function |: which takes the transpose of a matrix.

\$\endgroup\$
  • 1
    \$\begingroup\$ I used a similar approach, but avoided boxing. 45 bytes: |:@(-@i.@-@#|."0 1((,@,.~' '#~#)@$"0~1+i.@#)). it could surely be golfed further. this part -@i.@-@# is some low hanging fruit, most likely \$\endgroup\$ – Jonah Aug 15 '17 at 2:34
  • \$\begingroup\$ @Jonah I can't quickly decipher how your answer works, so I'll leave it to you to post it if you would like to, as I like to include an explanation of my answer. I guess J for me right now is a write-only language. \$\endgroup\$ – cole Aug 15 '17 at 3:01
  • \$\begingroup\$ helpful for quickly deciphering: f=. <some tacit expression>, then 5!:2 <'f'gives a boxed visualization and 5!:4 <'f' gives a tree visualization. in my case, try running $"0~1+i.@# with some string first, then run everything to the right of |."0 1, then understand that |."0 1 and everything to the left, save the final transpose, is just doing the necessary rotations. \$\endgroup\$ – Jonah Aug 15 '17 at 3:13
  • 1
    \$\begingroup\$ oh i wasn't expecting you to update your answer. it was more of a "hey, you might find this interesting." i could have posted it, but felt the high-level approaches were similar enough it wasn't worth it. \$\endgroup\$ – Jonah Aug 15 '17 at 3:21
  • 2
    \$\begingroup\$ Just remembered the complex arguments of # helps here, 26 bytes with |:@((-#)|."_1(1j1##)"0)~#\ \$\endgroup\$ – miles Aug 15 '17 at 10:59
1
\$\begingroup\$

JavaScript (ECMAScript 6), 161 bytes

(s,n=console.log)=>s.split("").map((q,i,a,p)=>n(p=" ".repeat(q=a.length-++i)+a.map((v,j)=>j>=q&&j%2==q%2?a[j]+' ':'').join(''))||p).reverse().map((v,i)=>i&&n(v))

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Perl 5, 86 + 2 (-F) = 88 bytes

Used @Dom's suggestions and a few of my own tweaks to lower the byte count.

for$k(0..$#F){$i=1;$a[$#F+$k]=$a[$#F-$k]=[map$i++<$k|($i+$k)%2?$":$_,@F]}say"@$_"for@a

Try it online!

\$\endgroup\$
  • \$\begingroup\$ Sorry about that, glad you got your answer out there though! Had a little bit of a play when I was trying to fix mine and coudnl't get a solution down, yours was a better approach! It is possible to drop a couple of bytes using -aF to put all the letters in @F and a couple of little tweaks (-F counts as 3 since it needs a space after): Try it online! \$\endgroup\$ – Dom Hastings Aug 18 '17 at 4:46
  • 1
    \$\begingroup\$ Why does -F count as 3? Shouldn't it be 2 at the most? Isn't it the difference between perl -e'code...' and perl -eF 'code...'. Also -a is unnecessary when using -F, so that byte can be cut. \$\endgroup\$ – Xcali Aug 18 '17 at 5:06
  • \$\begingroup\$ That's it exactly. So -F accepts an argument, but we don't wanna pass one in (-F allows us to control what -a splits on, no argument, split each char on its own) so it's the difference between perl -ae '...' and perl -aF -e '...'. By default -a splits on /\s+/. Hope that helps clarify! \$\endgroup\$ – Dom Hastings Aug 18 '17 at 5:42
  • \$\begingroup\$ Also, nice use of $#F! Always forget about that! \$\endgroup\$ – Dom Hastings Aug 18 '17 at 5:43
0
\$\begingroup\$

q/kdb+, 55 bytes

Solution:

-1(+){{1_a,((2*y)#" ",z),a:x#" "}'[(|)c;1+c:(!)(#)x]x};

Example:

q)-1(+){{1_a,((2*y)#" ",z),a:x#" "}'[(|)c;1+c:(!)(#)x]x}"Hello World";
          d
         l
        r d
       o l
      W r d
       o l
    o W r d
   l   o l
  l o W r d
 e l   o l
H l o W r d
 e l   o l
  l o W r d
   l   o l
    o W r d
       o l
      W r d
       o l
        r d
         l
          d

Explanation:

TODO. ungolfed version is 66 bytes:

-1 flip{{1_a,((2*y)#" ",z),a:x#" "}'[reverse c;1+c:til count x]x};

Bonus:

To get the same output as the example (74 bytes):

q)-1(+){1_'raze{(a,((2*y)#" ",z),a:x#" ";(2*y+x)#" ")}'[(|)c;1+c:(!)(#)x]x}"Hello World";
                    d
                  l
                r   d
              o   l
            W   r   d
              o   l
        o   W   r   d
      l       o   l
    l   o   W   r   d
  e   l       o   l
H   l   o   W   r   d
  e   l       o   l
    l   o   W   r   d
      l       o   l
        o   W   r   d
              o   l
            W   r   d
              o   l
                r   d
                  l
                    d
\$\endgroup\$

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