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If $f:\{0,1\}^m\rightarrow \{0,1\}^n$ with $n\geq m,$ then of course there are, the set of one-to-one functions, but such a function is not a cryptographic hash function, since it lacks the compression property. If $n<m,$ (or more generally if $|X|>|Y|$ for $f:X\rightarrow Y$), collisions will happen.


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Hexadecimal is traditional -- by this, I mean that there first were command-line tools that used hexadecimal for output, then other people using the hash functions found it fit to stick to hexadecimal, if only to be able to compare their values with the output of the aforementioned tools. That's how traditions get established: a more-or-less random choice at ...


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Is there any possibility to make a successful verification with just modifying message and signature, and without modifying public-key ? One would certainly hope not. If you can, then you've just shown that the signature algorithm used is broken, and not to be trusted. For a signature method to be considered secure, then it is required that someone ...


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I'm not a Java programmer, and I didn't compile this, but I modified answer to this question to achieve a hash chain of length equal to four. byte[] bytesOfMessage = yourString.getBytes("UTF-8"); MessageDigest md = MessageDigest.getInstance("MD5"); byte[] thedigest = md.digest(md.digest(md.digest(md.digest(bytesOfMessage)))); PS. Don't use MD5 for ...



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