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03-12 Key Exchange Solution

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    The answer is the first one.
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    Bob should compute yA to the xB power modulo q.
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    The second one would compute the same thing.
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    This is in fact exactly what Alice computed,
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    but Bob can't do this because he doesn't know xA.
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    The third one wouldn't compute the same key,
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    so the correctness property is that Alice and Bob obtain the same key,
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    and we can show this by just plugging in the values.
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    The key Alice computed was yB to the xA.
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    The value of yB is g to the xB, so that's equivalent
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    to g to the xB xA mod q.
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    The key that Bob would compute--and we'll write that as key BA
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    since we haven't yet shown that they're equivalent using this equation.
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    Well, yA is g to the xA, so this is g to the xA xB mod q.
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    And we already showed that powers of powers are commutative,
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    so these two are equivalent.
Title:
03-12 Key Exchange Solution
Team:
Udacity
Project:
CS387 - Applied Cryptography
Duration:
0:50
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