Substitute a string with itself!

Question 1

Substitute a string with itself

Your goal is to substitute a string with itself by replacing each character in the original string with the one before it, starting with the first character and wrapping around. Here are some examples to show what I mean:

1st example:

Input: program
Output: apgopra
How:
Program -> mrogram (replace p by m in program)
-> mpogpam (replace r by p in mrogram)
-> mprgpam (replace o by r in mpogpam)
-> mpropam (replace g by o in mprgpam)
-> mpgopam (replace r by g in mpropam)
-> mpgoprm (replace a by r in mpgopam)
-> apgopra (replace m by a in mpgoprm)

2nd example:

Input: robot
Output: orbro
How:
Robot -> tobot (replace r by t in robot)
-> trbrt (replace o by r in tobot)
-> trort (replace b by o in trbrt)
-> trbrt (replace o by b in trort)
-> orbro (replace t by o in trbrt)

3rd example:

Input: x
Output: x
How:
x -> x (replace x by x in x)

4th example:

Input: xy
Output: xx
How:
xy -> yy (replace x by y in xy)
-> xx (replace y by x in yy)

Sidenotes:

The string x will only contain lowercase alphanumeric characters and spaces
This is code-golf so shortest code in bytes wins!

Question 2

Do my edits match your original idea?

Question 3

Seems fine, I hope people understand that every round they basically encrypt an encrypted string by replacing characters every round. The examples make this clear, so I think they will.

Question 4

CJam, 11 bytes

q__1m>.{er}

Test it here.

Explanation

q__ e# Read input and make two copies.
1m> e# Rotate the second copy one character to the right.
.{er} e# For each pair of characters from the second and third string,
 e# replace occurrences of the first with the second.

Question 5

TeaScript, 17 bytes (削除) 19 (削除ここまで) (削除) 21 (削除ここまで) (削除) 24 (削除ここまで)

TeaScript is JavaScript for golfing

xd(#lg(i,xC(1#a))

Nice and short

Try it online (watch for trailing whitespace in the input)

Ungolfed & Explanation

x.reduce // Reduce over input
 (# // Anonymous function expands to ((l,i,a)=>
 l.g( // global replace...
 i // replace var i with...
 x.cycle(1) // Cycle x 1
 [a] // At position a
 )
 )

Question 6

JavaScript ES6, 69 bytes

s=>[...s].reduce((p,c,i)=>p.replace(RegExp(c,'g'),s.slice(i-1)[0]),s)

F=s=>[...s].reduce((p,c,i)=>p.replace(RegExp(c,'g'),s.slice(i-1)[0]),s)
input.oninput=()=>output.innerHTML=F(input.value)

#input, #output {
 width: 100%;
}

<textarea id='input' rows="5">
</textarea>
<div id="output"></div>

Question 7

You can skip the F= in your byte count.

Question 8

@Ypnypn Thanks never know what to do when they don't specify this stuff

Question 9

s.slice(i-1)[0] is not equal to s.slice(i-1,i) ?

Question 10

@edc65 No not when i=0

Question 11

There's now a replaceAll method that can save some bytes. I'm not sure which ES version introduced it, so it might not be ES6 then. s=>[...s].reduce((p,c,i)=>p.replaceAll(c,s.slice(i-1)[0]),s), 60 bytes.

Question 12

Ruby, (削除) 50 (削除ここまで) 48 bytes

->s{t=s.dup;t.size.times{|i|t.tr!s[i],s[i-1]};t}

Test:

f=->s{t=s.dup;t.size.times{|i|t.tr!s[i],s[i-1]};t}
f["program"]
=> "apgopra"

Question 13

Mathematica, (削除) 89 (削除ここまで) (削除) 75 (削除ここまで) (削除) 74 (削除ここまで) 57 bytes

""<>Fold[#/.#2&,c=Characters@#,Thread[c->RotateRight@c]]&

Question 14

""<>Fold[#/.#2&,c=Characters@#,Thread[c->RotateRight@c]]&

Question 15

@alephalpha Thanks, I tried that with Transpose and failed.

Question 16

Pyth, 13 bytes

u:G.*HC.>Bz1z

Try it online. Test suite.

Question 17

k2 - 17 char

Function taking 1 argument.

{_ssr/[x;x;-1!x]}

k2 has a builtin called _ssr for String Search and Replace. _ssr[x;y;z] will find y in x and replace it with z. So we use / to fold this functionality over each replacement we want to make. For those unfamiliar with folding (as in functional programming), essentially _ssr/[x; (y1; y2; y3); (z1; z2; z3)] becomes _ssr[_ssr[_ssr[x; y1; z1]; y2; z2]; y3; z3]. Strings are lists of their characters, so we may simply rotate the input back a step and get the replacements, and plug right in.

 {_ssr/[x;x;-1!x]} "program"
"apgopra"
 {_ssr/[x;x;-1!x]} "robot"
"orbro"
 {_ssr/[x;x;-1!x]} (,"x") / one-letter strings are ,"x" and parens are required
,"x"
 {_ssr/[x;x;-1!x]} "xy"
"xx"

Question 18

Haskell, 76 bytes

[]#_=[];(x:y)#g@(a,b)|x==a=b:y#g|2>1=x:y#g;h x=foldl(#)x$zip x$last x:init x

Too bad, Haskell doesn't even have a build-in substitution function.

Question 19

PHP, 76 bytes

function($s){$f=str_split;echo str_replace($f($s),$f(substr($s,-1).$s),$s);}

Here is the ungolfed version:

function selfSubstitute($originalString)
{
 $shiftedString = substr($originalString, -1) . $originalString;
 $splitOriginalString = str_split($originalString);
 $splitShiftedString = str_split($shiftedString);
 echo str_replace($splitOriginalString, $splitShiftedString, $originalString);
}

Question 20

Python, (削除) 67 (削除ここまで) (削除) 64 (削除ここまで) (削除) 62 (削除ここまで) 57 Bytes

Straightforward solution, will look into something to shorten this. Thanks to @RandyC for saving 5 bytes.

c=input()
for x in zip(c,c[-1]+c):c=c.replace(*x)
print c

Input should be in quotes.

Question 21

You can save a few bytes dropping the [:-1] since zip truncates to the shortest iterable.

Question 22

@RandyC Wow, good call! Thanks.

Question 23

Haskell, 58 bytes

r(x,y)c|x==c=y|0<1=c;f s=foldl(flip$map.r)s.zip s$last s:s

Pretty similar to Christian's solution, but using map and the fact that zip ignores superfluous elements if the lists are of unequal length. It folds through the list of replacements (on the form (from,to)), updating the string by mapping the hand written replacement function r on each letter.

The expression flip$map.r was derived using LambdaBot's "Pointless" plugin.

Question 24

Jelly, (削除) 10 (削除ここまで) (削除) 9 (削除ここまで) 7 bytes

ṙ-ż@y@ƒ

Try it online!

Having seen caird's y-based solution, I spent several hours trying to use matrix multiplication instead, but eventually I took a crack at y and arrived at this.

Speaking of caird's solution, I was about to pose the question to JHT of how to outdo J_.ịƲU for "obtaining pairs of elements from the input followed by the elements preceding them, wrapping around", but as I wrote that whole thing out I realized I never did actually try using ż!

The edit history contains some nice old explanations if anyone wants to read them.

ṙ Rotate the input left
 - by -1.
 ż@ Zip the input with the rotated input.
 ƒ Starting with the input, reduce the resulting pairs by
 y@ substitution.

Question 25

J, 18 bytes

rplc~/@,~|.,.1|.|.

Try it online!

Question 26

Jelly, 15 bytes

żṙ-,ドルμ,Ḣy\\/ḷ/¿

Try it online!

Jelly sucks at strings

How it works

żṙ-,ドルμ,Ḣy\\/ḷ/¿ - Main link. Takes a string S on the left
 $ - Group the previous two links together:
 - - -1
 ṙ - Rotate S 1 unit to the right
ż - Zip S with rotated S; Call this M
 , - Pair with S; [M, S]
 μ - Begin a new link with this pair as the argument
 \ - Group the previous two links together as a dyad f(p, s):
 \ - Group the previous two links together as a dyad g(p, s):
 Ḣ - Remove the first pair in p and use it; [c, t]
 y - Replace c with t in s
 , - Yield a pair with the modified p and the replaced s
 ¿ - While:
 ḷ/ - The first element is truthy
 / - Run f(M, S)

Question 27

Doesn't Jelly have transliterate built-in?

Question 28

@Shaggy Yeah, that’s the y. Unfortunately the issue is that unlike e.g. the k answer, Jelly isn’t a fan of reducing by y to perform repeated transliteration, so we need to go through the whole charade of a while loop instead

Question 29

Japt, 5 bytes

dUíUé

Try it

dUíUé :Implicit input of string U
d :For each pair XY in the following string, recursively replace X with Y
 Uí : Interleave U with
 Ué : U rotated right

Question 30

Perl 5 `-pF`, 34 bytes

for$a(0..$#F){s/$F[$a]/$F[$a-1]/g}

Try it online!

Question 31

Brachylog, 29 bytes

⟨tc≡⟩⟨gcs2f⟩{gtz{tv&th|h}m}lb

Try it online!

 c Concatenate
 t the last item of the input
⟨ ≡⟩ with the input.
 c Concatenate
 g that result in a singleton list
 ⟨ s2f⟩ with a list of its length-2 substrings.
 ... b Remove the extra element from the final result.
{ }l Left fold:
 gtz zip each element of the first element with the entire second element,
 { }m and for each of those pairs:
 tv the last element of both is the same
 &th and the first element of the last element of the pair
 | is the output, or
 h the first element of the pair is the output.

Martin Ender 198k67 gold badges455 silver badges998 bronze badges · Accepted Answer · 2015-10-25 12:59:09Z

CJam, 11 bytes

q__1m>.{er}

Test it here.

Explanation

q__ e# Read input and make two copies.
1m> e# Rotate the second copy one character to the right.
.{er} e# For each pair of characters from the second and third string,
 e# replace occurrences of the first with the second.

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Substitute a string with itself!