Vigenère cipher in Ruby Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Encrypter - Double Vigenere Cipher in PythonVigenère cipher in CVigenère cipher 2Vigenère Cipher in HaskellSimple Caesar cipher cryptanalysis in RubyVigenere cipher with random saltVigenère cipher in PythonBrute-force Vigenere Cipher using multiple threadsVigenere Cipher in CCracking Vigenere and Caesar Ciphered Text in Python

Suing a Police Officer Instead of the Police Department

Seek and ye shall find

Would reducing the reference voltage of an ADC have any effect on accuracy?

Is accepting an invalid credit card number a security issue?

std::is_constructible on incomplete types

What is the term for a person whose job is to place products on shelves in stores?

Justification for leaving new position after a short time

What is the best way to deal with NPC-NPC combat?

Implementing 3DES algorithm in Java: is my code secure?

Why did C use the -> operator instead of reusing the . operator?

What is /etc/mtab in Linux?

Is it acceptable to use working hours to read general interest books?

What ability score does a Hexblade's Pact Weapon use for attack and damage when wielded by another character?

Is there any hidden 'W' sound after 'comment' in : Comment est-elle?

What was Apollo 13's "Little Jolt" after MECO?

Is it OK if I do not take the receipt in Germany?

What's the difference between using dependency injection with a container and using a service locator?

How to not starve gigantic beasts

Passing args from the bash script to the function in the script

Prove the alternating sum of a decreasing sequence converging to 0 is Cauchy.

How can I wire a 9-position switch so that each position turns on one more LED than the one before?

How to open locks without disable device?

Align column where each cell has two decimals with siunitx

My bank got bought out, am I now going to have to start filing tax returns in a different state?



Vigenère cipher in Ruby



Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Encrypter - Double Vigenere Cipher in PythonVigenère cipher in CVigenère cipher 2Vigenère Cipher in HaskellSimple Caesar cipher cryptanalysis in RubyVigenere cipher with random saltVigenère cipher in PythonBrute-force Vigenere Cipher using multiple threadsVigenere Cipher in CCracking Vigenere and Caesar Ciphered Text in Python



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








3












$begingroup$


What I'm trying to do: implement the Vigenère cipher in Ruby. I already have a working version, but I want to make sure it is efficient and well-designed.



module Crypto
# Vigenère cipher encryption and decryption abstraction
module Vigenere
LETTERS = ('a'..'z').to_a.freeze
private_constant :LETTERS

module_function

# Encrypts a string
#
# @param string [String] the string that will be encrypted
# @param key [String] the key that will be used to encrypt the string
#
# @return [String] the encrypted string
def encrypt(string:, key:)
key = make_key(length: string.length, key: key)

string.length.times.map .join
end

# Decrypts an encrypted string
#
# @param string [String] the encrypted string that will be decrypted
# @param key [String] the key that will be used to decrypt the string
#
# @return [String] the decrypted string
def decrypt(string:, key:)
key = make_key(length: string.length, key: key)

string.length.times.map .join
end

# Repeats a word until it matches a certain length
#
# @param length [Integer] the length of the word being encrypted/decrypted
# @param key [String] the word that will be repeated
#
# @return [String] the word in its new form
def make_key(length:, key:)
i = 0
length.times do
i = 0 if i == key.length
break if key.length == length

key << key[i]
i += 1
end

key
end

private_class_method :make_key
end
end


I do have some specific questions:



1. private_class_method



This is the best way I found to define a private method in a module, but it feels weird to me. Isn't there a better way to do that? My first implementation was this:



module Crypto
class Vigenere
class << self
def encrypt # ...
def decrypt # ...

private

def make_key # ...
end
end
end


which was fine for me. But then I read this rule on the Ruby Style Guide repository. So I switched to using module, but it doesn't feel right to use private methods in this structure. Am I wrong?



2. reseting a counter (index)



Take a look at this snippet of code:



def make_key(length:, key:)
i = 0
length.times do
i = 0 if i == key.length
break if key.length == length

key << key[i]
i += 1
end

key
end


Defining a counter (i) and manually incrementing it... looks awkward, doesn't it (at least in the Ruby world)? Is there a better way to do this?



3. valid multi-line block with curly braces?



Now take a look at this snippet:



string.length.times.map i.join


I know most Ruby developers tend to use curly braces only for one-line blocks and do-end for multi-line blocks, but this time it seems okay using curly braces with a multi-line block, since I'm chaining #join right after. What would you do:



1. use do-end, store it in a variable and invoke #join after that



new_letters = string.length.times.map do |i|
p = LETTERS.find_index(string[i])
k = LETTERS.find_index(key[i])

LETTERS[(p + k) % 26]
end

new_letters.join


2. what I did (use curly braces even with multi-line block and chain #join)



And of course, if you have any other observations, please share.










share|improve this question







New contributor




sPaGhEtTiCaSe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 1




    $begingroup$
    If you take the time to spell Vigenère correctly then that's enough for me to tell that the code is going to be all right :)
    $endgroup$
    – Maarten Bodewes
    4 hours ago


















3












$begingroup$


What I'm trying to do: implement the Vigenère cipher in Ruby. I already have a working version, but I want to make sure it is efficient and well-designed.



module Crypto
# Vigenère cipher encryption and decryption abstraction
module Vigenere
LETTERS = ('a'..'z').to_a.freeze
private_constant :LETTERS

module_function

# Encrypts a string
#
# @param string [String] the string that will be encrypted
# @param key [String] the key that will be used to encrypt the string
#
# @return [String] the encrypted string
def encrypt(string:, key:)
key = make_key(length: string.length, key: key)

string.length.times.map .join
end

# Decrypts an encrypted string
#
# @param string [String] the encrypted string that will be decrypted
# @param key [String] the key that will be used to decrypt the string
#
# @return [String] the decrypted string
def decrypt(string:, key:)
key = make_key(length: string.length, key: key)

string.length.times.map .join
end

# Repeats a word until it matches a certain length
#
# @param length [Integer] the length of the word being encrypted/decrypted
# @param key [String] the word that will be repeated
#
# @return [String] the word in its new form
def make_key(length:, key:)
i = 0
length.times do
i = 0 if i == key.length
break if key.length == length

key << key[i]
i += 1
end

key
end

private_class_method :make_key
end
end


I do have some specific questions:



1. private_class_method



This is the best way I found to define a private method in a module, but it feels weird to me. Isn't there a better way to do that? My first implementation was this:



module Crypto
class Vigenere
class << self
def encrypt # ...
def decrypt # ...

private

def make_key # ...
end
end
end


which was fine for me. But then I read this rule on the Ruby Style Guide repository. So I switched to using module, but it doesn't feel right to use private methods in this structure. Am I wrong?



2. reseting a counter (index)



Take a look at this snippet of code:



def make_key(length:, key:)
i = 0
length.times do
i = 0 if i == key.length
break if key.length == length

key << key[i]
i += 1
end

key
end


Defining a counter (i) and manually incrementing it... looks awkward, doesn't it (at least in the Ruby world)? Is there a better way to do this?



3. valid multi-line block with curly braces?



Now take a look at this snippet:



string.length.times.map i.join


I know most Ruby developers tend to use curly braces only for one-line blocks and do-end for multi-line blocks, but this time it seems okay using curly braces with a multi-line block, since I'm chaining #join right after. What would you do:



1. use do-end, store it in a variable and invoke #join after that



new_letters = string.length.times.map do |i|
p = LETTERS.find_index(string[i])
k = LETTERS.find_index(key[i])

LETTERS[(p + k) % 26]
end

new_letters.join


2. what I did (use curly braces even with multi-line block and chain #join)



And of course, if you have any other observations, please share.










share|improve this question







New contributor




sPaGhEtTiCaSe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 1




    $begingroup$
    If you take the time to spell Vigenère correctly then that's enough for me to tell that the code is going to be all right :)
    $endgroup$
    – Maarten Bodewes
    4 hours ago














3












3








3





$begingroup$


What I'm trying to do: implement the Vigenère cipher in Ruby. I already have a working version, but I want to make sure it is efficient and well-designed.



module Crypto
# Vigenère cipher encryption and decryption abstraction
module Vigenere
LETTERS = ('a'..'z').to_a.freeze
private_constant :LETTERS

module_function

# Encrypts a string
#
# @param string [String] the string that will be encrypted
# @param key [String] the key that will be used to encrypt the string
#
# @return [String] the encrypted string
def encrypt(string:, key:)
key = make_key(length: string.length, key: key)

string.length.times.map .join
end

# Decrypts an encrypted string
#
# @param string [String] the encrypted string that will be decrypted
# @param key [String] the key that will be used to decrypt the string
#
# @return [String] the decrypted string
def decrypt(string:, key:)
key = make_key(length: string.length, key: key)

string.length.times.map .join
end

# Repeats a word until it matches a certain length
#
# @param length [Integer] the length of the word being encrypted/decrypted
# @param key [String] the word that will be repeated
#
# @return [String] the word in its new form
def make_key(length:, key:)
i = 0
length.times do
i = 0 if i == key.length
break if key.length == length

key << key[i]
i += 1
end

key
end

private_class_method :make_key
end
end


I do have some specific questions:



1. private_class_method



This is the best way I found to define a private method in a module, but it feels weird to me. Isn't there a better way to do that? My first implementation was this:



module Crypto
class Vigenere
class << self
def encrypt # ...
def decrypt # ...

private

def make_key # ...
end
end
end


which was fine for me. But then I read this rule on the Ruby Style Guide repository. So I switched to using module, but it doesn't feel right to use private methods in this structure. Am I wrong?



2. reseting a counter (index)



Take a look at this snippet of code:



def make_key(length:, key:)
i = 0
length.times do
i = 0 if i == key.length
break if key.length == length

key << key[i]
i += 1
end

key
end


Defining a counter (i) and manually incrementing it... looks awkward, doesn't it (at least in the Ruby world)? Is there a better way to do this?



3. valid multi-line block with curly braces?



Now take a look at this snippet:



string.length.times.map i.join


I know most Ruby developers tend to use curly braces only for one-line blocks and do-end for multi-line blocks, but this time it seems okay using curly braces with a multi-line block, since I'm chaining #join right after. What would you do:



1. use do-end, store it in a variable and invoke #join after that



new_letters = string.length.times.map do |i|
p = LETTERS.find_index(string[i])
k = LETTERS.find_index(key[i])

LETTERS[(p + k) % 26]
end

new_letters.join


2. what I did (use curly braces even with multi-line block and chain #join)



And of course, if you have any other observations, please share.










share|improve this question







New contributor




sPaGhEtTiCaSe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




What I'm trying to do: implement the Vigenère cipher in Ruby. I already have a working version, but I want to make sure it is efficient and well-designed.



module Crypto
# Vigenère cipher encryption and decryption abstraction
module Vigenere
LETTERS = ('a'..'z').to_a.freeze
private_constant :LETTERS

module_function

# Encrypts a string
#
# @param string [String] the string that will be encrypted
# @param key [String] the key that will be used to encrypt the string
#
# @return [String] the encrypted string
def encrypt(string:, key:)
key = make_key(length: string.length, key: key)

string.length.times.map .join
end

# Decrypts an encrypted string
#
# @param string [String] the encrypted string that will be decrypted
# @param key [String] the key that will be used to decrypt the string
#
# @return [String] the decrypted string
def decrypt(string:, key:)
key = make_key(length: string.length, key: key)

string.length.times.map .join
end

# Repeats a word until it matches a certain length
#
# @param length [Integer] the length of the word being encrypted/decrypted
# @param key [String] the word that will be repeated
#
# @return [String] the word in its new form
def make_key(length:, key:)
i = 0
length.times do
i = 0 if i == key.length
break if key.length == length

key << key[i]
i += 1
end

key
end

private_class_method :make_key
end
end


I do have some specific questions:



1. private_class_method



This is the best way I found to define a private method in a module, but it feels weird to me. Isn't there a better way to do that? My first implementation was this:



module Crypto
class Vigenere
class << self
def encrypt # ...
def decrypt # ...

private

def make_key # ...
end
end
end


which was fine for me. But then I read this rule on the Ruby Style Guide repository. So I switched to using module, but it doesn't feel right to use private methods in this structure. Am I wrong?



2. reseting a counter (index)



Take a look at this snippet of code:



def make_key(length:, key:)
i = 0
length.times do
i = 0 if i == key.length
break if key.length == length

key << key[i]
i += 1
end

key
end


Defining a counter (i) and manually incrementing it... looks awkward, doesn't it (at least in the Ruby world)? Is there a better way to do this?



3. valid multi-line block with curly braces?



Now take a look at this snippet:



string.length.times.map i.join


I know most Ruby developers tend to use curly braces only for one-line blocks and do-end for multi-line blocks, but this time it seems okay using curly braces with a multi-line block, since I'm chaining #join right after. What would you do:



1. use do-end, store it in a variable and invoke #join after that



new_letters = string.length.times.map do |i|
p = LETTERS.find_index(string[i])
k = LETTERS.find_index(key[i])

LETTERS[(p + k) % 26]
end

new_letters.join


2. what I did (use curly braces even with multi-line block and chain #join)



And of course, if you have any other observations, please share.







ruby vigenere-cipher






share|improve this question







New contributor




sPaGhEtTiCaSe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




sPaGhEtTiCaSe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




sPaGhEtTiCaSe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 5 hours ago









sPaGhEtTiCaSesPaGhEtTiCaSe

183




183




New contributor




sPaGhEtTiCaSe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





sPaGhEtTiCaSe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






sPaGhEtTiCaSe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 1




    $begingroup$
    If you take the time to spell Vigenère correctly then that's enough for me to tell that the code is going to be all right :)
    $endgroup$
    – Maarten Bodewes
    4 hours ago













  • 1




    $begingroup$
    If you take the time to spell Vigenère correctly then that's enough for me to tell that the code is going to be all right :)
    $endgroup$
    – Maarten Bodewes
    4 hours ago








1




1




$begingroup$
If you take the time to spell Vigenère correctly then that's enough for me to tell that the code is going to be all right :)
$endgroup$
– Maarten Bodewes
4 hours ago





$begingroup$
If you take the time to spell Vigenère correctly then that's enough for me to tell that the code is going to be all right :)
$endgroup$
– Maarten Bodewes
4 hours ago











2 Answers
2






active

oldest

votes


















1












$begingroup$

The code is rather clear, so consider the following to be nitpicks.




I'd say that you are generating a new key stream from the key. I'd certainly not reuse the key variable.




The first i = 0 before the loop seems spurious.



Using i as a counter is well understood, and I'd not worry overly much on the style of it. You are probably the only one who cares if it is really Ruby-esk; developers down the line will understand it.



What I wonder though is that you run your loop length times, but there is a break that seems to trigger before that. That's not all too clear to me.



I wonder what happens if you supply it an "empty" key string. Some guard statements may be in order.




Same for the curly braces. It's clear as it is, choose whatever you want. Personally I slightly favor the braces.




You could consider creating a mod function, however since % is already the modulus, which will never return a negative value if the right value is positive, it seems to me that removing the + 26 is probably the only thing you need to change (during decryption).



Instead of using 26 as unexplained magic value, you should get the size of the LETTERS range instead. That way you can also expand your ciphertext later.




I've got no opinion on the private_class_method as I'm not a Ruby developer (I'm specialized in knowing many languages / constructs and of course applied crypto).






share|improve this answer











$endgroup$












  • $begingroup$
    Thanks, that's really helpful! Sadly, I don't have enough reputation to upvote your answer. :(
    $endgroup$
    – sPaGhEtTiCaSe
    2 hours ago


















1












$begingroup$

Looping using string.length.times.map … and length.times do … is OK, but slightly on the awkward side. I recommend adhering to the convention of writing same-line blocks using and line-spanning blocks using do … end.



To extend the key, you can use the string multiplication operator. (Note that extending the key longer than necessary doesn't do much harm.)



You can also factor out more of the commonality between the encrypt and decrypt functions.



Instead of searching the LETTERS array, I recommend performing arithmetic on ASCII codes.



module Crypto
module Vigenere
module_function
def encrypt(plaintext, key)
vigenere(plaintext, key) (p + k) % 26
end

def decrypt(ciphertext, key)
vigenere(ciphertext, key) c, k
end

# Implementation of Vigenere cipher. The combiner block accepts
# one character from the text and the corresponding character from
# the key (encoded as a=0, b=1, ..., z=25), and returns the
# result using the same numerical scheme.
def vigenere(text, key, &combiner)
a = 'a'.ord
ext_key = key * (text.length / key.length + 1)
text.chars.zip(ext_key.chars).collect do |t, k|
(a + combiner.call(t.ord - a, k.ord - a)).chr
end.join
end
private_class_method :vigenere
end
end





share|improve this answer











$endgroup$













    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
    );



    );






    sPaGhEtTiCaSe is a new contributor. Be nice, and check out our Code of Conduct.









    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f219053%2fvigen%25c3%25a8re-cipher-in-ruby%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The code is rather clear, so consider the following to be nitpicks.




    I'd say that you are generating a new key stream from the key. I'd certainly not reuse the key variable.




    The first i = 0 before the loop seems spurious.



    Using i as a counter is well understood, and I'd not worry overly much on the style of it. You are probably the only one who cares if it is really Ruby-esk; developers down the line will understand it.



    What I wonder though is that you run your loop length times, but there is a break that seems to trigger before that. That's not all too clear to me.



    I wonder what happens if you supply it an "empty" key string. Some guard statements may be in order.




    Same for the curly braces. It's clear as it is, choose whatever you want. Personally I slightly favor the braces.




    You could consider creating a mod function, however since % is already the modulus, which will never return a negative value if the right value is positive, it seems to me that removing the + 26 is probably the only thing you need to change (during decryption).



    Instead of using 26 as unexplained magic value, you should get the size of the LETTERS range instead. That way you can also expand your ciphertext later.




    I've got no opinion on the private_class_method as I'm not a Ruby developer (I'm specialized in knowing many languages / constructs and of course applied crypto).






    share|improve this answer











    $endgroup$












    • $begingroup$
      Thanks, that's really helpful! Sadly, I don't have enough reputation to upvote your answer. :(
      $endgroup$
      – sPaGhEtTiCaSe
      2 hours ago















    1












    $begingroup$

    The code is rather clear, so consider the following to be nitpicks.




    I'd say that you are generating a new key stream from the key. I'd certainly not reuse the key variable.




    The first i = 0 before the loop seems spurious.



    Using i as a counter is well understood, and I'd not worry overly much on the style of it. You are probably the only one who cares if it is really Ruby-esk; developers down the line will understand it.



    What I wonder though is that you run your loop length times, but there is a break that seems to trigger before that. That's not all too clear to me.



    I wonder what happens if you supply it an "empty" key string. Some guard statements may be in order.




    Same for the curly braces. It's clear as it is, choose whatever you want. Personally I slightly favor the braces.




    You could consider creating a mod function, however since % is already the modulus, which will never return a negative value if the right value is positive, it seems to me that removing the + 26 is probably the only thing you need to change (during decryption).



    Instead of using 26 as unexplained magic value, you should get the size of the LETTERS range instead. That way you can also expand your ciphertext later.




    I've got no opinion on the private_class_method as I'm not a Ruby developer (I'm specialized in knowing many languages / constructs and of course applied crypto).






    share|improve this answer











    $endgroup$












    • $begingroup$
      Thanks, that's really helpful! Sadly, I don't have enough reputation to upvote your answer. :(
      $endgroup$
      – sPaGhEtTiCaSe
      2 hours ago













    1












    1








    1





    $begingroup$

    The code is rather clear, so consider the following to be nitpicks.




    I'd say that you are generating a new key stream from the key. I'd certainly not reuse the key variable.




    The first i = 0 before the loop seems spurious.



    Using i as a counter is well understood, and I'd not worry overly much on the style of it. You are probably the only one who cares if it is really Ruby-esk; developers down the line will understand it.



    What I wonder though is that you run your loop length times, but there is a break that seems to trigger before that. That's not all too clear to me.



    I wonder what happens if you supply it an "empty" key string. Some guard statements may be in order.




    Same for the curly braces. It's clear as it is, choose whatever you want. Personally I slightly favor the braces.




    You could consider creating a mod function, however since % is already the modulus, which will never return a negative value if the right value is positive, it seems to me that removing the + 26 is probably the only thing you need to change (during decryption).



    Instead of using 26 as unexplained magic value, you should get the size of the LETTERS range instead. That way you can also expand your ciphertext later.




    I've got no opinion on the private_class_method as I'm not a Ruby developer (I'm specialized in knowing many languages / constructs and of course applied crypto).






    share|improve this answer











    $endgroup$



    The code is rather clear, so consider the following to be nitpicks.




    I'd say that you are generating a new key stream from the key. I'd certainly not reuse the key variable.




    The first i = 0 before the loop seems spurious.



    Using i as a counter is well understood, and I'd not worry overly much on the style of it. You are probably the only one who cares if it is really Ruby-esk; developers down the line will understand it.



    What I wonder though is that you run your loop length times, but there is a break that seems to trigger before that. That's not all too clear to me.



    I wonder what happens if you supply it an "empty" key string. Some guard statements may be in order.




    Same for the curly braces. It's clear as it is, choose whatever you want. Personally I slightly favor the braces.




    You could consider creating a mod function, however since % is already the modulus, which will never return a negative value if the right value is positive, it seems to me that removing the + 26 is probably the only thing you need to change (during decryption).



    Instead of using 26 as unexplained magic value, you should get the size of the LETTERS range instead. That way you can also expand your ciphertext later.




    I've got no opinion on the private_class_method as I'm not a Ruby developer (I'm specialized in knowing many languages / constructs and of course applied crypto).







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 4 hours ago

























    answered 4 hours ago









    Maarten BodewesMaarten Bodewes

    582212




    582212











    • $begingroup$
      Thanks, that's really helpful! Sadly, I don't have enough reputation to upvote your answer. :(
      $endgroup$
      – sPaGhEtTiCaSe
      2 hours ago
















    • $begingroup$
      Thanks, that's really helpful! Sadly, I don't have enough reputation to upvote your answer. :(
      $endgroup$
      – sPaGhEtTiCaSe
      2 hours ago















    $begingroup$
    Thanks, that's really helpful! Sadly, I don't have enough reputation to upvote your answer. :(
    $endgroup$
    – sPaGhEtTiCaSe
    2 hours ago




    $begingroup$
    Thanks, that's really helpful! Sadly, I don't have enough reputation to upvote your answer. :(
    $endgroup$
    – sPaGhEtTiCaSe
    2 hours ago













    1












    $begingroup$

    Looping using string.length.times.map … and length.times do … is OK, but slightly on the awkward side. I recommend adhering to the convention of writing same-line blocks using and line-spanning blocks using do … end.



    To extend the key, you can use the string multiplication operator. (Note that extending the key longer than necessary doesn't do much harm.)



    You can also factor out more of the commonality between the encrypt and decrypt functions.



    Instead of searching the LETTERS array, I recommend performing arithmetic on ASCII codes.



    module Crypto
    module Vigenere
    module_function
    def encrypt(plaintext, key)
    vigenere(plaintext, key) (p + k) % 26
    end

    def decrypt(ciphertext, key)
    vigenere(ciphertext, key) c, k
    end

    # Implementation of Vigenere cipher. The combiner block accepts
    # one character from the text and the corresponding character from
    # the key (encoded as a=0, b=1, ..., z=25), and returns the
    # result using the same numerical scheme.
    def vigenere(text, key, &combiner)
    a = 'a'.ord
    ext_key = key * (text.length / key.length + 1)
    text.chars.zip(ext_key.chars).collect do |t, k|
    (a + combiner.call(t.ord - a, k.ord - a)).chr
    end.join
    end
    private_class_method :vigenere
    end
    end





    share|improve this answer











    $endgroup$

















      1












      $begingroup$

      Looping using string.length.times.map … and length.times do … is OK, but slightly on the awkward side. I recommend adhering to the convention of writing same-line blocks using and line-spanning blocks using do … end.



      To extend the key, you can use the string multiplication operator. (Note that extending the key longer than necessary doesn't do much harm.)



      You can also factor out more of the commonality between the encrypt and decrypt functions.



      Instead of searching the LETTERS array, I recommend performing arithmetic on ASCII codes.



      module Crypto
      module Vigenere
      module_function
      def encrypt(plaintext, key)
      vigenere(plaintext, key) (p + k) % 26
      end

      def decrypt(ciphertext, key)
      vigenere(ciphertext, key) c, k
      end

      # Implementation of Vigenere cipher. The combiner block accepts
      # one character from the text and the corresponding character from
      # the key (encoded as a=0, b=1, ..., z=25), and returns the
      # result using the same numerical scheme.
      def vigenere(text, key, &combiner)
      a = 'a'.ord
      ext_key = key * (text.length / key.length + 1)
      text.chars.zip(ext_key.chars).collect do |t, k|
      (a + combiner.call(t.ord - a, k.ord - a)).chr
      end.join
      end
      private_class_method :vigenere
      end
      end





      share|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        Looping using string.length.times.map … and length.times do … is OK, but slightly on the awkward side. I recommend adhering to the convention of writing same-line blocks using and line-spanning blocks using do … end.



        To extend the key, you can use the string multiplication operator. (Note that extending the key longer than necessary doesn't do much harm.)



        You can also factor out more of the commonality between the encrypt and decrypt functions.



        Instead of searching the LETTERS array, I recommend performing arithmetic on ASCII codes.



        module Crypto
        module Vigenere
        module_function
        def encrypt(plaintext, key)
        vigenere(plaintext, key) (p + k) % 26
        end

        def decrypt(ciphertext, key)
        vigenere(ciphertext, key) c, k
        end

        # Implementation of Vigenere cipher. The combiner block accepts
        # one character from the text and the corresponding character from
        # the key (encoded as a=0, b=1, ..., z=25), and returns the
        # result using the same numerical scheme.
        def vigenere(text, key, &combiner)
        a = 'a'.ord
        ext_key = key * (text.length / key.length + 1)
        text.chars.zip(ext_key.chars).collect do |t, k|
        (a + combiner.call(t.ord - a, k.ord - a)).chr
        end.join
        end
        private_class_method :vigenere
        end
        end





        share|improve this answer











        $endgroup$



        Looping using string.length.times.map … and length.times do … is OK, but slightly on the awkward side. I recommend adhering to the convention of writing same-line blocks using and line-spanning blocks using do … end.



        To extend the key, you can use the string multiplication operator. (Note that extending the key longer than necessary doesn't do much harm.)



        You can also factor out more of the commonality between the encrypt and decrypt functions.



        Instead of searching the LETTERS array, I recommend performing arithmetic on ASCII codes.



        module Crypto
        module Vigenere
        module_function
        def encrypt(plaintext, key)
        vigenere(plaintext, key) (p + k) % 26
        end

        def decrypt(ciphertext, key)
        vigenere(ciphertext, key) c, k
        end

        # Implementation of Vigenere cipher. The combiner block accepts
        # one character from the text and the corresponding character from
        # the key (encoded as a=0, b=1, ..., z=25), and returns the
        # result using the same numerical scheme.
        def vigenere(text, key, &combiner)
        a = 'a'.ord
        ext_key = key * (text.length / key.length + 1)
        text.chars.zip(ext_key.chars).collect do |t, k|
        (a + combiner.call(t.ord - a, k.ord - a)).chr
        end.join
        end
        private_class_method :vigenere
        end
        end






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 4 mins ago

























        answered 53 mins ago









        200_success200_success

        131k17157422




        131k17157422




















            sPaGhEtTiCaSe is a new contributor. Be nice, and check out our Code of Conduct.









            draft saved

            draft discarded


















            sPaGhEtTiCaSe is a new contributor. Be nice, and check out our Code of Conduct.












            sPaGhEtTiCaSe is a new contributor. Be nice, and check out our Code of Conduct.











            sPaGhEtTiCaSe is a new contributor. Be nice, and check out our Code of Conduct.














            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f219053%2fvigen%25c3%25a8re-cipher-in-ruby%23new-answer', 'question_page');

            );

            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







            Popular posts from this blog

            Invision Community Contents History See also References External links Navigation menuProprietaryinvisioncommunity.comIPS Community ForumsIPS Community Forumsthis blog entry"License Changes, IP.Board 3.4, and the Future""Interview -- Matt Mecham of Ibforums""CEO Invision Power Board, Matt Mecham Is a Liar, Thief!"IPB License Explanation 1.3, 1.3.1, 2.0, and 2.1ArchivedSecurity Fixes, Updates And Enhancements For IPB 1.3.1Archived"New Demo Accounts - Invision Power Services"the original"New Default Skin"the original"Invision Power Board 3.0.0 and Applications Released"the original"Archived copy"the original"Perpetual licenses being done away with""Release Notes - Invision Power Services""Introducing: IPS Community Suite 4!"Invision Community Release Notes

            Canceling a color specificationRandomly assigning color to Graphics3D objects?Default color for Filling in Mathematica 9Coloring specific elements of sets with a prime modified order in an array plotHow to pick a color differing significantly from the colors already in a given color list?Detection of the text colorColor numbers based on their valueCan color schemes for use with ColorData include opacity specification?My dynamic color schemes

            Ласкавець круглолистий Зміст Опис | Поширення | Галерея | Примітки | Посилання | Навігаційне меню58171138361-22960890446Bupleurum rotundifoliumEuro+Med PlantbasePlants of the World Online — Kew ScienceGermplasm Resources Information Network (GRIN)Ласкавецькн. VI : Літери Ком — Левиправивши або дописавши її