The domain server requires that all employees have a strong, random password conforming to the following rules:

  • Exactly 15 characters long.
  • Keyboard-typeable characters only (as shown in code-type below). Teaching the sales to use ALT+NUMPAD codes is not permitted.
  • At least 1 lower case letter: abcdefghijklmnopqrstuvwxyz
  • At least 1 upper case letter: ABCDEFGHIJKLMNOPQRSTUVWXYZ
  • At least 1 numeric digit: 0123456789
  • At least 1 symbol: `~!@#$%^&*()_+-={}|[]\:";'<>?,./

For this purpose IT have commissioned and will be distributing a Random Password Generator to all employees. All employees will be required to use the Random Password Generator. The requirements for the Random Password Generator are, in addition to the password restrictions above:

  • It must be able to generate all permutations of all allowable characters.
  • It must display the generated password on the screen.
  • The code is required to be as small as possible (in bytes).

Please submit your proposed solution within the next week.

  • 10
    \$\begingroup\$ You should also demand that all passwords which are allowed appear with the same probability (otherwise I can simply make a 30 characters long list with allowed characters, shuffle it, and give the first 15 ones) \$\endgroup\$ Commented Jan 5, 2014 at 22:36
  • \$\begingroup\$ @moose, agreed. I've added a new rule. \$\endgroup\$ Commented Jan 5, 2014 at 22:53
  • 22
    \$\begingroup\$ The IT guys should be fired, or at least better educated: If you do generate passwords randomly, then restricting the set of permissible passwords to those which include at least one character of each category in fact weakens the passwords, since it reduces the size of the permissible set. And our programs would be that much easier if we didn't have to check for that… OK, don't modify the contest after so many submissions have arrived; it's fine as a challenge. \$\endgroup\$
    – MvG
    Commented Jan 6, 2014 at 5:15
  • 10
    \$\begingroup\$ @MvG Indeed: correcthorsebatterystaple \$\endgroup\$ Commented Jan 6, 2014 at 5:46
  • 1
    \$\begingroup\$ You haven't really answered @moose by requiring that all passwords be generatable. They should appear with equal probability. \$\endgroup\$ Commented Jan 6, 2014 at 14:53

37 Answers 37


Mathematica (18)

Let me a little cheat

= 15char ASCII pwd

P.S. not safety :)

  • 7
    \$\begingroup\$ Where's the code? \$\endgroup\$
    – DavidC
    Commented Jan 5, 2014 at 23:38
  • 11
    \$\begingroup\$ Is that guaranteed to satisfy the at least one of each character class requirement? \$\endgroup\$ Commented Jan 5, 2014 at 23:42
  • 3
    \$\begingroup\$ @Hand-E-Food Yes, it is! If you look at the interpretation you will see: password length 15, lower-case letters required, upper-case letters required, numbers required, special characters required. \$\endgroup\$
    – ybeltukov
    Commented Jan 5, 2014 at 23:53
  • 6
    \$\begingroup\$ +1 Clever, but sneaky. \$\endgroup\$
    – DavidC
    Commented Jan 5, 2014 at 23:54
  • 11
    \$\begingroup\$ I knew that Mathematica has a function for everything, but this? \$\endgroup\$
    – null
    Commented Jan 6, 2014 at 8:23

Ruby, 74 69 bytes

Just randomly sample from the ascii range 33 - 126 until all classes of characters are present:


Ruby, 39 bytes

Using moose's clever discovery:


Edit to satisfy the mob:

Note that the rules changed after I first posted this. At the time both the previous entries applied to the rules. I would also like to point out that the rules are still not too well defined:

(..) all permutations of all allowable characters

"Permutations". There are no permutations of the allowable characters that complies with the rest of the rules, because any permutation of the set of allowable characters is as long as the set of allowable characters itself (while the password is supposed to be 15 characters long). And there are no repetitions in a permutation. However my first entry is still more "random" than many of the other well upvoted answers here.

Nevertheless, here you have it. Allows repetitions of characters and underscore:

Ruby, 77 bytes


I also used puts instead of p in this one because p prints out the string enclosed in "quotation marks" and some characters escaped with a backslash.

Ruby, 70 bytes

As Ventero points out, ~ can be skipped in front of the regexes, and print can replace puts$_. But with the ugly output this causes you might as well print all the rejected passwords too, squeezing it into a one-liner:



As requested. $_ is a semi-magical variable that contains the last line read from input - so you don't always need to store it, like this. Here however we use it because of another property, namely that the ~ operator applies a regex directly to it, a trick I first learned by chron. I replaced the usage of all, but it should be quite easy to understand if you get the rest (see the docs).

  • 2
    \$\begingroup\$ Could you please explain your code a bit? What does .all?{|r|~r} do? What does $_= do? \$\endgroup\$ Commented Jan 5, 2014 at 23:41
  • 3
    \$\begingroup\$ The sample line is clever and all, but I think it violates "It must be able to generate all permutations of all allowable characters." Nowhere does it says that the password may only contain a-d as far as letters are concerned. If z is an allowed character, there should be a chance >0 that z is in the password. \$\endgroup\$
    – nitro2k01
    Commented Jan 6, 2014 at 5:07
  • 1
    \$\begingroup\$ Does \W in Ruby include the underscore _? In most regex dialects I know it does not. And if your code could not generate passwords where a single _ was the only non-alphanumeric symbol, then it would violate one requirement. The second approach very obviously violates that requirement, but I guess it wasn't phrased correctly at that time. \$\endgroup\$
    – MvG
    Commented Jan 6, 2014 at 5:09
  • 1
    \$\begingroup\$ @MvG: You're correct. \W does not contain an underscore in Perl-compatible RegEx (source). \$\endgroup\$ Commented Jan 6, 2014 at 8:42
  • 1
    \$\begingroup\$ Furthermore, your solution is affected by the same problem @moose and I had with Python: sample does not repeat elements, so passwords with repeated elements cannot be generated by your code.Can you fix these two issues to make your answer conform to the question? Seeing how yours is the leading solution, excepting Wolfram Alpha, it would be nice to see whether you can conform and still keep the lead. I guess it shouldn't be too hard. \$\endgroup\$
    – MvG
    Commented Jan 6, 2014 at 10:26

Java 8 - 354 329 319 275 267 characters

Just for fun, using lambdas with Java 8 - each possible output has the same probability of being found.

It uses the fact that the allowed characters have consecutive ascii codes, from 33 to 126.

class A {
    //flags for, respectively, small caps, large caps, digits, punctuation
    static int a, A, d, p;

    public static void main(String[] x) {
        String s;
        do {
            //Using special String constructor that takes an int[]
            s = new String(new java.util.Random().ints(15, 33, 127)
                           0, 15);
            a = A = d = p = 0;
                .map(c ->
                      c > 96 & c < 123 ? a = 1
                    : c > 64 & c < 90  ? A = 1
                    : c > 47 & c < 58  ? d = 1
                    : (p = 1))
        } while (a + A + d + p < 4);

Sample output:


Compressed program:

class A{static int a,A,d,p;public static void main(String[]x){String s;do{s=new String(new java.util.Random().ints(15,33,127).toArray(),0,15);a=A=d=p=0;s.chars().map(c->c>96&c<123?a=1:c>64&c<90?A=1:c>47&c<58?d=1:(p=1)).min();}while(a+A+d+p<4);System.out.println(s);}}

  • \$\begingroup\$ How about while(a+A+d+p<4) together with a|=1 instead of a++? Or use bitmasks, i.e. stuff like a|=1 through a|=8, with a<15 as loop condition. This saves another 13 chars if I counted correctly. \$\endgroup\$
    – MvG
    Commented Jan 6, 2014 at 5:04
  • \$\begingroup\$ @MvG good point - did something similar, saving a couple extra chars I believe. \$\endgroup\$
    – assylias
    Commented Jan 6, 2014 at 9:33
  • \$\begingroup\$ @MvG And using new String(int[],int,int) saves another 40-odd chars! \$\endgroup\$
    – assylias
    Commented Jan 6, 2014 at 9:41

Python 2.X + 3.X (229 characters): Generate and replace


  1. First make a list with 15 allowed symbols
  2. Replace a random position r by a random digit
  3. Replace a random position s, with s != r, by an upper case letter
  4. The same for lower case letter and symbol as in 2 and 3.


from random import randint as r, shuffle as s
for i in range(15):

Python 2.X + 3.X (194 characters): Generate and check

import random
from re import search as s
while not all([s("\d",p),s("[a-z]",p),s("[A-Z]",p),s("[\W_]",p)]):
 p=str(map(chr,[random.choice(list(range(33,127))) for i in range(15)]))
  • Thanks to MvG who told me that \u and \l does not exist in Python regex.
  • Thanks to grc who told me that random.sample is without replacement, be to get every possible allowed password we need sampling with replacement.

Using flaw in the problem description

Currently, the problem description does not demand that every symbol / digit appears with the same probability. With the following solution, you cannot make any assumption about a single symbol and/or position. But you can do it with multiple ones.

Python 2.X+ 3.X (62 characters)

from random import sample

Thanks to daniero for the idea to use sample.

  • \$\begingroup\$ Very smooth finding the flaw! I've plugged that one, but bonus points for identifying it. :-) \$\endgroup\$ Commented Jan 5, 2014 at 22:59
  • \$\begingroup\$ Your gen&check is similar to my approach. Out of curiosity:where is this \l and so on for python regexes documented? Don't see it in the reference. My Python 3.3.3 won't even accept "\u". The str(…) does not join the letters in either 3.3.3 or 2.7.6. One suggestion for optmization: all(s("\\"+i,p)for i in "dluW"). \$\endgroup\$
    – MvG
    Commented Jan 6, 2014 at 4:09
  • \$\begingroup\$ random.sample chooses elements without replacement, so not all passwords are possible. \$\endgroup\$
    – grc
    Commented Jan 6, 2014 at 7:15
  • \$\begingroup\$ @MvG: Thank you. I've just seen that \u and \l is vim-only. \$\endgroup\$ Commented Jan 6, 2014 at 8:36

Bash on *nix (109)

while ! grep -Pq [A-Z].*[a-z].*[0-9].*[\\W_]<<<$a$a$a$a
do a=`tr -dc !-~</dev/urandom|head -c15`
echo $a

To work correctly, $a must not be set to a valid but non-random password up front. If you want to include a= and a line break up front, that's three more characters but it allows you to run the thing repeatedly. You can obviously also replace all newlines with ; so you have a one-liner which you can execute as often as you whish.

Furthermore, you should have set LC_ALL=C or not set any locale-specific environment variables (LANG and LC_CTYPE in particular), since the character ranges depend on collation order being equal to ascii order.

/dev/urandom is the a source of random bytes. !-~ is the range of all permissible characters, as specified in the question. tr -dc removes all characters not listed in its next argument. head takes 15 of the remaining characters. grep checks whether each of the required kinds does occur at least once. Its input consists of four copies of the candidate, so order of the symbols does not matter, hence all possible passwords stand a chance of getting selected. The -q to grep suppresses output.

For reasons unknown, /dev/random instead of /dev/urandom takes ages. It seems like entropy got exhausted pretty quickly. If you cd into /dev, you can avoid some more bytes, but that feels a bit like cheating.

Python 2 (138)

import re,random
while not re.search('[A-Z].*[a-z].*[0-9].*[\W_]',a*4):
print a

To make the code readable I added a newline and indentation after the loop which is not neccessary and which I did not count.

This is essentially the same idea as in the bash version. The random source here is random.sample, which will not repeat elements. To counter this fact, we use 15 copies of the list of permissible letters. That way, every combination can still occur, although those with repeated letters will occur less often. But I decide to consider this a feature, not a bug, since the question did not require equal probability for all permutations, only the possibility.

Python 3 (145)

import re,random
while not re.search('[A-Z].*[a-z].*[0-9].*[\W_]',a*4):

One newline and one indent again not counted. Apart from some Python-3-specific syntax overhead this is the same solution as for Python 2.

JavaScript (161)


I added the newlines for readability, but did not count them.

R (114)


Linebreak and indentation inside loop added but not counted. If you feel like it, you can again move this to a single ;-separated line.

  • \$\begingroup\$ Ha! I was just about to point out that you could have used grepl in your R code. If only I'd thought of repeating the test password four times so you could do all the checks in one. And you know, if only I'd thought about sample and intToUtf8. However, you need to add replace=TRUE (or more concisely, you just need to add ,T) to your sample method to make sure you're getting all possible passwords. \$\endgroup\$
    – AmeliaBR
    Commented Jan 6, 2014 at 6:26
  • \$\begingroup\$ @AmeliaBR: You are right, fixed that replace=T mistake, thanks for pointing this out. Finding intToUtf8 by guessing likely names with tab completion took me quite a while; I knew such a function has to exist, but the more common names like chr and so on were not used. \$\endgroup\$
    – MvG
    Commented Jan 6, 2014 at 6:33
  • \$\begingroup\$ @MvG: I don't understand why your Python code terminates at all. Why do you need that *4? I thought your regex would match any string, that ad first one uppercase letter, then anything, then one lowercase letter, than anything ... what did I get wrong? \$\endgroup\$ Commented Jan 6, 2014 at 8:49
  • \$\begingroup\$ @moose: As you already noticed, my regex checks for the required categories in a specific order. But by taking the concatenation of four copies of the current candidate, I can ensure that the order no longer matters: Even if my password is symbols followed by digits followed by lowercase followed by uppercase, then there would still be a match. The only way a match can fail is if a category is missing altogether. Also note that I do re.search not re.match, so the regex might match anywhere in the candidate password. Does this explain why it will terminate eventually? \$\endgroup\$
    – MvG
    Commented Jan 6, 2014 at 8:55
  • \$\begingroup\$ Ah, I did not notice that you use re.search instead of re.match. That explains it. But I still think you don't need the *4. Thank you for the explanation (+1) \$\endgroup\$ Commented Jan 6, 2014 at 9:09

C# (123 - 139 103 - 127 characters compacted):

Using a perfectly adequate framework method in System.Web.dll:

class P
    static void Main()
        Console.WriteLine(System.Web.Security.Membership.GeneratePassword(15, 1));


class P{static void Main()



Alternatively, take the value of the second parameter (int numberOfNonAlphanumericCharacters) from the command line:

class P
    static void Main(string[] a)
        Console.WriteLine(System.Web.Security.Membership.GeneratePassword(15, int.Parse(a[0])));
  • 3
    \$\begingroup\$ GeneratePassword does not support the full set of symbols specified in the question. Neither did I find any guarantees about the minimum number of occurrences of each character category. \$\endgroup\$
    – MvG
    Commented Jan 6, 2014 at 8:39
  • 2
    \$\begingroup\$ You could compact further by using class P and string[] a. \$\endgroup\$
    – d3dave
    Commented Jan 6, 2014 at 14:45
  • \$\begingroup\$ @MvG, that's interesting. It looks like it excludes any symbol that's commonly used to write accented characters in languages such as French. Probably a smart move. Changing the keyboard language would be enough to stuff up your password. \$\endgroup\$ Commented Jan 6, 2014 at 21:19

R (301 322 characters)

Correction forgot to check for digits.


(whitespace added for clarity only).

Generates all possible 15-character permutations of the 94 characters. Then randomly selects one until it matches the criteria.

The magic is in the q:q operation, which generates a new factor data type that is the interaction of all the factors in the first q list with all the factors in the second list, with every possible combination of those two lists being included in the list of "levels" of that factor. Interact 15 copies of the list of allowed characters, and you get (94^15) possible levels.

Please do not try this at home. The code takes a couple seconds to figure out all the three-character permutations, I really can't imagine how long it would take to figure out all the 15-character permutations, if your computer didn't just run out of memory in the meantime. When I ran the finished (three-character password) script to check it, the first password it spit out was "oO=", which I think about sums up the reaction you should have to this code.

  • \$\begingroup\$ @MvG has an R script that is both much more practical and much shorter, if much less awesome: codegolf.stackexchange.com/a/17700/12413 \$\endgroup\$
    – AmeliaBR
    Commented Jan 6, 2014 at 6:28
  • \$\begingroup\$ Nevertheless, I like your idea. Many code-golf snippets I saw let specific features of the language do the hard work. And your code certainly does that for R, with those interactions. \$\endgroup\$
    – MvG
    Commented Jan 7, 2014 at 13:13

Mathematica 170





Python 2.7 (182)

import random as r,string as s
print ''.join(j)
  • \$\begingroup\$ You can get 9 digits less by removing the join as it is not required by the problem descript. Another 2 less by removing spaces. \$\endgroup\$ Commented Jan 5, 2014 at 22:47
  • \$\begingroup\$ @moose I took out the spaces right before you commented :-) I feel like the join kinda has to be there: Would users be expected to understand python list syntax from the output: ['q', 'u', ...]? \$\endgroup\$ Commented Jan 5, 2014 at 22:50
  • 1
    \$\begingroup\$ I thought about removing print at all. When size in bytes is important, they might live in the punch card time. That way, they might be able to read the memory ... just by looking at it. Or they are "real programmers": xkcd.com/378 \$\endgroup\$ Commented Jan 5, 2014 at 22:53
  • 1
    \$\begingroup\$ If I'm reading the code correctly, this does not fulfill the all permutations requirement, it will always have 12 lowercase characters, making passwords with more than one of the other groups (like aA$bc1111111111) impossible. \$\endgroup\$
    – IQAndreas
    Commented Jan 5, 2014 at 23:31
  • 1
    \$\begingroup\$ In Johnathon's defence, I think the permutations rule was added 5 minutes after his post. \$\endgroup\$ Commented Jan 5, 2014 at 23:36

Golfscript (60)

Since the obl. golfscript is missing and as a noob I need the practice anyway :)

[48 10{rand}:r~+65 26r+97 26r+33 15r+11,{;32 96r+}%~]{r}$''+

It just builds an array with the 4 required + 11 random characters and sorts in random order.

  • \$\begingroup\$ +1 for {r}$. That's a pretty dirty way to shuffle a list — I like it! ;-) \$\endgroup\$ Commented Jan 7, 2014 at 2:47
  • \$\begingroup\$ ...however, I don't think this can ever output e.g. 0Aa~~~~~~~~~~~~. :-( \$\endgroup\$ Commented Jan 7, 2014 at 13:43

JavaScript 258 240 233 225


Using a rule where:

function(x){return x*x} can be re-written as function(x)x*x. Only seems to work for functions returning a value.

Next iteration, reduced x.sort(function().5-R()) to x.sort(y=>.5-R())

Next iteration, reduced further with fat arrow notation, which sadly only works for Firefox 22 and above.

  • \$\begingroup\$ Nice compacting! :D \$\endgroup\$
    – IQAndreas
    Commented Jan 6, 2014 at 9:01

JavaScript (269 characters compacted)

For clarity, this is the code before I compacted it down JS-Fiddle of it:

var lowerLetters = "abcdefghijklmnopqrstuvwxyz";
var numbers = "0123456789";
var symbols = "`~!@#$%^&*()_+-={}|[]\\:\";'<>?,./";
var allCharacters = lowerLetters + upperLetters + numbers + symbols;

String.prototype.randomChar = function() {
    return this[Math.floor(this.length * Math.random())];

var minLength = 15;
var result = [];

// Start off by picking one random character from each group
// Next, pick a random character from all groups until the desired length is met
while(result.length < minLength) {
result.shuffle(); // Finally, shuffle the items (custom function; doesn't actually exist in JavaScript, but is very easy to add) -> http://stackoverflow.com/questions/2450954/how-to-randomize-shuffle-a-javascript-array

Here it is compacted down to 269 characters (JS-Fiddle of it):


function r(t){
    return t[~~(t.length*R())]

x.sort(function(){return .5-R()});
  • \$\begingroup\$ Since I'm ending lines with semi-colons, all removable whitespace was ignored for character counting, but left in for clarity. \$\endgroup\$
    – IQAndreas
    Commented Jan 5, 2014 at 23:57
  • \$\begingroup\$ What do you mean by shuffle() being a "custom function". Is it part of JavaScript or code you would have to write it yourself? \$\endgroup\$ Commented Jan 6, 2014 at 1:51
  • \$\begingroup\$ @Hand-E-Food I meant it is not built into JavaScript, and since any developers here should know how to shuffle an array, I felt including the function in the code was unnecessary. It is available in the JS-Fiddle though (line 16). \$\endgroup\$
    – IQAndreas
    Commented Jan 6, 2014 at 1:58
  • 1
    \$\begingroup\$ My point is, it counts towards your byte-count. But I see now you have implemented it in the compacted version, so please ignore me. :-) \$\endgroup\$ Commented Jan 6, 2014 at 4:39

Clojure (63):

(->> (map char (range 33 127)) (shuffle) (take 15) (apply str))

But need to be improved to ensure that containing at least 1 character of each category (Upper, Lower, Digit, Symbol).


In sql-server

declare @a nvarchar(28)
set @a='abcdefghijklmnopqrstuvwxyz'
declare @b nvarchar(max)
declare @c nvarchar(max)
set @c='0123456789'
declare @d nvarchar(max)
set @d='~!@#$%^&*()_+-={}|[]\:";<>?,./'

select left(substring(@a,cast(rand()*10 as int),3)+substring(@b,cast(rand()*10 as int),6)+substring(@c,cast(rand()*10 as int),3)+substring(@d,cast(rand()*10 as int),5),15)

See it in action--1

see it in Action--2

  • 1
    \$\begingroup\$ I'm having trouble following the last line, but the code does not seem to fulfill the all permutations requirement. \$\endgroup\$
    – IQAndreas
    Commented Jan 6, 2014 at 9:15
  • \$\begingroup\$ Your code will never generate any password starting with ~0Aa, nor any password where b is followed by a. \$\endgroup\$
    – Heinzi
    Commented Jan 6, 2014 at 9:39
  • \$\begingroup\$ @Heinzi:yes i agree all permutations required r not taken into account it just displays a 15 length..characters randomly choosen from a...z,A..Z,0..9,!...+ :(... \$\endgroup\$
    – vhadalgi
    Commented Jan 6, 2014 at 9:47

SAS (191)

%macro c(p);compress(p,,"&p")ne''%mend;data x;length p$15;do x=1by 1;do t=1to 15;substr(p,t,1)=byte(ranuni(7)*94+33);end;if %c(kd)and %c(kl)and %c(ku)and %c(ad)then do;put p;stop;end;end;run;



%macro c(p); /*compress removes or keeps certain classes of characters*/
data x;
length p$15;
do x=1by 1;
    do t=1to 15;
        substr(p,t,1)=byte(ranuni(7)*94+33); /*give it a 33-126, byte discards the noninteger portion rounding down*/
    if %c(kd)and %c(kl)and %c(ku)and %c(ad)then do; /*k=keep d=digit l/u=lower/upper ad=remove digits and alphas*/
        put p;
        stop;  /*met our requirement, head home*/

PowerShell: 119

Gofled Code


Un-golfed and Commented

# Start for loop definition.
    # Skip variable initialization, start definition of run condition.
    # Loop runs if $x does not meet complexity requirements.
    # Length requirement is not tested here because it is enforced by the generator later.
    # Much thanks to @VasiliSyrakis for the RegEx help.
    # Reset $x in case the script block has already run.
    # Use ForEach-Object, via the % alias, to run a loop 15 times.
        # Append a random ASCII character from 33-126 to $x.
        # Note: Use get-random instead of random for faster performance.
# Display $x.
# Variable cleanup - not included in golfed code.
rv x
  • \$\begingroup\$ I think this regex may make it ever so slightly shorter: ^.*(?=.{15,})(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[!#$%&? "]).*$, you can do one match with this, which will only match when there is one Upper,Lower,Digit,Symbol. \$\endgroup\$ Commented Jan 9, 2014 at 6:01
  • \$\begingroup\$ @VasiliSyrakis Ok, You might have to walk me through that a bit. Feel free to start up a chat room if you think it will take awhile. A couple things I'm confused on: 1.) I see the number 15 included there. Is that to make sure the string is exactly 15 characters? If so, this can be omitted, since the script naturally only generates 15-character strings. 2.) What do you mean "will only match when there is one Upper, Lower, Digit, Symbol"? Does that mean it will only match when there is exactly one of each, or at least one of each? The former will break things. \$\endgroup\$
    – Iszi
    Commented Jan 9, 2014 at 6:07
  • \$\begingroup\$ Also, does your RegEx ignore ordering of characters? For example, if tuned down to match 4-character strings, would both 1aZ% and (p3R match? I had some difficulty finding ways to do that online. \$\endgroup\$
    – Iszi
    Commented Jan 9, 2014 at 6:08
  • \$\begingroup\$ Tested the new RegEx against output from my current script. It doesn't seem quite reliable. Code: $x-cmatch'^.*(?=.{15,})(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[!#$%&? "]).*$' Good matches: C>suD1?hTwbDx(z j%4O]HyeG|u[U$5 O/rGeD0$hJk=GO/ Failed matches: 3evthX3_X^nBrR` .nA~uYzrR4YV-r.` u-IjZE48ntQ;HxV \$\endgroup\$
    – Iszi
    Commented Jan 9, 2014 at 6:17
  • \$\begingroup\$ How do I open a chat room? \$\endgroup\$ Commented Jan 9, 2014 at 6:17

Python 2.7 (149)

from random import*
''.join(map(lambda x:chr(randint(*(x[1]or(32,126)))),sorted(map(None,sample(range(15),15),((48,57),(65,90),(97,122),(33,47))))))

Written out in a more readable (and not executable) way;

from random import *
''.join(                                          # Concatenate characters to string
  map(                                            # Map all characters using below lambda
    lambda x:chr(randint(*(x[1] or (32, 126)))),  # Map a single range to a random character
                                                  # within a specific range if supplied,
                                                  # otherwise the default "all" range.
    sorted(                                       # After distributing ranges, sort
      map(None,                                   # zip_longest alternative, distributes the
                                                  # required ranges over 4 random positions
        sample(range(15), 15),                    # 0-14 in random order
        ((48, 57), (65, 90), (97, 122), (33, 47)) # The 4 required ranges

Fairly straight forward and surprisingly not much longer than a "generate, retry on match fail" version.

  • \$\begingroup\$ Are you sure this can really generate all eligible passwords, including e.g. 0Aa~~~~~~~~~~~~? (Note that '~' == chr(126).) \$\endgroup\$ Commented Jan 7, 2014 at 13:39

PSQL (189)

Feels like PSQL is a bit verbose... :)

SELECT ARRAY_TO_STRING(ARRAY_AGG(CHR((TRUNC((b-a)*RANDOM()+a))::int)ORDER BY RANDOM()),'')FROM(SELECT 32 a,127 b FROM generate_series(1,11)UNION ALL VALUES(48,58),(65,91),(97,123),(33,48))a

SQLfiddle demo.


PHP, 235 225

This script shuffles characters around and then is checked via RegEx to make sure the password is strong (or it is regenerated).

while(!preg_match('/^(?=.*[A-Z])(?=.*[^A-Za-z])(?=.*[0-9])(?=.*[a-z]).{15}$/',$p)){ $p = substr(str_shuffle('ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789`~!@#$%^&*()_+-={}|[]\:";\'<>?,./'),0,15); }
echo $p;
  • 1
    \$\begingroup\$ Clever, but doesn't allow duplicate characters. \$\endgroup\$ Commented Jan 6, 2014 at 1:44
  • 1
    \$\begingroup\$ Instead of while(true) ... if (condition) break you can use while (!condition) \$\endgroup\$
    – exussum
    Commented Jan 6, 2014 at 8:12

Javascript (209)

r=Math.random;function t(x,y){return String.fromCharCode(Math.floor(y*r()+x))};x=[t(33,14),t(48,10),t(65,26),t(97,26)];for(i=0;i<11;i++)x.push(t(32,95));console.log(x.sort(function(){return r()-0.5}).join(''))


// Return a character in the range [x,x+y)
function t(x,y) { return String.fromCharCode(Math.floor(y*Math.random()+x)) }
// Prefill required ranges
x=[ t(33,14), t(48,10), t(65,26), t(97,26)]
// Push 11 totally random (valid) characters
for(i=0; i<11; i++)
// Shuffle and output as string
console.log(x.sort(function(){return Math.random()-0.5})

Perl, 92

Not as concise as the Ruby answer, but I'm sure a Perl wizard could make this even shorter... I'm not too happy with all the m//s at the end, but seems to work and should satisfy the conditions to eventually generate all permutations.

do{$_=join"",map{(map{chr}33..127)[rand 94]}0..14}while!(/[A-Z]/&/[a-z]/&/\d/&/[\W_]/);print

Sample usage:

perl -e 'do{$_=join"",map{(map{chr}33..127)[rand 94]}0..14}while!(/[A-Z]/&/[a-z]/&/\d/&/[\W_]/);print'

Edited to fix validation and change [[:punct:]] to [\W_] after MvGs comments.

  • 1
    \$\begingroup\$ Your generation part is nice, but your selection criterion in the loop condition is plain wrong: e.g. a password of aaaaaaaaaaaaaa would cause the loop to terminate. You should test criteria with non-random passwords to make sure they do what you want them to. \$\endgroup\$
    – MvG
    Commented Jan 7, 2014 at 13:10
  • \$\begingroup\$ Indeed, you are correct, I've fixed this and saved some bytes! Thanks! \$\endgroup\$ Commented Jan 7, 2014 at 13:50
  • 1
    \$\begingroup\$ Are you sure about this [[:punct:]]? I guess I'd prefer '[\W_], which is shorter and of which I'm even more sure that it is correct, at least combined with your 33..127` range. \$\endgroup\$
    – MvG
    Commented Jan 7, 2014 at 15:04
  • \$\begingroup\$ A good point, I think I was concerned that \W didn't include _, however you're absolutely right, it's not needed: gist.github.com/anonymous/8301237. Thank you! \$\endgroup\$ Commented Jan 7, 2014 at 15:41

Java 7 (270 234 characters)

The premise is the same used by @assylias with java 8 (generate random passwords until valid password). However, instead of using lambdas, the password is generated by iterating a char array and validated by matching a regex.

class A {
  public static void main(String [] a) {
    byte[] b = new byte[15];
    String s;
    do {
      new java.util.Random().nextBytes(b);
      s = new String(b);
    } while(!s.matches("(?=.*?[a-z])(?=.*?[A-Z])(?=.*?\\d)(?=.*?[!-/:-@\\[-`]).*"));

Minified Code:

class A {public static void main(String[] a){byte[] b=new byte[15];String s;do{new java.util.Random().nextBytes(b);s=new String(b);}while(!s.matches("(?=.*?[a-z])(?=.*?[A-Z])(?=.*?\\d)(?=.*?[!-/:-@\\[-`]).*"));System.out.println(s);}}


One Liner version (143 bytes)

sal g random;1..11|%{$P+=[char](33..126|g)};(65..90|g),(97..122|g),(48..57|g),(33..47+58..64+123..126|g)|%{$P=$P.insert((1..11|g),[char]$_)};$P

Mini version (146 bytes)

sal g random

Readable version (860 bytes)

function pwgen {

    # Fulfill Upper,Lower,Digit,Symbol requirement by predefining ASCII ranges for each
    # These will be added into the string beginning at line 24

    [array[]]$symbolrange = (33..47),(58..64),(123..126)

    [char]$upper = (get-random (65..90))
    [char]$lower = (get-random (97..122))
    [char]$digit = (get-random (48..57))
    [char]$symbol = $symbolrange | get-random

    [char[]]$requirement = $upper + $lower + $digit + $symbol

    # Create the first 11 characters using any ASCII character between 32 - 126

    foreach ($number in (1..11)) {
        [string]$pass += [char](get-random (33..126))

    # Insert each requirement character at a random position in the string

    foreach ($char in $requirement) {
        [string]$pass = $pass.insert((Get-Random (1..11)),$char)

    return $pass

Credit to Iszi for various tips to shorten the code.

  • 1
    \$\begingroup\$ This doesn't cover all permutations. As an example, abcd1234ABCD{|}~ will never come up because $symbol forces at least one of the symbols to be between ASCII 33 and 47. \$\endgroup\$ Commented Jan 6, 2014 at 4:43
  • \$\begingroup\$ Dangit! Must you point out my laziness!? Just kidding... I've edited it now. I also made each "requirement" character go to a separate index within the string, instead of clumping the four in together at the same index. Now if only I could shorten this... \$\endgroup\$ Commented Jan 6, 2014 at 6:23
  • \$\begingroup\$ Is there any reason you can't shave a couple more characters by shortening $SR to, perhaps $Q? \$\endgroup\$
    – Iszi
    Commented Jan 8, 2014 at 23:54
  • \$\begingroup\$ You should also be able to cut stuff like (g(65..90)) down to 65..90|g'. And change the foreach` statements to foreach-object loops using the % alias. Example: foreach($N in (1..11)){...} should be doable as 1..11|%{...}. I'm pretty sure there's other optimizations which are possible, but I've actually got a completely different implementation in mind that I'm planning to try later. \$\endgroup\$
    – Iszi
    Commented Jan 9, 2014 at 0:03
  • \$\begingroup\$ Nice tips :) I shortened it to 213 bytes if I take out the carriage returns and replace with semicolons. \$\endgroup\$ Commented Jan 9, 2014 at 0:20

Factor, 196 characters

Same algorithm as MvG and moose's. It is not the shortest but should satisfy all the (current) criteria in the question:

USING: io kernel math pcre random sequences sequences.repeating ;
[ 15 94 random-integers [ 33 + ] "" map-as dup 60 cycle
"[A-Z].*[a-z].*[0-9].*[\\W_]" findall { } = not ] [ drop ] until print
  • \$\begingroup\$ I might be misinterpreting the regex, but I think something like ~{}|1234abcdABCD will fail the regex. \$\endgroup\$ Commented Jan 8, 2014 at 23:41
  • 1
    \$\begingroup\$ No it will work: "~{}|1234abcdABCD" 60 cycle "[A-Z].*[a-z].*[0-9].*[\\W_]" findall empty? not => t \$\endgroup\$ Commented Jan 9, 2014 at 8:41
  • \$\begingroup\$ I'll take your word for it. :-) \$\endgroup\$ Commented Jan 9, 2014 at 22:51

C - 154 characters

char p[16],c,f,w;main(){srand(time());while(f^15){c=p[15]=f=0;while(c^15){w=33+rand()%94;f|=w

How do I hate srand()? Let me count the ways.


Haskell, 192

import System.Random
main=getStdGen>>= \g->(print.(take 15))$until((\x->all(any(`elem`x))[['a'..'z'],['A'..'Z'],['0'..'9'],['!'..'/']++":;<=>?@[\\]^_`{|}~"]).(take 15))tail$randomRs('"','~')g

The printed string has quotes around it and escapes the backslash and quote characters; if that's unacceptable, print can be replaced with putStrLn for 3 more bytes. Here's a more readable version:

import System.Random

main = do
    g <- getStdGen
    let chars = randomRs ('"', '~') g
    let password = take 15 $ until (hasChars.(take 15)) tail chars
    print password

hasChars :: String -> Bool
hasChars x = all (any (`elem` x)) $
    [ ['a'..'z']
    , ['A'..'Z']
    , ['0'..'9']
    , "!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~"

It's quite straightforward, it just creates an infinite/lazy list of random ASCII characters in the range '!' to '~', then tosses out the first element until the first 15 characters have at least one character from each string of required characters.


Excel VBA, 209 bytes

For i = 1 To 15
x = x + Chr(Int(90 * Rnd + 33))
p = "^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*(_|[^\w])).+$"
With New RegExp
.Pattern = p
Set m = .Execute(x)
If m.Count = 0 Then
MsgBox "redo"
MsgBox x
End If
End With

Randomly generates 15 ASCII characters so all possible combinations are possible. Then uses a regular expression pattern to check if it contains at least one of each criteria.

If it does then the password is displayed, if not "redo" is displayed.

Credit to Bart Kiers for the Regular Expression pattern: https://stackoverflow.com/questions/1559751/regex-to-make-sure-that-the-string-contains-at-least-one-lower-case-char-upper


Python 3 + passgen, 64 bytes

import passgen as p

If you have any suggestions for improvement, let me now in the comments!

Try it online! (or rather don't, as the passgen module isn't on TIO)

Github: https://github.com/EacyCoding/Password_Generator

or in one line:

import passgen as p;print(p.passgen(length=15,punctuation=True))

Sample Outputs:

  • \$\begingroup\$ Welcome to the site! Both of these solutions are 117 bytes. You can remove the spaces before the * and after the ) ()for x) and be sure to check out our Tips for golfing in Python page for more ways you could golf your answer. Also, the challenge requires that you include "at least 1" for 4 different categories. It's unlikely, but this program does have a chance to return e.g. abcdefghijklmno, which doesn't meet the challenge specs. \$\endgroup\$ Commented Jan 7, 2021 at 13:04
  • \$\begingroup\$ @cairdcoinheringaahing thank you very much for your feedback, I see the problem with the chance to return abcdefghijklmno, so I have a new solution (: \$\endgroup\$
    – EacyCoding
    Commented Jan 7, 2021 at 13:22
  • \$\begingroup\$ Feel free to edit the post and link to the try it yourself page. I don't know how to automatically insert my code in the link, would love if you coul explain me. \$\endgroup\$
    – EacyCoding
    Commented Jan 7, 2021 at 13:27
  • \$\begingroup\$ I've edited the title slightly to indicate the use of a non-standard library (perfectly fine, so long as it's made clear). To get the answer format from Try It Online (TIO), click the link button at the top of the page and copy paste the contents of the "Code Golf submission" text area \$\endgroup\$ Commented Jan 7, 2021 at 13:32
  • \$\begingroup\$ @cairdcoinheringaahing thank you \$\endgroup\$
    – EacyCoding
    Commented Jan 8, 2021 at 16:20

AutoHotkey 352

global o:={}
loop 4
loop 11
loop 15
msgbox % s
return z
return mod(z,r)
while o[t:=m(r(),15)]!=""
return t
r := chr(r%f%+m(r(),l%f%))
if f=3
return r

Using - Just run the script


Python (121 characters)

Makes use of the fact that you can multiply lists in Python [1,2,3] * 2 gives [1,2,3,1,2,3]. Imports random. Numbers in a list multiplied by three are borders between ranges in ascii table for needed characters, e.g. [65, 90] maps to uppercase letters.

print "".join([random.choice([chr(i) for i in range(z[0],z[1])]) for z in [[33,48],[48,58],[58,65],[65,90],[90,123]]* 3])
  • 1
    \$\begingroup\$ "It must be able to generate all permutations of all allowable characters.". I don't think it does that since the ranges are always applied in the same order...? \$\endgroup\$ Commented Jan 6, 2014 at 13:01
  • \$\begingroup\$ You are right, thank you. Indeed I didn't notice that ranges should be applied in random order, they should be shuffled, I'll edit that in a moment. \$\endgroup\$
    – Pawelmhm
    Commented Jan 6, 2014 at 13:57
  • \$\begingroup\$ This needs to actually include import random in the code. \$\endgroup\$
    – user45941
    Commented May 21, 2016 at 11:06

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