Dual_Ec_Drbg backdoor: a proof of concept

Dual_EC_DRBG backdoor: a proof of concept

What’s this ?

Dual_EC_DRBG is an pseudo-random number generator promoted by NIST in NIST SP 800-90A and created by NSA. This algorithm is problematic because it has been made mandatory by the FIPS norm (and should be implemented in every FIPS approved software) and some vendors even promoted this algorithm as first source of randomness in their applications. edit: I’ve been told it’s not the case anymore in FIPS-140-2 but the cat is already out of the bag

If you still believe Dual_EC_DRBG was not backdoored on purpose, please keep reading.
In 2007 already, Dan Shumow and Niels Ferguson from Microsoft showed that Dual_EC_DRBG algorithm could be backdoored. Twitter also uncovered recently that this algorithm was even patented in 2004 by Dan Brown (Not the Da Vinci guy, the Certicom one) as a “key escrow mechanism” (government jargon/lingo for trapdoor/backdoor).
I will go a little bit further in explaining how it works and give a proof-of-concept code, based on OpenSSL FIPS. This is in the best of my knowledge the only public proof of concept published today. (correct me if I’m wrong).

Dual_EC_DRBG in a nutshell

The PRNG works as following: it takes a seed that goes through a hashing algorithm. This data is then “fed” into two Elliptic Curve points, P and Q, before being slightly transformed and output.

In order to understand how the algorithm (and the backdoor) works, let’s see the relevant maths from Elliptic Curves:

Elliptic curves

Many types of elliptic curves exist. They are classified in function of their properties and equations. We’re going to see here the curve NIST-p256 which is one of the three curves being used by Dual_EC_DRBG. NIST-p384 and NIST-p521 have very similar characteristics. I’ll try to (poorly) give you the basic theoretics for EC, but I’m no mathematician. Please forgive me and correct me if I’m wrong.

A curve is a set of points that follow a group structure. The curve is defined over several parameters (for NIST GFp curves):

  • Equation: all the members of that group (called points) must satisfy this equation y^2 = x^3 + ax + b (mod p)
  • Prime modulus p: a prime number used to define the finite field Z/pZ in which the equation elements are defined.
  • The order r: this is the order of the EC and the total number of points into the group.
  • a and b: fixed integers used in curve’s equation. a is set to -3 in NIST GF(p) curves.
  • A Generator point defined by Gx and Gy: This point is considered as the base element of the group.

Curve members are points. A point is a pair of coordinates X and Y that satisfy the curve’s equation. They are written as capital letters such as G, P or Q. Points have some characteristics from groups:

  • They have an addition operation (+) defined between two points. (eg. P + Q).
  • This addition is commutative and associative.
  • Since you can add a point with itself as many time as you want, it has a scalar multiplication, which is the multiplication of a scalar (0..n) with a point and results in another point.
  • That scalar multiplication is associative/commutative (a(bP) = b(aP)).
  • This scalar should be in the group of integers modulo r (the order of the curve).

The group of Elliptic Curves is very useful in cryptography, because an equation such as iQ = P is very easy to resolve for Q or P (if you know i) but very hard to resolve for i (if all you know is P and Q). This is the Discrete logarithm problem in the EC group. That’s why most of the time the points are used as public parameters/keys and scalars as private counterparts.

All NIST curves have different parameters p, r, a, b and points G. These parameters have been constructed from a sha1 hash of a public seed but nobody knows how the seed itself has been chosen.

Enough on the theory, let’s study the PRNG !

Dual_EC_DRBG internals

Dual_EC_DRBG is defined in NIST SP800-90A page 60. It is an algorithm generating an infinite number of pseudo-random sequences from a single seed, taken in the first step or after an explicit reseed.
It is unfortunate that SP800-90A and the presentation from Microsoft use conflicting terminology (variable names). So I will use these variables:
i_n: Internal seed value.
o_n: External (output) value.
You can also see in the document two functions: phi and x(). phi is a function that does a mapping between a large integer and binary data. It doesn’t do much, in fact we can happily ignore it. x(P) simply gets the X coordinate of a point, as the X and Y coordinates are mostly redundant (as we’re going to see later). If we unroll the inductive feedback loop on the first two generated outputs, we get this:

i_0 = randomseed()
i_1 = phi(x(i_0 P))
o_0 = phi(x(i_1 Q))
output(30 least significant bytes of o_0)
i_2 = phi(x(i_1 P))
o_1 = phi(x(i_2 Q))
output(30 least significant bytes of o_1)

An attack

Let’s begin working on o_0 and look if we can guess o_1 from its content. We can see that o_0 is the X coordinate of a point, with 16 bits missing (we lost the 2 most significant bytes in the output process). In a NIST GFp curve, There are for every value of X, zero, one or two points on the curve. So we have at most 17 bits of bruteforce to do to recover the original point A. Let’s work on the hypothesis that we know the point A and it is equal (by definition) to A = i_1 Q.

Then the equation:

i_1 Q = A
but if we have a (secret) relation between P and Q such as dQ = P with d = secret number (our backdoor !):
i_1 d Q = d A multiplicating each side by d
i_1 P = d A (replacing dQ with P)

If you look carefully at the unrolled algorithm, you will notice that if we know i_1 P we can calculate i_2 = phi(x(d A)) and we have all the required information to calculate subsequent o_1 ..o_n and i_2 ..i_n. All we need to do is to guess a value of A (based on a bruteforce approach), multiply it by the secret value d, then multiply the resulting scalar with Q, strip two bytes and publish the output. It is also very interesting that if we learn (in a practical attack) the first 32 bytes generated by this PRNG, the 30 first bytes give us candidates for A and the remaining two bytes can be used to validate our findings. If the X value had been output on 32 bytes, we would have an one over two chance of success because of the two coexisting points on same coordinate X. (Remember from high school, second degree equations can have two solutions).

Generating the constants

As you have seen before, for our backdoor to work we need to choose the P and Q points in order to have the secret key to the backdoor. We have chosen to define P = dQ, however, that can’t work, because P is a generator for the EC group and its value has already been fixed. So, we have to find a value e such as Q = eP. This value can be calculated :

P = dQ
eP = edQ (mult. by e)

We need to find a value e such as ed = 1 for the curve C. The equation to resolve is
ed = 1 (mod r) where r is the order of the EC.
The value of e is the inverse of d modulo r. We can then use that value to generate Q.

/* 256 bits value randomly generated */
unsigned char d[]=
d_bn = BN_new();
assert(d_bn != NULL);

BN_bin2bn(d, 32, d_bn);
/* ensure d is well inside the group order */
EC_GROUP_get_order(curve, r, bn_ctx);
BN_mod(d_bn, d_bn, r, bn_ctx);
/* calculate Q = d * Generator + (NULL * NULL) */
ret = EC_POINT_mul(curve, my_Q, d_bn, NULL, NULL, bn_ctx);
assert(ret == 1);

/* calculate e = d^-1 (mod r) */
e_bn = BN_new();
assert(e_bn != NULL);

/* invert d to get the value of e */
assert(NULL != BN_mod_inverse(e_bn, d_bn, r, bn_ctx));

(note: I know I mixed up e with d between the code and blog post but that doesn’t change anything at all.)


You can find the proof of concept code on my github. I’ll comment how it works:

Install OpenSSL/FIPS

Most of the work needed for this POC actually was fighting with OpenSSL FIPS mode (getting it to compile at first) and finding the good APIs to use. OpenSSL FIPS and OpenSSL are two different software that share some codebase. I had to fetch a specific commit of OpenSSL FIPS (one that would compile) and patch it a little to have a few functions from Bignums usable from inside my application. I haven’t been able to mix FIPS and regular libcrypto, because of header incompatibilities (or a bug in my code I though was caused by incompatibilities). The README explains the steps to take (please read it).

Recover point A

If you remember, we have the 30 least significant bytes of the X coordinate, that means we need to bruteforce our way into A point candidates. This can be easily done in a loop over the 2^16 possibilities. OpenSSL doesn’t provide any way of recovering a point from a single coordinate (there exists a point compression algorithm, but it is so badly patented that it’s not implemented anywhere). We have to resolve the curve’s equation:
y^2 = x^3 - 3x + b (mod p)
Resolving such an equation for y is not so different from the equation resolving you learned at high school:

for (prefix = 0; prefix <= 0x10000 ; ++prefix){
    x_bin[0] = prefix >> 8;
    x_bin[1] = prefix & 0xff;
    BN_bin2bn(x_bin, 32, x_value);
    //bnprint("X value", x_value);

    /* try to find y such as */
    /* y^2 = x^3 - 3x + b (mod p) */
    /* tmp1 = x^2 */
    ret = BN_mod_mul(tmp1, x_value, x_value, &curve->field, bn_ctx);
    assert(ret == 1);

    ret = BN_set_word(tmp2, 3);
    assert(ret == 1);

    /* tmp1 = x^2 - 3 */
    ret = BN_mod_sub(tmp1, tmp1, tmp2, &curve->field, bn_ctx);
    assert(ret == 1);

    /* tmp1 = (x^2 -3) * x */
    ret = BN_mod_mul(tmp1, x_value, tmp1, &curve->field, bn_ctx);
    assert(ret == 1);

    /* tmp1 = x^3 - 3x + b */
    ret = BN_mod_add(tmp1, tmp1, b_bn, &curve->field, bn_ctx);
    assert(ret == 1);

    //bnprint("Y squared", tmp1);
    if (NULL != BN_mod_sqrt(y_value, tmp1, &curve->field, bn_ctx)) {
        //printf("value %x match !\n", prefix);
            bnprint("calculated Y", y_value);

        BN_mod_sub(y_value, zero_value, y_value, &curve->field, bn_ctx);
            bnprint("calculated Y opposite", y_value);
        test_candidate(buffer + 30, x_value, y_value, bn_ctx);
        valid_points += 2;

I mentioned that for every X, there were zero, one or two solutions: zero if the square root fails (not all elements of Z/pZ are quadratic residues), one if the result is 0 and two for all other answers. There are then two valid points (A_x, A_y) and (A_x, -A_y) where -A_y is the opposite of the first value modulo p. Explanation (thanks Rod):

y^2 = a^2
y^2 - a^2 = 0
(y-a) (y + a) = 0
y = a or y = -a

Recover PRNG state and generate next block

This part is pretty straightforward. We import the estimated x and y values, verify that they are in the curve (they should !), then multiply that point with the secret value. We then multiply Q with the resulting scalar and we get 30 bytes of the next output. If the two first bytes match, we have successfully guessed the 28 remaining bytes. That attack can recover everything that’s output by that PRNG till a reseed.

/* create the point A based on calculated coordinates x and y */
ret = EC_POINT_set_affine_coordinates_GFp(curve, point, x, y, bn_ctx);
assert(ret == 1);
/* Normally the point should be on curve but we never know */
if (!EC_POINT_is_on_curve(curve, point, bn_ctx))
    goto end;

/* calculates i2 = phi(x(e.A)) */
ret = EC_POINT_mul(curve, point, NULL, point, e_bn, bn_ctx);
assert(ret ==1);

ret = EC_POINT_get_affine_coordinates_GFp(curve, point, i2x, NULL, bn_ctx);
assert(ret ==1);
    bnprint ("i2_x", i2x);

/* calculate o1 = phi(x(i2 * Q)) */
ret = EC_POINT_mul(curve, point, NULL, my_Q, i2x, bn_ctx);
assert(ret == 1);
ret = EC_POINT_get_affine_coordinates_GFp(curve, point, o1x, NULL, bn_ctx);
    bnprint ("o1_x", o1x);
BN_bn2bin(o1x, o1x_bin);
if (o1x_bin[2] == buffer[0] && o1x_bin[3] == buffer[1]){
    printf("Found a match !\n");
    bnprint("A_x", x);
    bnprint("A_y", y);
    print_hex("prediction", o1x_bin + 4, 28);

Let’s run it !

aris@kalix86:~/dualec$ ./dual_ec_drbg_poc
s at start of generate:
[omitted: many output from openssl]
y coordinate at end of mul:
r in generate is:
Random bits written:
y coordinate at end of mul:
s in generate:
[omitted: many output from openssl]
PRNG output: ebc0d507c39f3b5ed8c96e789cc3e6861e1ddfb9d4170d3d5ff68e242437449e
Found a match !
A_x: 96e8ebc0d507c39f3b5ed8c96e789cc3e6861e1ddfb9d4170d3d5ff68e242437
A_y: 0663bc78276a258d2f422be407f881aa51b8d2d82ece31481db69dfbc6c4d010
prediction: a3cbc223507c197ec2598e6cff61cab0d75f89a68ccffcb7097c09d3
Reviewed 65502 valid points (candidates for A)
PRNG output: a3cbc223507c197ec2598e6cff61cab0d75f89a68ccffcb7097c09d3

Yikes !


It is quite obvious in light of the recent revelations from Snowden that this weakness was introduced by purpose by the NSA. It is very elegant and leaks its complete internal state in only 32 bytes of output, which is very impressive knowing it takes 32 bytes of input as a seed. It is obviously complete madness to use the reference implementation from NIST and I’m not the only one to be angry about it. You could use it with custom P and Q values, but that’s very seldom possible. Nevertheless having a whole EC point parameter leaked in the output makes it too easy to distinguish from real random and should never have been made into any specs at all. Let’s all bury that PRNG and the “security” companies bribed by NSA to enable backdoors by default for thirty silver coins.

edits: fixed Dan Brown’s employer name, changed a variable name to avoid confusion, fixed counting error 28 bytes to 30 bytes
note: I did not break the official algorithm. I do not know the secret value used to compute the Q constant, and thus cannot break the default implementation. Only NSA (and people with access to the key) can exploit the PRNG weakness.
note2: Some questions were raised about the security of that secret value. The secret value I speak of is the scalar used to generate Q. As EC is a cyclicgroup, it is proven that for every P and Q there is a value such as eP = Q.
How long would it be to bruteforce that value ? Finding the value e is equivalent to resolving the discrete logarithm problem in NIST-p256. ANSSI tells that GF(p) 256 bits has the same security as a 128bits symmetric cipher encryption key. Keys up to 80 bits are believed to be impossible to crack today, so we can say it’s more likely to be stolen by an NSA employee than being cracked.

edit3: NIST-p256 is a cyclic group and that’s what makes the note2 relation possible
edit4: There’s an awesome video explaining elliptic curves and Dual-ec-drbg.

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The Conversation {30 comments}

  1. Kyhwana {Wednesday January 1, 2014 @ 10:52 am}

    However, using your own Q and P parameters is patented, so you’re not allowed to do that..

  2. phi² {Wednesday January 1, 2014 @ 1:36 pm}

    Well Done, Aris. Happy New Year . Phi² 😉

  3. Victor {Wednesday January 1, 2014 @ 8:48 pm}

    Do you know that the link to your blog is being blocked by Facebook? We found out today when trying to share it. Google+ is cool with it. Very strange.

  4. me {Wednesday January 1, 2014 @ 9:20 pm}

    Awesome work! A pleasure to read!

  5. Aris Adamantiadis {Wednesday January 1, 2014 @ 9:26 pm}

    Victor: bad news :( try http://www.badcode.be/ instead, it redirects here.

  6. ehrichweiss {Wednesday January 1, 2014 @ 9:44 pm}

    Victor. I can confirm this. They’re refusing to give a preview of it and so I haven’t yet submitted it. I suspect when I do I’m going to get an error message.

  7. Aris Adamantiadis {Wednesday January 1, 2014 @ 9:49 pm}

    I got the error message :( I wrote a little request in the feedback form but have little hope that it works.

  8. infojunkie {Wednesday January 1, 2014 @ 9:53 pm}

    Also confirmed, had to type www badcode be to get it past facebook’s filter.

  9. victor {Wednesday January 1, 2014 @ 10:32 pm}

    It appears to work through donotlink.com, but through tinyurl it is now being blocked (it was not at first…). The plot thickens.

  10. Aris Adamantiadis {Wednesday January 1, 2014 @ 10:35 pm}

    Facebook even tell me now that my email address is invalid and I have to change it. it seems convinced that my domain name is malicious

  11. Pizza {Wednesday January 1, 2014 @ 11:51 pm}

    Facebook blocks url shortners. I’ve made a 301 redirect with a subdomain to your site and posted it on Facebook. This works, but for how long?

  12. Pizza {Wednesday January 1, 2014 @ 11:56 pm}

    After 5 minutes, my subdomain was also blocked by Facebook.

  13. Aris {Thursday January 2, 2014 @ 12:32 am}

    I think the best you can do is to post the link to the slashdot article http://it.slashdot.org/story/14/01/01/1830238/dualecdrbg-backdoor-a-proof-of-concept

  14. Victor {Thursday January 2, 2014 @ 12:39 am}

    Did you find out why they are blocking this blog?

  15. Sandra {Thursday January 2, 2014 @ 2:34 am}

    Very interesting read!

    Why can the secret value not be calculated?

    And what key are you talking about in “Only NSA (and people with access to the key) can exploit the PRNG weakness.” ?

  16. Dave {Thursday January 2, 2014 @ 3:26 am}

    Interesting to note that 32 bytes is the length of the SSL/TLS client/server random, so if you use the EC-DRBG for SSL/TLS then you’ll hand over its state to an attacker in your Hello message, and they can then get the next piece of output, the premaster secret, and decrypt your SSL session.

    (In practice it’s not quite that easy since only 28 bytes are RNG output, so you’d have to brute-force the remaining 4 bytes).

  17. Kell {Thursday January 2, 2014 @ 3:51 am}

    brute forcing 4 bytes is essentially nothing anymore. An 8 core 3GHz processor can run that many guesses in under 1 second.

  18. joev {Thursday January 2, 2014 @ 3:55 am}

    Yes can you explain why we can’t bruteforce d?

  19. Dave {Thursday January 2, 2014 @ 9:48 am}

    It’s a bit more than that since you need to simulate the SSL handshake to find out whether you’ve got the right premaster -> master -> crypto keys, but other than that it’s pretty easily do-able.

  20. Aris Adamantiadis {Thursday January 2, 2014 @ 10:36 am}

    Sandra: I added a note at end of paper over this question.
    Dave: Initially breaking a TLS session was my first target but did not have the time to go further, and I am not sure it would be worthwhile as there are many other ways of leaking session keys into compromised TLS sessions. The bruteforcing of the 4 remaining bytes would definitively be a problem because the POC already takes a minute to run on my i7 macbook. It’s definitively not doable on the fly.

  21. Matt Caswell {Thursday January 2, 2014 @ 11:34 am}

    “OpenSSL doesn’t provide any way of recovering a point from a single coordinate”

    I think you can use the following function for this:

  22. Aris Adamantiadis {Thursday January 2, 2014 @ 11:47 am}

    Hi Matt,
    It may have changed but last time (roughly 1 yr ago) when I tried to use that function I got error codes because it was hardcoded to reply back with an error code. It’s possible that this code had not made into the distribution I used because it is patented. (see http://www.google.com/patents/US6141420)

    Openssl code:
    You are right it exists ! and the good thing (for me) is that the code looks a lot like mine. My code may be more optimized for our use case because I generate two points from a single X and avoid calling that function two times (or I would have to extract the Y component and set its opposite). Please take a moment to think about how flawed the patent system is if I could “reinvent” it in an hour of my free time.

  23. Dave {Thursday January 2, 2014 @ 12:58 pm}

    You don’t need to do it on the fly, since most of the net still uses RSA key transport you can just record the session and decrypt it later (assuming it’s the client using EC_DRBG, since the server doesn’t contribute anything to the premaster secret). Once more of the net moves to applying PFS (via DH/ECDH) this will become more difficult to do, but that’ll take years to achieve (I know that Google does it, but that assumes you’re using Chrome to connect to a Google-run site, it doesn’t help the net in general). For things like banks, healthcare, and other industries doing X.12 (EDI, XML, etc) over SSL, it could take halfway to eternity before the problematic crypto is phased out.

  24. Dave {Thursday January 2, 2014 @ 12:59 pm}

    Sorry, by “doable” I meant doable for a nation-state adversary, not necessarily that you could throw together an attack in an afternoon.

  25. Manuel {Tuesday January 7, 2014 @ 11:06 am}

    Being nitpicky here, but “As EC is a group, it is proven that for every P and Q there is a value such as eP = Q.” is not correct. The result is true, but only because it’s a *cyclic* group of prime order. (The group being cyclic means that there exists a Q such that for any P, there is a d such that P = dQ. The fact the the group’s order is prime implies that it works in fact for any Q.) The NIST prime curves are cyclic groups of prime order, but this is not true of any curve.

    Of course this point is mainly irrelevant here, because the NSA doesn’t need this property: even it were not true, the NSA would just pick a point Q, an integer d, compute dQ and call than P, et voilà!

    Oh, and I nearly forgot: thanks and congratulations for this excellent article!

  26. Aris Adamantiadis {Tuesday January 7, 2014 @ 11:13 am}

    Hi Manual, I missed the cyclic term and you are right. It’s fixed now. And you’re right, this property is not needed to create the backdoor, it only proves this value exists even if no one knows it.
    Thanks for feedback :)

  27. nymble {Wednesday January 8, 2014 @ 7:32 am}

    Used by Cisco and MIcrosoft … in fact the only certified DRNG in the certification for MS: http://goo.gl/jj1sGC

  28. drakes00 {Friday February 28, 2014 @ 11:08 am}

    Hi, thanks a lot for this article :-)

    Just for me to understand right, what you’re basically telling is “Knowing d such that Q = dP, we are able to predict the next random bit sequences” ?

    For my culture, is there a way to perform the opposite deduction, like, for a given state (and maybe additional information), would we be able to predict former generated bit sequences ?

  29. Trey {Wednesday April 23, 2014 @ 12:53 pm}

    Hey, very interesting read. has this got anything to do with the heartbleed vulnerability in OPENSSL? my uneducated guess tells me that it does

  30. Guarana Dong {Sunday May 4, 2014 @ 3:40 pm}


    This doesn’t have anything to do with the “heart bleed” coined memory leak vulnerability attributed to openssl.

    There however is a pattern of similar/closely related vulnerabilities related to openssl/dual_ec_drb vulnerabilities.

    Look at the attributes of the dual_ec_drbg vulnerability, and openssl vulnerabilities, and you will see patterns, and preferred vectors of exploitation by the nsa. There are probably a LOT more undetected exploits out there that utilize same methods/behaviors as this article covers/openssl vulnerability, but they haven’t been discovered yet.

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