1. Smoothing using only the two hi order bits (aggregation byK 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U V 1 3 7 1 2 4 2 3 1 2 1 3 5
Smoothing using
only the two hi
order bits
(aggregation by
counts within
2-hi grid cells)
x y z A B
C D
E N F
G H
I J K L M
t u
r s q
n o
l m
j k
h i
p O
P Q R
S T
U V v
d e f g
9 a b c w

2. Smoothing with count aggregation by using the top level LMs:K 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U 1 3 7 1 2 3 2 3 1 2 1 3 4
x y z A B
C D
E N F
G H
I J K L M
t u
r s q
n o
l m
j k
h i
p O
P Q R
S T U v
d e f g
9 a b c w 1 13 1 12 1 23 1 22 13 12 11 10 23 22 21 20 1 2 3 4 5 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 01 1 1 01 01 01 1 1 1 01 1 1 01 01 01 1 01 1 1 1 1 1 1 01 01 01 1 1 01 1 1 10 1 1 01 1 01 1 1 01 1 1 1 1 1 1 10 01 01 01 10 10 1 1 1 1 1 01 1 1 1 1 01 10 10 10 10 01 1 1 10 1 1 1 1 1 1 1 01 1 1 10 10 10 01 01 10 10 01 10 1 01 1 1 10
0100 1 01 1
1011 01 10 10 10 01 10 01 10

3. Smoothing using only the two hi order bits (existentialK 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U V 1 3 7 1 2 4 2 3 1 2 1 3 5
Smoothing using
only the two hi
order bits
(existential
aggregation within
2-hi grid cells)
x y z A B
C D
E N F
G H
I J K L M
t u
r s q
n o
l m
j k
h i
p O
P Q R
S T
U V v
d e f g
9 a b c w

4. Smoothing with existential aggregation by using the top level LMs:K 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U 1 3 7 1 2 3 2 3 1 2 1 3 4
x y z A B
C D
E N F
G H
I J K L M
t u
r s q
n o
l m
j k
h i
p O
P Q R
S T U v
d e f g
9 a b c w 1 13 1 12 1 23 1 22 13 12 11 10 23 22 21 20 1 2 3 4 5 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 01 1 1 01 01 01 1 1 1 01 1 1 01 01 01 1 01 1 1 1 1 1 1 01 01 01 1 1 01 1 1 10 1 1 01 1 01 1 1 01 1 1 1 1 1 1 10 01 01 01 10 10 1 1 1 1 1 01 1 1 1 1 01 10 10 10 10 01 1 1 10 1 1 1 1 1 1 1 01 1 1 10 10 10 01 01 10 10 01 10 1 01 1 1 10
0100 1 01 1
1011 01 10 10 10 01 10 01 10

5. Smoothing using three hi order bits (aggregation by counts withinK 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U V 1 3 7 1 2 4 1 3 5 2 3 1 2 1 3 5 2 1 3 5
Smoothing using
three hi order bits
(aggregation by
counts within
3-hi grid cells)
n o
l m
j k
h i
t u
r s v w
x y z A B
C D
E N F
G H
I J K L M q
p O
P Q R
S T
U V
d e f g
9 a b c

6. Smoothing using three hi order bits (aggregation by counts withinK 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U V 1 3 7 1 2 4 1 3 5 2 3 1 2 1 3 5 2 1 3 5
Smoothing using
three hi order bits
(aggregation by
counts within
3-hi grid cells)
x y z A B
C D
E N F
G H
I J K L M
t u
r s q
n o
l m
j k
h i
p O
P Q R
S T
U V v
d e f g
9 a b c w

7. Smoothing with exitential aggregation by using the both levels of LMs:
(the red! When there is no read one must
interpret from the black and blue).
x y z A B
C D
E N F
G H
I J K L M
t u
r s q
n o
l m
j k
h i
p O
P Q R
S T
U V v
d e f g
9 a b c w 1 13 1 12 11 1 1 23 1 22 1 21 13 12 11 10 23 22 21 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 01 1 1 01 01 01 1 1 1 01 1 1 01 01 01 1 01 1 1 1 1 1 1 01 01 01 1 1 01 1 1 10 1 1 01 1 01 1 1 01 1 1 1 1 1 1 10 01 01 01 10 10 1 1 1 1 1 01 1 1 1 1 01 10 10 10 10 01 1 1 10 1 1 1 1 1 1 1 01 1 1 10 10 10 01 01 10 10 01 10 1 01 1 1 10
0100 1 01 1
1011 01 10 10 01 10 01 10 10

8. resLeaf exists iff resLM=1 and then resLeaf=^TresLeaf^T'^PresLeaf'
resLM = ^TLM ^T'^PPM'
resLeaf exists iff resLM=1 and then resLeaf=^TresLeaf^T'^PresLeaf'
(if no operands, install pure1 or create a PM) K 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U 1 3 7 1 2 3 1 3 4 1 22 2 3 1 2 1 3 4 2 1 3 5 2 1 3 1 32 5 1 31 7 1 30 3 1 21 3 1 20 7 1 10 2 2 32 1 7 2 31 1 7 2 30 1 5 2 21 1 2 20 1 9 2 10 1
e.g., 13^12: P= T= , resLM = LM13^LM12 =
resLeaf(3): same P and T. LMs in red and PMs in blue below.
so resLM ^ = PMs show that the 2 middle leaves are pure1 (rc=4 already)
and that the last leaf of 13 is pure1 so just retrieve last leaf of 12 (01) and accumulate 1-count into rc (=5) and
ANDing first leaves, 01 ^ 10 = 00, so rc=5 1 1 1 13 12 11 10 23 22 21 20 1 2 3 4 5 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 01 1 1 01 01 01 1 1 1 01 1 1 01 01 01 1 01 1 1 1 1 1 1 01 01 01 1 1 01 1 1 10 1 1 01 1 01 1 1 01 1 1 1 1 1 1 10 01 01 01 10 10 1 1 1 1 1 01 1 1 1 1 01 10 10 10 10 01 1 1 10 1 1 1 1 1 1 1 01 1 1 10
Note that PM(B')=LM'(B) LM(B')= PM'(B
P = AND-input-pattern (vertical slices involved in AND)
T = AND=truth-pattern (truth value of thos inolved) 10 10 01 01 10 10 01 10 1 01 1 1 10
0100 1 01 1
1011 01 10 10 10 01 10 01 10

9. resLM = ^TLM ^T'^PPM' resPM unnecessary - must be construct.
resLeaf exists iff resLM=1 and then resLeaf=^TresLeaf^T'^PresLeaf' K 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U 1 3 7 1 2 3 1 3 4 1 22 2 3 1 2 1 3 4 2 1 3 5 2 1 3 1' 3 1 2 1 2 2 1
13'^12^20: P= T= , resLM = LM12^LM20 ^PM'13 = resLeaf(3456):
^ ^ = rc=12 1 1
1000
1000 10 1 1 1 01 1 11 10 00 1 1 1 1 10 00 01 11
1011 1 10 1 11 01 10 13 12 11 10 23 22 21 20 1 2 3 4 5 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 01 1 1 01 01 01 1 1 1 01 1 1 01 01 01 1 01 1 1 1 1 1 1 01 01 01 1 1 01 1 1 10 1 1 01 1 01 1 1 01 1 1 1 1 1 1 10 01 01 01 10 10 1 1 1 1 1 01 1 1 1 1 01 10 10 10 10 01 1 1 10 1 1 1 1 1 1 1 01 1 1 10
Note that PM(B')=LM'(B) LM(B')= PM'(B
P = AND-input-pattern (vertical slices involved in AND)
T = AND=truth-pattern (truth value of thos inolved) 10 10 01 01 10 10 01 10 1 01 1 1 10
0100 1 01 1
1011 01 10 10 01 10 01 10 10

10. APPENDIX Leaf Maps Concatenated Leaf Map LeafSizes= 8, 4 p12 10111101 11 1
p11 10 1
p10 K 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U 1 3 7 1 2 3 1 3 4 1 22 2 3 1 2 1 3 4 2 1 3 5 2 1 3 13 12 11 10 13 1
p13 12 1
p22 21 1
p21 20 1
p20 23 22 21 20 23 1
p23 22 1
Leaf Maps 13 1 12 1 11 1 10 1 23 1 22 1 21 1 20 1
Concatenated Leaf Map
LeafSizes= 8, 4

11. P-tree operation: AND 0 1 2 3 4 5 6 7 8 9 positions Even better:
P-tree operation: AND
0p2
0p4
0p1
0p6
1p3
1p5 1
0-Leaf Map1
0Pure Map1
0-Leaf Map2
0Pure Map2
0-Leaf Map4
0Pure Map4
1-Leaf Map5
1Pure Map5
1-Leaf Map3
1Pure Map3
0-Leaf Map6
0Pure Map6
positions
Assumemixed leaves are clustered by Leaf Offset ( vertically down the cube), and the collection of LMs (and PMs) are stored separately on one additional extent.
1. AND all 0-LMs --> A
2. scan l-to-r across A for next 1 bit,
if that position in any 1-PM=1,
then GOTO 2
else fetch & AND nonpure leaves --> B; GOTO 2
3. A forms the LM of the result and the Bs are the nonpure leaves.
2. pos=1, PM3(1)=0 so fetch & AND p p p
res
E.g., p1 ^ p3 ^ p6 ^ ^ =
3. Result Ptree:
0-Leaf_Map:
0-Pure_Map:
impure leaves:
root-count = 1
Even better:
^{0LM} ^{1PM'}  0LM (result is always type0)
Fetch & AND leaves corresp. to 1-bits in 0LM.
Set Purity Map.
In ASM, is there an operation, AND and COUNT? to count 1-bits as they are produced?

12. Leaf Maps (red are type-1) K 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U 1 3 7 1 2 3 1 3 4 1 22 2 3 1 2 1 3 4 2 1 3 5 2 1 3 13 12 11 10 13 1
0-p13 12 1 12 1
0-p12 11 1
1-p11 10 1
0-p10 23 22 21 20 23 1 23 1
0-p23 22 1 22 1
1-p22 21 1 21 1
1-p21 20 1
0-p20
LeafOff=0;
LeafOff=1;
Leaf Maps (red are type-1)
LeafOff=2; 13 1 12 1 11 1 10 1 23 1 22 1 21 1 20 1
LeafOff=3;
LeafOff=4;
Purity Maps
LeafOff=5; 12 1 23 1 22 1 21 1
LeafSize=8, NOPL=7
LeafOff=6;

13. P13^p12 P11^p10 ^{0LM} ^{1PM'}  0LM Fetch & AND lo=0,1,2,3,4,5,61 12 1
0LM 1
P11^p10
^{0LM} ^{1PM'}  0LM
Fetch & AND lo=0,1,2,3,4,5,6 10 1 LM 1
^{0LM} ^{1PM'}  0LM
lo=0;
 rc=2
Fetch & AND lo=3
lo3; 13 12
rc=5
lo=4;
 rc=1
lo=1;
 rc=2
lo=5;
 rc=1
lo=2;
 rc=2
lo=6;
 rc=1
lo=3;
 rc=3
Total rc=12
lo0;
LeafMaps; (red=type1) 13 1 12 1 11 1 10 1 23 1 22 1 21 1 20 1
lo1;
lo2;
lo3;
lo4;
PureMaps; 12 1 23 1 22 1 21 1
lo5;
lo6;

14. P22^p21
^{0LM} ^{1PM'}  0LM
Fetch & AND lo of 0LM 1-bit positions (i.e., 2,3,4,5,6) for P21, p22 (those that exits)
22' 1
21' 1 LM 1
rc=4
rc=7
rc=0
rc=4
lo2;
lo3;
lo4;
lo5;
lo6;
Total rc=22
lo0;
LeafMaps; (red=type1) 13 1 12 1 11 1 10 1 23 1 22 1 21 1 20 1
lo1;
lo2;
lo3;
lo4;
PureMaps; 12 1 23 1 22 1 21 1
lo5;
lo6;

15. P22^p21^p13^p12 ^{0LM} ^{1PM'}  0LM
Fetch & AND lo of 0LM 1-bit positions (i.e., lo3) for 21, 22 13, 12 (those that exits)
22' 1
21' 13 12 LM 1
lo3;
rc=4
lo0;
LeafMaps; (red=type1) 13 1 12 1 11 1 10 1 23 1 22 1 21 1 20 1
lo1;
lo2;
lo3;
lo4;
PureMaps; 12 1 23 1 22 1 21 1
lo5;
lo6;

16. Then it makes sense to just use 0-type (next slide).
P22^p21
^{0LM} ^{1PM'}  0LM
Fetch & AND lo of 0LM 1-bit positions (i.e., 2,3,4,5,6) for P21, p22 (those that exits)
22' 1
21' 1 LM 1
We can note that we only use PM of 1-types and LM of 0-types (except complement). Therefore, if we always precompute and store the complement of every Ptree, we don't need 1LMs and 0PMs at all (and then just use 1PM'i,j=Mi,j and 0LMi,j=Mi,j )
Then it makes sense to just use 0-type (next slide).
rc=4
rc=7
rc=0
rc=4
lo2;
lo3;
lo4;
lo5;
lo6;
Total rc=22
lo0;
LeafMaps; (red=type1) 13 1 12 1 11 1 10 1 23 1 22 1 21 1 20 1
lo1;
lo2;
lo3;
lo4;
PureMaps; 12 1 23 1 22 1 21 1
lo5;
lo6;

17. P22^p21^p13^p12
^M resultM
Fetch & AND lo of resultM 1-bit positions (i.e., lo3) for 21, 22 13, 12 (those that exits) 22 1 21 13 12
resM 1
lo3;
rc=4 13 1
p13 12 1
p12 11 1
p11 10 1
p10
First Level Maps: 13 1 12 1 11 1 10 1 23 1 22 1 21 1 20 1 23 1
p23 22 1
p22 21 1
p21 20 1
p20
LeafOff=0;
LeafOff=1;
LeafOff=2;
LeafOff=3;
LeafOff=4;
LeafOff=5;
LeafOff=6;

18. A further simplification is to store LeafOffset PtreeList in map form
P22^p21
^M  M
Fetch & AND lo of M 1-bit positions (i.e., 2,3,4,5,6) for PtreeMap 22 1 21 1 M 1
A further simplification is to store LeafOffset PtreeList in map form
rc=4
rc=7
rc=0
rc=4
lo2;
lo3;
lo4;
lo5;
lo6;
Total rc=22
PtreeMap(lo):
LeafMaps(pt): lo 13 1 12 1 11 1 10 1 23 1 22 1 21 1 20 1 lo lo lo lo lo lo

19. We note taking a = (1, 0) and a = (0, 2)
=(7.4, 4.3)
We note taking a = (1, 0) and a = (0, 2)
the 4 resulting TV-Xa-contours nearly partition a thick ring.
Thickening the Xa-contours even more, gives better coverage without increasing the neighborhood much.
Does this hold true in higher dimensions?
Do we need to consider other diagonals, e.g., (0,0.., i,0,.., j,0,..,0) etc. ?
d e f g
9 a b c
n o
l m
j k
h i
t u
r s v w
x y z A B
C D
E N F
G H
I J K L M q
p O
P Q R
S T
U V
1111
1110
1101
1100
1011
1010
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000

20. x y z A B t u C D r s E N F q G H I J K L M n o p O v l m P Q R j k1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J L M N O P Q R S T U V 1 3 7 1 2 4 1 3 5 1 23 2 3 1 2 1 3 5 2 1 3 5 2 1 3 1 32 5 1 31 7 1 30 3 1 21 4 1 20 8 1 10 3 2 32 1 7 2 31 1 7 2 30 1 5 2 21 1 2 20 1 9 2 10 1
x y z A B
C D
E N F
G H
I J K L M
t u
r s q
n o
l m
j k
h i
p O
P Q R
S T
U V v
d e f g
9 a b c w

21. Leaf Maps (red are type-1)
LeafSize=8 NOPL=7
LeafSize=2 NOPL=4
LM lo=0
LM lo=1
LM lo=2
LM lo=3
Leaf Maps (red are type-1) 11 1 10 1 20 1 11 1 10 1 21 1 20 1 11 1 10 1 22 1 21 1 20 1 13 1 10 1 22 1 21 1 20 1 13 1 12 1 11 1 10 1 23 1 22 1 21 1 20 1
PM lo=0
PM lo=1
PM lo=2
PM lo=2 11 1 20 1 11 1 21 1 20 1 11 1 22 1 21 1 20 1 13 1 22 1 21 1 20 1 1 01
Purity Maps
lo0 10
lo1 10
lo2 11
lo3 13 12 1 23 1 22 1 21 1 01 13 1
0-p13 12 1 12 1
0-p12 11 1
1-p11 10 1
0-p10 23 1 23 1
0-p23 22 1 22 1
1-p22 21 1 21 1
1-p21 20 1
0-p20

22. 13 1 3 4 12 1 2 12 11 1 19 10 1 21 23 2 3 1 12 22 2 1 18 21 1 2 19 20 1 2 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 01 1 1 01 01 01 1 1 1 01 1 1 01 01 01 1 01 1 1 1 1 1 1 01 01 01 1 1 1 1 10 1 1 01 1 1 01 1 1 01 1 1 1 1 1 1 10 01 01 01 10 10 1 1 10 1 1 1 01 1 1 1 1 01 10 01 10 01 1 1 10 1 1 1 1 1 1 1 1 1 10 10 10 01 01 10 10 01 10 1 01 1 1 10
0100 1 01 1
1011 01 10 10 01 10 01 10 10

23. 1 1 1 1 1 13 23 12 22 1 1 11 21 10 20 1 01 01 01 01 1 1 1 01 01 01 01 1 01 1 1 01 01 01 1 1 1 1 10 1 1 01 1 1 01 01 01 01 1 1 10 1 1 1 01 01 01 1 1 10 1 1 1 10 01 10 10 01 10 1 01 1 01 10 10 10 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 01 1 1 1 1 1 1 10 10 10 1 1 1 1 10 10 01 1 1 1 1 1 1 10 10 01 1 10
0100 1 01 1
1011 10 01 10

24. 1 10 01 13 23 12 22 11 21 20
1011
0100

25. µ If we plot the 1st level LM 2-lo cells as a smoothing--> µK 1 2 3 4 5 6 7 1 3 1 2 4 2 3 1 4 2 1 5
1111
1110
1101
1100
1011
1010
1001
1000
x y z A B
C D
E N F
G H
I J K L M
t u
r s q 1 7 1 7 2 1 6 2 1 7
If we plot the 1st level LM 2-lo cells as a smoothing-->
0111
0110
0101
0100
0011
0010
0001
0000
n o
l m
j k
h i
p O
P Q R
S T
U V v
µi = (xX xi) / |X|
=(1/7)x k 2kxi,
(1/7)k 2kxxi,k
d e f g
9 a b c
=(1/7)k 2krcPi,k w
for i=1 =(1/7)*(1*23
+4*22) =3.4 (rather than 4.3)
for i=2 =(1/7)*(4*23
+5*22) =7.4 (instead of 6.4)
Doesn't move the mean much! This is the right way to compute the mean (since only 2-bits are used).
In general, there is some shifting of concentration due to the "existential" aggregation for smoothing.

26. µ If we plot the 1st level LM 2-lo cells as a smoothing-->K 1 2 3 4 5 6 7 1 3 1 2 4 2 3 1 2 1 5
If we use majority
smoothing? (changes
shown in red).
Note there is additional processing involved.
1111
1110
1101
1100
1011
1010
1001
1000
x y z A B
C D
E N F
G H
I J K L M
t u
r s q
If we plot the 1st level LM 2-lo cells as a smoothing-->
0111
0110
0101
0100
0011
0010
0001
0000
n o
l m
j k
h i
p O
P Q R
S T
U V v
µi = (xX xi) / |X|
=(1/7)x k 2kxi,
(1/7)k 2kxxi,k
d e f g
9 a b c
=(1/7)k 2krcPi,k w
for i=1 =(1/7)*(1*23
+4*22) =3.4 (rather than 4.3)
for i=2 =(1/7)*(3*23
+5*22) =6.3 (instead of 6.4)
Moves hardly at all! However, in this example, majority smoothing doesn't seem much better than
"existential" smoothing. Both are pretty good smoothings!! Existential is no added work!

27. Nest we do smoothing only at the lowest level (agglomerting pairs).K 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s 28 1 3 4 1 2 1 9 2 3 1 2 1 8 1 2 9
111
110
101
100
x y z A B
C D
E N F
G H
I J K L M p i
h j k
d e f
t u
r s q
l m
n o
011
010
001
000
n o
l m
j k
h i
b c
q r
p O
P Q R
S T U v g
9 a
d e f g
9 a b c
5 6 7 8 s
1 2 3 4 w
µi =
=(1/7)k 2krcPi,k
i=1 (1/28)(423+(12)22+19) =3.5
i=2 (1/28)(12*23+18*22+19) =6.7
Nest we do smoothing only at the lowest level (agglomerting pairs).