PRBHA-10: A hashing algorithm in PythonHashing passwords for a websiteHow easy is it to crack this encryption algorithm?PHP password encryption algorithmMy images have secrets A.K.A. the making of aesthetic passwordsMy images have secrets A.K.A. the making of aesthetic passwords V.2Python image encryption security analysis
How bad would a partial hash leak be, realistically?
What are the words for people who cause trouble believing they know better?
Do manufacturers try make their components as close to ideal ones as possible?
Traffic law UK, pedestrians
Why is c4 bad when playing the London against a King's Indian?
Avoiding cliches when writing gods
Pronoun introduced before its antecedent
Can a magnetic field of an object be stronger than its gravity?
How would you say “AKA/as in”?
Can a 2nd-level sorcerer use sorcery points to create a 2nd-level spell slot?
How do photons get into the eyes?
Completing the square to find if quadratic form is positive definite.
Building a road to escape Earth's gravity by making a pyramid on Antartica
Whats the next step after commercial fusion reactors?
What happened to all the nuclear material being smuggled after the fall of the USSR?
How is it possible that Gollum speaks Westron?
Do any instruments not produce overtones?
Company did not petition for visa in a timely manner. Is asking me to work from overseas, but wants me to take a paycut
PhD student with mental health issues and bad performance
How hard would it be to convert a glider into an powered electric aircraft?
Is it legal in the UK for politicians to lie to the public for political gain?
On the Twin Paradox Again
What's the correct term describing the action of sending a brand-new ship out into its first seafaring trip?
C SIGINT signal in Linux
PRBHA-10: A hashing algorithm in Python
Hashing passwords for a websiteHow easy is it to crack this encryption algorithm?PHP password encryption algorithmMy images have secrets A.K.A. the making of aesthetic passwordsMy images have secrets A.K.A. the making of aesthetic passwords V.2Python image encryption security analysis
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
So I've been working on a simple little hashing algorithm for Python that I like to call PRBHA-10 for short, or Pseudo-Random-Based Hash Algorithm. This is a hash algorithm in Python that is meant to be simplistic and secure. As you could see from the title, the algorithm implements a system with seeding the pseudo-random number generator to be able to produce the correct integer values when provided the correct password, therefore enabling the decryption of an encrypted byte sequence or string.
Here's the advantages of my method:
The functions are tiny and simple, unlike some hashing methods out there.
Despite the simplicity of the algorithm, it is secure due to the generation of large random numbers that scale with the password size.
There is no chance of revealing even part of the encrypted string without the full password due to the huge numbers created by the seeded PRNG and multiplication that completely messes up the string if the wrong password is entered.
It is fast. I have not done
timeit
s yet, but the functions run in a trivial amount of time to generate their results.
The disadvantages:
A somewhat large password is required to ensure security - usually anything over seven characters is completely cryptographically secure.
It is vulnerable to the same brute-force attacks that any cryptographic encryption method is vulnerable to.
With larger passwords comes greater security, but also a larger hash. The hash size slightly increases with each character added onto the password.
Here's the whole module program, prhba.py
:
"""Pseudo-Random Based Hash Algorithm (PRHBA-10) is a simple
Python cipher that uses the integer representation of a
string, along with taking advantage of the built-in
pseudo-random number generator to be able to securely encrypt
strings and bytearrays.
The algorithm uses the integer representation of byte sequences,
seeding the random number generator using the password to
create a predictable random number for later use, and a special
encoded form to turn any byte sequence into an ASCII string
using only the letters 'aceilnorst'. This is secure due to the
extremely large random number generated ensuring the security
of the result.
The result can be later decrypted using a special algorithm to
decipher the text into a large integer number. Then, the
password is used as a seed to once again get the random number
used earlier in the process. That integer is used to modify
the large integer to be able to get the integer form of the
original string, which can then be converted to a string.
This cipher currently only works in Python due to the specific
random number generation implementation in the `random` module.
"""
import math
import random
__all__ = ["encrypt", "decrypt"]
LETTERS = "aceilnorst"
def to_num(s):
return int.from_bytes(s.encode(), 'little')
def from_num(n):
return n.to_bytes(math.ceil(n.bit_length() / 8), 'little')
def encrypt(data, password):
assert len(password) > 1, "Password length cannot be less than two"
random.seed(to_num(password))
unique_id = to_num(data) * random.getrandbits(len(password))
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)
def decrypt(encrypted_data, password):
random.seed(to_num(password))
partnum_digits = []
for char in encrypted_data:
partnum_digits.append(str(LETTERS.index(char)))
partnum = int("".join(partnum_digits))
return from_num(partnum // random.getrandbits(len(password)))
Here's the questions I have:
Is it PEP-8 compliant?
Are the variable names confusing?
Is there any parts that could be improved or further reduced?
Do you have any other suggestions?
Thanks in advance!
python python-3.x random cryptography
$endgroup$
add a comment |
$begingroup$
So I've been working on a simple little hashing algorithm for Python that I like to call PRBHA-10 for short, or Pseudo-Random-Based Hash Algorithm. This is a hash algorithm in Python that is meant to be simplistic and secure. As you could see from the title, the algorithm implements a system with seeding the pseudo-random number generator to be able to produce the correct integer values when provided the correct password, therefore enabling the decryption of an encrypted byte sequence or string.
Here's the advantages of my method:
The functions are tiny and simple, unlike some hashing methods out there.
Despite the simplicity of the algorithm, it is secure due to the generation of large random numbers that scale with the password size.
There is no chance of revealing even part of the encrypted string without the full password due to the huge numbers created by the seeded PRNG and multiplication that completely messes up the string if the wrong password is entered.
It is fast. I have not done
timeit
s yet, but the functions run in a trivial amount of time to generate their results.
The disadvantages:
A somewhat large password is required to ensure security - usually anything over seven characters is completely cryptographically secure.
It is vulnerable to the same brute-force attacks that any cryptographic encryption method is vulnerable to.
With larger passwords comes greater security, but also a larger hash. The hash size slightly increases with each character added onto the password.
Here's the whole module program, prhba.py
:
"""Pseudo-Random Based Hash Algorithm (PRHBA-10) is a simple
Python cipher that uses the integer representation of a
string, along with taking advantage of the built-in
pseudo-random number generator to be able to securely encrypt
strings and bytearrays.
The algorithm uses the integer representation of byte sequences,
seeding the random number generator using the password to
create a predictable random number for later use, and a special
encoded form to turn any byte sequence into an ASCII string
using only the letters 'aceilnorst'. This is secure due to the
extremely large random number generated ensuring the security
of the result.
The result can be later decrypted using a special algorithm to
decipher the text into a large integer number. Then, the
password is used as a seed to once again get the random number
used earlier in the process. That integer is used to modify
the large integer to be able to get the integer form of the
original string, which can then be converted to a string.
This cipher currently only works in Python due to the specific
random number generation implementation in the `random` module.
"""
import math
import random
__all__ = ["encrypt", "decrypt"]
LETTERS = "aceilnorst"
def to_num(s):
return int.from_bytes(s.encode(), 'little')
def from_num(n):
return n.to_bytes(math.ceil(n.bit_length() / 8), 'little')
def encrypt(data, password):
assert len(password) > 1, "Password length cannot be less than two"
random.seed(to_num(password))
unique_id = to_num(data) * random.getrandbits(len(password))
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)
def decrypt(encrypted_data, password):
random.seed(to_num(password))
partnum_digits = []
for char in encrypted_data:
partnum_digits.append(str(LETTERS.index(char)))
partnum = int("".join(partnum_digits))
return from_num(partnum // random.getrandbits(len(password)))
Here's the questions I have:
Is it PEP-8 compliant?
Are the variable names confusing?
Is there any parts that could be improved or further reduced?
Do you have any other suggestions?
Thanks in advance!
python python-3.x random cryptography
$endgroup$
add a comment |
$begingroup$
So I've been working on a simple little hashing algorithm for Python that I like to call PRBHA-10 for short, or Pseudo-Random-Based Hash Algorithm. This is a hash algorithm in Python that is meant to be simplistic and secure. As you could see from the title, the algorithm implements a system with seeding the pseudo-random number generator to be able to produce the correct integer values when provided the correct password, therefore enabling the decryption of an encrypted byte sequence or string.
Here's the advantages of my method:
The functions are tiny and simple, unlike some hashing methods out there.
Despite the simplicity of the algorithm, it is secure due to the generation of large random numbers that scale with the password size.
There is no chance of revealing even part of the encrypted string without the full password due to the huge numbers created by the seeded PRNG and multiplication that completely messes up the string if the wrong password is entered.
It is fast. I have not done
timeit
s yet, but the functions run in a trivial amount of time to generate their results.
The disadvantages:
A somewhat large password is required to ensure security - usually anything over seven characters is completely cryptographically secure.
It is vulnerable to the same brute-force attacks that any cryptographic encryption method is vulnerable to.
With larger passwords comes greater security, but also a larger hash. The hash size slightly increases with each character added onto the password.
Here's the whole module program, prhba.py
:
"""Pseudo-Random Based Hash Algorithm (PRHBA-10) is a simple
Python cipher that uses the integer representation of a
string, along with taking advantage of the built-in
pseudo-random number generator to be able to securely encrypt
strings and bytearrays.
The algorithm uses the integer representation of byte sequences,
seeding the random number generator using the password to
create a predictable random number for later use, and a special
encoded form to turn any byte sequence into an ASCII string
using only the letters 'aceilnorst'. This is secure due to the
extremely large random number generated ensuring the security
of the result.
The result can be later decrypted using a special algorithm to
decipher the text into a large integer number. Then, the
password is used as a seed to once again get the random number
used earlier in the process. That integer is used to modify
the large integer to be able to get the integer form of the
original string, which can then be converted to a string.
This cipher currently only works in Python due to the specific
random number generation implementation in the `random` module.
"""
import math
import random
__all__ = ["encrypt", "decrypt"]
LETTERS = "aceilnorst"
def to_num(s):
return int.from_bytes(s.encode(), 'little')
def from_num(n):
return n.to_bytes(math.ceil(n.bit_length() / 8), 'little')
def encrypt(data, password):
assert len(password) > 1, "Password length cannot be less than two"
random.seed(to_num(password))
unique_id = to_num(data) * random.getrandbits(len(password))
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)
def decrypt(encrypted_data, password):
random.seed(to_num(password))
partnum_digits = []
for char in encrypted_data:
partnum_digits.append(str(LETTERS.index(char)))
partnum = int("".join(partnum_digits))
return from_num(partnum // random.getrandbits(len(password)))
Here's the questions I have:
Is it PEP-8 compliant?
Are the variable names confusing?
Is there any parts that could be improved or further reduced?
Do you have any other suggestions?
Thanks in advance!
python python-3.x random cryptography
$endgroup$
So I've been working on a simple little hashing algorithm for Python that I like to call PRBHA-10 for short, or Pseudo-Random-Based Hash Algorithm. This is a hash algorithm in Python that is meant to be simplistic and secure. As you could see from the title, the algorithm implements a system with seeding the pseudo-random number generator to be able to produce the correct integer values when provided the correct password, therefore enabling the decryption of an encrypted byte sequence or string.
Here's the advantages of my method:
The functions are tiny and simple, unlike some hashing methods out there.
Despite the simplicity of the algorithm, it is secure due to the generation of large random numbers that scale with the password size.
There is no chance of revealing even part of the encrypted string without the full password due to the huge numbers created by the seeded PRNG and multiplication that completely messes up the string if the wrong password is entered.
It is fast. I have not done
timeit
s yet, but the functions run in a trivial amount of time to generate their results.
The disadvantages:
A somewhat large password is required to ensure security - usually anything over seven characters is completely cryptographically secure.
It is vulnerable to the same brute-force attacks that any cryptographic encryption method is vulnerable to.
With larger passwords comes greater security, but also a larger hash. The hash size slightly increases with each character added onto the password.
Here's the whole module program, prhba.py
:
"""Pseudo-Random Based Hash Algorithm (PRHBA-10) is a simple
Python cipher that uses the integer representation of a
string, along with taking advantage of the built-in
pseudo-random number generator to be able to securely encrypt
strings and bytearrays.
The algorithm uses the integer representation of byte sequences,
seeding the random number generator using the password to
create a predictable random number for later use, and a special
encoded form to turn any byte sequence into an ASCII string
using only the letters 'aceilnorst'. This is secure due to the
extremely large random number generated ensuring the security
of the result.
The result can be later decrypted using a special algorithm to
decipher the text into a large integer number. Then, the
password is used as a seed to once again get the random number
used earlier in the process. That integer is used to modify
the large integer to be able to get the integer form of the
original string, which can then be converted to a string.
This cipher currently only works in Python due to the specific
random number generation implementation in the `random` module.
"""
import math
import random
__all__ = ["encrypt", "decrypt"]
LETTERS = "aceilnorst"
def to_num(s):
return int.from_bytes(s.encode(), 'little')
def from_num(n):
return n.to_bytes(math.ceil(n.bit_length() / 8), 'little')
def encrypt(data, password):
assert len(password) > 1, "Password length cannot be less than two"
random.seed(to_num(password))
unique_id = to_num(data) * random.getrandbits(len(password))
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)
def decrypt(encrypted_data, password):
random.seed(to_num(password))
partnum_digits = []
for char in encrypted_data:
partnum_digits.append(str(LETTERS.index(char)))
partnum = int("".join(partnum_digits))
return from_num(partnum // random.getrandbits(len(password)))
Here's the questions I have:
Is it PEP-8 compliant?
Are the variable names confusing?
Is there any parts that could be improved or further reduced?
Do you have any other suggestions?
Thanks in advance!
python python-3.x random cryptography
python python-3.x random cryptography
asked 8 hours ago
connectyourchargerconnectyourcharger
1908
1908
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Don’t roll your own encryption. It takes a team of experts to develop new secure encryption methods, and even they can get it wrong from time to time.
Huge hole in your DIY encryption:
If I use a 2-character password, I might naïvely expect I’d have $62^2$ possible passwords that I can encrypt the data with. I’d be really shocked when it turns out there are only 4.
random.getrandbits(len(password))
generates 2 random bits, for $2^2$ possible values to multiply to_num(data)
by. Only 4 possibilities is a lot easier to attack than $62^2$ different possibilities!
And one of those possibilities ... all bits zero ... destroys all the data you want to encode. So we’re down to actually only 3 possible numbers to test to reverse the encryption.
Any encryption mechanism worth its salt uses a salt (initial random data) to prevent the same message with the same password from being encrypted as the same text.
Code improvements: use generator expressions. Eg)
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)
would be more efficient rewritten as:
return "".join(LETTERS[int(digit)] for digit in str(unique_id))
The chunk
list is never created in memory; instead, the join
method pulls items one at a time from the generator expression.
Finally, don’t roll your own encryption.
$endgroup$
$begingroup$
Thanks for the suggestion. I agree, it is lacking in the department of small passwords. That's why I said it needs sufficiently large passwords in the post. This was just something I was playing with - I don't actually expect it to compete with any standard cryptographic methods.
$endgroup$
– connectyourcharger
6 hours ago
$begingroup$
You missed the point. A two character pass phrase creates a two bit encryption key; an 10 character pass phrase creates a 10 bit encryption key!!!
$endgroup$
– AJNeufeld
4 hours ago
$begingroup$
No, yes, I understand that. It's a limitation I've been trying to get past.
$endgroup$
– connectyourcharger
4 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "196"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f221499%2fprbha-10-a-hashing-algorithm-in-python%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Don’t roll your own encryption. It takes a team of experts to develop new secure encryption methods, and even they can get it wrong from time to time.
Huge hole in your DIY encryption:
If I use a 2-character password, I might naïvely expect I’d have $62^2$ possible passwords that I can encrypt the data with. I’d be really shocked when it turns out there are only 4.
random.getrandbits(len(password))
generates 2 random bits, for $2^2$ possible values to multiply to_num(data)
by. Only 4 possibilities is a lot easier to attack than $62^2$ different possibilities!
And one of those possibilities ... all bits zero ... destroys all the data you want to encode. So we’re down to actually only 3 possible numbers to test to reverse the encryption.
Any encryption mechanism worth its salt uses a salt (initial random data) to prevent the same message with the same password from being encrypted as the same text.
Code improvements: use generator expressions. Eg)
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)
would be more efficient rewritten as:
return "".join(LETTERS[int(digit)] for digit in str(unique_id))
The chunk
list is never created in memory; instead, the join
method pulls items one at a time from the generator expression.
Finally, don’t roll your own encryption.
$endgroup$
$begingroup$
Thanks for the suggestion. I agree, it is lacking in the department of small passwords. That's why I said it needs sufficiently large passwords in the post. This was just something I was playing with - I don't actually expect it to compete with any standard cryptographic methods.
$endgroup$
– connectyourcharger
6 hours ago
$begingroup$
You missed the point. A two character pass phrase creates a two bit encryption key; an 10 character pass phrase creates a 10 bit encryption key!!!
$endgroup$
– AJNeufeld
4 hours ago
$begingroup$
No, yes, I understand that. It's a limitation I've been trying to get past.
$endgroup$
– connectyourcharger
4 hours ago
add a comment |
$begingroup$
Don’t roll your own encryption. It takes a team of experts to develop new secure encryption methods, and even they can get it wrong from time to time.
Huge hole in your DIY encryption:
If I use a 2-character password, I might naïvely expect I’d have $62^2$ possible passwords that I can encrypt the data with. I’d be really shocked when it turns out there are only 4.
random.getrandbits(len(password))
generates 2 random bits, for $2^2$ possible values to multiply to_num(data)
by. Only 4 possibilities is a lot easier to attack than $62^2$ different possibilities!
And one of those possibilities ... all bits zero ... destroys all the data you want to encode. So we’re down to actually only 3 possible numbers to test to reverse the encryption.
Any encryption mechanism worth its salt uses a salt (initial random data) to prevent the same message with the same password from being encrypted as the same text.
Code improvements: use generator expressions. Eg)
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)
would be more efficient rewritten as:
return "".join(LETTERS[int(digit)] for digit in str(unique_id))
The chunk
list is never created in memory; instead, the join
method pulls items one at a time from the generator expression.
Finally, don’t roll your own encryption.
$endgroup$
$begingroup$
Thanks for the suggestion. I agree, it is lacking in the department of small passwords. That's why I said it needs sufficiently large passwords in the post. This was just something I was playing with - I don't actually expect it to compete with any standard cryptographic methods.
$endgroup$
– connectyourcharger
6 hours ago
$begingroup$
You missed the point. A two character pass phrase creates a two bit encryption key; an 10 character pass phrase creates a 10 bit encryption key!!!
$endgroup$
– AJNeufeld
4 hours ago
$begingroup$
No, yes, I understand that. It's a limitation I've been trying to get past.
$endgroup$
– connectyourcharger
4 hours ago
add a comment |
$begingroup$
Don’t roll your own encryption. It takes a team of experts to develop new secure encryption methods, and even they can get it wrong from time to time.
Huge hole in your DIY encryption:
If I use a 2-character password, I might naïvely expect I’d have $62^2$ possible passwords that I can encrypt the data with. I’d be really shocked when it turns out there are only 4.
random.getrandbits(len(password))
generates 2 random bits, for $2^2$ possible values to multiply to_num(data)
by. Only 4 possibilities is a lot easier to attack than $62^2$ different possibilities!
And one of those possibilities ... all bits zero ... destroys all the data you want to encode. So we’re down to actually only 3 possible numbers to test to reverse the encryption.
Any encryption mechanism worth its salt uses a salt (initial random data) to prevent the same message with the same password from being encrypted as the same text.
Code improvements: use generator expressions. Eg)
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)
would be more efficient rewritten as:
return "".join(LETTERS[int(digit)] for digit in str(unique_id))
The chunk
list is never created in memory; instead, the join
method pulls items one at a time from the generator expression.
Finally, don’t roll your own encryption.
$endgroup$
Don’t roll your own encryption. It takes a team of experts to develop new secure encryption methods, and even they can get it wrong from time to time.
Huge hole in your DIY encryption:
If I use a 2-character password, I might naïvely expect I’d have $62^2$ possible passwords that I can encrypt the data with. I’d be really shocked when it turns out there are only 4.
random.getrandbits(len(password))
generates 2 random bits, for $2^2$ possible values to multiply to_num(data)
by. Only 4 possibilities is a lot easier to attack than $62^2$ different possibilities!
And one of those possibilities ... all bits zero ... destroys all the data you want to encode. So we’re down to actually only 3 possible numbers to test to reverse the encryption.
Any encryption mechanism worth its salt uses a salt (initial random data) to prevent the same message with the same password from being encrypted as the same text.
Code improvements: use generator expressions. Eg)
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)
would be more efficient rewritten as:
return "".join(LETTERS[int(digit)] for digit in str(unique_id))
The chunk
list is never created in memory; instead, the join
method pulls items one at a time from the generator expression.
Finally, don’t roll your own encryption.
answered 6 hours ago
AJNeufeldAJNeufeld
7,8871725
7,8871725
$begingroup$
Thanks for the suggestion. I agree, it is lacking in the department of small passwords. That's why I said it needs sufficiently large passwords in the post. This was just something I was playing with - I don't actually expect it to compete with any standard cryptographic methods.
$endgroup$
– connectyourcharger
6 hours ago
$begingroup$
You missed the point. A two character pass phrase creates a two bit encryption key; an 10 character pass phrase creates a 10 bit encryption key!!!
$endgroup$
– AJNeufeld
4 hours ago
$begingroup$
No, yes, I understand that. It's a limitation I've been trying to get past.
$endgroup$
– connectyourcharger
4 hours ago
add a comment |
$begingroup$
Thanks for the suggestion. I agree, it is lacking in the department of small passwords. That's why I said it needs sufficiently large passwords in the post. This was just something I was playing with - I don't actually expect it to compete with any standard cryptographic methods.
$endgroup$
– connectyourcharger
6 hours ago
$begingroup$
You missed the point. A two character pass phrase creates a two bit encryption key; an 10 character pass phrase creates a 10 bit encryption key!!!
$endgroup$
– AJNeufeld
4 hours ago
$begingroup$
No, yes, I understand that. It's a limitation I've been trying to get past.
$endgroup$
– connectyourcharger
4 hours ago
$begingroup$
Thanks for the suggestion. I agree, it is lacking in the department of small passwords. That's why I said it needs sufficiently large passwords in the post. This was just something I was playing with - I don't actually expect it to compete with any standard cryptographic methods.
$endgroup$
– connectyourcharger
6 hours ago
$begingroup$
Thanks for the suggestion. I agree, it is lacking in the department of small passwords. That's why I said it needs sufficiently large passwords in the post. This was just something I was playing with - I don't actually expect it to compete with any standard cryptographic methods.
$endgroup$
– connectyourcharger
6 hours ago
$begingroup$
You missed the point. A two character pass phrase creates a two bit encryption key; an 10 character pass phrase creates a 10 bit encryption key!!!
$endgroup$
– AJNeufeld
4 hours ago
$begingroup$
You missed the point. A two character pass phrase creates a two bit encryption key; an 10 character pass phrase creates a 10 bit encryption key!!!
$endgroup$
– AJNeufeld
4 hours ago
$begingroup$
No, yes, I understand that. It's a limitation I've been trying to get past.
$endgroup$
– connectyourcharger
4 hours ago
$begingroup$
No, yes, I understand that. It's a limitation I've been trying to get past.
$endgroup$
– connectyourcharger
4 hours ago
add a comment |
Thanks for contributing an answer to Code Review Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f221499%2fprbha-10-a-hashing-algorithm-in-python%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown