Attack on DSA modification with bad hash function

Given a (bad) modification of DSA where the hash function is the identity ($H(m)=m$), I am now supposed to design an attack. So $(m, (r,s))$ are given and I should be able to create a legit signature $(r',s')$ for a chosen $m'$ (of course $m'\neq m$).

Since retrieving the private key $x$ doesn't work. My new approach is to choose $m'$ as a modification of $m$ such that $m'=t \cdot m$. Then I want to calculate $(r',s')$ to be a legit signature of $m'$. So I was doing the verification step with $m'$ in order to see how to modify $r'$:

$v'=g^{m' \cdot w} y^{r \cdot w}=g^{m' \cdot w + xrw} \mod q$ now $v'$ is supposed to equal $r'$. This means $g^{m' \cdot w + xrw}=g^k$ so $m'w+xrw=k$. But I can't just choose k, can I (because it's already $k=mw+xrw$ ? And I don't know to modify $r$ in order to make this work...

• Does the statement allow the attacker to choose $m$ or/and $m'$? With what constraints? – fgrieu Jul 15 '14 at 9:36
• Apparantly, m is given and fix, but m' can be chosen. I updated the question. – CGFoX Jul 15 '14 at 15:46
• "Apparantly, I'll have to calculate the private key $x$" -- nope, recovering the private key is not required. – poncho Jul 15 '14 at 17:23
• Ok, new approach. I updated the question, please have a look. – CGFoX Jul 16 '14 at 9:26
• @CGFoX: If for any $t$ you could forge the signature for $m'=t\cdot m\bmod q$ in the weakened system, that would also break the real DSA [by choosing $t=H(m')\cdot H(m)^{-1}\bmod q$ and using the same attack]. You want to exhibit a narrower class of transformations $m'=f_t(m,r,s,p,q,g,y)$ for which an acceptable signature $(r',s')$ can be forged. – fgrieu Jul 16 '14 at 13:19

With my new idea I seem to solve the problem and answer my question, so I'll go ahead and post it as the answer. I choose the new $m'$ as $m'=t+m$ with $t>0$. Now the verification works like this:
$v'=g^{m'w} y^{rw}=g^{m'w+xrw}=g^{tw+mw+xrw}=g^{tw} \cdot g^k=r' \mod q$
So my new $r'=g^{tw}r=g^{ts^{-1}}r$
This means I can create a legit signature to any $m'=t+m$ which is $(r',s')=(g^{ts^{-1}}r, s)$. Correct?
• Again, if for any $t$ you could forge the signature for $m′=t+m\bmod q$ in the weakened system, that would also break the real DSA [by choosing $t=H(m′)-H(m)\bmod q$ and using the same attack]. So no this does not cut it, and you need a narrower choice of $m$, of the form $m'=f(m,r,s,p,q,g,y,t)$ for some $f$; and it won't be possible to find $t$ to obtain a chosen $m'$. The mistake in the argument given is that it is assumed $s$ does not change, rather than proven that with $s$ unchanged the verification procedure will pass with the $r'$ that you propose; indeed the verification will fail. – fgrieu Jul 17 '14 at 5:50
• Thanks for your comment, but I'm afraid I don't know what to do now. How am I supposed to find a $m'=f(m,r,s,p,q,y,t)$? Also I will have to hand this in in a few hours, so I would be extra thankful for some quick help. (I will still appreciate it later, too.) – CGFoX Jul 17 '14 at 7:46
• Write down the main equation used by the verifier for testing that $(m,r,s)$ is an acceptable signature in the weak system. The valid signature gives known values satisfying that equation. Your goal is finding $(m',r',s')$ with $m'\not\equiv m\pmod q$ which keeps the equation satisfied. What's $g^q\bmod p$? $y^q\bmod p$? What kind of changes does that allow while maintaining the equation satisfied? Perhaps replacing $s$ with $w=s^{-1}\bmod q$ in the equation (and $s'$ with $w'=s'^{-1}\bmod q$) will help you finding the appropriate changes. – fgrieu Jul 17 '14 at 9:34