New Password Idea: Word-walker

Question

I thought of a new way to generate my passwords, and even though it's probably not very clever in the long run, it could still make for a fun code-golf.

Taking a string of words, the password is generated thus:

• Pick the nth character in the nth word
• If n is larger than the word, continue counting backwards

Example:

This is a fun task!
T     s a  u      !

T is the first character
s is the second
a is the first, but going back and forth it is also the third
u is the second but because of counting backwards it's also the fourth
'!' is the fifth character in 'task!' and thus will be included in the final password, Tsau!

Rules

• Input will be a string
• Separate the string on spaces, all other characters must be included
• Uppercase letters must remain uppercase, same with lowercase
• You take n steps in each word, where n is the number of words that have come before plus one
• If n is larger than the word, you must step backwards through the word, if you hit the start, you go forward again until you have stepped n times
• The first and last character is only stepped once, so 'fun' on the seventh position as an example goes 'funufun' and ends on n, not 'funnuff' and ends on f
• Output must be a string

Examples:

Input              Output
Once Upon A Time   OpAe
There was a man    Taaa
Who made a task    Waak
That was neat!     Taa
This is a long string to display how the generator is supposed to work  Tsagnoyotoipto

The shortest code in bytes wins!

Answer 1

Python 2, 72 bytes

lambda s:''.join((-~i*(w+w[-2:0:-1]))[i]for i,w in enumerate(s.split()))

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Answer 2

05AB1E, 11 bytes

#vyN©Fû}®è?

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Explanation

#             # split input on spaces
 vy           # for each word in input
   N©F        # N times do, where N is the current iteration
      û}      # palendromize the word
        ®è    # use N to index into the resulting word
          ?   # print

Answer 3

Jelly, 12 bytes

ḲJị"ŒBṖȯ$€$Ɗ

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Answer 4

Java 10, 148 117 114 110 bytes

s->{int i=-1,j;for(var a:s.split(" "))System.out.print(a.charAt((j=a.length()-1)>0*i++?i/j%2<1?i%j:j-i%j:0));}

-31 bytes thanks to @SamYonnou by creating a port of @user71546's JavaScript answer.
-4 bytes thanks to @SamYonnou again, optimizing the algorithm for Java.

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Explanation:

s->{                            // Method with String parameter and no return-type
  int i=-1,                     // Step integer, starting at -1
      j;                        // Temp integer
  for(var a:s.split(" "))       // Loop over the parts split by spaces
    System.out.print(           // Print:
     a.charAt((j=a.length()-1)  //  Set `j` to the the length of the part minus 1
               >0               //  If the length of the part is larger than 1 (`j` > 0)
                 *i++?          //  (and increase `i` by 1 in the process with `i++`)
                i/j%2<1?        //   If `i` integer-divided by `j` is even:
                 i%j            //    Print the character at index `i` modulo-`j`
                :               //   Else:
                 j-i%j          //    Print the character at index `j` minus `i` modulo-`j`
               :                //  Else:
                0));}           //   Print the first (and only) character
                                //   (the >0 check is added to prevent divided-by-0 errors)

Answer 5

Charcoal, 16 bytes

⭆⪪Ｓ §⁺ι✂ι±²¦⁰±¹κ

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

  Ｓ                 Input string
 ⪪                  Split on spaces
⭆                   Map over words and join
      ι ι           Current word
       ✂ ±²¦⁰±¹     Slice backwards from 2nd last character to start exclusive
     ⁺              Concatenate
    §          κ    Cyclically index on current word index
                    Implicitly print

I don't often get to use Slice's last parameter.

Answer 6

Stax, 9 bytes

éñ~╗D¡┤Gq

Run and debug it

Unpacked, ungolfed, and commented, it looks like this.

j       split into words
{       start block for mapping
  cDrD  copy word; remove first and last character; reverse
  +     concatenate with original word
  i@    modularly (wrap-around) index using map iteration index
m       perform map

Run this one

Answer 7

JavaScript (Node.js), 78 70 69 68 bytes

-1 byte @Arnauld

x=>x.split` `.map((y,i)=>y[a=i%(l=y.length-1)|0,i/l&1?l-a:a]).join``

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Explanation

x=>
 x.split` `                    // Split the words by spaces
 .map((y,i)=>                  // For each word:
  y[                           //  Get the character at index:
                               //   A walk has cycle of length (2 * y.length - 2)
   a=i%(l=y.length-1)|0,       //   Calculate index a = i % (y.length - 1)
   i/l&1                       //   Check in which half the index i in
   ?l-a                        //   If in the second half of cycle, use y.length - 1 - a
   :a                          //   If in the first half of cycle, use a                  
  ]
 ).join``                      // Join back the letters

Answer 8

Red, 135 bytes

func[s][n: 0 p: copy""foreach w split s" "[append/dup r: copy""append w
reverse copy/part next w back tail w n: n + 1 append p r/(n)]p]

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Readable:

f: func[s][
    n: 0
    p: copy ""
    foreach w split s "  "[
        r: copy ""
        append/dup r append w reverse copy/part next w back tail w n: n + 1
        append p r/(n)
    ]
    p
]

Answer 9

Perl 5, 76 bytes

print map{$l=length;substr$_.reverse,$i++%(2*$l-2||1)*(1+1/$l),1}split" ",<>

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Answer 10

k, 31 30 28 bytes

{x{*|y#x,1_|1_x}'1+!#x}@" "\

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Answer 11

Husk, 14 11 bytes

SzS!ö!…¢ŀŀw

Try it online! Inspiration was taken from this program by Zgarb for the forward-and-reverse stepping.

Answer 12

Pyth, 12 bytes

s.e@+b_Ptbkc

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s.e@+b_PtbkcQ   Final Q (input) implicit

           cQ   Split on spaces
 .e             Map the above with b=element, k=index
       Ptb        Remove 1st and last character
      _           Reverse
    +b            Prepend the unaltered element ('abcd' -> 'abcdcb')
   @      k       Get the kth character (0 indexed, wrapping)
s               Join on empty string, implicit output

Answer 13

Japt, -P, 11 bytes

¸Ëê ŪD gEÉ

Try it

¸Ë+s1J w)gE

Try it

---

Explanations

¸Ëê ŪD gEÉ
¸               :Split on spaces
 Ë              :Map over each element D at index E
  ê             :  Palindromise
    Å           :  Slice off the first character
     ªD         :  Logical OR with the original element (the above will return an empty string for single character words)
        g       :  Get the character at index
         EÉ     :  E-1

¸Ë+s1J w)gE
¸               :Split on spaces
 Ë              :Map over each element D at index E
   s1J          :  Slice off the first and last characters
       w        :  Reverse
  +     )       :  Append to D
         gE     :  Get the character at index E

Answer 14

C (gcc), 148 bytes (string version), 114 bytes (print version)

If I must return a string (long version):

char c[99]={0};char*f(s,t,u,i,j,k)char*s,*t,*u;{for(u=c,i=0;t=strtok(s," ");s=0,i++)*u++=t[j=strlen(t),k=2*j-(j>1)-1,(i%k<j?i%k:k-i%k)%j];return c;}

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Otherwise, I just print and don't worry about a buffer (short version):

f(s,t,i,j,k)char*s,*t;{for(i=0;t=strtok(s," ");s=0,i++)putchar(t[j=strlen(t),k=2*j-(j>1)-1,(i%k<j?i%k:k-i%k)%j]);}

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Answer 15

AWK, 79 bytes

Mostly because I'm curious to see any better awk or bash solutions!

{for(i=1;i<=NF;i++){l=length($i);k=int(i/l)%2?l-i%l:k%l;printf substr($i,k,1)}}

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Answer 16

C# (.NET Core), 111 bytes

s=>{int i=-1,j;return String.Concat(s.Split(' ').Select(a=>a[++i>(j=a.Length-1)?j>0?i/j%2<1?i%j:j-i%j:0:i]));};

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Answer 17

Haskell, 65 62 61 bytes

zipWith(\i->(!!i).cycle.(id<>reverse.drop 1.init))[0..].words

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It requires the latest version of Prelude which features the <> function.

                   words    -- split the input string into a list of words
zipWith(\i->     )[0..]     -- zip the elements i of [0..] and the words pairwise
                            -- with the function      
      ... <> ...            --   call the functions with a word and concatenate
                            --   the results. The functions are
        id                  --     id: do nothing
        reverse.drop 1.init --     drop last and first element and reverse
    cycle                   --   repeat infinitely
(!!i)                       -- take the ith elemnt of  

Edit: -3 bytes thanks to @user28667, -1 byte thanks to @B. Mehta

Answer 18

PHP, 77 bytes

while(ord($w=$argv[++$i]))echo($w.=strrev(substr($w,1,-1)))[~-$i%strlen($w)];

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• -3 bytes thanks to Kevin
• -10 bytes thanks to Titus

Answer 19

Vyxal, 11 bytes

⌈₅(ʁ):ẏ¨£i∑

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⌈           # Input split on spaces
 ₅( )       # repeat len(that) times...
   ʁ        # Palindromise each word
       ¨£   # Zipping...
     :      # Words
      ẏ     # range(0, len(words))
         i  # Index the numbers into the words
          ∑ # Stringify

Answer 20

Powershell 208 186 170 bytes

$args|%{$i=0;-join($_.Split()|%{$l=($b=($a=$_)).Length;if($l-gt2){$b=($a|%{-join$a[($l-2)..1]})}for($j=0;$a.Length-le$i;$j++){$a+=($b,$_)[$j%2]}$a.Substring($i,1);$i++})}

Ungolfed:

$args|%{
   $i=0;
    -join($_.Split()|%{
        $l=($b=($a=$_)).Length;
        if($l-gt2){
            $b=($a|%{-join$a[($l-2)..1]})
        }
        for($j=0;$a.Length-le$i;$j++){
            $a+=($b,$_)[$j%2]
        }
        $a.Substring($i,1);
        $i++
    })
}

Test cases below or try it online

@(
    "This is a fun task!",
    "Once Upon A Time",
    "There was a man",
    "Who made a task",
    "That was neat",
    "This is a long string to display how the generator is supposed to work"
)|%{$i=0;-join($_.Split()|%{$l=($b=($a=$_)).Length;if($l-gt2){$b=($a|%{-join$a[($l-2)..1]})}for($j=0;$a.Length-le$i;$j++){$a+=($b,$_)[$j%2]}$a.Substring($i,1);$i++})}

Answer 21

J, 43 bytes

[:(>{~"_1#@>|i.@#)[:(,}.@}:)&.>[:<;._1' '&,

ungolfed

[: (> {~"_1 #@> | i.@#) [: (, }.@}:)&.> [: <;._1 ' '&,

• <;._1 ' '&, split on spaces
• (, }.@}:)&.> for each word, kill the first and last elm and append to word
• #@> | i.@# take the remainder of each word's length divided into its index
• > {~"_1 take that result and pluck it from each word.

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Answer 22

CJam, 22 bytes

qS/{_);1>W%+T_):T@*=}%

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Answer 23

Ruby -pl, 60 bytes

gsub(/(\S+) ?/){w=$1+$1.reverse[1..-2]
w[(-2+$.+=1)%w.size]}

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