WEBVTT

1
00:00:00.000 --> 00:00:05.267
This overview explains the retained EPFL
comparison between CM and CSE-flat,

2
00:00:05.267 --> 00:00:09.333
separating kernel execution time from preparation cost.

3
00:00:09.333 --> 00:00:17.155
This matters because similar execution times do not show whether
extra preparation is repaid over the actual pattern of reuse.

4
00:00:17.155 --> 00:00:17.167
This matters because similar execution times do not show whether
extra preparation is repaid over the actual pattern of reuse.

5
00:00:17.467 --> 00:00:22.300
Use a small circuit with an and gate, an inversion, and a final and.

6
00:00:22.300 --> 00:00:26.400
The output cone contains the gates that can feed this output.

7
00:00:26.400 --> 00:00:31.033
We will trace values through that cone before interpreting what a benchmark family name

8
00:00:31.033 --> 00:00:34.667
says about implementation cost.

9
00:00:34.667 --> 00:00:39.333
For one one zero, the first and gives one and not C gives one.

10
00:00:39.333 --> 00:00:41.967
The final and therefore gives one.

11
00:00:41.967 --> 00:00:46.900
Each intermediate belongs to a specific gate, so the signal path is visible rather than

12
00:00:46.900 --> 00:00:50.500
hidden in one compound expression.

13
00:00:50.500 --> 00:00:56.033
Changing C to one leaves the first gate unchanged but makes its inverted signal zero.

14
00:00:56.033 --> 00:00:58.233
The output becomes zero.

15
00:00:58.233 --> 00:01:03.167
Circuit structure and input values together determine this result; a graph's size alone

16
00:01:03.167 --> 00:01:07.484
does not describe every execution cost.

17
00:01:07.484 --> 00:01:07.500
does not describe every execution cost.

18
00:01:07.500 --> 00:01:10.800
The accepted EPFL AND/INV study

19
00:01:10.800 --> 00:01:18.533
found equal instruction and executed-operation
counts for CM and CSE-flat on its selected cones.

20
00:01:18.533 --> 00:01:23.667
That structural equality is consistent with the observed kernel parity.

21
00:01:23.667 --> 00:01:29.933
The retained kernel-time ratio is CM divided by CSE-flat: about one.

22
00:01:29.933 --> 00:01:34.767
Its circuit-clustered 95% interval crosses one.

23
00:01:34.767 --> 00:01:40.567
Preparation is separate: CM took 4.11 times as much.

24
00:01:40.567 --> 00:01:48.100
Under the retained reuse calculation, 55 of 129 cases never broke even.

25
00:01:48.100 --> 00:01:55.333
These historical results describe selected AND/INV
cones and their retained implementation.

26
00:01:55.333 --> 00:02:03.833
They do not contradict the different B2/B4 workload, and they
do not show that every future circuit gives identical timings.

27
00:02:03.833 --> 00:02:10.967
Equal structural counts support a mechanism
interpretation without measuring hardware traffic directly.

28
00:02:11.267 --> 00:02:17.133
Suppose two methods have the same per-use execution cost but unequal preparation.

29
00:02:17.133 --> 00:02:21.467
These are illustrative units rather than EPFL timings.

30
00:02:21.467 --> 00:02:26.800
The candidate spends eight preparing, while the control spends two.

31
00:02:26.800 --> 00:02:31.867
For any reuse count n, both totals grow by two per use.

32
00:02:31.867 --> 00:02:34.933
The preparation difference stays six.

33
00:02:34.933 --> 00:02:40.200
No crossover exists under these costs, even though the candidate can be correct and its

34
00:02:40.200 --> 00:02:43.667
execution can be at parity.

35
00:02:43.667 --> 00:02:48.500
The retained evidence determines what happens on the actual circuit set.

36
00:02:48.500 --> 00:02:53.700
A never-crossing result under a declared model means the stated preparation premium was

37
00:02:53.700 --> 00:02:55.567
not repaid there.

38
00:02:55.567 --> 00:03:02.533
It does not prove an impossibility for every future implementation or workload.

39
00:03:02.533 --> 00:03:02.543
On these selected cones, near-equal kernel time and higher
preparation cost do not establish an end-to-end advantage.

40
00:03:02.543 --> 00:03:10.800
On these selected cones, near-equal kernel time and higher
preparation cost do not establish an end-to-end advantage.

41
00:03:10.800 --> 00:03:13.667
The practical comparison includes preparation,

42
00:03:13.667 --> 00:03:20.264
execution, and the actual reuse pattern, within the retained study's scope.