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If P+K < 26 then key will remain same and if P+K > 26 then key = P+K - 26.
U made the MOD work so easy, don't even need a calculator :)
Not >26, it's >25* Because we start assigning "a"=0If it's 26, you don't need to calculate.. the answer is simply "a"
Love u bro u made a new formula love u
Woww kya bhaari sikhaya hein bhaiya thank you so much 🎉🎉🎉🎉🎉
The answer to the question is ZICVTWQNGRZGVTW
Correct
Wrong
0:42
Correct answer
@@Theudaan-y3t its right
Thanks A LOT FOR THIS VIDEO, HELPFUL, EXAM IN 10 DAYS
Best explanation ever 🎉
Thank you brother ❤
Bro u are GOD. Thank you so much
Perfect and too the point love from PK ❤
Vernam and vigenere cipher techniques me kya differences hai
The homework questions answers is:ZICVTWQNGRZGVTW 🙏IF p+k < 26 no need to calculate write the same value as it is...IF p+k > 26 then 26 - (p+k) 👍
provide link for the watch you worn!!
This js vernam cipher (one time pad )
Thank you sir ❤
ab decrypt kese hoga
😊😊
best ever
Thanks ❤
Thank you
what if key's length greater than the plain text's length ?
Double the plain text until it matches the length of key but most of time key is shorter the pt
voice is just like Amir Khan.
mera kaal exam hain 😅🤣
Final Answer Of the Homework Question : ZICVSBAXSYITIS
ans to correct hai lekin aolve krne me aaadha ghnta lag gyaaaa
Home work question solution:ZICVTWQNYRZGVTW
Ink
ZICVTWQNGRZGVTW Answer to HW question.
Before last TW there is "Y" also
import stringdef alphabet_table(): alphabets_ = {} alphabets = string.ascii_lowercase for i, alphabet in zip(range(26), alphabets): alphabets_[alphabet] = i return alphabets_def key_table(): keys = {} alphabets = string.ascii_lowercase for i, alphabet in zip(range(26), alphabets): keys[i] = alphabet return keysdef get_keys_for_alphabets(text): return [alphabet_table()[char] for char in text]def vigenere_cipher(text, key): len_of_text = len(text) len_of_key = len(key) if len_of_text > len_of_key: desired_length = len_of_text - len_of_key padded_key = (key * ((desired_length // len(key)) + 1))[:desired_length] new_key = key + padded_key cipher_keys_list = get_keys_for_alphabets(new_key) else: cipher_keys_list = get_keys_for_alphabets(key) plain_list = get_keys_for_alphabets(text) cipher_text_list = [(p + k) % 26 for (p, k) in zip(plain_list, cipher_keys_list)] cipher_chars = [key_table()[val] for val in cipher_text_list] cipher_text = "".join(cipher_chars) return cipher_textif __name__ == "__main__": print(vigenere_cipher("wearediscovered", "deceptive"))
Cipher : z I c g x l d w f s x t I m y sir this correcet ???
nope
Z I C O T W W N G R Z H V T W hope i am right
no you are not
Z I C V T W Q N G R Z G V T W This is the correct answer bro
@@omegajain3570 yes bro
Z I C V T W Q N G R Z G V Y T W
thank you
If P+K < 26 then key will remain same and if P+K > 26 then key = P+K - 26.
U made the MOD work so easy, don't even need a calculator :)
Not >26, it's >25*
Because we start assigning "a"=0
If it's 26, you don't need to calculate.. the answer is simply "a"
Love u bro u made a new formula love u
Woww kya bhaari sikhaya hein bhaiya thank you so much 🎉🎉🎉🎉🎉
The answer to the question is
ZICVTWQNGRZGVTW
Correct
Wrong
0:42
Correct answer
@@Theudaan-y3t its right
Thanks A LOT FOR THIS VIDEO, HELPFUL, EXAM IN 10 DAYS
Best explanation ever 🎉
Thank you brother ❤
Bro u are GOD. Thank you so much
Perfect and too the point love from PK ❤
Vernam and vigenere cipher techniques me kya differences hai
The homework questions answers is:
ZICVTWQNGRZGVTW 🙏
IF p+k < 26 no need to calculate write the same value as it is...
IF p+k > 26 then 26 - (p+k) 👍
provide link for the watch you worn!!
This js vernam cipher (one time pad )
Thank you sir ❤
ab decrypt kese hoga
😊😊
best ever
Thanks ❤
Thank you
what if key's length greater than the plain text's length ?
Double the plain text until it matches the length of key but most of time key is shorter the pt
voice is just like Amir Khan.
mera kaal exam hain 😅🤣
Final Answer Of the Homework Question :
ZICVSBAXSYITIS
ans to correct hai lekin aolve krne me aaadha ghnta lag gyaaaa
Home work question solution:
ZICVTWQNYRZGVTW
Ink
ZICVTWQNGRZGVTW Answer to HW question.
Before last TW there is "Y" also
import string
def alphabet_table():
alphabets_ = {}
alphabets = string.ascii_lowercase
for i, alphabet in zip(range(26), alphabets):
alphabets_[alphabet] = i
return alphabets_
def key_table():
keys = {}
alphabets = string.ascii_lowercase
for i, alphabet in zip(range(26), alphabets):
keys[i] = alphabet
return keys
def get_keys_for_alphabets(text):
return [alphabet_table()[char] for char in text]
def vigenere_cipher(text, key):
len_of_text = len(text)
len_of_key = len(key)
if len_of_text > len_of_key:
desired_length = len_of_text - len_of_key
padded_key = (key * ((desired_length // len(key)) + 1))[:desired_length]
new_key = key + padded_key
cipher_keys_list = get_keys_for_alphabets(new_key)
else:
cipher_keys_list = get_keys_for_alphabets(key)
plain_list = get_keys_for_alphabets(text)
cipher_text_list = [(p + k) % 26 for (p, k) in zip(plain_list, cipher_keys_list)]
cipher_chars = [key_table()[val] for val in cipher_text_list]
cipher_text = "".join(cipher_chars)
return cipher_text
if __name__ == "__main__":
print(vigenere_cipher("wearediscovered", "deceptive"))
Cipher : z I c g x l d w f s x t I m y
sir this correcet ???
nope
Z I C O T W W N G R Z H V T W
hope i am right
no you are not
Z I C V T W Q N G R Z G V T W
This is the correct answer bro
@@omegajain3570 yes bro
Z I C V T W Q N G R Z G V Y T W
thank you