Finding the private key for this signature scheme
Let $m'$ be an arbitrary message. Then, the triple $(m', a', b')$ with:
\begin{align}
a' &amp;= -\mathcal{H}(m') \pmod{p-1} \\
b' &amp;= g^{-\mathcal{H}(m')} \cdot y^{-1} \pmod{p}
\end{align}
is a valid signature on $m'$. Since we forged a signature on an arbitrary message, we've achieved total break for this scheme.