Shortest acceptance versus clocked all-branch halting

On either a bit or the blank scanned for empty x, N's first option enters the accepting state in one transition. Its start state is not accepting, so no length-zero acceptance exists and ${\tau }_N(x)=1$. On the other branch it reaches cell one in state $q_{\mathrm{\infty }}$ after the first transition and cell t in that state after t transitions, for every $t\ge 1$. Hence this branch never halts and N has no finite all-branch running-time function. Its language is nevertheless all binary words.

Take a clock whose first transition on every symbol enters its halt state, so T(n)=1 exactly. Here its transition can be incorporated directly into the finite control: make the same two initial choices as N, but replace the $q_{\mathrm{\infty }}$ target by rejection. The resulting machine C has precisely two one-step branches, one accepting and one rejecting, on every input. Thus every branch halts after one step and $L(C)=\{x:{\tau }_N(x)\le 1\}=\{0,1{\}}^{\ast }$. This is an explicit optimized instance of the inclusive product-clock construction of F2, with no copying needed.

If one tests the cutoff zero instead, its language is empty because the initial state is nonaccepting. A decider can recognize it by a single rejecting transition, but it is not a zero-transition physical exact clock. At cutoff one, testing acceptance before expiry matters: rejecting immediately at the clock's last transition without testing N's simultaneous acceptance would incorrectly remove all inputs. C keeps that acceptance, as its two explicit branches show. This separates raw shortest and worst-branch times while verifying the correct clocked language and endpoints without AC.