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As far as I know (which, admittedly, might be limited; I do not claim to possess encyclopedic knowledge of attacks on KDFs), there are no known practical attacks against KDF0 or KDF1 (which this page, following ISO-18033-2, seems to call KDF1 and KDF2 respectively) when instantiated with a secure hash function. Regarding the relative security of these KDFs ...


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In general this is not the case. Consider a PRF F which ignores the first bit of the key. Then you can distinguish F'(k,.)=F(.,k) from a random function (and thus break its security as a PRF) by querying it on inputs x and x', where with x' we denote x with the first bit flipped. Note that the outputs of F'(k,.) will collide on inputs x and x' as ...


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$\;\;\;$ No. $\:$ A PRF involves a secret key; a Random Oracle doesn't. $\;\;\;$ One can construct a PRF from a random oracle.


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Yes, you're misinterpretting the PRF. It's not just a hash function (and when you hit the end of the function function, start back at the beginning). Instead, if is a function that generates a rather long (actually, infinite) output; we use the first $N$ bits of that output to populate the various key values. See section 5 of RFC5246; we have: TLS's ...


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It is unclear if you wanted to compare TLS 1.1 PRF or TLS 1.2 PRF. Different TLS versions have different PRFs. Assuming you meant TLS 1.1 PRF although you linked TLS 1.2 RFC. TLS 1.1 PRF Short: HKDF is commonly a better choice than TLS 1.1 PRF, but not always. Consider these aspects: HKDF is a generic construct. HKDF is extract and expand. TLS1.1 PRF ...


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The answer given by Henrick is good, but I try to give a explanation with more details in security area. When you think about PRF (Pseudo Random Function), you will think that there are three elements with PRF, which is $K, X, Y$. $K$ means the key, $X$ means the message and $Y$ means the output. PRF is a function, when you give this function $K$ and $X$, ...



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