Is it possible to reverse a SHA-1?
I'm thinking about using a SHA-1 to create a simple lightweight system to authenticate a small embedded system that communicates over an unencrypted connection.
Let's say that I create a sha1 like this with input from a "secret key" and spice it with a timestamp so that the SHA-1 will change all the time.
sha1("My Secret Key"+"a timestamp")
Then I include this SHA-1 in the communication and the server, which can do the same calculation. And hopefully, nobody would be able to figure out the "secret key".
But is this really true?
If you know that this is how I did it, you would know that I did put a timestamp in there and you would see the SHA-1.
Can you then use those two and figure out the "secret key"?
secret_key = bruteforce_sha1(sha1, timestamp)
Note1:
I guess you could brute force in some way, but how much work would that actually be?
Note2:
I don't plan to encrypt any data, I just would like to know who sent it.
No, you cannot reverse SHA-1, that is exactly why it is called a Secure Hash Algorithm.
What you should definitely be doing though, is include the message that is being transmitted into the hash calculation. Otherwise a man-in-the-middle could intercept the message, and use the signature (which only contains the sender's key and the timestamp) to attach it to a fake message (where it would still be valid).
And you should probably be using SHA-256 for new systems now.
sha("My Secret Key"+"a timestamp" + the whole message to be signed)
You also need to additionally transmit the timestamp in the clear, because otherwise you have no way to verify the digest (other than trying a lot of plausible timestamps).
If a brute force attack is feasible depends on the length of your secret key.
The security of your whole system would rely on this shared secret (because both sender and receiver need to know, but no one else). An attacker would try to go after the key (either but brute-force guessing or by trying to get it from your device) rather than trying to break SHA-1.
SHA-1 is a hash function that was designed to make it impractically difficult to reverse the operation. Such hash functions are often called one-way functions or cryptographic hash functions for this reason.
However, SHA-1's collision resistance was theoretically broken in 2005. This allows finding two different input that has the same hash value faster than the generic birthday attack that has 280 cost with 50% probability. In 2017, the collision attack become practicable as known as shattered.
As of 2015, NIST dropped SHA-1 for signatures. You should consider using something stronger like SHA-256 for new applications.
Jon Callas on SHA-1:
It's time to walk, but not run, to the fire exits. You don't see smoke, but the fire alarms have gone off.
The question is actually how to authenticate over an insecure session.
The standard why to do this is to use a message digest, e.g. HMAC.
You send the message plaintext as well as an accompanying hash of that message where your secret has been mixed in.
So instead of your:
sha1("My Secret Key"+"a timestamp")
You have:
msg,hmac("My Secret Key",sha(msg+msg_sequence_id))
The message sequence id is a simple counter to keep track by both parties to the number of messages they have exchanged in this 'session' - this prevents an attacker from simply replaying previous-seen messages.
This the industry standard and secure way of authenticating messages, whether they are encrypted or not.
(this is why you can't brute the hash:)
A hash is a one-way function, meaning that many inputs all produce the same output.
As you know the secret, and you can make a sensible guess as to the range of the timestamp, then you could iterate over all those timestamps, compute the hash and compare it.
Of course two or more timestamps within the range you examine might 'collide' i.e. although the timestamps are different, they generate the same hash.
So there is, fundamentally, no way to reverse the hash with any certainty.
In mathematical terms, only bijective functions have an inverse function. But hash functions are not injective as there are multiple input values that result in the same output value (collision).
So, no, hash functions can not be reversed. But you can look for such collisions.
Edit
As you want to authenticate the communication between your systems, I would suggest to use HMAC. This construct to calculate message authenticate codes can use different hash functions. You can use SHA-1, SHA-256 or whatever hash function you want.
And to authenticate the response to a specific request, I would send a nonce along with the request that needs to be used as salt to authenticate the response.
It is not entirely true that you cannot reverse SHA-1 encrypted string.
You cannot directly reverse one, but it can be done with rainbow tables.
Wikipedia:
A rainbow table is a precomputed table for reversing cryptographic hash functions, usually for cracking password hashes. Tables are usually used in recovering a plaintext password up to a certain length consisting of a limited set of characters.
Essentially, SHA-1 is only as safe as the strength of the password used. If users have long passwords with obscure combinations of characters, it is very unlikely that existing rainbow tables will have a key for the encrypted string.
You can test your encrypted SHA-1 strings here:
http://sha1.gromweb.com/
There are other rainbow tables on the internet that you can use so Google reverse SHA1.
Note that the best attacks against MD5 and SHA-1 have been about finding any two arbitrary messages m1 and m2 where h(m1) = h(m2) or finding m2 such that h(m1) = h(m2) and m1 != m2. Finding m1, given h(m1) is still computationally infeasible.
Also, you are using a MAC (message authentication code), so an attacker can't forget a message without knowing secret with one caveat - the general MAC construction that you used is susceptible to length extension attack - an attacker can in some circumstances forge a message m2|m3, h(secret, m2|m3) given m2, h(secret, m2). This is not an issue with just timestamp but it is an issue when you compute MAC over messages of arbitrary length. You could append the secret to timestamp instead of pre-pending but in general you are better off using HMAC with SHA1 digest (HMAC is just construction and can use MD5 or SHA as digest algorithms).
Finally, you are signing just the timestamp and the not the full request. An active attacker can easily attack the system especially if you have no replay protection (although even with replay protection, this flaw exists). For example, I can capture timestamp, HMAC(timestamp with secret) from one message and then use it in my own message and the server will accept it.
Best to send message, HMAC(message) with sufficiently long secret. The server can be assured of the integrity of the message and authenticity of the client.
You can depending on your threat scenario either add replay protection or note that it is not necessary since a message when replayed in entirety does not cause any problems.
Hashes are dependent on the input, and for the same input will give the same output.
So, in addition to the other answers, please keep the following in mind:
If you start the hash with the password, it is possible to pre-compute rainbow tables, and quickly add plausible timestamp values, which is much harder if you start with the timestamp.
So, rather than use
sha1("My Secret Key"+"a timestamp")
go for
sha1("a timestamp"+"My Secret Key")
I believe the accepted answer is technically right but wrong as it applies to the use case: to create & transmit tamper evident data over public/non-trusted mediums.
Because although it is technically highly-difficult to brute-force or reverse a SHA hash, when you are sending plain text "data & a hash of the data + secret" over the internet, as noted above, it is possible to intelligently get the secret after capturing enough samples of your data. Think about it - your data may be changing, but the secret key remains the same. So every time you send a new data blob out, it's a new sample to run basic cracking algorithms on. With 2 or more samples that contain different data & a hash of the data+secret, you can verify that the secret you determine is correct and not a false positive.
This scenario is similar to how Wifi crackers can crack wifi passwords after they capture enough data packets. After you gather enough data it's trivial to generate the secret key, even though you aren't technically reversing SHA1 or even SHA256. The ONLY way to ensure that your data has not been tampered with, or to verify who you are talking to on the other end, is to encrypt the entire data blob using GPG or the like (public & private keys). Hashing is, by nature, ALWAYS insecure when the data you are hashing is visible.
Practically speaking it really depends on the application and purpose of why you are hashing in the first place. If the level of security required is trivial or say you are inside of a 100% completely trusted network, then perhaps hashing would be a viable option. Hope no one on the network, or any intruder, is interested in your data. Otherwise, as far as I can determine at this time, the only other reliably viable option is key-based encryption. You can either encrypt the entire data blob or just sign it.
Note: This was one of the ways the British were able to crack the Enigma code during WW2, leading to favor the Allies.
Any thoughts on this?
SHA1 was designed to prevent recovery of the original text from the hash. However, SHA1 databases exists, that allow to lookup the common passwords by their SHA hash.
Is it possible to reverse a SHA-1?
SHA-1 was meant to be a collision-resistant hash, whose purpose is to make it hard to find distinct messages that have the same hash. It is also designed to have preimage-resistant, that is it should be hard to find a message having a prescribed hash, and second-preimage-resistant, so that it is hard to find a second message having the same hash as a prescribed message.
SHA-1's collision resistance is broken practically in 2017 by Google's team and NIST already removed the SHA-1 for signature purposes in 2015.
SHA-1 pre-image resistance, on the other hand, still exists. One should be careful about the pre-image resistance, if the input space is short, then finding the pre-image is easy. So, your secret should be at least 128-bit.
SHA-1("My Secret Key"+"a timestamp")
This is the pre-fix secret construction has an attack case known as the length extension attack on the Merkle-Damgard based hash function like SHA-1. Applied to the Flicker. One should not use this with SHA-1 or SHA-2. One can use
HMAC-SHA-256 (HMAC doesn't require the collision resistance of the hash function therefore SHA-1 and MD5 are still fine for HMAC, however, forgot about them) to achieve a better security system. HMAC has a cost of double call of the hash function. That is a weakness for time demanded systems. A note; HMAC is a beast in cryptography.
KMAC is the pre-fix secret construction from SHA-3, since SHA-3 has resistance to length extension attack, this is secure.
Use BLAKE2 with pre-fix construction and this is also secure since it has also resistance to length extension attacks. BLAKE is a really fast hash function, and now it has a parallel version BLAKE3, too (need some time for security analysis). Wireguard uses BLAKE2 as MAC.
Then I include this SHA-1 in the communication and the server, which can do the same calculation. And hopefully, nobody would be able to figure out the "secret key".
But is this really true?
If you know that this is how I did it, you would know that I did put a timestamp in there and you would see the SHA-1. Can you then use those two and figure out the "secret key"?
secret_key = bruteforce_sha1(sha1, timestamp)
You did not define the size of your secret. If your attacker knows the timestamp, then they try to look for it by searching. If we consider the collective power of the Bitcoin miners, as of 2022, they reach around ~293 double SHA-256 in a year. Therefore, you must adjust your security according to your risk. As of 2022, NIST's minimum security is 112-bit. One should consider the above 128-bit for the secret size.
Note1: I guess you could brute force in some way, but how much work would that actually be?
Given the answer above. As a special case, against the possible implementation of Grover's algorithm ( a Quantum algorithm for finding pre-images), one should use hash functions larger than 256 output size.
Note2: I don't plan to encrypt any data, I just would like to know who sent it.
This is not the way. Your construction can only work if the secret is mutually shared like a DHKE. That is the secret only known to party the sender and you. Instead of managing this, a better way is to use digital signatures to solve this issue. Besides, one will get non-repudiation, too.
Any hashing algorithm is reversible, if applied to strings of max length L. The only matter is the value of L. To assess it exactly, you could run the state of art dehashing utility, hashcat. It is optimized to get best performance of your hardware.
That's why you need long passwords, like 12 characters. Here they say for length 8 the password is dehashed (using brute force) in 24 hours (1 GPU involved). For each extra character multiply it by alphabet length (say 50). So for 9 characters you have 50 days, for 10 you have 6 years, and so on. It's definitely inaccurate, but can give us an idea, what the numbers could be.
I'm using (CryptoKit) to use AES-GCM to encrypt some data and authenticate it as well.
However, I was wondering how I would generate an AES-GCM key from a plain text password. Normally, you would use a KDF function for that, like PBKDF2.
In CryptoKit, there is a HKDF class which does about what I want: https://developer.apple.com/documentation/cryptokit/hkdf
However, I am wondering what KDF algorithm the DeriveKey function uses. Does it use PBKDF2? Does it use bcrypt? If so, how do I specify settings, or are the settings automatically determined?
HKDF is defined in RFC5869. It is intended to generate keys from some cryptographically secure "keying material" (IKM). It is not intended for stretching a human-generated password. As discussed in section 4 Applications of HKDF:
On the other hand, it is anticipated that some applications will not
be able to use HKDF "as-is" due to specific operational requirements,
or will be able to use it but without the full benefits of the
scheme. One significant example is the derivation of cryptographic
keys from a source of low entropy, such as a user's password. The
extract step in HKDF can concentrate existing entropy but cannot
amplify entropy. In the case of password-based KDFs, a main goal is
to slow down dictionary attacks using two ingredients: a salt value,
and the intentional slowing of the key derivation computation. HKDF
naturally accommodates the use of salt; however, a slowing down
mechanism is not part of this specification. Applications interested
in a password-based KDF should consider whether, for example, [PKCS5]
meets their needs better than HKDF.
I don't believe that CryptoKit offers a PBKDF of any kind (PBKDF2, scrypt, bcrypt, argon2). It's a very limited framework (I have yet to find a situation where it was useful). You will likely need to continue to use CommonCrypto for this, or implement it yourself (or use something like CryptoSwift, which I believe implements several).
I need to compare hashed data from different platforms. The hash function used is Keccak's SHA3.
I started with the assumption that the hash generated values were platform independent; but then these two online tools proved me the opposite:
http://sha3calculator.appspot.com/
http://emn178.github.io/online-tools/sha3_256.html
Here is the respective output when I hashed the word data (sha3 256bits):
8f54f1c2d0eb5771cd5bf67a6689fcd6eed9444d91a39e5ef32a9b4ae5ca14ff
efda893aa850b0c0e61f33325615b9d93bcf6b42d60d8f5d37ebc720fd4e3daf
My knowledge on the subject is quite limited, but I have always taken for granted that hash functions were platform independent (that's why we use MD5 checksum, right ?).
Any light on the subject will be highly appreciated.
Best.
Yes a hash algorithm should return the same value everywhere, aside from encoding issues.
You will get the same hash if you choose the algorithm Keccak-256 instead, i'm not sure what they use for SHA3-256:
http://emn178.github.io/online-tools/keccak_256.html
SHA-3, specifically (and distinctly from Keccak) is specified in FIPS 202 as of August 2015.
All implementations should product the same results with the same inputs.
Official test Vectors are available at the NIST CAVP Testing: Secure Hashing page.
I've never taken any classes on encryption or security and I'm trying to teach myself some basics, so forgive me if this is a silly question (don't worry, I'm not working on anything sensitive)
So, I'm playing around with Crypto++ so that I can make a signature of a file to see if the file has been edited by someone other than me. The test application that comes with the library looks like it has options (rs and rv) that do exactly what I want to do in my own program (verify the integrity of the signature of a file). Of course, before doing that I need to generate a public and private key. When doing so with the test application's g option it asks me to specify the key length in bits. What difference does the key length make?
The key length determines how hard it is for someone to break your cryptography. For digital signatures, that means how hard is it for someone to generate a fake signature.
For RSA a key length of 1024 bits is sufficient for non-sensitive information, but it should only be used for a few years and then replaced with a new key. 2048 bits is stronger and 4096 is stronger still.
For a naive brute-force attacker, adding a single bit to the key length doubles the amount of work they need to do to compromise your key. However, algorithms like RSA do not scale in this way: a 2048-bit RSA key is not 2^1024 times as hard to break as a 1024-bit key (unless the attacker is really stupid).
Generally public key algorithms (e.g. RSA) need much larger keys than symmetric key algorithms (e.g. AES) because they rely on different mathematical properties.
For a good primer on cryptography you should check out Peter Gutmann's godzilla crypto tutorial. It's pretty readable and gives you a good overview of how crypto works in its various forms.
I'm currently experimenting with both public-key and personal file encryption. The programs I use have 2048 bit RSA and 256 bit AES level encryption respectively. As a newbie to this stuff (I've only been a cypherpunk for about a month now - and am a little new to information systems) I'm not familiar with RSA algorithms, but that's not relevant here.
I know that unless some secret lab or NSA program happens to have a quantum computer, it is currently impossible to brute force hack the level of security these programs provide, but I was wondering how much more secure it would be to encrypt a file over and over again.
In a nutshell, what I would like to know is this:
When I encrypt a file using 256-bit AES, and then encrypt the already encrypted file once more (using 256 again), do I now have the equivalent of 512-bit AES security? This is pretty much a question of whether or not the the number of possible keys a brute force method would potentially have to test would be 2 x 2 to the 256th power or 2 to the 256th power squared. Being pessimistic, I think it is the former but I was wondering if 512-AES really is achievable by simply encrypting with 256-AES twice?
Once a file is encrypted several times so that you must keep using different keys or keep putting in passwords at each level of encryption, would someone** even recognize if they have gotten through the first level of encryption? I was thinking that perhaps - if one were to encrypt a file several times requiring several different passwords - a cracker would not have any way of knowing if they have even broken through the first level of encryption since all they would have would still be an encrypted file.
Here's an example:
Decrypted file
DKE$jptid UiWe
oxfialehv u%uk
Pretend for a moment that the last sequence is what a cracker had to work with - to brute-force their way back to the original file, the result they would have to get (prior to cracking through the next level of encryption) would still appear to be a totally useless file (the second line) once they break through the first level of encryption. Does this mean that anyone attempting to use brute-force would have no way of getting back to the original file since they presumably would still see nothing but encrypted files?
These are basically two questions that deal with the same thing: the effect of encrypting the same file over and over again. I have searched the web to find out what effect repeated encryption has on making a file secure, but aside from reading an anecdote somewhere that the answer to the first question is no, I have found nothing that pertains to the second spin on the same topic. I am especially curious about that last question.
**Assuming hypothetically that they somehow brute-forced their way through weak passwords - since this appears to be a technological possibility with 256-AES right now if you know how to make secure ones...
In general, if you encrypt a file with k-bit AES then again with k-bit AES, you only get (k+1) bits of security, rather than 2k bits of security, with a man-in-the-middle attack. The same holds for most types of encryption, like DES. (Note that triple-DES is not simply three rounds of encryption for this reason.)
Further, encrypting a file with method A and then with method B need not be even as strong as encrypting with method B alone! (This would rarely be the case unless method A is seriously flawed, though.) In contrast, you are guaranteed to be at least as strong as method A. (Anyone remembering the name of this theorem is encouraged to leave a comment; I've forgotten.)
Usually you're much better off simply choosing a single method as strong as possible.
For your second question: Yes, with most methods, an attacker would know that the first layer had been compromised.
More an opinion here...
First, when computer are strong enough to do a brute-force attack on AES-256 for example, it will be also for iterations of the same... doubling or tripling the time or effort is insignificant at that level.
Next, such considerations can be void depending on the application you are trying to use this encryption in... The "secrets" you will need to carry become bigger (number of iterations and all the different keys you will need, if in fact they are different), the time to do the encryption and the decryption will also need to increase.
My hunch is that iterating the encryption does not help much. Either the algorithm is strong enough to sustain a brute-force attach or it is not. The rest is all in the protection of the keys.
More practically, do you think your house is more protected if you have three identical or similar locks on your front door ? (and that includes number of keys for you to carry around, don't loose those keys, make sure windows and back door are secured also...)
Question 1:
The size of the solution space is going to be the same for two passes of the 256-bit key as the 512-bit key, since 2^(256+256) = 2^512
The actual running time of each decrypt() may increase non-linearly as the key-size grows (it would depend on the algorithm), in this case I think brute forcing the 256+256 would run faster than the 2^512, but would still be infeasible.
Question 2:
There are probably ways to identify certain ciphertext. I wouldn't be surprised if many algorithms leave some signature or artifacts that could be used for identification.