Written answers: Python Code

Written answers: Python Code

A. Monoalphabetic vs. polyalphabetic (3 points) Explain briefly how to analyze a long ciphertext to determine whether it was the result of a monoalphabetic cipher or the result of a polyalphabetic one.

B. One-time pad (3 points) Provide three reasons why the one-time pad (OTP) system is unbreakable.

C. Breaking Vigenère with a crib (8 points) You are given the ciphertext below that you know to be the result of applying the Vigenère cipher with a key of no more than 10 letters. You suspect that the crib word “quantum” occurs in the last 20 letters. Use that to break it.


In your answer, it is not enough to provide the plaintext.

· Explain how you used or would use the crib word (“quantum”) to find at least part of the key

· Give the plaintext with spaces and punctuation inserted

· Provide the key

· Provide the author of the quote

it might help to determine the likely length of the key using the program you wrote in assignment 1.

D. Finding 𝑝 and 𝑞 in RSA (8 points) We discussed several times that knowing any one of the secret keys in RSA would lead to knowing the others. Remember that we always know the two public values 𝑛 and e. But there are four secret values:

1. 𝑝

2. 𝑞

3. (𝑝−1)(𝑞−1), which is also called 𝜙(𝑛)

4. 𝑑

Clearly if one knows 𝑝, one can find 𝑞 and, from those two values, 𝜙(𝑛) and 𝑑.

But what if we only know 𝜙(𝑛)? As it turns out, algebra will help us find the values 𝑝+𝑞 and 𝑝−𝑞:

· One can see that 𝜙(𝑛)=(𝑝−1)(𝑞−1)=𝑝𝑞−𝑝−𝑞+1. Show how to use the last formula to find 𝑝+𝑞.

· One can also see that (𝑝−𝑞)2=𝑝2−2𝑝𝑞+𝑞2. Show how to use the last formula to find 𝑝−𝑞.

· Show how to use the formulas 𝑝+𝑞 and 𝑝−𝑞 to find 𝑝 and 𝑞.

Now, apply this technique to the values:

· 𝑛= 908445664156710023209119550968341886822411188352723705601669

· 𝜙(𝑛)= 908445664156710023209119550966411338125651051061875208447520

Provide the values of 𝑝 and 𝑞.

E. ElGamal (9 points) Please complete exercise 16 in section 10.6.

F. RSA (9 points) Please complete exercise 2 in section 14.4.

Programming problem

Implementing one round of AES encryption

(15 points) Implement (almost) the first round of AES with a 128-bit key. Given a 128-bit plaintext and a 128-bit key, this involves:

1. Performing a bitwise XOR with the plaintext and the key

2. Applying byte substitution for each byte in the state

3. Applying shift rows on the state table rows

4. Applying mix columns on the state table columns

Output the state table at this point. The “almost” above means that you don’t have to perform the XOR with the state and the first sub-key.

Please name the program AESEncryptRound.java or aesencryptround.py. You may hard code the inputs.

Please use the following inputs, which are the same as given in Appendix B of the FIPS197 document (located in the Notes module):

plaintext = 0x3243f6a8885a308d313198a2e0370734

key = 0x2b7e151628aed2a6abf7158809cf4f3c

The byte substitution, multiply-by-2, and multiply-by-3 tables are available in the text file AEStables.txt, which is in the D2L module with this exam. Each table is given as a list where the 𝑖𝑡ℎ entry is the value of applying the given operation to 𝑖.

Please print the state table after each step. Using the inputs given in Appendix B, my program outputs:

Original plaintext

32 88 31 e0

43 5a 31 37

f6 30 98 07

a8 8d a2 34

After ARK with original key

19 a0 9a e9

3d f4 c6 f8

e3 e2 8d 48

be 2b 2a 08

After SB

d4 e0 b8 1e

27 bf b4 41

11 98 5d 52

ae f1 e5 30

After SR

d4 e0 b8 1e

bf b4 41 27

5d 52 11 98

30 ae f1 e5

After MC

04 e0 48 28

66 cb f8 06

81 19 d3 26

e5 9a 7a 4c

My suggestion is to work on the layers in the order given so that it is easy to get partial credit. I found the mix columns layer the most challenging, which is another reason to leave it to the end. Also, be careful with byte ordering. Note that the state table is filled starting at the leftmost byte and in column-major order. It’s a small detail but getting it wrong results in outputs that are not close to what is correct.


In the following arrays, the value at index i is found by applying the operation mentioned above the array.  For example, in the ByteSub array, the entry at index 3 is 0x7b, which means that applying ByteSub to the value 0x03 returns the value 0x7b.

Multiplied by 2 in GF(256)

[0x00, 0x02, 0x04, 0x06, 0x08, 0x0a, 0x0c, 0x0e,

 0x10, 0x12, 0x14, 0x16, 0x18, 0x1a, 0x1c, 0x1e,

 0x20, 0x22, 0x24, 0x26, 0x28, 0x2a, 0x2c, 0x2e,

 0x30, 0x32, 0x34, 0x36, 0x38, 0x3a, 0x3c, 0x3e,

 0x40, 0x42, 0x44, 0x46, 0x48, 0x4a, 0x4c, 0x4e,

 0x50, 0x52, 0x54, 0x56, 0x58, 0x5a, 0x5c, 0x5e,

 0x60, 0x62, 0x64, 0x66, 0x68, 0x6a, 0x6c, 0x6e,

 0x70, 0x72, 0x74, 0x76, 0x78, 0x7a, 0x7c, 0x7e,

 0x80, 0x82, 0x84, 0x86, 0x88, 0x8a, 0x8c, 0x8e,

 0x90, 0x92, 0x94, 0x96, 0x98, 0x9a, 0x9c, 0x9e,

 0xa0, 0xa2, 0xa4, 0xa6, 0xa8, 0xaa, 0xac, 0xae,

 0xb0, 0xb2, 0xb4, 0xb6, 0xb8, 0xba, 0xbc, 0xbe,

 0xc0, 0xc2, 0xc4, 0xc6, 0xc8, 0xca, 0xcc, 0xce,

 0xd0, 0xd2, 0xd4, 0xd6, 0xd8, 0xda, 0xdc, 0xde,

 0xe0, 0xe2, 0xe4, 0xe6, 0xe8, 0xea, 0xec, 0xee,

 0xf0, 0xf2, 0xf4, 0xf6, 0xf8, 0xfa, 0xfc, 0xfe,

 0x1b, 0x19, 0x1f, 0x1d, 0x13, 0x11, 0x17, 0x15,

 0x0b, 0x09, 0x0f, 0x0d, 0x03, 0x01, 0x07, 0x05,

 0x3b, 0x39, 0x3f, 0x3d, 0x33, 0x31, 0x37, 0x35,

 0x2b, 0x29, 0x2f, 0x2d, 0x23, 0x21, 0x27, 0x25,

 0x5b, 0x59, 0x5f, 0x5d, 0x53, 0x51, 0x57, 0x55,

 0x4b, 0x49, 0x4f, 0x4d, 0x43, 0x41, 0x47, 0x45,

 0x7b, 0x79, 0x7f, 0x7d, 0x73, 0x71, 0x77, 0x75,

 0x6b, 0x69, 0x6f, 0x6d, 0x63, 0x61, 0x67, 0x65,

 0x9b, 0x99, 0x9f, 0x9d, 0x93, 0x91, 0x97, 0x95,

 0x8b, 0x89, 0x8f, 0x8d, 0x83, 0x81, 0x87, 0x85,

 0xbb, 0xb9, 0xbf, 0xbd, 0xb3, 0xb1, 0xb7, 0xb5,

 0xab, 0xa9, 0xaf, 0xad, 0xa3, 0xa1, 0xa7, 0xa5,

 0xdb, 0xd9, 0xdf, 0xdd, 0xd3, 0xd1, 0xd7, 0xd5,

 0xcb, 0xc9, 0xcf, 0xcd, 0xc3, 0xc1, 0xc7, 0xc5,

 0xfb, 0xf9, 0xff, 0xfd, 0xf3, 0xf1, 0xf7, 0xf5,

 0xeb, 0xe9, 0xef, 0xed, 0xe3, 0xe1, 0xe7, 0xe5]

Multiplied by 3 in GF(256)

[0x00, 0x03, 0x06, 0x05, 0x0c, 0x0f, 0x0a, 0x09,

 0x18, 0x1b, 0x1e, 0x1d, 0x14, 0x17, 0x12, 0x11,

 0x30, 0x33, 0x36, 0x35, 0x3c, 0x3f, 0x3a, 0x39,

 0x28, 0x2b, 0x2e, 0x2d, 0x24, 0x27, 0x22, 0x21,

 0x60, 0x63, 0x66, 0x65, 0x6c, 0x6f, 0x6a, 0x69,

 0x78, 0x7b, 0x7e, 0x7d, 0x74, 0x77, 0x72, 0x71,

 0x50, 0x53, 0x56, 0x55, 0x5c, 0x5f, 0x5a, 0x59,

 0x48, 0x4b, 0x4e, 0x4d, 0x44, 0x47, 0x42, 0x41,

 0xc0, 0xc3, 0xc6, 0xc5, 0xcc, 0xcf, 0xca, 0xc9,

 0xd8, 0xdb, 0xde, 0xdd, 0xd4, 0xd7, 0xd2, 0xd1,

 0xf0, 0xf3, 0xf6, 0xf5, 0xfc, 0xff, 0xfa, 0xf9,

 0xe8, 0xeb, 0xee, 0xed, 0xe4, 0xe7, 0xe2, 0xe1,

 0xa0, 0xa3, 0xa6, 0xa5, 0xac, 0xaf, 0xaa, 0xa9,

 0xb8, 0xbb, 0xbe, 0xbd, 0xb4, 0xb7, 0xb2, 0xb1,

 0x90, 0x93, 0x96, 0x95, 0x9c, 0x9f, 0x9a, 0x99,

 0x88, 0x8b, 0x8e, 0x8d, 0x84, 0x87, 0x82, 0x81,

 0x9b, 0x98, 0x9d, 0x9e, 0x97, 0x94, 0x91, 0x92,

 0x83, 0x80, 0x85, 0x86, 0x8f, 0x8c, 0x89, 0x8a,

 0xab, 0xa8, 0xad, 0xae, 0xa7, 0xa4, 0xa1, 0xa2,

 0xb3, 0xb0, 0xb5, 0xb6, 0xbf, 0xbc, 0xb9, 0xba,

 0xfb, 0xf8, 0xfd, 0xfe, 0xf7, 0xf4, 0xf1, 0xf2,

 0xe3, 0xe0, 0xe5, 0xe6, 0xef, 0xec, 0xe9, 0xea,

 0xcb, 0xc8, 0xcd, 0xce, 0xc7, 0xc4, 0xc1, 0xc2,

 0xd3, 0xd0, 0xd5, 0xd6, 0xdf, 0xdc, 0xd9, 0xda,

 0x5b, 0x58, 0x5d, 0x5e, 0x57, 0x54, 0x51, 0x52,

 0x43, 0x40, 0x45, 0x46, 0x4f, 0x4c, 0x49, 0x4a,

 0x6b, 0x68, 0x6d, 0x6e, 0x67, 0x64, 0x61, 0x62,

 0x73, 0x70, 0x75, 0x76, 0x7f, 0x7c, 0x79, 0x7a,

 0x3b, 0x38, 0x3d, 0x3e, 0x37, 0x34, 0x31, 0x32,

 0x23, 0x20, 0x25, 0x26, 0x2f, 0x2c, 0x29, 0x2a,

 0x0b, 0x08, 0x0d, 0x0e, 0x07, 0x04, 0x01, 0x02,

 0x13, 0x10, 0x15, 0x16, 0x1f, 0x1c, 0x19, 0x1a]


[0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5,

 0x30, 0x01, 0x67, 0x2b, 0xfe, 0xd7, 0xab, 0x76,

 0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59, 0x47, 0xf0,

 0xad, 0xd4, 0xa2, 0xaf, 0x9c, 0xa4, 0x72, 0xc0,

 0xb7, 0xfd, 0x93, 0x26, 0x36, 0x3f, 0xf7, 0xcc,

 0x34, 0xa5, 0xe5, 0xf1, 0x71, 0xd8, 0x31, 0x15,

 0x04, 0xc7, 0x23, 0xc3, 0x18, 0x96, 0x05, 0x9a,

 0x07, 0x12, 0x80, 0xe2, 0xeb, 0x27, 0xb2, 0x75,

 0x09, 0x83, 0x2c, 0x1a, 0x1b, 0x6e, 0x5a, 0xa0,

 0x52, 0x3b, 0xd6, 0xb3, 0x29, 0xe3, 0x2f, 0x84,

 0x53, 0xd1, 0x00, 0xed, 0x20, 0xfc, 0xb1, 0x5b,

 0x6a, 0xcb, 0xbe, 0x39, 0x4a, 0x4c, 0x58, 0xcf,

 0xd0, 0xef, 0xaa, 0xfb, 0x43, 0x4d, 0x33, 0x85,

 0x45, 0xf9, 0x02, 0x7f, 0x50, 0x3c, 0x9f, 0xa8,

 0x51, 0xa3, 0x40, 0x8f, 0x92, 0x9d, 0x38, 0xf5,

 0xbc, 0xb6, 0xda, 0x21, 0x10, 0xff, 0xf3, 0xd2,

 0xcd, 0x0c, 0x13, 0xec, 0x5f, 0x97, 0x44, 0x17,

 0xc4, 0xa7, 0x7e, 0x3d, 0x64, 0x5d, 0x19, 0x73,

 0x60, 0x81, 0x4f, 0xdc, 0x22, 0x2a, 0x90, 0x88,

 0x46, 0xee, 0xb8, 0x14, 0xde, 0x5e, 0x0b, 0xdb,

 0xe0, 0x32, 0x3a, 0x0a, 0x49, 0x06, 0x24, 0x5c,

 0xc2, 0xd3, 0xac, 0x62, 0x91, 0x95, 0xe4, 0x79,

 0xe7, 0xc8, 0x37, 0x6d, 0x8d, 0xd5, 0x4e, 0xa9,

 0x6c, 0x56, 0xf4, 0xea, 0x65, 0x7a, 0xae, 0x08,

 0xba, 0x78, 0x25, 0x2e, 0x1c, 0xa6, 0xb4, 0xc6,

 0xe8, 0xdd, 0x74, 0x1f, 0x4b, 0xbd, 0x8b, 0x8a,

 0x70, 0x3e, 0xb5, 0x66, 0x48, 0x03, 0xf6, 0x0e,

 0x61, 0x35, 0x57, 0xb9, 0x86, 0xc1, 0x1d, 0x9e,

 0xe1, 0xf8, 0x98, 0x11, 0x69, 0xd9, 0x8e, 0x94,

 0x9b, 0x1e, 0x87, 0xe9, 0xce, 0x55, 0x28, 0xdf,

 0x8c, 0xa1, 0x89, 0x0d, 0xbf, 0xe6, 0x42, 0x68,

 0x41, 0x99, 0x2d, 0x0f, 0xb0, 0x54, 0xbb, 0x16]

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Institute of Agricultural Policy and Market Research Chair of Agricultural and Food Market Analysis TAKE-HOME ASSIGNMENT MK102 Global Food Markets (WiSe 2021-22) Student name:             **please fill in your name here** Student ID-number:     **please fill in your student ID here** Start time:                    Thursday, February 17, 2022 at 10:00 am (CET) End

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Ayesha Khan is the daughter of Salman and Huma. Until they retired, Salman and Huma operated a successful business in Melbourne importing engineering components from Karachi. Thirty years ago, Salman and Huma’s accountant advised them that they should set up a discretionary trust for tax purposes, which they did. This

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The transborder challenge or threat

Your essay should be structured as follows: (excluding the reference list). o What are the main debates surrounding the causes of this challenge, which voices dominate in this debate, and why? This part of the essay involves: (a) identifying the main debates surrounding the causes of this challenge/threat, the key

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LWZ223 Property Law: Assessment One

Word Limit 2 000 words AGLC referencing and bibliography are required. Refer to rubric for marking guide. Relevant Unit Learning Outcomes for task, 1, 4, 5 and 6: advice on disputes in hypothetical problem and actual case law situations. Tip: Remember to support your response with law. This can be

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LAW5211 Constitutional Law Essay

LAW/CRI ASSESSMENT TASK SHEET – LAW5211 Assessment # 1 Assessment Name and description: Constitutional Law Essay Assessment purpose/scope and relevance:    The purpose of this assessment is to:    assess students’ ability to research issues of constitutional law; assess students’ ability to analyse and apply the results of that research

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SOC201A Mediation and Conflict Management

  Assessment Brief   Program   Bachelor of Applied Social Science   Subject   Mediation and Conflict Management   Subject code   SOC201A   Name of assessment   Assessment 3: Case Study   Length   2000 words   Learning outcomes addressed by this assessment:   A, B, C, D, E,

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This assignment addresses market demand in census tract neighbourhoods in the Toronto Census Metropolitan Area (CMA). You will first collect and present demographic data on two census tracts. Then you will compare and contrast the two census tract neighbourhoods. Finally, you will briefly identify census variables from your own neighbourhood.

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