Since everyone until now has simply defined what a hash function was, I will bite. A one-way function is not just a hash function — a function that loses information — but a function f for which, given an image y (“SE” or 294 in existing answers), it is difficult to find a pre-image x … Read more
You need to hash about 2^64 values to get a single collision among them, on average, if you don’t try to deliberately create collisions. Hash collisions are very similar to the Birthday problem. If you look at two arbitrary values, the collision probability is only 2-128. The problem with md5 is that it’s relatively easy … Read more
This has very little to do with hash functions. A cryptographic hash function may be part of the implementation, but it’s not required. Actually, it generates the digits on a time-based interval, if I press the button for it to generate the digits, it generates the digits and after about 25 seconds, and I press … Read more
There’s an article on the Code Project that has all the code you need to do this. It’s just a couple of classes so it’s a light-weight solution. To get the bytes for either a certificate or a key from the PEM file the following method will work, regardless of the order of the key … Read more
MD5 is designed to be cryptographically irreversible. In this case, the most important property is that it is computationally unfeasible to find the reverse of a hash, but it is easy to find the hash of any data. For example, let’s think about just operating on numbers (binary files after all, could be interpreted as … Read more
The weak collision resistance property is sometimes also referred to as second preimage resistance: Given an arbitrary x there exists no x’ with x’ != x so that h(x) = h(x’) Strong collision resistance on the other hand is defined as: There exist no x and x’ with x != x’ so that h(x) = … Read more
In practice and with respect to security, no, mathematically, yes. If you have a private key (N, D), there is algebraically an infinite number of solutions to the equation 1 = E*D (mod Phi(N)). However, if you make two such solutions (E, N) and (E’, N) that both satisfy the equation public, you will have … Read more
An IV is essential when the same key might ever be used to encrypt more than one message. The reason is because, under most encryption modes, two messages encrypted with the same key can be analyzed together. In a simple stream cipher, for instance, XORing two ciphertexts encrypted with the same key results in the … Read more
The IV has been thoroughly covered by the other answers, so I’ll focus just on storing the key. First… I can’t except it could not be done on a single server at software level. Anything done in software can be undone in software. You can encrypt, hide, and lock it in as many safes as … Read more
Do you have PKCS#12 or just PFX-file? In the Microsoft world it is the same, but other think another (see this archived page). You can try just following X509Certificate2 cert = X509Certificate2(byte[] rawData, “password”); X509Certificate2 cert2 = X509Certificate2(byte[] rawData, “password”, X509KeyStorageFlags.MachineKeySet | X509KeyStorageFlags.PersistKeySet | X509KeyStorageFlags.Exportable); (X509Certificate2(Byte[])) or X509Certificate2 cert = X509Certificate2(“C:\Path\my.pfx”, “password”); (see X509Certificate2(String, String) … Read more