Epfl Parity

This overview explains the retained EPFL comparison between CM and CSE-flat, separating kernel execution time from preparation cost. This matters because similar execution times do not show whether extra preparation is repaid over the actual pattern of reuse. Use a small circuit with an and gate, an inversion, and a final and. The output cone contains the gates that can feed this output. We will trace values through that cone before interpreting what a benchmark family name says about implementation cost. For one one zero, the first and gives one and not C gives one. The final and therefore gives one. Each intermediate belongs to a specific gate, so the signal path is visible rather than hidden in one compound expression. Changing C to one leaves the first gate unchanged but makes its inverted signal zero. The output becomes zero. Circuit structure and input values together determine this result; a graph's size alone does not describe every execution cost. The accepted EPFL AND/INV study found equal instruction and executed-operation counts for CM and CSE-flat on its selected cones. That structural equality is consistent with the observed kernel parity. The retained kernel-time ratio is CM divided by CSE-flat: about one. Its circuit-clustered 95% interval crosses one. Preparation is separate: CM took 4.11 times as much. Under the retained reuse calculation, 55 of 129 cases never broke even. These historical results describe selected AND/INV cones and their retained implementation. They do not contradict the different B2/B4 workload, and they do not show that every future circuit gives identical timings. Equal structural counts support a mechanism interpretation without measuring hardware traffic directly. Suppose two methods have the same per-use execution cost but unequal preparation. These are illustrative units rather than EPFL timings. The candidate spends eight preparing, while the control spends two. For any reuse count n, both totals grow by two per use. The preparation difference stays six. No crossover exists under these costs, even though the candidate can be correct and its execution can be at parity. The retained evidence determines what happens on the actual circuit set. A never-crossing result under a declared model means the stated preparation premium was not repaid there. It does not prove an impossibility for every future implementation or workload. On these selected cones, near-equal kernel time and higher preparation cost do not establish an end-to-end advantage. The practical comparison includes preparation, execution, and the actual reuse pattern, within the retained study's scope.