Solutions Manual for an Introduction to Cryptography with Coding Theory (2nd Edition)
Wade Trappe, Lawrence C. Washington
Language: English
Pages: 181
ISBN: 2:00196939
Format: PDF / Kindle (mobi) / ePub
The accompanying solutions manual to an Introduction to Cryptography with Coding Theory (2nd Edition) by Wade Trappe, Lawrence C. Washington (Pearson).
Systems Analysis and Design in a Changing World (5th Edition)
Exam Ref 70-410: Installing and Configuring Windows Server 2012
Bioinformatics Data Skills: Reproducible and Robust Research with Open Source Tools (1st Edition)
Practical Distributed Processing (Undergraduate Topics in Computer Science)
codewords. Therefore, M ≤ q. An example with M = q is obtained by repeating each letter of the alphabet n times. Therefore, M = q is the largest possible M , so Aq (n, n) = q. 8. A vector (x1 , . . . , xn ) is in C ⊥ if and only if (x1 , . . . , xn ) · (1, . . . , 1) = 0, which is exactly the equation defining the parity check code of length n. 9. Suppose u−v ∈ C. Let u+c ∈ u+C. Then u+c = v+(c+(u−v)) ∈ v+C, since c + (u − v) is the sum of two elements of C, hence is in C. Therefore u + C ⊆ v +
Out[19]= 14 In[20]:= coinc hdsf, 6 Out[20]= 14 The key length is probably 4. Look at frequencies of letters in positions 1 mod 4: In[21]:= vigvec hdsf, 4, 1 Out[21]= 0.151163, 0.0465116, 0.0116279, 0, 0.0348837, 0.127907, 0.0697674, 0.0465116, 0, 0.0232558, 0, 0.0348837, 0, 0.0581395, 0.0232558, 0.0232558, 0, 0, 0.0465116, 0.0697674, 0.0465116, 0.0813953, 0, 0.0116279, 0.0465116, 0.0465116 Find the dot products with the alphabet frequency vector: In[22]:= corr % Out[22]= 0.0448023, 0.0458953,
-48382 mod 83987: In[122]:= Mod 48382, 83987 Out[122]= 35605 65 Therefore, we found the square root of -48382 mod 83987, rather than the square root of 48382. This is because 48382 does not have a square root mod 83987. Chapter 6 - Mathematica Note: There are two commands that can be used to change text to numbers: num1 and txt2num1. They are the same function. Similarly, both alph1 and num2txt1 change numbers back to text and are the same function. Problem 1. Encrypt each of the two messages
values. 8. We first calculate the coincidences to determine the most likely key length: >> coinc(ocwy,1) ans = 13 >> coinc(ocwy,2) ans = 13 >> coinc(ocwy,3) ans = 11 >> coinc(ocwy,4) ans = 136 6 >> coinc(ocwy,5) ans = 15 >> coinc(ocwy,6) ans = 23 >> coinc(ocwy,7) ans = 8 Observe that the key length is most likely 6. >> vigvec(ocwy,6,1) ans = 0.0577 0.0769 0.0385 0 0 0.0192 0 0.0962 0.0385 0.0192 0.0577 0.0769 0 0.0577 0.0769 0.0769 0 0.0192 0.0192 0.0385 0.1346 0.0192 0.0385 0 137 0.0192
has 2 matches. Thus, the most likely key length is 2. (b) To decrypt, we use the fact that the key length is 2 and extract off every odd letter to get BBBAB, and then every even letter to get AAAAA. Using a frequency count on each of these yields that a shift of 0 is the most likely scenario for the first character of the Vigenere key, while a shift of 1 is the most likely case for the second character of the key. Thus, the key is AB. Decrypting each subsequence yields BBBAB and BBBBB. Combining
