46 lines
2.4 KiB
Markdown
46 lines
2.4 KiB
Markdown
1. **Choose a short phrase of between 50-60 characters. You could choose a website headline, song title, etc. Convert this phrase using Caesar shift encryption and a key of 5 to create your ciphertext. Carry out a cryptanalytic attack to try to work out the decryption key and plaintext. (10 marks)**
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F1 veteran to retire and take up new role in 2024.
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| Original String | F1 veteran to retire and take up new role in 2024. |
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| Key = 1 | G1 wfufsbo up sfujsf boe ublf vq ofx spmf jo 2024. |
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| Key = 2 | H1 xgvgtcp vq tgvktg cpf vcmg wr pgy tqng kp 2024. |
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| Key = 3 | I1 yhwhudq wr uhwluh dqg wdnh xs qhz uroh lq 2024. |
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| Key = 4 | J1 zixiver xs vixmvi erh xeoi yt ria vspi mr 2024. |
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| Key = 5 | K1 ajyjwfs yt wjynwj fsi yfpj zu sjb wtqj ns 2024. |
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| Character | Count |
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| j | 7 |
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| y | 4 |
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| w | 4 |
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| s | 4 |
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| f | 3 |
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The most common English letter is E, so I am going to substitute J for E, since I assume it will be the most frequent character.
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E <- F <- G <- H <- I <- J
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This results in a key of 5, so translating the 2nd word: ajyjwfs -> veteran. This looks promising as it resulted in an English word. Trying the 4th word: wjynwj -> retire. Since this word is also in English, I can assume this candidate key is likely to work, so I will translate the entire string.
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`K1 ajyjwfs yt wjynwj fsi yfpj zu sjb wtqj ns 2024.` -> `F1 veteran to retire and take up new role in 2024.`
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1. Computer A sends 5 packets of data to computer B using Sliding Windows Flow Control.
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- The transmission time (time to put on the network) for a packet of data is 1 'time units'
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- Transmission time for an acknowledgement is 0 ‘time units’ (they are very small)
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- The propagation time (time to travel through network) for any transmission is random (between 3 and 5 ‘time units’, you choose a random time for each packet and acknowledgement sent).
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- B's packet processing time is 2. B cannot process multiple packets simultaneously.
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- The initial window size is 2
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Draw a diagram to show how flow will be controlled while the data is being sent. (10 marks)
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Transmission Time = 1tu
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Acknowledgement = 0tu
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Propagation Time = 3-5tu
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Processing Time (B) = 2tu
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Window Size = 2
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For propagation time, I am going to use the following values:
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( 0 = 3, 1 = 3, 2 = 4, 3 = 4, 4 = 5), and for simplicity's sake, I will mirror the propagation time there with the propagation time back for the acknowledgement.
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