Encryption Warm-Up Exercise

Spend 5 minutes responding to the prompt below:

Prompt: Why might people hide messages?

1. List two reasons why someone would conceal information:
   
   
   
2. For each reason, provide a real-world example where encryption is used:
   
   
   

Cipher Wheel Activity

Objective: Practice encoding and decoding messages using the Caesar cipher and a cipher wheel.

Materials:

• One pre-assembled cipher wheel per student or pair
  - Blank message cards or slips of paper
  - Pens or pencils
  
  Instructions:

1. Pair up and secretly agree on a shift key between 1 and 25. Record your key.
   
   2. Player A writes a short message (3–5 words) on a message card. Using the cipher wheel and your agreed key, encode the message into ciphertext.
   
   3. Exchange the ciphertext card with Player B. Player B uses the cipher wheel and the same key to decrypt the message back into plaintext. Verify correctness.
   
   4. Switch roles: Player B creates a new message and key, encodes it, and Player A decodes.
   
   5. Repeat as time allows, trying different shift keys each round.

Reflection Questions:

1. Which shift keys were easiest to work with? Which were most challenging, and why?
   
   
   
2. How did the cipher wheel help you understand the mechanics of the Caesar cipher?
   
   
   
3. What strategies did you develop to avoid errors in encoding or decoding?
   
   
   
   
   

Session 1 Practice Worksheet

Use the Caesar cipher with the given shift key to complete the exercises below. You may use Cipher Wheel Activity Materials to help with encoding and decoding.

Part A: Encoding

1. Plaintext: SECRET
   Key = 4
   Ciphertext: ____________________________
   
   
   
2. Plaintext: HELLO WORLD
   Key = 5
   Ciphertext: ____________________________
   
   
   
3. Plaintext: DIGITAL
   Key = 2
   Ciphertext: ____________________________
   
   
   

Part B: Decoding

1. Ciphertext: KHOOR
   Key = 3
   Plaintext: ____________________________
   
   
   
2. Ciphertext: YMJBTWQI
   Key = 5
   Plaintext: ____________________________
   
   
   
3. Ciphertext: GCUA VJCV KC C EQOG
   Key = 2
   Plaintext: ____________________________
   
   
   

Part C: Create & Exchange

1. Write your own short message (3–5 words), choose a key (1–25), and encode it. Then swap with a partner and decode their message.Your message (plaintext): ____________________________
   
   
   
   
   Key used: ______
   Ciphertext: ____________________________
   
   
   
   
   Partner’s decoded message: ____________________________
   
   
   

Part D: Reflection

1. Which shift keys felt easiest to work with? Why?
   
   
   
   
   
2. Describe one strategy you used to avoid errors when encoding or decoding.
   
   
   
   
   

Good luck, Codebreakers! Keep track of your process and be prepared to discuss your strategies in class.

Session 1 Quick Quiz Answer Key

This answer key provides the correct responses and the reasoning steps for each question.

Question 1

Prompt: What do we call the original readable message before encryption?
Correct Answer: plaintext

Explanation / Reasoning:

• Plaintext is the standard term for any message or data in its original, human-readable form.
• After encryption, plaintext becomes ciphertext, which is unreadable without the proper key.

Question 2

Prompt: Using a Caesar cipher with a shift key of 4, encode the word “HELLO”. Write the resulting ciphertext.
Correct Answer: LIPPS

Step-by-Step Encoding:

1. Identify the shift: +4 positions in the alphabet.
2. Encode each letter:
   • H → move forward 4 → I (1), J (2), K (3), L (4)
   • E → F (1), G (2), H (3), I (4)
   • L → M (1), N (2), O (3), P (4)
   • L → P (same as previous L)
   • O → P (1), Q (2), R (3), S (4)
3. Combine the results: L I P P S → LIPPS

Note: Students must show understanding of the shift process. Full credit for the correct ciphertext LIPPS.

Question 3

Prompt: Provide one real-world example where encryption is important and briefly explain why.
Sample Model Answer:

• Example: Online banking transactions.
• Explanation: When you log in to your bank’s website or mobile app, your username, password, and account data are encrypted. This prevents hackers or eavesdroppers from intercepting and reading your personal financial information.

Rubric / Scoring Guidelines:

• 2 points: Student names a valid context where encryption is used and provides a clear, relevant explanation.
• 1 point: Student names a valid context but gives only a vague or incomplete rationale.
• 0 points: Response does not identify a correct encryption use case or explanation is incorrect.
   
   
  Good work, Codebreakers! Ensure you review these steps before moving on to the Session 1 mini-test and reflection prompt.

Session 1 Mini Test Answer Key

This answer key provides the correct responses, step-by-step solutions, and scoring guidelines for each question on the Session 1 Mini Test.

Question 1

Prompt: What is the process of converting readable information into an unreadable form called?
Correct Answer: Encryption

Explanation / Reasoning:

• Encryption is the standard term for taking plaintext (readable data) and transforming it into ciphertext (unreadable form) using an algorithm and key.

Scoring Guidelines:

• 1 point for selecting Encryption.
• 0 points for any other choice.

Question 2

Prompt: What do we call the process of converting ciphertext back to plaintext?
Correct Answer: Decryption

Explanation / Reasoning:

• Decryption reverses encryption, transforming ciphertext back into its original readable form using the correct key.

Scoring Guidelines:

• 1 point for selecting Decryption.
• 0 points for any other choice.

Question 3

Prompt: Define ‘plaintext’ in your own words.
Model Answer:
Plaintext is the original, human-readable message or data before it is encrypted into ciphertext.

Explanation / Reasoning:

• Emphasize that plaintext is the starting (unencrypted) form of any message.

Scoring Guidelines:

• 2 points: Clear, accurate definition mentioning “readable” or “original message” before encryption.
• 1 point: Partial description (e.g., “unencrypted data”) without clarity on readability.
• 0 points: Incorrect or missing definition.

Question 4

Prompt: In the context of a Caesar cipher, what is a ‘key’?
Model Answer:
A key is the number of positions each letter in the plaintext is shifted in the alphabet to produce the ciphertext.

Explanation / Reasoning:

• The key determines how far forward (or backward) you move each letter.

Scoring Guidelines:

• 2 points: Correctly identifies the key as the shift amount in the alphabet.
• 1 point: Mentions that the key is needed to shift letters but lacking detail on shift amount.
• 0 points: Incorrect or no answer.

Question 5

Prompt: Using a Caesar cipher with a shift key of 4, encode the message ‘CLASS’. Write the ciphertext.
Correct Answer: G P E W W → GPEWW

Step-by-Step Encoding:

1. C → move forward 4 → D (1), E (2), F (3), G (4)
2. L → M (1), N (2), O (3), P (4)
3. A → B (1), C (2), D (3), E (4)
4. S → T (1), U (2), V (3), W (4)
5. S → W (same as previous S)
6. Combine: G P E W W

Scoring Guidelines:

• 2 points: Correct ciphertext GPEWW and evidence of correct shifting.
• 1 point: Minor arithmetic or transcription error but clear understanding of shifting.
• 0 points: Incorrect answer without demonstration of shift process.

Question 6

Prompt: Decode the following ciphertext using a shift key of 2: EKUVN. Write the plaintext.
Correct Answer: C I S T L → CISTL

Step-by-Step Decoding:

1. E → move backward 2 → D (–1), C (–2)
2. K → J (–1), I (–2)
3. U → T (–1), S (–2)
4. V → U (–1), T (–2)
5. N → M (–1), L (–2)
6. Combine: C I S T L

Scoring Guidelines:

• 2 points: Correct plaintext CISTL with clear backward shifting.
• 1 point: Minor error but correct approach to shifting.
• 0 points: Incorrect answer without proper use of the key.

Total Points Available: 10

Ensure students review these solutions before proceeding to deeper encryption methods in Session 2.

Session 1 Cool Down

Take 5 minutes to reflect on today’s work with the Caesar cipher. Respond to the prompts below:

1. What was one key insight you gained about how the Caesar cipher transforms plaintext into ciphertext?
   
   
   
2. Which step in encoding or decoding did you find most challenging, and why?
   
   
   
3. What is one question or idea about encryption you’d like to explore in Session 2?
   
   
   
   
   

Session 2 Warm-Up Exercise

Spend 5 minutes responding to the prompts below as you prepare to explore modern encryption methods:

1. List two differences you predict between symmetric and asymmetric encryption:
   
   
   
2. For each difference, provide a real-world example of where that type of encryption is used:
   
   
   

Diffie–Hellman Key Exchange Activity

Objective: Simulate the Diffie–Hellman key exchange to understand how two parties can agree on a shared secret over an insecure channel without directly sharing their private keys.

Materials:

• Pre-printed parameter cards (one per pair) with prime p and base g (e.g., p = 23, g = 5; alternate set: p = 17, g = 3)
• Secret exponent cards (one per student) with a private number between 1 and p–2 (keep this face-down)
• Scratch paper or whiteboards for calculations
• Optional: calculators for modular arithmetic

Instructions:

1. Form pairs. Each pair receives one parameter card listing your agreed public values: prime p and base g.
2. Each student secretly selects a private exponent (secret) a (or b) from their card—do not reveal this to your partner.
3. Compute your public value:
   • Student A computes A = g^a mod p
   • Student B computes B = g^b mod p
4. Exchange your public values (swap only A and B).
5. Compute the shared secret key s:
   • Student A computes s = B^a mod p
   • Student B computes s = A^b mod p
     Note: Both calculations yield the same number s.
6. Verify with your partner that you both arrived at the same shared secret s (reveal only this number).

Extension:

• Repeat with the alternate parameter set (p = 17, g = 3) and compare results.
• Try choosing different secret exponents and repeat the process to see how the shared secret changes.

Reflection Questions:

1. Did both participants calculate the same shared secret? Why does the math guarantee this?
   
   
   
2. Why is it secure that each party’s private exponent was never sent over the channel?
   
   
   
3. If an eavesdropper knows p, g, A, and B, what information would they need to compute the shared secret? Why is this difficult?
   
   
   
   
   

Public-Key Practice Worksheet

Use your Diffie–Hellman Key Exchange Activity materials and RSA formulas from the slides to complete the exercises below. Show all calculations and be prepared to discuss your results.

Part A: Diffie–Hellman Key Exchange Simulation

Parameters Set 1: p = 23, g = 5

1. Your secret exponent (a): ________________________
2. Compute your public value: A = gᵃ mod p = ________________________
   
   
3. Partner’s public value (B): ________________________
4. Compute shared secret: s = Bᵃ mod p = ________________________
   
   

Parameters Set 2: p = 17, g = 3
5. Your secret exponent (a): ________________________
6. Compute your public value: A = gᵃ mod p = ________________________


7. Partner’s public value (B): ________________________
8. Compute shared secret: s = Bᵃ mod p = ________________________

Part B: Basic RSA Practice

Use small primes p = 3 and q = 11.

1. Compute the modulus: n = p × q = ________________________
2. Compute Euler’s totient: φ(n) = (p–1)(q–1) = ________________________
   
   
3. Given public exponent e = 3, find the private exponent d such that e · d ≡ 1 (mod φ(n)): d = ________________________
   
   
4. Encryption: For message M = 4, compute ciphertext C = Mᵉ mod n = ________________________
   
   
5. Decryption: For ciphertext C = 31, compute plaintext M = Cᵈ mod n = ________________________
   
   
   

Part C: Reflection Questions

1. In your own words, explain why two parties end up with the same shared secret in Diffie–Hellman even though they never exchange their private exponents.
   
   
   
   
   
   
2. Describe how the choice of a large prime modulus (p) in Diffie–Hellman and RSA contributes to the security of each system.
   
   
   
   
   
   

Good work, Codebreakers! Be ready to share your calculations and insights in our next discussion.

Session 2 Quick Quiz Answer Key

This answer key provides the correct responses and the reasoning steps for each question on the Session 2 Quick Quiz.

Question 1

Prompt: Which statement correctly describes the difference between symmetric and asymmetric encryption?

Correct Answer: Symmetric encryption uses the same secret key for encryption and decryption, while asymmetric encryption uses a public/private key pair.

Explanation / Reasoning:

• Symmetric encryption relies on one shared secret key for both encrypting and decrypting data.
• Asymmetric encryption employs two mathematically related keys: a public key for encryption (shared openly) and a private key for decryption (kept secret).

Scoring Guidelines:

• 1 point: Selects the exact statement above.
• 0 points: Any other selection.

Question 2

Prompt: In a Diffie–Hellman exchange with p = 23, g = 5, your secret exponent a = 6, and your partner’s public value B = 19, what shared secret s will you compute? Show your work.

Correct Answer: s = 2

Step-by-Step Calculation:

1. Identify values: p = 23, g = 5, a = 6, B = 19.
2. Compute s = Bᵃ mod p = 19⁶ mod 23.
   • 19² mod 23 = 361 mod 23 = 16
   • 19⁴ mod 23 = (19²)² = 16² = 256 mod 23 = 3
   • 19⁶ mod 23 = (19⁴ · 19²) mod 23 = (3 · 16) mod 23 = 48 mod 23 = 2
3. Result: s = 2

Scoring Guidelines:

• 2 points: Correctly computes s = 2 with clear modular arithmetic steps.
• 1 point: Correct s = 2 but with incomplete or minor errors in the work.
• 0 points: Incorrect result or missing calculation.

Question 3

Prompt: Using the RSA parameters p = 3, q = 11, e = 3, and d = 7, decrypt the ciphertext C = 31. What is the plaintext message M? Show your calculation.

Correct Answer: M = 4

Step-by-Step Calculation:

1. Compute modulus: n = p · q = 3 · 11 = 33
2. Decrypt using M = Cᵈ mod n = 31⁷ mod 33.
   • 31² mod 33 = 961 mod 33 = 4
   • 31⁴ mod 33 = (31²)² = 4² = 16
   • 31⁶ mod 33 = (31⁴ · 31²) mod 33 = (16 · 4) mod 33 = 64 mod 33 = 31
   • 31⁷ mod 33 = (31⁶ · 31) mod 33 = (31 · 31) mod 33 = 961 mod 33 = 4
3. Result: M = 4

Scoring Guidelines:

• 2 points: Correctly finds M = 4 with modular exponentiation steps.
• 1 point: Correct M but with incomplete or minor errors in work.
• 0 points: Incorrect result or missing calculation.
   
   
  Great work, Codebreakers! Review these steps to solidify your understanding before the Session 2 test and next unit activities.

Session 2 Test Answer Key

This answer key provides the correct responses, step-by-step solutions, and scoring guidelines for each question on the Session 2 Test.

Question 1

Prompt: Which statement correctly describes asymmetric encryption?
Correct Answer: Involves a public key to encrypt and a private key to decrypt

Explanation / Reasoning:

• Asymmetric encryption uses a pair of keys: the public key is shared openly for encryption, and the private key is kept secret for decryption.

Scoring Guidelines:

• 1 point for selecting the correct statement.
• 0 points for any other choice.

Question 2

Prompt: Which values are exchanged publicly during the Diffie–Hellman key exchange?
Correct Answer: Public values (A and B)

Explanation / Reasoning:

• Each party computes a public value (A = g^a mod p, B = g^b mod p) and shares only these in the clear.
• The private exponents (a and b) remain secret, and the shared secret is derived from A and B plus each party’s own private exponent.

Scoring Guidelines:

• 1 point for selecting Public values (A and B).
• 0 points for any other selection.

Question 3

Prompt: In a Diffie–Hellman exchange with p = 23, g = 5, your secret exponent a = 10, and your partner’s public value B = 19, what shared secret s will you compute? Show your work.
Correct Answer: s = 6

Step-by-Step Calculation:

1. Identify values: p = 23, g = 5, a = 10, B = 19.
2. Compute powers of 19 modulo 23:
   • 19² mod 23 = 361 mod 23 = 16
   • 19⁴ = (19²)² = 16² = 256 mod 23 = 3
   • 19⁸ = (19⁴)² = 3² = 9
3. Combine to find 19¹⁰ = 19⁸ · 19² = 9 · 16 = 144 mod 23 = 144 – 23·6 = 6
4. Result: s = 6

Scoring Guidelines:

• 2 points: Correctly computes s = 6 with clear modular arithmetic steps.
• 1 point: Correct result but incomplete or minor errors in the work.
• 0 points: Incorrect result or missing calculation.

Question 4

Prompt: Using RSA with parameters p = 3, q = 11, e = 3, and d = 7, encrypt the message M = 5. What is the ciphertext C? Show your calculation.
Correct Answer: C = 26

Step-by-Step Calculation:

1. Compute modulus: n = p · q = 3 · 11 = 33
2. Encrypt: C = Mᵉ mod n = 5³ mod 33
   • 5² = 25 mod 33 = 25
   • 5³ = 5² · 5 = 25 · 5 = 125 mod 33 = 125 – 33·3 = 26
3. Result: C = 26

Scoring Guidelines:

• 2 points: Correct ciphertext 26 with evidence of modular exponentiation.
• 1 point: Correct result but missing or flawed steps.
• 0 points: Incorrect result or no work shown.

Question 5

Prompt: Using the same RSA parameters (p = 3, q = 11, e = 3, d = 7), decrypt the ciphertext C = 26. What is the plaintext message M? Show your calculation.
Correct Answer: M = 5

Step-by-Step Calculation:

1. Compute modulus: n = 33
2. Decrypt: M = Cᵈ mod n = 26⁷ mod 33
   • 26² mod 33 = 676 mod 33 = 16
   • 26⁴ = (26²)² = 16² = 256 mod 33 = 25
   • Combine for 26⁷ = 26⁴ · 26² · 26 = 25 · 16 = 400 mod 33 = 4; then 4 · 26 = 104 mod 33 = 104 – 33·3 = 5
3. Result: M = 5

Scoring Guidelines:

• 2 points: Correct plaintext 5 with clear modular exponentiation.
• 1 point: Correct result but partial or minor errors in work.
• 0 points: Incorrect result or missing calculation.

Question 6

Prompt: Explain one advantage and one disadvantage of symmetric versus asymmetric encryption.
Sample Model Answer:

• Advantage of symmetric encryption: It is computationally efficient and fast, making it ideal for encrypting large amounts of data.
• Disadvantage of symmetric encryption: Securely sharing the single secret key between parties can be challenging, requiring a secure channel.

Explanation / Reasoning:

• Symmetric encryption uses one shared key (speed, efficiency) but suffers from a key distribution problem.
• Asymmetric encryption (by contrast) simplifies key distribution (public key can be shared openly) but is slower and less efficient for large data volumes.

Scoring Guidelines:

• 2 points: Clearly states one relevant advantage and one relevant disadvantage.
• 1 point: Mentions only one of the two or provides an incomplete explanation.
• 0 points: Incorrect or no response.

Total Points Available: 10