 Solution

For the exact difference, the representability condition becomesWithin the same operand square $[-H,U{]}^2$, the closed diagonal band between $y=x-U$ and $y=x+H$ is representable. Above the upper line, $x-y<-H$ gives negative overflow; below the lower line, $x-y>U$ gives positive overflow. The right-hand panel of the figure shows these regions, including their representable boundary points.

Positive overflow can occur only when $x\ge 0$ and $y<0$; negative overflow can occur only when $x<0$ and $y\ge 0$. The corresponding exact inequalities are still required: operands with different signs do not always overflow. Same-sign subtraction is always representable. In sign-bit terms, subtraction overflows exactly when the operands have different signs and the wrapped result's sign differs from that of $x$.

One must test subtraction directly or use its correct overflow rule: forming -y in the same $n$ bits fails at $y=-H$, even though some differences with that operand are representable. For example $(-H)-(-H)=0$ is valid.