Helloellolloloo Worldorldrldldd

Question

Make a program that takes the word you input, and adds that word to the back of itself minus its first letter, then repeats until all letters are gone. For example, cat would become catatt, and hello would become helloellolloloo.

Input
Any of the 26 letters of the English alphabet. There may be multiple words separated by spaces, and the change should be applied to every word.

Output
The word(s) inputted, with each word put after itself with its first letter missing, and then with its second letter missing, and so on until there are no more letters to add.

More examples:

ill eel outputs illlll eelell

laser bat outputs laserasersererr batatt

darth vader outputs dartharthrththh vaderaderdererr

This is code golf, so the shortest code wins.

Clarification:
You can treat the input or output as a list. You can separate words using newline instead of space. You can add a trailing space to the input.

Answer 1

Japt -m, 6 3 bytes

Input and output are arrays of words.

£sY

Try it

---

Explanation

        :For each word in the input array
£       :Map each letter at index Y
 sY     :  Slice the current word from index Y

Answer 2

brainfuck, 60 56 bytes

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

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Requires a trailing space and prints a leading space. Both of these could be circumvented, but that ends up at 112 bytes.

Explanation

,[  Loop over each byte of input
  Tape: 32 w o r-32 d'
  >++++++++[-<----<++++>>]   Subtract 32 from the character and add 32 to the previous char
  Tape: 32 w o r d-32 0'
  <[>>]<   If the last character was a space
  Tape: 32 w o r d-32 0'
  or
  Tape: 32 w o r d' space-32
  [
    [<]>.   Move to the end of the word and print out the space
    [   Loop over each letter
      [-]    Remove the first letter (initially space)
      >[.>]  Print the rest of the word
      <[<]>  Move back to the first letter
    ]
    Tape: clear
  ]
,]  Get the next byte of input

Answer 3

Haskell, 36 21 bytes

map$concat.scanr(:)""

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Edit: -15 bytes, because of new IO format (list of words instead of space separated words)

Answer 4

Perl -p, 36 25 23 bytes

s!\b|\S!$'=~s/ .*//r!eg

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This is a single regsub. First, it matches all word boundaries or non-space characters:

[][H][e][l][l][o] [][W][o][r][l][d]

Note that each of these matches should be replaced with the rest of the word:

[→Hello][H→ello][e→llo][l→lo][l→o][o→] (...)

We can accomplish this with the special variable $', which stores the part of the string after the match. However, we need to apply the nested regsub s/ .*// to it, which removes everything past the first space in $', in order to get rid of the remaining words in the input.

Thanks to @nwellnhof for 2 bytes.

Answer 5

Jelly, 3 bytes

ḊƬ€

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Don’t need the Ks anymore since array input/output is now allowed.

ḊƬ€
  €   For each word:
Ḋ       Remove the first letter
 Ƭ      until there are none left.

Answer 6

Python 3, 49 bytes

d=lambda s:' '.join(n+d(n[1:])for n in s.split())

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This takes advantage of the fact that "".split() returns an empty array so that acts as the check for the base case in the recursion.

Answer 7

JavaScript (ES6), 33 bytes

Saved 1 byte thanks to @ShieruAsakoto

I/O format: array of words.

a=>a.map(g=w=>w&&w+g(w.slice(1)))

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---

JavaScript (ES6), 35 bytes

I/O format: array of words.

a=>a.map(w=>w.replace(/./g,"$&$'"))

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Answer 8

APL(Dyalog), 19 9 bytes

{⌽∊,\⌽⍵}¨

thanks to @H.PWiz for jogging my brain

This works because all strings in APL are character arrays.

{⌽∊,\⌽⍵}¨ 
        ¨ - for each string
      ⍵} - string argument - ex. "hello"
     ⌽ - reverse - "olleh"
   ,\ - scan magic - "o" "ol" "oll" "olle" "olleh"
  ∊ - enlist(join together) "oolollolleolleh"
{⌽ - reverse - "helloellolloloo"

TIO

Answer 9

R, 82 75 67 bytes

write(sapply(x<-scan(,""),substring,1:(y=max(nchar(x))),y),1,y,,"")

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Several bytes saved thanks to JayCe

Separates output with newlines.

The sapply(...) expression generates a matrix/column vector of the appropriate substrings, padding with "" as needed. write then prints the elements of the matrix, y per line, separating them with "".

Answer 10

brainfuck, 94 93 bytes

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

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• Saved one byte thanks to Nitrodon -- golfing .[-]>[.>]<[<]> to [.>]<[<]>[-]>.

Explanation

[[[ (dynamic) tape layout: ... NUL STR ... STR CHR FLG BUF SPC NUL ... ]]]

load a 32 into SPC
-[-[-<]>>+<]>-

while FLG
<<+[

 read a word
 [
  clear FLG; read CHR
  [-]<,
  copy CHR to BUF (using FLG as a temporary)
  [->+>+<<]>[-<+>]
  subtract SPC from BUF and save SPC
  >>[-<->>+<]
  move tape layout one to the right
  <
 ]

 strip trailing space; set FLG to true
 << [>>+<<[-]]
 to STR's first character
 <[<]>
 print word in all reduced forms
 [
  [.>]<[<]>[-]>
 ]

 print SPC; move to FLG
 >>>>.<<
]

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Attribution

Esolang's brainfuck constant collection was used for the initial space load.

Answer 11

Husk, 6 4 bytes

^{-2 bytes thanks to Jonathan Allan (taking input as a list)!}

moΣṫ

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Explanation

Takes input as a list of strings and maps the following function:

Σṫ  -- example argument: "abc"
 ṫ  -- tails: ["abc","bc","c"]
Σ   -- concat: "abcbcc"

Answer 12

05AB1E, 5 bytes

€.síJ

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Explanation

€.s        # push suffixes of each
   í       # reverse each
    J      # join suffixes

Answer 13

Vim, 47 bytes (38 key strokes)

Start with your input as the sole line in a Vim buffer.

:s/<Space>/\r/g<CR>ggqaywPlxqqb99@aj0q99@bgg99J

Explanation

This puts each word on its own line, iterates over each line, then rejoins them all. Breaks if words are longer than 99 characters or if your input has more than 99 words.

1. :s/<Space>/\r/g<CR> replaces spaces with new lines (\r)
2. gg positions the cursor at the beginning of the first line
3. qa begins recording macro a:
   • yw yanks the rest of the word
   • P puts it behind the cursor
   • lx removes the first letter of the latter word
   • q stops recording macro a
4. qb begins recording macro b:
   • 99@a executes macro a ninety-nine times (introduces the character limit)
   • j0 positions the cursor at the start of the next line
   • q stops recording macro b
5. 99@b executes macro b ninety-nine times (introduces the word limit)
6. gg positions the cursor at the first line
7. 99J joins the following ninety-nine lines with spaces (word limit again)

For another 2 bytes (2 key strokes) you could extend the word limit to 999. Another 4 bytes, 9999, etc.

Answer 14

Retina 0.8.2, 15 bytes

 
¶
.
$&$%'
¶
 

Try it online! Note: trailing spaces. Explanation:

 
¶

Split on spaces.

.
$&$%'

Append its suffix to each letter. The % means that we only get the word's suffix.

¶
 

Join with spaces.

Answer 15

Pepe, 167 153 bytes

REEerEeeEeeeeeRrEEEEerEEEEEeerEErEEeerreErEEeErreEREEEEEEEreereErEerEEEErEEeerrEEreRRErEEEEreREEreeereReeRerEEEEEErEEEeerreEerEEeerEEEEerEEeEreereErEeree

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Answer 16

16-bit x86 assembly code, 24 bytes

     47             inc    di
     B020           mov    al,20h
l1:  3806           cmp    [si],al
     7212           jb     l5 ;less means end of string
     7401           je     l2  ;equal means space was seen
     4F             dec    di ;overwrite extra space
l2:  E80300         call   l3
     46             inc    si ;move to next character in word
     75F1           jne    l1
l3:  56             push   si
l4:  3806           cmp    [si],al
     A4             movsb      ;copy character
     77FB           ja     l4  ;until either zero or space is seen
     5E             pop    si
l5:  C3             ret

Call with si = pointer to source string, di = pointer to output buffer.
The source string requires a zero byte to end it.
The code is the same in 16- or 32- or 64-bit (si/di become either esi/edi or rsi/rdi).
32-bit code is two bytes larger because of the expanded call.
64-bit code is three bytes larger still because the inc/dec of rsi/rdi attracts a prefix (but if it is known that they are within 32-bit memory space, then they can be esi/edi again to avoid that penalty).

Answer 17

MATL, 18 16 bytes

"@gXH"HX@Jh)]0&h

Input is a cell array of words. Try it online!

Explanation

"         % Implicit input: cell array of strings. For each cell
  @g      %   Push content of current cell, that is, a word
  XH      %   Copy into clipboard H
  "       %   For each letter
    H     %     Push word
    X@    %     Push iteration index
    Jh)   %     Index from that until the end into current word
  ]       %   End
  0       %   Push 0. Will be cast to char. Char(0) is displayed as space
  &h      %   Concatenate horizontally all elements so far. Implicit display

Answer 18

K4 / K (oK), 9 bytes

Solution:

,/'(1_)\'

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Explanation:

,/'(1_)\' / the solution
        ' / apply to each
       \  / scan
   (  )   / do this together
    1_    / drop first
,/'       / flatten (,/) each (')

Answer 19

C++ (clang), 174 bytes

#include<map>
#include<string.h>
std::string r(std::string w){while(auto x=strchr(w.c_str(),32))return r(w.substr(0,x-w.c_str()))+" "+r(x+1);return w!=""?w+r(w.substr(1)):w;}

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It's my first submission, and i didn't know if returning string instead of printing it is okay :)

Answer 20

Stax, 3 bytes

m|]

Run and debug it

Explanation:

m   Map over the lines
 |] Get all suffixes (suffices?)
    Implicit flatten and output

Answer 21

Charcoal, 14 bytes

⪫Ｅ⪪Ｓ ⭆ι✂ιμＬι¹ 

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

   Ｓ            Input string
  ⪪             Split on spaces
 Ｅ              Map over each word
      ι         Current word
     ⭆          Map over each character and join
        ι       Current word
         μ      Current index
           ι    Current word
          Ｌ     Length
            ¹   Literal 1
       ✂        Slice
⪫               Join with spaces
                Implicitly print

Answer 22

C (gcc), 79 bytes

f(s,t)char*s,*t;{for(;*s;printf("%.*s",*s^32?t?t-s:~0:1,s),s++)t=strchr(s,32);}

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Answer 23

Pip -s, 11 bytes

J_@>,#_Mq^s

Takes the space-separated list of words from stdin. Try it online!

Explanation

             s is space (implicit)
        q    Read a line of stdin
         ^s  Split it on spaces
       M     Map this lambda function to each word:
 _            The word...
  @>          sliced starting at index...
    ,#_       range(len(word))
              This creates len(word) slices ["word" "ord" "rd" "d"]
J             Join those into a single string
             The resulting list of modified words is printed; the -s flag uses space
             as the separator

Answer 24

Ruby, 42 bytes

->s{s.map{|a|(w=a.b).chars{a[0]='';w<<a}}}

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Answer 25

Python 2, 63 bytes

lambda s:' '.join(map(g,s.split()))
g=lambda s:s and s+g(s[1:])

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Answer 26

Canvas, 6 bytes

±［±］⇵］

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5 bytes with a crazy output format

Answer 27

C#, 111 90 bytes

b=>string.Join(" ",(b.Split(' ').Select(x=>string.Concat(x.Select((y, i)=>x.Substring(i))))))

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By Changing input and output to arrays, I saved a few bytes:

b=>b.Select(x=>string.Concat(x.Select((y,i)=>x.Substring(i)))).ToArray()

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Answer 28

K (oK), 17 13 bytes

{,/|:'|,\|x}'

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Prefix anonymous function; Input is taken as a list of strings, which in turn are lists of characters.

Thanks @streetster for 4 bytes.

How:

{,/|:'|,\|x}' //Main function, argument x → ("ill";"eel")
            ' // For each element of the argument
         |x}  // Flip it. x → ("lli";"lee")
       ,\     // Concatenate each element, keeping intermediates. x → (("l";"ll";"lli");("l";"le";"lee")
      |       // Flip it again. x → (("lli";"ll";"l");("lee";"le";"l"))
   |:'        // Now flip each element. x → (("ill";"ll";"l");("eel";"el";"l"))
{,/           // Concatenation scan. x → ("illlll";"eelell")

Answer 29

Common Lisp, 179 bytes

(defun r(s)(cond((endp s)nil)((eql(first s)#\Space)(princ " ")(r(rest s)))(t(q s)(r(rest s)))))(defun q (l)(cond((eql(first l)#\Space)t)((endp l)t)(t(princ(first l))(q(rest l)))))

Try it online!

This is my first try at golfing any edits are welcome

Answer 30

Lua, 70 bytes

for i=1,#arg do x=arg[i]for i=1,#x do io.write(x:sub(i))end print()end

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Explanation

The arguments in Lua are stored in the table arg starting at index 1. The unary operator # returns the size of the table and function s:sub(a,b) returns a substring based on string s delimited by integers a and b, if b is not passed it will return the rest of the string.

I had to use io.write() instead of print() to avoid line breaking, and added print() at the end for the opposite reason.

Answer 31

Python 3, 79 74 bytes

-5 bytes thanks to mypetlion

print(*map(lambda x:''.join(x[n:]for n in range(len(x))),input().split()))

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A full program that takes input from stdin and outputs to stdout.

Answer 32

Perl 6, 44 32 bytes

-12 bytes thanks to nwellnhof!

~*.words>>.&{[~] $_,{S/.//}...0}

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An anonymous Whatever lambda that takes a string and returns a string.

Explanation:

 *.words   # Split the given word by spaces
        >>.&{                  }  # Map each word to
                           ...  # A list composed of
                 $_              # The initial string
                   ,{S/.//}      # Remove the first character
                              0  # Until it's empty
             [~]   # All joined together
~   # Convert the list to a string, which joins by spaces

Answer 33

JavaScript (Node.js), 33 bytes

s=>s.replace(/\B(?=(\S+))/g,"$1")

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I don't really know JavaScript, but I just copied Arnauld's syntax with an idea I had for the regex substitution :P

Answer 34

MathGolf, 6 bytes

ÆÅ_╞↑ 

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Outputs a trailing space, though you can add another byte to remove this.

Hopefully 6 bytes when implicit input is implemented. Yay.

Explanation:

        Implicit input
Æ       Implicit for loop over the next 5 instructions
 Å  ↑   While true without popping (empty string is false)
  _╞    Duplicate the top of stack and remove the first letter
        (space) Append a space to the stack
        Implicitly output the stack joined together

Answer 35

Java (JDK 10), 90 bytes

s->{var r="";for(var x:s){for(int i=0;i<x.length();)r+=x.substring(i++);r+=" ";}return r;}

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Credits

• -11 bytes thanks to Kevin Cruijssen, notifying me of the rule change about the input.

Answer 36

Powershell, 29 23 bytes

Port of Javascript by @Arnauld

$args-replace'.',"$&$'"

Input and output are a list of word. Test script:

$f = {

$args-replace'.',"$&$'"

}

@(
    ,('illlll eelell', 'ill','eel')
    ,('laserasersererr batatt', 'laser','bat')
    ,('dartharthrththh vaderaderdererr', 'darth','vader')
) | % {
    $e,$s = $_
    $r = &$f @s
    $r = "$r"
    "$($r-eq$e): $r"
}

Output:

True: illlll eelell
True: laserasersererr batatt
True: dartharthrththh vaderaderdererr

Answer 37

Scala, 97 bytes

Scala main that takes a single String argument containing all the words.

args(0).split(" ").foreach(x=>{for(i<-0 to x.length)print(x.substring(i,x.length));print(" ")})

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Answer 38

Python3, 86 bytes

Reduced a lot, thanks to @ElPedro and @manatwork

def f(a):return a+f(a[1:])if''<a else'' 
for x in input().split():print(f(x),end=' ')

P.S. I tried removing split but then nothing is printing, its probably required in python3

Answer 39

Python3, 58 bytes

Takes a space-separated string of words from stdin and returns stdout likewise.

f=lambda x:x and x+f(x[1:])
print(*map(f,input().split()))

It works by splitting the string into a list of words and using map() to call the recursive function f on each element. Honestly, I don't know exactly how f works. When the stack reaches the end of the word, x becomes the empty string (''), so the and statement returns False, which I suspect acts as the base case to end recursion.

Here's the same idea but without the functional aspects:

def f(x):
    return x and x + f(x[1:])

print(*( f(word) for word in input().split() ))

Answer 40

q 78 38 bytes

" "sv{r::x;{x,r::1_r}/[count x;x]}each

Answer 41

Attache, 28 bytes

Join&sp##Sum@Suffixes=>Split

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Alternatives

ReplaceF&(Sum@Suffixes)&/"\\w+"               ?? 31 bytes
ReplaceF«_,/"\\w+",Sum@Suffixes»              ?? 34 bytes, 32 chars
ReplaceF<~_,/"\\w+",Sum@Suffixes~>            ?? 34 bytes

Explanation

Join&sp##Sum@Suffixes=>Split    input, e.g.: "abc defg"
                       Split    split the input on spaces
                                e.g.: ["abc", "defg"]
                     =>         on each word:
                                  e.g.: "abc"
             Suffixes             take the suffixes of that word
                                  e.g.: ["abc", "bc", "c"]
         Sum@                     join them together
                                  e.g.: "abcbcc"
       ##                       then
Join&                           join the result by:
     sp                           spaces

Answer 42

Scala, 81 bytes

def e(s:List[String])=s.map(_.foldRight(("","")){(a,b)=>(a+b._1,a+b._1+b._2)}._2)

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As per comments, this can receive and output a List[String]. This builds the output from right to left, starting with the rightmost letter, then prepending the two rightmost letters, and so on down the line.

Answer 43

Pyth, 8 6 bytes

m_s.__

_{Decreased byte(s?) thanks to @Steven H.}

Try it online! Input and output are a list.

Expalanation:

        - implicit output
m       - map function d:
      _ -   ...d reversed...
    ._  -   ...get all prefixes of it...
  _s    -   ...joined and reversed
        - ...over implicit Q (input)

Answer 44

Pascal (FPC), 129 bytes

var s:string;i,j:int32;begin read(s);repeat j:=Pos(' ',s);for i:=1to j do write(s[i..j-1]);write(' ');Delete(s,1,j)until s=''end.

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Requires that the input ends in space, allowed in this comment.

Answer 45

Twig, 73 bytes

Creates a macro with the function f() that receives an array of words, displaying the words separated by new lines.

{%macro f(a)%}{%for v in a%}{%for i in 0..v|length%}{{v[i:]}}{%endfor%}
{%endfor%}{%endmacro%}

To use it, just import:

{% import 'macro.twig' as a %}

{{ a.f(['a','sentence','is','a','list','of','words']) }}

{# to pass a string, you can split it #}
{{ a.f('a sentence is a list of words'|split(' ')) }}

You can try it on https://twigfiddle.com/8xech1
Warning: due to the way that output is handled in the link, I was forced to add a simple period (.) at the end of the line. This is NOT needed for normal operation.

Answer 46

Brachylog, 7 bytes

{a₁ᶠc}ᵐ

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Explanation

{    }ᵐ          Map for each word:
 a₁ᶠ               Find all suffixes
    c              Concatenate into a single string

Answer 47

Haskell, 24 bytes

This solution is not optimal in two ways: 1) the other Haskell answer by nimi is 3 bytes shorter and 2) the input is limited to 9223372036854775807 characters (~36 exabytes), though I think it's an interesting find:

map.mconcat$drop<$>[0..]

Try it online! (sets a limit of 255 characters for TIO would time out otherwise)

Explanation

Since drop has type Int -> [a] -> [a] it forces [0..] to be [0..9223372036854775807] which is finite and thus the program terminates (eventually).

And mconcat takes the list of [drop 0, drop 1..] and gives us a function String -> String which is equivalent to:

\str-> drop 0 str ++ drop 1 str ++ .. ++ drop 9223372036854775807 str

_{Note: Once the argument of drop is greater or equal to the length of str it just appends empty strings.}

Practical solution, 40 bytes

This won't take as long as the shorter solution but at a great cost of 16 bytes:

map$mconcat=<<map drop.zipWith pure[0..]

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Answer 48

C (gcc), 94 bytes

f(char*s){char*e=s+strlen(s),*i=s;for(;i<e;*i++=*i-32?*i:0);for(i=s;i<e;i++)printf(*i?i:" ");}

Try it online!

Saves a pointer to the end of the string in char *e, then replaces spaces with null characters, then prints the string starting with each character and ending at the next null character, but printing a space if the current character is a null character.

Answer 49

JavaScript (Node.js), 88 bytes

f=s=>s.split(' ').map(e=>[...e].map((_,i,a)=>a.slice(i,a.length).join``).join``).join` `

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Answer 50

C (gcc), 64 bytes

f(s,t)char*s,*t;{for(;*s;t+=*s<33)for(strcpy(t,s++);*t&31;t++);}

Take two char* parameters: The first one for input, and the second one for output. The caller is responsible for malloc and free. Like most string functions in C, this one is designed to be vulnerable by overflow the buffer.

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Answer 51

Gema, 24 characters

<L>=@x{$0}
x:?*=?*@x{$2}

Sample run:

bash-4.4$ gema '<L>=@x{$0};x:?*=?*@x{$2}' <<< 'darth vader'
dartharthrththh vaderaderdererr

Answer 52

Red, 67 66 bytes

func[s][foreach w split s" "[until[prin w take w tail? w]prin" "]]

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Answer 53

Kotlin, 87 bytes

fun f(s:String):String=if(s=="")s else s.split(" ").joinToString(" "){it+f(it.drop(1))}

Answer 54

Python 2, 55 bytes

for x in input():
 y=""
 while x:y+=x;x=x[1:]
 print y,

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Takes input as a list. output is separated by spaces.

Answer 55

Java, 131 bytes

char[]e(char[]a){int b=0,l=a.length,c,d=0;char[]n=new char[l*(l+1)/2];for(;b<l;b++){for(c=l-b;c>0;c--){n[d]=a[l-c];d++;}}return n;}

I know there is a better Java answer but, wanted to do it just with char arrays

char[] e(char[]a){
    int b=0,l=a.length,c,d=0;      //l to keep length, b for each iteration, c for each letter in iteration 
    char[] n= new char[l*(l+1)/2]; //Create new array with length based in triangular number sequence
    for(;b<l;b++) {
        for(c=l-b;c>0;c--) {
            n[d]=a[l-c];          //Fill it
            d++;                   //d keeps position in new array
        }
    }
    return n;
}

Answer 56

Julia 0.7, 75 65 bytes

p(a)=for i in a for j=1:endof(i) print(i[j:end])end;print(" ")end

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First codegolf. Yay v0.7 becase it it just deprecates the endof method and 1.0 replaces it with lastindex.

Update: As mentioned in the comments I am not sure if an array of words as input is permitted. So I included the bytes to take a string as input

Update2: Apparently an array of strings is ok so here we go down to 65 bytes. Thx to @JonathanFrech for removing some more whitespace

Answer 57

Kotlin, 48 bytes

Lambda takes a List<String> as input and returns a List<String> as output, which is allowed. The extra code in the footer is just so the input can be tested easily.

{it.map{it.indices.fold(""){a,v->a+it.drop(v)}}}

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Answer 58

GolfScript, 23 bytes

" "/{{""=!}{(;}/}/]" "*

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Answer 59

Multi User Forth(121 bytes), Protomuck variant.

" " explode_array { swap foreach nip { swap  begin dup 1 strcut nip dup not until }cat repeat }list " " array_join .tell

I guess if you want to require that I use entirely standard stuff, then .tell it becomes me @ swap notify which brings it up to 132 bytes. If you want to go the other direction, then I guess that swap could become .s and nip .n, dup .d etc, would turn it into

" " explode_array { .s foreach .n { .s  begin .d 1 strcut .n .d not until }cat repeat }list " " array_join .tell

Which has 113 bytes.

Luckily we can continue further, begin turns into .b, until .u and explode_array .e, if we use our thinking caps we can turn foreach into .f by defining that into foreach nip (who needs the index of the array anyway?) and we end at 94 bytes.

" " .e { .s .f { .s  begin .d 1 strcut .n .d not until }cat repeat }list " " array_join .tell

But, this is getting silly, and probably only hurting any chance of this being seen as a meaningful first post on codegolf.

Answer 60

C (clang), -DT=t=strtok -DZ=" ") 178 113 106 105 95 91 bytes

i;f(*a){char*t;for(T(a,Z;t;printf(Z,T(0,Z)for(i=0;t[i];)printf(t+i++);}

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Answer 61

Scala, 81 bytes

def x(l:String):String=if(l=="")" "else l+x(l.tail)
println(args.map(x).mkString)

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Answer 62

Elixir, 122 bytes

a = ["Hello","World"]
f = fn n,f -> if(n != "") do "#{n}#{f.(String.slice(n, 1..-1), f)}" end end
Enum.map(a, &(f.(&1,f)))

# OUTPUT
["Helloellolloloo", "Worldorldrldldd"]

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Answer 63

Kotlin Android, 204 bytes

fun f(s:String){ val a=s.split(" ".toRegex()).dropLastWhile{it.isEmpty()}.toTypedArray();for(i in a.indices){var t=0;while(t<a[i].length){for(j in t until a[i].length){print(a[i][j]+"")};t++};print(" ")}

}

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This Program is written in Kotlin Android.

Answer 64

C#, 193 bytes

using System;class a{static void Main(string[]b){foreach(var c in b)d(c, 0);}static void d(string c,int e){if(e==c.Length-1){Console.Write(" ");return;}Console.Write(c.Substring(e++));d(c, e);}}

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Since .NET automatically splits the arguments at spaces, it's possible to loop over them without splitting them manually.
Other than that, this uses recursion to subtract one character at a time using the overload of string.Substring that takes only 1 argument which is the index to start at.

Answer 65

sed, 42 37+2 bytes

s:\b:_:g
:A
s:_(.(\w*_))|__:\1\2:g
tA

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Repeatedly match parts of words between underscores:

_hello_ _world_
hello_ello_ world_orld_
helloello_llo_ worldorld_rld_
helloellollo_lo_ worldorldrld_ld_
helloellollolo_o_ worldorldrldld_d_
helloellolloloo__ worldorldrldldd__
helloellolloloo worldorldrldldd

Answer 66

Burlesque, 8 bytes

{iS\[}ww

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{
 iS # All suffixes
 \[ # Concatenate
}ww # For each word