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CTR mode calculate messages before new key is required

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I am studying a cryptography video on Coursera here titled: Modes of Operation: Many Time Key (CTR).

I have just two simple questions:

  1. At around 4:30 in the video they show 2^48 without saying where this number came from, perhaps it forms part of the AES specification?

  2. They then go on to explain that they plugged in the value 2^48 into the underlined equation on the line above, to calculate how many messages and of what size could be sent before a new key is required. How did they do this and what was the output?

Here is a picture from the video: Screen shot from video

kelalaka avatar
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square-root square-root!
questioner avatar
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@kelalaka What did they square root?
Maarten Bodewes avatar
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Well, you have three numbers 128, 32 and 48 as powers of two... They have chosen 48 bit as margin, but remember the birthday bound!
questioner avatar
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@MaartenBodewes Would you mind posting a reply to the question, because I did the square root of 128 and got a different answer... Then there is the case of the second question.
Maarten Bodewes avatar
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I think I can explain the slide, but I'm currently unsure about the term $|X|$... The squared q is needed because the nonce can collide otherwise. Basically you use 128 - 32 = 96 bits, and the square root of that is 48 bits. However, you still have to deal with the counter as well. However, I have too little context to create a readable answer - these slides are not the best readable without context - i.e. the presentation.
questioner avatar
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@MaartenBodewes |x| means the block length of X, so in AES case 2^128. Why did you subtract 32 from 128? I took the square root of 96 but my answer was not 48. I did link the video on the first line of my question.
Maarten Bodewes avatar
in flag
The block length of AES is $128$, not $2^{128}$, which is what confused me. Maybe $|X|$ is the group size of the input / output - that would make more sense: there are $2^{128}$ possible input and output values after all. The square root of 96 bits is 48 bits. $\sqrt{2^{96}} = (2 ^ {96})^ {1 \over 2} = 2 ^ {96 / 2} = 2^{48}$. Bits are already exponential.
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