WEBVTT

1
00:00:00.000 --> 00:00:07.967
This overview explains how ordered row and column variables
place each assignment and its output in an explicit matrix.

2
00:00:07.967 --> 00:00:15.165
Without the axis key, a coordinate does not identify
the input case, even when the stored bit is correct.

3
00:00:15.165 --> 00:00:15.167
Without the axis key, a coordinate does not identify
the input case, even when the stored bit is correct.

4
00:00:15.467 --> 00:00:21.067
For this four-input layout, declare zero-first binary order on both axes.

5
00:00:21.067 --> 00:00:26.267
The row receives A and B, while the column receives C and D.

6
00:00:26.267 --> 00:00:31.367
This choice differs from the true-first convention used for the small operator matrices;

7
00:00:31.367 --> 00:00:36.133
the labels, rather than memory, determine the address.

8
00:00:36.133 --> 00:00:42.833
For assignment one zero one one, the row bits are one zero, which represent two.

9
00:00:42.833 --> 00:00:46.633
The column bits are one one, which represent three.

10
00:00:46.633 --> 00:00:53.167
A and B gives zero, C or D gives one, and exclusive-or gives one.

11
00:00:53.167 --> 00:00:57.467
Place that answer at row two, column three.

12
00:00:57.467 --> 00:01:01.733
Keep the row fixed and walk across all four columns.

13
00:01:01.733 --> 00:01:04.400
The first branch remains zero.

14
00:01:04.400 --> 00:01:12.833
The second branch gives zero, one, one, one, so exclusive-or returns that same sequence.

15
00:01:12.833 --> 00:01:17.000
Filling a row means evaluating every column assignment; it is more work

16
00:01:17.000 --> 00:01:20.526
than looking up one cell.

17
00:01:20.526 --> 00:01:20.533
than looking up one cell.

18
00:01:20.533 --> 00:01:24.267
First compute A AND B and C OR D.

19
00:01:24.267 --> 00:01:28.167
The function takes the exclusive-or of those two results.

20
00:01:28.167 --> 00:01:32.833
Assignment 1011 produces output one.

21
00:01:32.833 --> 00:01:38.567
Choose ordered row variables A, B and column variables C, D.

22
00:01:38.567 --> 00:01:42.200
Read each pair most-significant bit first.

23
00:01:42.200 --> 00:01:48.367
Row bits 10 give 1 × 2 + 0 × 1: row two.

24
00:01:48.367 --> 00:01:51.733
Column bits 11 give column three.

25
00:01:51.733 --> 00:01:55.400
A AND B is zero and C OR D is one.

26
00:01:55.400 --> 00:01:59.933
Their exclusive-or is one, so cell two, three stores one.

27
00:01:59.933 --> 00:02:06.567
Evaluate all 16 assignments under the same
rule to fill the complete 4×4 layout.

28
00:02:06.567 --> 00:02:14.333
The axis key lets us decode row two and
column three back into A-B 10 and C-D 11.

29
00:02:14.333 --> 00:02:19.400
The coordinate identifies the input; the stored bit gives its output.

30
00:02:19.400 --> 00:02:25.367
As another worked example, 1110 selects row three and column two.

31
00:02:25.367 --> 00:02:29.833
Both inner results are one, so the stored output is zero.

32
00:02:29.833 --> 00:02:35.400
For 0100, row one and column zero store zero.

33
00:02:35.400 --> 00:02:38.367
Both inner results are zero.

34
00:02:38.367 --> 00:02:43.367
This layout places A on the rows and B-C-D on the columns.

35
00:02:43.367 --> 00:02:48.400
One row bit gives two rows; three column bits give eight columns.

36
00:02:48.400 --> 00:02:55.433
The tracked assignment 1011 moves to row
one, column three and keeps output one.

37
00:02:55.433 --> 00:03:00.767
This layout places A-B-C on the rows and D on the columns.

38
00:03:00.767 --> 00:03:06.000
The same assignment is at row five, column one in an 8×2 layout.

39
00:03:06.000 --> 00:03:09.033
Each layout contains sixteen cells.

40
00:03:09.033 --> 00:03:14.867
Changing the partition moves assignment-output
pairs without changing their values.

41
00:03:14.867 --> 00:03:20.827
These rectangular repository layouts are
distinct from the paper's square construction.

42
00:03:20.827 --> 00:03:20.833
These rectangular repository layouts are
distinct from the paper's square construction.

43
00:03:21.133 --> 00:03:27.633
Work a second complete case with A 1, B 1, C 0, D 1.

44
00:03:27.633 --> 00:03:31.133
Keep this assignment fixed throughout the calculation.

45
00:03:31.133 --> 00:03:35.500
Each symbol retains the meaning already introduced; a new value does not

46
00:03:35.500 --> 00:03:38.700
create a new logical rule.

47
00:03:38.700 --> 00:03:44.933
The row address is three and the column address is one in the declared zero-first layout.

48
00:03:44.933 --> 00:03:50.167
Both inner branches give one, so exclusive-or gives zero.

49
00:03:50.167 --> 00:03:54.333
This cell stores zero even though both branches are true.

50
00:03:54.333 --> 00:04:00.000
A matrix lookup and a program trace must agree on that same assignment.

51
00:04:00.000 --> 00:04:03.567
This case establishes a specific contrast.

52
00:04:03.567 --> 00:04:07.233
Address (3,1) stores 0.

53
00:04:07.233 --> 00:04:10.200
The layout does not change the outer gate.

54
00:04:10.200 --> 00:04:14.800
Keep the input, intermediate values, and conclusion together when comparing

55
00:04:14.800 --> 00:04:16.867
it with the previous case.

56
00:04:16.867 --> 00:04:23.593
The calculated answer belongs to this complete assignment under the stated rule.

57
00:04:23.593 --> 00:04:23.600
The calculated answer belongs to this complete assignment under the stated rule.

58
00:04:23.900 --> 00:04:30.767
Now change the partition so that rows contain A and C and columns contain B and D.

59
00:04:30.767 --> 00:04:33.367
The Boolean rule has not changed.

60
00:04:33.367 --> 00:04:36.733
Only the address convention has changed, so every assignment

61
00:04:36.733 --> 00:04:40.300
must be relocated consistently.

62
00:04:40.300 --> 00:04:44.300
Track the same four-bit assignments through both layouts.

63
00:04:44.300 --> 00:04:50.333
One zero one one moves from row two, column three to row three, column one.

64
00:04:50.333 --> 00:04:55.667
Zero one zero zero moves from one, zero to zero, two.

65
00:04:55.667 --> 00:05:00.300
One one zero one moves from three, one to two, three.

66
00:05:00.300 --> 00:05:03.767
Each retains its own output.

67
00:05:03.767 --> 00:05:10.100
A matrix position is meaningful only with its declared partition and axis order.

68
00:05:10.100 --> 00:05:15.733
Two arrays can look different while encoding the same function under different layouts.

69
00:05:15.733 --> 00:05:21.433
Conversely, treating different conventions as identical can compare the wrong assignments

70
00:05:21.433 --> 00:05:25.367
and produce a false disagreement.

71
00:05:25.367 --> 00:05:25.378
The function determines the output.

72
00:05:25.378 --> 00:05:28.633
The function determines the output.

73
00:05:28.633 --> 00:05:34.400
The ordered axis lists and bit convention determine its position.