I'm curious as to how a backdoor can be built into a hash function. What might it take the form of? What's probability of getting one that can be backdoored in a way that can actually undermine its security? Could this backdoor be asymmetric?

  • $\begingroup$ Every backdoor is asymmetric in some fashion. That's what makes it a backdoor. $\endgroup$
    – cpast
    Apr 15 '15 at 4:42
  • $\begingroup$ Is it possible to backdoor a hash function and still have it compatible with other implementations? $\endgroup$
    – user9070
    May 15 '15 at 20:30
  • $\begingroup$ a paper for future readers who might be interested Malicious sha1 via modifying the constants used $\endgroup$
    – Ella Rose
    Aug 8 '16 at 2:49

Here is a "backdoored" hash function:

Let $p = 2q + 1$ be a big prime of length $2048$ bits, such that $q$ is also prime. Let $a$ be an integer of order $q$ modulo $p$, i.e. $a \neq 1$ but $a^q = 1 \pmod p$; it can be shown that $a = 4$ is always a valid solution. Let $s$ be a (secret) integer between $1$ and $q-1$, and let $b = a^s \pmod p$.

Then define operation $\phi(x, y)$ such that input $x$ is a sequence of $2048$ bits, and $y$ is a sequence of $2042$ bits (not $2048$); the output is a $2048$-bit value:

  • Concatenate $x$ and $y$ and split it back into $u$ and $v$, such that $u$ and $v$ both have length $2045$ bits each: $$u || v = x || y$$

  • Interpret $u$ and $v$ as unsigned big integers, using some convention (e.g. big-endian). Note that both $u$ and $v$ are lower than $2^{2045}$.

  • Compute: $$z = a^u b^v \pmod p$$

  • $\phi(x, y)$ is the encoding of $z$ over exactly $2048$ bits (there again, using a fixed convention).

Note that $\phi(x, y) \neq 0$ for all $x$ and $y$.

Now that you have $\phi$, define a hash function $h$ as follows:

  • For input $m$, pad it with some "removable" padding sequence (e.g. a Merkle-Damgård compliant padding) so that the total length of $m$ is now a multiple of $2042$.
  • Split $m$ into $n$ blocks of $2042$ bits (denoted $m_1$, $m_2$, ... $m_n$).
  • Define $x_0 = 0$ (a sequence of $2048$ bits, all of value $0$).
  • For $i = 1$ to $n$, define: $x_i = \phi(x_{i-1}, m_i)$
  • Define the hash output $h(m)$ to be SHA-256($x_n$).

Let's see the security of this construction:

  • Preimages: for a given $t$, finding $m$ such that $h(m) = t$ implies, in particular, finding some $x_n$ such that SHA-256($x_n$) $= t$, so our function $h$ is at least as much preimage-resistant as SHA-256.

  • Second preimages: second preimages are a special case of collisions, so $h$ will resist second preimages at least as well as it resists collisions. Ideally we would want something better (i.e. resistance up to $2^{256}$ for second preimages, even though collisions can be found in effort $2^{128}$), but since collisions should be infeasible anyway, this is probably acceptable.

  • Collisions: suppose that $m \neq m'$ and $h(m) = h(m')$; $m$ yields a sequence of $n$ words $x_1$ to $x_n$, while $m'$ yields $x'_1$ to $x'_{n'}$. We suppose, without loss of generality, that $n' \geq n$. Then one of the following holds:

    • $x_n \neq x'_{n'}$ but SHA-256($x_n$) $=$ SHA-256($x'_{n'}$). We have found a collision on SHA-256.

    • There is an index $i$ such that $x_{i-1} \neq x'_{i-1}$ but $x_i = x'_i$. This means that we know integers of at most $2045$ bits $u$, $v$, $u'$ and $v'$ such that $(u, v) \neq (u', v')$, but $a^u b^v = a^{u'} b^{v'} \pmod p$. This implies that $a^{u-u'} = b^{v'-v} \pmod p$. Since $a$ and $b$ have order $q$ (greater than $2^{2046}$) and $u-u'$ and $v-v'$ cannot be both $0$, then $u-u'$ and $v'-v$ are invertible modulo $q$, and we can write $b = a^g \pmod p$ with $g = (u-u')/(v'-v) \pmod q$ (i.e. $g$ is the secret value $s$). We just solved discrete logarithm.

    • There is an index $j$ such that $n \lt j \leq n'$ and $x'_j = x_n$. By "rewinding" the steps this implies that either we find an instance of the previous situation ($a^u b^v = a^{u'} b^{v'}$ for $(u,v) \neq (u',v')$), or we have $x'_{j-n} = x_0$, which is not possible since $x_0 = 0$ and $x'_{j-n} = \phi(x'_{j-n-1}, m_{j-n}) \neq 0$.

    Therefore, finding a collision on $h$ requires either finding a collision on SHA-256, or solving discrete logarithm modulo $p$. Since both are computationally infeasible (as far as we know), this function $h$ can be deemed "secure" as a hash function.

The backdoor is the knowledge of $s$. In all of the above, we used the $a$ and $b$ values "as is". The value of $s$ is not needed anywhere. However, if you know $s$, then you can compute collisions at will:

  • Generate two random blocks $m_1$ and $m'_1$. Compute the corresponding $x_1$ and $x'_1$.
  • Let $u$ be the first $2045$ bits of $x_1$, and $u'$ be the first $2045$ bits of $x'_1$.
  • Compute $k = (u-u')/s \pmod q$
  • If $k$ is less than $2^{2045}$, or greater than $q - 2^{2045}$, then you can find $v$ and $v'$ of $2045$ bits each, such that $v'-v = k \pmod q$, from which you can infer two blocks $m_2$ and $m'_2$ that match $(u,v)$ and $(u',v')$, respectively. You have your collision.
  • The condition on $k$ works with probability at least $1/2$. If you were unlucky, just try again with new random blocks $m_1$ and $m'_1$.

The above is thus a hash function with a backdoor. However, mind the fine print:

  • The backdoor is explicit. While the knowledge of $s$ is necessary to exploit the backdoor, everybody knows that it is there. Conversely, if you generate $a$ and $b$ as NUMS numbers (thus with $s$ convincingly unknown to everybody), then the hash function is "backdoorless" (but also useless).

  • When you exploit the backdoor, you produce two colliding messages $m$ and $m'$, from which anybody can recompute $s$. Thus, the backdoor is "one-shot".

  • The backdoor allows finding collisions, not preimages. A hash function designed to have a backdoor for preimages would need something else. Ideally you would use a one-way trapdoor permutation, but (to my knowledge) no such object is currently known, that would yield an output of suitable length (e.g. 256 bits).

  • $\begingroup$ Know of any way to construct a hash function with a multi-time backdoor? How did you come up with this one? $\endgroup$
    – Melab
    Apr 10 '16 at 8:54
  • $\begingroup$ I came up with that example by my own thinking (I was initially looking for a way to make a hash function based on elliptic curves and with a security that could reduced to that of discrete logarithm). I am not aware of any publication of that construction, other than this answer (whether a StackExchange answer counts as "publication" is another question). $\endgroup$ Apr 11 '16 at 12:54
  • $\begingroup$ How we can infer $m_2$ from $v$? please explain $\endgroup$
    – Lisbeth
    Aug 4 '17 at 11:47
  • $\begingroup$ By construction, $u||v = x_1 || m_2$. So it's mostly a matter of splitting the bit string at the right place to account for the slight differences in lengths (2042, 2045, 2048 bits). $\endgroup$ Aug 4 '17 at 12:12

I'm not very familiar with hash functions, but consider the following toy example: take any hash function $H$, and define a new function $H'$ which on input $x$ discards the last bit and applies $H$ to the result. $H'$ is not collision resistant because flipping the last bit of any input does not change the hash. Yet this may not be immediately obvious, and so could be considered a "backdoor".

Of course, the design of such a function would need to be kept secret for the "backdoor" to work, this is why you shouldn't trust any crypto primitive whose design is not public. And indeed, even if the design is public, the backdoor can of course be more subtle than that, this is why you shouldn't trust any crypto primitive which hasn't been extensively studied, even if its design is public.


The backdoor is possible. Imagine if we calculate the SHA256 of all combinations of a 256bit string, we already have a dictionary containing already a big subset of the SHA256 results. Once matched in the dictionary, the source can be replaced by the corresponding 256bit string. Now I agree and believe that bitcoin could be a conspiracy in that it helps to create SHA256 dictionaries!!! That could create messy conditions.

  • 4
    $\begingroup$ This is not a back door and even running the Bitcoin network at full power for centuries wouldn't help to find anything anywhere close to full preimages. $\endgroup$ Mar 31 '19 at 15:11
  • $\begingroup$ Have to upvote. It's not often that I get such a new and exciting conspiracy theory to fret over... $\endgroup$
    – Paul Uszak
    Mar 31 '19 at 16:50

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