I am very new to cryptography (so be kind), but I have a question that may seem silly.

If the one-time pad is the perfect cipher and impossible to crack, why would the following algorithm not be one of the strongest:

To encrypt:

  • generate a random bit array using a pseudo number generator with the key as the seed
  • The generated bit array is the size of the data
  • use a simple XOR to encrypt using the generated bit array

To decrypt:

  • pseudo-random number generator will generate the same bit array with the same seed (the key)
  • generate bit array the same size of the encrypted data
  • run XOR on encrypted data using bit array that's generated

This is very simplified, but you can imagine how for larger encryption you could break down the data into blocks, and in order to ensure that the block bit arrays are random a new seed for each can be derived from the key. So, if $f(k)$ is the pseudo random number generator, and b is the block number, you can encrypt each block with $f(k+b)$. ($k$ is key)

Instead of using a pseudo-random number generator, you could use some hash algorithm to generate the pad.

Ceteris paribus (assuming you can use equivalently strong keys) would this not be a very strong cipher method? Perhaps it may be strong and heavy on memory since the pad needs to be the size of the data. So during encryption and decryption you need to generate a pad that's equivalent to the size of the data.

Is there a flaw in this algorithm, or would something like this not be practical and much slower than current standards?

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    $\begingroup$ Stating the obvious: that's NOT the One-Time Pad; it's known under the name stream cipher; it ensures confidentiality (not security) of the message if the key is secret, not reused, and the Pseudo-Random Number Generator is cryptographically secure. $\endgroup$
    – fgrieu
    Commented Jan 29, 2014 at 16:23
  • $\begingroup$ A terminology note: The OTP is the pad, and the equivalent for a stream cipher is the keystream. $\endgroup$
    – forest
    Commented Jan 7, 2019 at 3:37

9 Answers 9


It's a good question. As pg1989 said, this is the basis behind stream ciphers, which are very fast in practice.

I thought I'd quickly expand upon your statement that "the one-time pad is the perfect cipher and impossible to crack." This is true, in a sense, but it's worth pointing out that sometimes an attacker wants to do something simpler than "cracking" the encryption scheme in order to recover all plaintexts.

For instance, an attacker may just want to change the message so that the honest recipient decrypts something else. As a simple example, suppose that the attacker knew for sure that the message was either "yes" or "no!" and just wants to change the message to the other value. Note that:

  1. This is simple to do for messages encrypted using a one-time pad. The linked Wikipedia page describes why this is true, but it's probably more fun to figure it out for yourself.

  2. The attacker can do this without learning the plaintext. Hence, the original claim of a one-time pad being uncrackable is still true.

There are some other issues with a one-time pad construction too, such as the fact that usually some state will be required to make sure that you don't re-use the seed during the next encryption. Hence, while your question is definitely not silly, and stream ciphers are very useful in certain circumstances, they also have their limits.

  • $\begingroup$ Awesome answer. Thank you. But if the key between the two parties is absolutely secret, how can the attacker change the message? The wiki article dealt with RSA encryption where there is a shared public key in addition to the private keys. I might be missing the connection though. $\endgroup$
    – dardawk
    Commented Sep 19, 2013 at 13:27
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    $\begingroup$ One Time Pads do not provide data integrity. So an attacker can change the message and the receiver won't know that the data has been changed. The attacker cannot recover the plain text, but they can change the cipher text. Mayank's yes/no is an example of this. The attacker can see Yes or No's cipher text and can change this cipher text so the plain text is changed when the receiver decrypts it. I learned this from the class pg1989 recommended. $\endgroup$
    – Ryan
    Commented Sep 19, 2013 at 13:41
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    $\begingroup$ Hmm, but if "Yes" is encrypted using the pad, the attacker does not know the message or the pad so if they try to send something else, decrypting an attacker's message with the original pad will surely produce garbage? Of course, this method is useless against a plain-text attack since the user gen infer the pad very easily. Is this what you mean? Again, I'm just at the beginning of the course so perhaps I just need to be patient. :) $\endgroup$
    – dardawk
    Commented Sep 19, 2013 at 14:35
  • $\begingroup$ @dardawk Well, lets imagine that you are sending single-bit yes/no messages that you are encrypting using a one-time pad shared between you and the recipient. By XORing the bit with a shared secret bit, you can ensure that, whatever the message you are sending, nobody listening in on the line will be able to tell whether the message means yes or no. However, by just inverting the message bit, the attacker can derive a valid cryptogram for the opposite value. Then they can send it to the recipient, who will see 'yes' if you sent 'no, and 'no' if you sent 'yes'. $\endgroup$ Commented Sep 19, 2013 at 15:03
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    $\begingroup$ Let $m \in \left\{ { \text{yes}, \text{no!} } \right\}$, $k$ be a random key, and the ciphertext $c = k \oplus m$. You wish to modify $c$ so that it decrypts to the other possible plaintext. Let $x = \text{yes} \oplus \text{no!}$. Calculate $c^\prime = c \oplus x$. This is your new ciphertext. Note that if $m = \text{yes}$, $c' = k \oplus \text{yes} \oplus \text{yes} \oplus \text{no!} = k \oplus \text{no!}$ and if $m = \text{no!}$, $c' = k \oplus \text{no!} \oplus \text{no!} \oplus \text{yes} = k \oplus \text{yes}$. $\endgroup$ Commented Sep 19, 2013 at 17:35

Yes, this is a widely-used cryptographic construction called a stream cipher. For more information about this and other encryption schemes, Coursera's cryptography class is a good resource.


Congratulations you just reinvented the stream cipher.

The main strength of the one-time pad is that the key space is as large as the message space. This means that any cipher-text only attacks always fail because all plaintexts are valid.

This automatically means that any construct that decreases the key space (like using a seed for a PRNG) severely weakens the one-time pad.

Besides this a one-time pad (and the XOR based stream cipher) is very vulnerable to repeated-key attacks while there are other ciphers which are (more) secure against it.


If using a cryptographically-secure random number generator then the result is a stream cipher. If using actual random numbers, then it's a one-time pad.

Any output you get from a random source needs to be run through a randomness extractor anyway in a 2:1 ratio (2 bits in, 1 bit out).

Don't forget to provide a MAC along with the ciphertext to prevent an attacker altering the message.

If you want to use CSPRNG don't use just one algorithm. There's potential for a powerful attacker to determine a pattern from it. Generate a few separate keys with quality randomness of 512 bits+ length each. Then with each key generate a separate key stream using a different algorithm. Then combine/mix the streams together, perhaps with XOR.

Don't forget you need a way to keep in sync with the other person even in network failure so you don't re-use part of the stream/pad.


From what I understand from your question, you are describing a stream cipher.

If the one-time pad is the perfect cipher and impossible to crack, why would the following algorithm not be one of the strongest ...

You're on the right track; a one-time pad is essentially a perfect (unbreakable) stream cipher.

Without going into (any) mathematical details, the main difference between your average stream cipher and a one-time pad (OTP) is that an ideal OTP requires a random key at least as long as the plaintext, whereas your average stream cipher requires a key length of only N, regardless of plaintext size. (Assuming keys for both cases are completely random).

In a perfect universe, your stream cipher would use a short key size to come up with an unguessable, unpredictable, completely random number of length equal to the plaintext, and XOR the two together.

What, however, would be the purpose of that? If the end result of the cipher is nondeterministic, there would be no way, given only the key, to recover the original message.

What we need then is a deterministic mathematical function where we can create a long string of predetermined digits given a particular input (the key), which are otherwise (very) unlikely to be guessed without the key.

Our system's security is now defined by the weaker of either our key size, or the flaws in our mathematical function.

  • $\begingroup$ Right. What I meant by the pseudo random number generator is the deterministic f'n that would output a long string unique to the key. I used the pseudo random number generator just because it was fresh in my mind. Thanks for your answer! $\endgroup$
    – dardawk
    Commented Sep 19, 2013 at 13:31

I also want to throw in a bit more on this - as why you would not want to do that. To use the analogy of sending secret message from a capitol to a diplomatic embassy, it is a great and secure, somewhat "foolproof" encryption method - but the practical use has some drawbacks:

  • You would need as much random data (i.e. the "codebook") as you would ever want to transmit. If you sent enough messages with sufficient size, you have exhausted your random pool. You cannot re-use the same random numbers ("codes") or it would be insecure.
  • You would now need a secure and reliable way to hand the codebook over. If there were any eavesdroppers, you couldn't do this. This means that you can't just email your key over to someone, you'd need to transport it in something akin to a locked attache case handcuffed to your wrist with a tamper proof (or tamper evident) seal.
  • Your local embassy has the codebook. Therefore, it is possible that a spy inside the embassy could use it to forge a message to the embassy, pretending it was from the capitol. (i.e. it is a symmetric cypher, the codebook can be used two-ways).

So when your random data pool ran out, you'd need to send out a new attache, and could not use encryption until this was done.

With other methods, a public key can be sent in the clear - if an eavesdropper were to receive it, it would do them little good, and there is no "pool" of random data that would be exhausted.

As awkward and restrictive as it is however, I still believe there are HIGH SECURITY applications in which this may provide the strongest encryption.

  • $\begingroup$ Brad, that's why I mention the pseudo-random-number generator (which I should have called a deterministic f'n that generates a long unique string--the pad--based on a much shorter input key). The keys still need to be agreed upon and the "spy" can get at they key just as they could at any (non RSA\public-key) cipher. Of course, I see your point that the one-time pad would not be useful if you want to re-use the key. I just wanted to compare this simple algorithm against against other private key ciphers: AES, serpent, twofish, etc. ... $\endgroup$
    – dardawk
    Commented Sep 19, 2013 at 14:41
  • $\begingroup$ Ah, then yes - I concur with the comments about "reinventing the stream cipher" ;-) The big keypoint (that I was going to say but thought it might be too obvious) was that the "security is only as good as the uncrackable randomness of your "codebook" data. $\endgroup$
    – Brad
    Commented Sep 20, 2013 at 15:09
  • $\begingroup$ Governments and diplomats also have the option of using a diplomatic bag to transfer items to their embassies in another country. These have diplomatic immunity from search or seizure under the Vienna Convention on Diplomatic Relations. So it's not as difficult as you're making it out to be. A locked attache case handcuffed to your wrist with a tamper proof (or tamper evident) seal would probably still be required however. Obviously there would be no panic when you ran out of random data. You would have a program which alerted you ahead of time. $\endgroup$
    – NDF1
    Commented Jan 14, 2015 at 10:25

This is indeed a good question; let me try to make it a bit more precise. Suppose:

  • Alice has a plaintext message of some number of bits, call it p.
  • Alice and Bob share a crypto-strength random number generator that generates n truly random bits.
  • Alice and Bob share a pseudo-random number generator that can take a seed of size n and produce one of 2n sequences of p bits.
  • Alice and Bob have an insecure channel and a secure channel.
  • Alice (or Bob) creates a truly random key KEY of size n and sends it to Bob (or Alice) over the secure channel.
  • Alice creates a pseudo-random sequence SEQ of size p and xors it with the plaintext to produce a ciphertext of size p
  • Alice sends the ciphertext to Bob over the insecure channel.
  • Bob decrypts the ciphertext in the obvious way.
  • Details of the system are not part of the key. That is, the attacker should be presumed to know precisely how the PRNG works.

Ok, so first of all, the obvious criticism is: if Alice and Bob have a secure channel then why are they messing with encryption at all? And the answer is: the secure channel may be more expensive or only available at inconvenient times.

The second obvious criticism is: if Alice and Bob have a device that can generate n truly random bits then why not use it to generate p truly random bits in the first place, and get the pseudo-random number generator out of the picture? And the answer is: because sending p bits over the secure channel may be too expensive.

So what attacks could be mounted against this system?

The accepted answer points out that this system only defends against one attack, namely, it makes it very hard for Eve to decrypt the intercepted ciphertext. It does not provide any mechanism for Bob to verify that the ciphertext received was actually produced by Alice. If bits of that ciphertext are flipped, they'll be flipped in the plaintext too and Bob has no way of knowing that.

Another attack is: suppose n is a relatively small number, say, 32. That means there are only four billion possible values for KEY and therefore only that many pseudo-random bit sequences SEQ. Eve can obtain the ciphertext and simply try all of them until one of them decrypts it into something sensible. This implies that n had better be pretty large; large enough that this brute-force attack is infeasible. (And of course, if n has to be so large to prevent this attack that it is a significant fraction of p, then again, take the PRGN out of the picture and just generate as many truly random bits as there are bits in the plaintext.)

And yet another attack is: suppose the attacker manages to obtain the whole ciphertext -- easy -- and manages to obtain or guess k bits of the plaintext. This is not as hard as you might think; lots of messages have common patterns in them. (WWII era allied codebreakers made good progress by assuming that a significant fraction of messages would end in a greeting to the Führer; and they were right.) If k bits of the plaintext and all of the ciphertext are known then k bits of SEQ are known. Now the question is: Can the attacker make a good guess as to the value of KEY given k bits of SEQ and knowledge of the internals of the PRNG? If they can then the task of generating SEQ just got a lot easier for the attacker. The PRNG had better be carefully designed to make this kind of key recovery infeasible.


You would need to generate random numbers with the same size and rate as your data. True randomness is curical, and you should not re-use them. Quantum RNGs may be the solution.

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    $\begingroup$ How would the recipient decrypt the data then? (Not my downvote.) $\endgroup$
    – user
    Commented Sep 19, 2013 at 9:31
  • $\begingroup$ @MichaelKjörling By obtaining a copy of the one-time pad by a separate channel. This is the problem with one-time pads: they're ultimately secure in some theoretical sense, but impractical. $\endgroup$ Commented Sep 19, 2013 at 11:53
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    $\begingroup$ @Gilles Exactly. If you have a secure channel to transmit the OTP, why not just use that channel to transmit the message in the first place? If on the other hand the second channel is no more secure than the first, you have solved nothing from a confidentiality point of view - anyone who can eavesdrop on one channel can also eavesdrop on the other. It may pose a problem to a garden variety attacker, but not to a determined adversary - and to protect against a garden variety attacker, a reasonable mode of AES-256 is more than enough and requires only the secure distribution of 32 bytes of key. $\endgroup$
    – user
    Commented Sep 19, 2013 at 12:23
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    $\begingroup$ @Gilles I did comment somewhere (might have been IT Security) that OTPs have a use where you have a secure channel now but need to transmit provably secure messages later. But even then, you still at the very least have the problem of storing the pad until it's needed. $\endgroup$
    – user
    Commented Sep 19, 2013 at 12:24
  • $\begingroup$ @MichaelKjörling The channel might be secure but high latency. For example consider a trusted courier who takes several hours to fly around the world. Or some conspirators who meet in person for the key-setup, and then later on exchange encrypted messages. A one-time-pad allows you to seperate the time of message exchange from the time you have the trusted channel. $\endgroup$ Commented Sep 19, 2013 at 15:52

I'm not an expert either, but I think an obvious flaw is that if you used a well-known PRNG algorithm with well-known constants, then it might be possible for an attacker to guess the pseudo-random number stream.

I'm using a method similar to the one you describe, but I have 4 PRNGs, each with "non-standard" constants I chose myself. I generate the pseudo-random key stream like this:

  1. using PRNG#4 generate a random seed#1 for PRNG#1, then use this to generate key stream 1, to the same length as the cleartext.

  2. using PRNG#4 generate a random seed#2 for PRNG#2, then use this to generate key stream 2, to the same length as the cleartext.

  3. using PRNG#4 generate a random seed#3 for PRNG#3, then use this to generate key stream 3, to the same length as the cleartext.

  4. XOR key streams 1, 2 & 3 together, to create the final key stream, which is XOR'd with the cleartext to create the cyphertext. Append seed#1, seed#2 & seed#3 to the cyphertext to enable decryption by the exact same process.

If each PRNG can generate (e.g.) 64000 PRNs before repeating, then there will be 64000 * 64000 * 64000 = 260000000000000 different final key streams. This will help ensure that any particular final key stream will very rarely get re-used. I tested my algorithm for 300 million encryptions and no final key stream ever got repeated. Because I know the number encryptions I perform per year (approx 1 million), I'm confident that no final key stream would get re-used within 300 years - well beyond the lifetime of my system or its data. And because I have chosen my own PRNG constants, it would be difficult for an attacker to guess them. An important point to make here is that should you decide to choose your own PRNG constants, you need to do some sanity checking of the randomness of the resulting PRN stream - badly chosen constants can result in a "PRN stream" that is completely predictable. My system depends on the encryption algorithm and implementation code being kept secure & private, and with only the cyphertext publicly visible (which is the case).

The usual warnings apply here... when there are so many theoretically proven encryption algorithms available, you need a good reason to devise your own, unless it's just for fun.

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    $\begingroup$ 1) Any decent stream cipher will only repeat after more than 2^60 output bytes, so this limitation is not of practical concern 2) The number of different streams and the length of one stream is unrelated. Typical stream ciphers will have at least 2^128 different streams. 3) It's far more important to use a good stream cipher than combining multiple. Very little reason to modify the constants yourself, just choose a random key and IV as seed. $\endgroup$ Commented Jan 29, 2014 at 10:19
  • $\begingroup$ Am I missing something? In step 4, you append the seeds to the ciphertext, and apparently send them in the clear. Why can't an attacker use those seeds to decrypt the message? If the answer is "because the attacker doesn't know the 'non-standard' constants in the key streams", well, that means that those non-standard constants are effectively the key to this system; using keys bolted into the system makes for, at least, very questionable security. $\endgroup$
    – poncho
    Commented Jul 28, 2014 at 17:29
  • $\begingroup$ To guess the PRN stream used to encrypt the plaintext, the attacker would need to guess:- (a) the PRNG algorithm I used, and (b) the exact "non-standard" constants I plugged into that algorithm. Admittedly, my resulting PRN stream is definitely not strictly cryptographically secure, but I think its secure enough for my purposes. And if I've made a mistake and someone breaks it, then I will have to take responsibility. $\endgroup$ Commented Jul 28, 2014 at 17:33
  • $\begingroup$ PS.. If anyone thinks they can break it, I would be very interested to know. $\endgroup$ Commented Jul 28, 2014 at 17:41
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    $\begingroup$ @MarkTaylor Regarding your …the attacker would need to guess:- (a) the PRNG algorithm I used, and (b) the exact "non-standard" constants I plugged into that algorithm… and your If anyone thinks they can break it, I would be very interested to know. – Maybe you’re not aware if it yet, but cryptographers assume Kerckhoff’s principle. Everything else would be “security by obscurity”, but not cryptography. In cryptography, the complete algorithm is assumed to be known by the attacker/enemy/Eve and only the key represents an unknown secret. $\endgroup$
    – e-sushi
    Commented Jul 28, 2014 at 18:51

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