Well, you could look it up in Wikipedia… But since you want an explanation, I’ll do my best here:

## Hash Functions

They provide a mapping between an arbitrary length input, and a (usually) fixed length (or smaller length) output. It can be anything from a simple crc32, to a full blown cryptographic hash function such as MD5 or SHA1/2/256/512. The point is that there’s a one-way mapping going on. It’s always a many:1 mapping (meaning there will always be collisions) since every function produces a smaller output than it’s capable of inputting (If you feed every possible 1mb file into MD5, you’ll get a ton of collisions).

The reason they are hard (or impossible in practicality) to reverse is because of how they work internally. Most cryptographic hash functions iterate over the input set many times to produce the output. So if we look at each fixed length chunk of input (which is algorithm dependent), the hash function will call that the current state. It will then iterate over the state and change it to a new one and use that as feedback into itself (MD5 does this 64 times for each 512bit chunk of data). It then somehow combines the resultant states from all these iterations back together to form the resultant hash.

Now, if you wanted to decode the hash, you’d first need to figure out how to split the given hash into its iterated states (1 possibility for inputs smaller than the size of a chunk of data, many for larger inputs). Then you’d need to reverse the iteration for each state. Now, to explain why this is VERY hard, imagine trying to deduce `a`

and `b`

from the following formula: `10 = a + b`

. There are 10 positive combinations of `a`

and `b`

that can work. Now loop over that a bunch of times: `tmp = a + b; a = b; b = tmp`

. For 64 iterations, you’d have over 10^64 possibilities to try. And that’s just a simple addition where some state is preserved from iteration to iteration. Real hash functions do a lot more than 1 operation (MD5 does about 15 operations on 4 state variables). And since the next iteration depends on the state of the previous and the previous is destroyed in creating the current state, it’s all but impossible to determine the input state that led to a given output state (for each iteration no less). Combine that, with the large number of possibilities involved, and decoding even an MD5 will take a near infinite (but not infinite) amount of resources. So many resources that it’s actually significantly cheaper to brute-force the hash if you have an idea of the size of the input (for smaller inputs) than it is to even try to decode the hash.

## Encryption Functions

They provide a 1:1 mapping between an arbitrary length input and output. And they are always reversible. The important thing to note is that it’s reversible using some method. And it’s always 1:1 for a given key. Now, there are multiple input:key pairs that might generate the same output (in fact there usually are, depending on the encryption function). Good encrypted data is indistinguishable from random noise. This is different from a good hash output which is always of a consistent format.

## Use Cases

Use a hash function when you want to compare a value but can’t store the plain representation (for any number of reasons). Passwords should fit this use-case very well since you don’t want to store them plain-text for security reasons (and shouldn’t). But what if you wanted to check a filesystem for pirated music files? It would be impractical to store 3 mb per music file. So instead, take the hash of the file, and store that (md5 would store 16 bytes instead of 3mb). That way, you just hash each file and compare to the stored database of hashes (This doesn’t work as well in practice because of re-encoding, changing file headers, etc, but it’s an example use-case).

Use a hash function when you’re checking validity of input data. That’s what they are designed for. If you have 2 pieces of input, and want to check to see if they are the same, run both through a hash function. The probability of a collision is astronomically low for small input sizes (assuming a good hash function). That’s why it’s recommended for passwords. For passwords up to 32 characters, md5 has 4 times the output space. SHA1 has 6 times the output space (approximately). SHA512 has about 16 times the output space. You don’t really care what the password *was*, you care if it’s the same as the one that was stored. That’s why you should use hashes for passwords.

Use encryption whenever you need to get the input data back out. Notice the word **need**. If you’re storing credit card numbers, you need to get them back out at some point, but don’t want to store them plain text. So instead, store the encrypted version and keep the key as safe as possible.

Hash functions are also great for signing data. For example, if you’re using HMAC, you sign a piece of data by taking a hash of the data concatenated with a known but not transmitted value (a secret value). So, you send the plain-text and the HMAC hash. Then, the receiver simply hashes the submitted data with the known value and checks to see if it matches the transmitted HMAC. If it’s the same, you know it wasn’t tampered with by a party without the secret value. This is commonly used in secure cookie systems by HTTP frameworks, as well as in message transmission of data over HTTP where you want some assurance of integrity in the data.

## A note on hashes for passwords:

A key feature of cryptographic hash functions is that they should be very fast to create, and **very** difficult/slow to reverse (so much so that it’s practically impossible). This poses a problem with passwords. If you store `sha512(password)`

, you’re not doing a thing to guard against rainbow tables or brute force attacks. Remember, the hash function was designed for speed. So it’s trivial for an attacker to just run a dictionary through the hash function and test each result.

Adding a salt helps matters since it adds a bit of unknown data to the hash. So instead of finding anything that matches `md5(foo)`

, they need to find something that when added to the known salt produces `md5(foo.salt)`

(which is very much harder to do). But it still doesn’t solve the speed problem since if they know the salt it’s just a matter of running the dictionary through.

So, there are ways of dealing with this. One popular method is called key strengthening (or key stretching). Basically, you iterate over a hash many times (thousands usually). This does two things. First, it slows down the runtime of the hashing algorithm significantly. Second, if implemented right (passing the input and salt back in on each iteration) actually increases the entropy (available space) for the output, reducing the chances of collisions. A trivial implementation is:

```
var hash = password + salt;
for (var i = 0; i < 5000; i++) {
hash = sha512(hash + password + salt);
}
```

There are other, more standard implementations such as PBKDF2, BCrypt. But this technique is used by quite a few security related systems (such as PGP, WPA, Apache and OpenSSL).

The bottom line, `hash(password)`

is not good enough. `hash(password + salt)`

is better, but still not good enough… Use a stretched hash mechanism to produce your password hashes…

## Another note on trivial stretching

**Do not under any circumstances feed the output of one hash directly back into the hash function**:

```
hash = sha512(password + salt);
for (i = 0; i < 1000; i++) {
hash = sha512(hash); // <-- Do NOT do this!
}
```

The reason for this has to do with collisions. Remember that all hash functions have collisions because the possible output space (the number of possible outputs) is smaller than then input space. To see why, let’s look at what happens. To preface this, let’s make the assumption that there’s a 0.001% chance of collision from `sha1()`

(it’s **much** lower in reality, but for demonstration purposes).

```
hash1 = sha1(password + salt);
```

Now, `hash1`

has a probability of collision of 0.001%. But when we do the next `hash2 = sha1(hash1);`

, **all collisions of hash1 automatically become collisions of hash2**. So now, we have hash1’s rate at 0.001%, and the 2nd

`sha1()`

call adds to that. So now, `hash2`

has a probability of collision of 0.002%. That’s twice as many chances! Each iteration will add another `0.001%`

chance of collision to the result. So, with 1000 iterations, the chance of collision jumped from a trivial 0.001% to 1%. Now, the degradation is linear, and the real probabilities are **far**smaller, but the effect is the same (an estimation of the chance of a single collision with

`md5`

is about 1/(2^{128}) or 1/(3×10

^{38}). While that seems small, thanks to the birthday attack it’s not really as small as it seems).

Instead, by re-appending the salt and password each time, you’re re-introducing data back into the hash function. So any collisions of any particular round are no longer collisions of the next round. So:

```
hash = sha512(password + salt);
for (i = 0; i < 1000; i++) {
hash = sha512(hash + password + salt);
}
```

Has the same chance of collision as the native `sha512`

function. Which is what you want. Use that instead.