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seen May 17 '13 at 14:46

Jun
25
awarded  Teacher
Apr
18
comment Why is $h(H, m) = E(m, H) \oplus m$ insecure?
I think you are right on the latter part, there do have various attacks on encryption algorithms.
Apr
18
answered Why is $h(H, m) = E(m, H) \oplus m$ insecure?
Apr
17
comment Why is $h(H, m) = E(m, H) \oplus m$ insecure?
It is true if E is 128 bit, but in the lecture slide [homework][1] the instructor did not specify encryption bits, consider a previous slide [example][2], is possible to construct a $H'$ that is not collision resistant to any random $m'$? [1]: i.imgur.com/NSuGUk3.jpg [2]: i.imgur.com/bX7mp2T.jpg
Apr
17
awarded  Student
Apr
17
awarded  Editor
Apr
17
revised Why is $h(H, m) = E(m, H) \oplus m$ insecure?
reformat in single $ LaTeX
Apr
17
asked Why is $h(H, m) = E(m, H) \oplus m$ insecure?