shortest code to encrypt a string

Question

Using a language of your choice write a program that takes a string as an input, and returns the string encrypted.

Algorithm:

You should assume that each character is always encoded in a byte and is ASCII-based. The extended ASCII codes are mapped to whatever is the default code page in use by the system.

You should encrypt the code in the following way: code of first character +1, code of second character +2, code of third character +3 .....

Extended ASCII codes are based on this table: http://www.asciitable.com/

Shortest code wins.

Answer 1

BrainFuck: 21

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

Assumes EOF=zero

echo -n 'My secret password is 123456!' | beef encrypt.bf
==> N{#wjiym}*{mv36HJLNPR>

Answer 2

C — 67 56 49 characters

During compilation it might warn that q is missing a type...

q;main(p){for(;p=~getchar();)putchar(q++%255-p);}

Thanks to Josh for the significantly shorter rewrites!

$ echo 'My secret password is 123456!' | ./_encrypt
N{#wjiym}*{m����v3}�6HJLNPR>(

Confirm it handles the rollover from 255 to 1 correctly by encrypting 343 xs and checking for no xs in the output:

$ echo xxxxxxx | sed -e 's/x/xxxxxxx/g' -e 's/x/xxxxxxx/g' | ./_encrypt | grep -o x | wc -l
0

Answer 3

Befunge 98 - 15 bytes

:~+'U3*%1+,1+#@

Keeps a counter (starting at 0) and adds that to the ~ input. Then, computes 255 (as 85 * 3 in the form of 'U3*), then mods (%). Then, it adds 1 and prints it, then adds one to the counter, then uses # to skip over the @. At the end of the input, the IP goes the other way and hits the @, ending the program.

Alternatively, if your interpreter supports unicode, this 14 byte (13 char) solution works:

:~+'ÿ%1+,1+#@

Sample run (in command prompt, so some of the unicode characters are actually some ascii value):

This string will be encrypted using the predetermined method!2.7182818284590452353602874713526624

Output:

Ujlw%y{zrxr,äw{|1tx4zäzèÆèÅüü>öôèÉèDÖÄîHÖ£ÉÉÆóöó₧¢íÖÖVñ¥¡ó¬á^pmwrzu|v~yÇ}⌂ä|üâüâåàëäçÄÄîÉïÄæÅöòÆò

Answer 4

GolfScript 20 19 15 12

{0)255%:0+}%

This assigns a variable to the character 0, which allows us to not define it on the first iteration of the mapping. I love that you can do that in GolfScript! Each iteration of the mapping increments the 0 variable, and adds it to subsequent characters of the string (which is treated like an array).

This alternate 20-character GolfScript solution using zip encounters an interesting issue:

.,,]zip{~)255%+}%''+

Try it here. This standard online interpreter, however, treats zip differently for an array containing a string and an array. See this vs this. I'm not sure if an alternate version of Ruby or GolfScript is being used, or which one is considered canonical, but the second interpreter can be resolved with 2 characters:

.,,[\]zip{~)255%+}%''+

I also like a different solution for 23 characters:

1\{(@.@255%+\)@.}do]''+

Answer 5

J, 21 characters

a.{~1+256|(+i.@#)a.i.

Usage (borrowing Yimin's test text):

   a.{~1+256|(+i.@#)a.i.'My secret password is 123456!'
N{#wjiym}*{m????v3}?6HJLNPR>

Answer 6

Python2.7 (67 chars)

import sys;print''.join([chr((ord(x)+1)%256)for x in sys.argv[1]])

Answer 7

JavaScript, 90c

t.split("").map(function(c,a){return String.fromCharCode(c.charCodeAt(0)+a+1)}).join("")

[].map.call(t,function(c,a){return String.fromCharCode(c.charCodeAt(0)+a%255+1)}).join("")

Fiddledee

Answer 8

Javascript/JScript: 72 to 105 chars

Here is a Javascript/JScript attempt at making a version of this code.

I'm aware that there is another Javascript code, but that one relies on recent Javascript implementations.

That is a bad thing!!!

That code is invalid in ie8< without polyfills.

Here is my version:

function $(c){for(var i=0,l=c.length,o='';i<l;o+=String.fromCharCode(c.charCodeAt(i++)+i&255));return o;}
//function call, 105 chars, absolutely valid!

function $(c){for(i=0,l=c.length,o='';i<l;o+=String.fromCharCode(c.charCodeAt(i++)+i&255));return o;}
//function call, no var keyword, 101 chars, completely valid!

function $(c){for(i=0,l=c.length,o='';i<l;o+=String.fromCharCode(c.charCodeAt(i++)+i));return o;}
//function call, 97 chars, most likely invalid!

c='';//here goes input, output is in var `o`, dont count this line
for(i=0,l=c.length,o='';i<l;o+=String.fromCharCode(c.charCodeAt(i++)+i&255));
//no function call, input goes to var c, 77 chars of pure working code! valid for sure!

c='';//here goes input, output is in var `o`, dont count this line
for(i=0,l=c.length,o='';i<l;o+=String.fromCharCode(c.charCodeAt(i++)+i));
//no function call, input goes to var c, no char wrapping after 255, 73 chars of probably invalid code

---

@toothbrush gave me this answer:

i=0;while(i<c.length)c[i]=String.fromCharCode(c.charCodeAt(i++)+i%255+1)

It changes the original string and is only 72 bytes!

I don't know why he has %255+1 in the end when &255 does the same job, is faster and shorter.

But I will keep the original answer intact.

---

Consider this and test it with a Javascript/JScript interpreter.

I know that the 73 char version and the 97 char function call version will be invalid, but its still here!

If you will be using the code in a production environment (website), use the 105 chars long and consider changing the function name to something else (or you might have problems with jQuery and MooTools).

Answer 9

Python3 (99 chars)

e=''
i=0
for c in __import__("sys").argv[1]:
        i=(i+1)%255
        e+=chr(ord(c)+i)
print(e)

Example

python encrypt.py password outputs qcvw|uyl

Answer 10

CoffeeScript, 73 64 chars

t.replace /./g,(c,a)->String.fromCharCode c.charCodeAt()+a%255+1

[].map.call(t,(c,a)->String.fromCharCode c.charCodeAt(0)+a%255+1).join ""

Example: "Test string".replace /./g,(c,a)->String.fromCharCode c.charCodeAt()+a%255+1 outputs Ugvx%y{zrxr

Coffee to Javascript converter (paste in right box): http://js2coffee.org/

As per Ismael Miguel's comments, this method should work in IE 5.5+.

Answer 11

Ruby functional, 67

puts ARGV[0].bytes.map.with_index(1){|b,i|b+i%255}.map(&:chr).join

Example:

$ ruby encrypt.rb password
=> qcvw|uyl

Answer 12

Julia 51 (-4 if in place is allowed)

g(s)=([s.data[i]+=(i-1)%255+1 for i=1:length(s)];s)
g(s)=[s.data[i]+=(i-1)%255+1 for i=1:length(s)] # in place version
julia> g("a"^256) # ^ operator repeats strings
"bcdefghijklmnopqrstuvwxyz{|}~\x7f��������������������������������������������������������������������������������������������������������������������������������\0\x01\x02\x03\x04\x05\x06\a\b\t\n\v\f\r\x0e\x0f\x10\x11\x12\x13\x14\x15\x16\x17\x18\x19\x1a\e\x1c\x1d\x1e\x1f !\"#\$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`b"

Answer 13

GolfScript, 27 characters

[.,,\]zip[{{+}*255%1+}/]''+

Hmmm...seems a bit long. Maybe I should stick to J.

Answer 14

Haskell, 77C 59C

import Data.Char
e s=h s 1
h""_=""
h(x:s)i=(chr(mod((ord x)+i)128)):h s(i+1)

v2

import Data.Char
e s = map chr(zipWith(+)(map ord s)[1..])

I am assuming that input doesn't need to be on STDIN.

Example run:

*Main> e "hello"
"igopt"

Answer 15

Erlang: 60C

Not the smallest or prettiest, but this one can be run standalone in the command line. Set T to your text.

{S,_}=lists:mapfoldl(fun(C,I)->{C+I,(I+1)rem 255}end,1,T),S.

Yields: "qcvw|uyl"

Answer 16

Rebol, 58 chars

s: input forall s[s/1: mod(index? s)+(to-integer s/1)256]s

Answer 17

Groovy, 53 Characters

i=0;args[0].chars.collect{(char)it+(++i%255)}.join()

Answer 18

Ruby, 45 39 bytes

gets.chars{|c|putc(c.ord+$.)%256;$.+=1}

Answer 19

Perl, 54 char

$i=1;print map{chr((ord()+$i++)%256)}split//,$ARGV[0];

Example:

perl encrypt.pl password
> qcvw|uyl

Answer 20

Dart, 68

f(s,{i=1})=>new String.fromCharCodes(s.runes.map((c)=>(c+i++)%255));

Example:

f('My secret password is 123456!');
> N{#wjiym}*{mv3}6HJLNPR>