#StackBounty: #matlab #random Transforming draws in Matlab from Gaussian mixture to uniform

Bounty: 100

Consider the following draws for a 2x1 vector in Matlab with a probability distribution that is a mixture of two Gaussian components.

P=10^3; %number draws

%First component
mu_a = [0,0.2806];
sigma_a = [v,0;0,v];

%Second component
mu_b = [0,-1.6806];
sigma_b = [v,0;0,v];

MU = [mu_a;mu_b];
SIGMA = cat(3,sigma_a,sigma_b);
w = ones(1,2)/2; %equal weight 0.5
obj = gmdistribution(MU,SIGMA,w);

RV_temp = random(obj,P);%Px2

% Transform each component of RV_temp into a uniform in [0,1] by estimating the cdf.
RV1=ksdensity(RV_temp(:,1), RV_temp(:,1),'function', 'cdf');
RV2=ksdensity(RV_temp(:,2), RV_temp(:,2),'function', 'cdf'); 

Now, if we check whether RV1 and RV2 are uniformly distributed on [0,1] by doing


we can see that RV1 is uniformly distributed on [0,1] (the empirical cdf is close to the 45 degree line) while RV2 is not.

Could you help me to understand why?

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#StackBounty: #python #python-2.7 #random #steganography Python PRNG_Steganography lsb method

Bounty: 50

This is my implementation of a PRNG_Steganography tool for Python. You can also find the code on GitHub.


from PIL import Image
import numpy as np
import sys
import os
import getopt
import base64
import random
from cryptography.fernet import Fernet
from cryptography.hazmat.primitives import hashes
from cryptography.hazmat.backends import default_backend

# Set location of directory we are working in to load/save files
__location__ = os.path.realpath(os.path.join(os.getcwd(), os.path.dirname(__file__)))

Encryption and decryption of the text methods via the Cryptography module

def get_key(password):
    digest = hashes.Hash(hashes.SHA256(), backend=default_backend())
    return base64.urlsafe_b64encode(digest.finalize())

def encrypt_text(password, token):
    f = Fernet(get_key(password))
    return f.encrypt(bytes(token))

def decrypt_text(password, token):
    f = Fernet(get_key(password))
    return f.decrypt(bytes(token))

Main encryption method

def encrypt(filename, text, magic):
    # check whether the text is a file name
    if len(text.split('.')[1:]):
        text = read_files(os.path.join(__location__, text))
    t = [int(x) for x in ''.join(text_ascii(encrypt_text(magic, text)))] + [0]*7  # endbit
        # Change format to png
        filename = change_image_form(filename)

        # Load Image
        d_old = load_image(filename)

        # Check if image can contain the data
        if d_old.size < len(t):
            print '[*] Image not big enough'

        # get new data and save to image
        d_new = encrypt_lsb(d_old, magic, t)
        save_image(d_new, 'new_'+filename)
    except Exception, e:
        print str(e)

Main decryption method

def decrypt(filename, magic):
        # Load image
        data = load_image(filename)

        # Retrieve text
        text = decrypt_lsb(data, magic)
        print '[*] Retrieved text: n%s' % decrypt_text(magic, text)
    except Exception, e:
        print str(e)

Random methods

def text_ascii(text):
    return map(lambda char: '{:07b}'.format(ord(char)), text)

def ascii_text(byte_char):
    return chr(int(byte_char, 2))

def next_random(random_list, data):
    next_random_number = random.randint(0, data.size-1)
    while next_random_number in random_list:
        next_random_number = random.randint(0, data.size-1)
    return next_random_number

def generate_seed(magic):
    seed = 1
    for char in magic:
        seed *= ord(char)
    print '[*] Your magic number is %d' % seed
    return seed

Encrypt via lsb_method

def encrypt_lsb(data, magic, text):
    print '[*] Starting Encryption'

    # We must alter the seed but for now lets make it simple

    random_list = []
    for i in range(len(text)):
        next_random_number = next_random(random_list, data)
        data.flat[next_random_number] = (data.flat[next_random_number] & ~1) | text[i]

    print '[*] Finished Encryption'
    return data

Decrypt via lsb_method

def decrypt_lsb(data, magic):
    print '[*] Starting Decryption'

    random_list = []
    output = temp_char = ''

    for i in range(data.size):
        next_random_number = next_random(random_list, data)
        temp_char += str(data.flat[next_random_number] & 1)
        if len(temp_char) == 7:
            if int(temp_char) > 0:
                output += ascii_text(temp_char)
                temp_char = ''
                print '[*] Finished Decryption'
                return output

Image handling methods

def load_image(filename):
    img = Image.open(os.path.join(__location__, filename))
    data = np.asarray(img, dtype="int32")
    return data

def save_image(npdata, outfilename):
    img = Image.fromarray(np.asarray(np.clip(npdata, 0, 255), dtype="uint8"), "RGB")
    img.save(os.path.join(__location__, outfilename))

def change_image_form(filename):
    f = filename.split('.')
    if not (f[-1] == 'bmp' or f[-1] == 'BMP' or f[-1] == 'PNG' or f[-1] == 'png'):
        img = Image.open(os.path.join(__location__, filename))
        filename = ''.join(f[:-1]) + '.png'
        img.save(os.path.join(__location__, filename))
    return filename

def read_files(filename):
    if os.path.exists(filename):
        with open(filename, 'r') as f:
            return ''.join([i for i in f])
    return filename.replace(__location__, '')[1:]

Usage and main

def usage():
    print "Steganography prng-Tool @Ludisposed & @Qin"
    print ""
    print "Usage: prng_stego.py -e -m magic filename text "
    print "-e --encrypt              - encrypt filename with text"
    print "-d --decrypt              - decrypt filename"
    print "-m --magic                - encrypt/decrypt with password"
    print ""
    print ""
    print "Examples: "
    print "prng_stego.py -e -m pass test.png howareyou"
    print 'python prng_stego.py -e -m magic test.png tester.sh'
    print 'python prng_stego.py -e -m magic test.png file_test.txt'
    print 'prng_stego.py --encrypt --magic password test.png "howareyou  some other text"'
    print ''
    print "prng_stego.py -d -m password test.png"
    print "prng_stego.py --decrypt --magic password test.png"

if __name__ == "__main__":
    if not len(sys.argv[1:]):
        opts, args = getopt.getopt(sys.argv[1:], "hedm:", ["help", "encrypt", "decrypt", "magic="])
    except getopt.GetoptError as err:
        print str(err)

    magic = to_encrypt = None
    for o, a in opts:
        if o in ("-h", "--help"):
        elif o in ("-e", "--encrypt"):
            to_encrypt = True
        elif o in ("-d", "--decrypt"):
            to_encrypt = False
        elif o in ("-m", "--magic"):
            magic = a
            assert False, "Unhandled Option"

    if magic is None or to_encrypt is None:

    if not to_encrypt:
        filename = args[0]
        decrypt(filename, magic)
        filename = args[0]
        text = args[1]
        encrypt(filename, text, magic) 

Any general Coding tips are welcome!

I would also love some tips on how the use the magic to create a wierd seed.

It works by seeding the random module and getting the next unique random integer with that seed. At decryption it knows when to stop looking for bits if it finds the endbit [0]*7

Furthermore I’m interested in how Random this is? I think this is harder to decrypt then just normal lsb_stego, but can not prove anything.

Kind Regards: Ludisposed

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Best way to select random rows PostgreSQL

Given, you have a very large table with 500 Million rows, and you have to select some random 1000 rows out of the table and you want it to be fast.

Given the specifications:

  • You assumed to have a numeric ID column (integer numbers) with only few (or moderately few) gaps.
  • Ideally no or few write operations.
  • Your ID column should have been indexed! A primary key serves nicely.

The query below does not need a sequential scan of the big table, only an index scan.

First, get estimates for the main query:

SELECT count(*) AS ct              -- optional
     , min(id)  AS min_id
            , max(id)  AS max_id
            , max(id) - min(id) AS id_span
FROM   big;

The only possibly expensive part is the count(*) (for huge tables). You will get an estimate, available at almost no cost (detailed explanation here):

SELECT reltuples AS ct FROM pg_class WHERE oid = 'schema_name.big'::regclass;

As long as ct isn’t much smaller than id_span, the query will outperform most other approaches.

WITH params AS (
    SELECT 1       AS min_id           -- minimum id <= current min id
         , 5100000 AS id_span          -- rounded up. (max_id - min_id + buffer)
    SELECT p.min_id + trunc(random() * p.id_span)::integer AS id
    FROM   params p
          ,generate_series(1, 1100) g  -- 1000 + buffer
    GROUP  BY 1                        -- trim duplicates
    ) r
JOIN   big USING (id)
LIMIT  1000;                           -- trim surplus
  • Generate random numbers in the id space. You have “few gaps”, so add 10 % (enough to easily cover the blanks) to the number of rows to retrieve.
  • Each id can be picked multiple times by chance (though very unlikely with a big id space), so group the generated numbers (or use DISTINCT).
  • Join the ids to the big table. This should be very fast with the index in place.
  • Finally trim surplus ids that have not been eaten by dupes and gaps. Every row has a completely equal chance to be picked.

Short version

You can simplify this query. The CTE in the query above is just for educational purposes:

    SELECT DISTINCT 1 + trunc(random() * 5100000)::integer AS id
    FROM   generate_series(1, 1100) g
    ) r
JOIN   big USING (id)
LIMIT  1000;

Refine with rCTE

Especially if you are not so sure about gaps and estimates.

WITH RECURSIVE random_pick AS (
   FROM  (
      SELECT 1 + trunc(random() * 5100000)::int AS id
      FROM   generate_series(1, 1030)  -- 1000 + few percent - adapt to your needs
      LIMIT  1030                      -- hint for query planner
      ) r
   JOIN   big b USING (id)             -- eliminate miss

   UNION                               -- eliminate dupe
   SELECT b.*
   FROM  (
      SELECT 1 + trunc(random() * 5100000)::int AS id
      FROM   random_pick r             -- plus 3 percent - adapt to your needs
      LIMIT  999                       -- less than 1000, hint for query planner
      ) r
   JOIN   big b USING (id)             -- eliminate miss
FROM   random_pick
LIMIT  1000;  -- actual limit

We can work with a smaller surplus in the base query. If there are too many gaps so we don’t find enough rows in the first iteration, the rCTE continues to iterate with the recursive term. We still need relatively few gaps in the ID space or the recursion may run dry before the limit is reached – or we have to start with a large enough buffer which defies the purpose of optimizing performance.

Duplicates are eliminated by the UNION in the rCTE.

The outer LIMIT makes the CTE stop as soon as we have enough rows.

This query is carefully drafted to use the available index, generate actually random rows and not stop until we fulfill the limit (unless the recursion runs dry). There are a number of pitfalls here if you are going to rewrite it.

Wrap into function

For repeated use with varying parameters:

CREATE OR REPLACE FUNCTION f_random_sample(_limit int = 1000, _gaps real = 1.03)
   _surplus  int := _limit * _gaps;
   _estimate int := (           -- get current estimate from system
      SELECT c.reltuples * _gaps
      FROM   pg_class c
      WHERE  c.oid = 'big'::regclass);

   WITH RECURSIVE random_pick AS (
      SELECT *
      FROM  (
         SELECT 1 + trunc(random() * _estimate)::int
         FROM   generate_series(1, _surplus) g
         LIMIT  _surplus           -- hint for query planner
         ) r (id)
      JOIN   big USING (id)        -- eliminate misses

      UNION                        -- eliminate dupes
      SELECT *
      FROM  (
         SELECT 1 + trunc(random() * _estimate)::int
         FROM   random_pick        -- just to make it recursive
         LIMIT  _limit             -- hint for query planner
         ) r (id)
      JOIN   big USING (id)        -- eliminate misses
   FROM   random_pick
   LIMIT  _limit;
$func$  LANGUAGE plpgsql VOLATILE ROWS 1000;


SELECT * FROM f_random_sample();
SELECT * FROM f_random_sample(500, 1.05);

You could even make this generic to work for any table: Take the name of the PK column and the table as polymorphic type and use EXECUTE … But that’s beyond the scope of this post. See:

Possible alternative

IF your requirements allow identical sets for repeated calls (and we are talking about repeated calls) I would consider a materialized view. Execute above query once and write the result to a table. Users get a quasi random selection at lightening speed. Refresh your random pick at intervals or events of your choosing.

Postgres 9.5 introduces TABLESAMPLE SYSTEM (n)

It’s very fast, but the result is not exactly random. The manual:

The SYSTEM method is significantly faster than the BERNOULLI method when small sampling percentages are specified, but it may return a less-random sample of the table as a result of clustering effects.

And the number of rows returned can vary wildly. For our example, to get roughly 1000 rows, try:

SELECT * FROM big TABLESAMPLE SYSTEM ((1000 * 100) / 5100000.0);

Where n is a percentage. The manual:

The BERNOULLI and SYSTEM sampling methods each accept a single argument which is the fraction of the table to sample, expressed as a percentage between 0 and 100. This argument can be any real-valued expression.

Bold emphasis mine.



Simple way to generate a random password in PHP

When creating web apps, there’s often a need to generate a random password for your users. There are a number of ways to do this, but in needing to do it recently I came up with this very simple function that will generate a password (or other random string) of whatever length you wish. It’s particularly useful when generating passwords for users that they will then change in the future. It uses PHP’s handy str_shuffle() function:

    function random_password( $length = 8 ) {
         $chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*()_-=+;:,.?";
            $password = substr( str_shuffle( $chars.$password ), 0, $length );
         return $password;

<?php $password = random_password(8); ?>

NOTE: This method is better than the on posted in the Source. It randomizes the password generation process enough to be safe from brute force or predictability.