please answer all and show steps. write neat please!! 4. Decrypt the following m
ID: 3714698 • Letter: P
Question
please answer all and show steps. write neat please!!
4. Decrypt the following message: "Cher zngurzngvpf vf, va vgf jnl, gur cbrgel bs ybtvpny vqrnf." Nyoreg Rvafgrva. 10 points] 5. Encrypt the famous quote "Once we accept our limits, we go beyond them." by Albert Einstein using Vigenere cipher with key "MATH". 10 points] 6. Let p -23, 17,e 59, d 179 as in Module 3 Lesson 4. Encrypt the message m 31 (You must show all your steps as in the lecture videos to receive full credit, including the work done for exponentiation with square and multiply algorithm!)30 points)Explanation / Answer
4.The above cipher text is generated by CAESAR CIPHER with an offset of 13.i.e.,each letter in the
alphabets is shifted by 13 characters.Thus a=1,becomes a=a+13=14=n.
So,the decrypted text that I have obtained is
purt mathematics is in its way poetry of logical ideas
albert einstein
5.The VIGENERE Cipher works by repeating the key a number of times by running over the entire plain text
.Thus ,when the key is MATH and the plain text is as specified ,the cipher text that is obtained is
ANVLIETJOEIAAUKSUMBAEPLSOULKOGKTHEM
We have positin of m=13(first letter of key) and position of o=14,thus the cipher text obtained is
13+14=27mod26=1(Since there are only 26 letters in the alphabets)
i.e. a.This procedure is repeated for the entire plaintext and by repeating the key over it.For example,the key length ends at only.so,for w,the same key repeats starting from m.
6.For encryption and decryption using the RSA algorithm where
p=23,q=17 e=59 and d=179.
We have product of p and =n=23*17=391.
Now,the public key is given by : pu(e,n)
and the private key is given by:pr(d,n)
Thus,the public key becomes (59,391)
and private key becomes (179,31)
So,the encrypted text is given by:
plaintext^e mod n;
and the decrypted text is given by:
ciphertext^d mod n;
In this case,we have,plaintext=31.
Hence,by following the above procedure,we can obtain ciphertext=163
and you see that when you decrypt it,you will get back the plain text =31.
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