Space Filling Curves
Hierarchical Basis
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Agenda
1. Motivation
2. Space filling curves
a.
b.
c.
d.
Definition
Hilbert’s space filling curve
Peano’s space filling curve
Usage in numerical simulations
3. Hierarchical basis and generating systems
a. Standard nodal basis for FEM-analysis
b. Hierarchical basis
c. Generating systems
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1. Motivation
Time needed for simulating phenomena depends on both performing the
corresponding calculation steps and transferring data to and from the processor(s).
–
The latter can be optimised by arranging the data in an effective way to avoid cache
misses and support optimisation techniques (e.g. pre-fetching). Standard organization
in matrices / arrays leads to bad cache behaviour.
 Space-Filling Curves
–
Use (basis) functions for FEM-analysis with which LSE solvers show faster
convergence
 Hierarchical Basis
 Generating System (for higher dimensional problems)
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2.a Definition of Space-Filling Curves
We can construct a mapping from a 1-D interval to a n-D interval where n is finite,
i.e. we can construct mappings as follow:
  [0,1]    [0,1]2
  [0,1]    [0,1]3
  [0,1]    [0,1]n , n  4,5,...
If the curve of this mapping passes through every point of the target space we call
this a “Space-Filling Curve”.
We are interested in a continuous mapping. This cannot be bijective, it’s only
surjective.
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2.a Geometric Interpretation
0
First we partition the interval I into n
subintervals and the square  into n sub
squares. Now one can easily establish a
continuous mapping between these areas.
1/
2/
9
9
1
1,1
This idea holds for any finite-dimensional
target space. It’s not necessary to partition
into equally sized areas!
Remark:
When moving a certain distance on the unit
interval we can easily estimate the possible
position in the target space (~ radius).
0,0
1/
3,
1,0
0
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2.b Hilbert’s Space-Filling Curve
The German mathematician David Hilbert (1862-1943) was the first one to give the so far only
algebraically described space-filling curves a geometric interpretation.
In the 2 dimensional case he splat I into four subintervals and  into four sub squares. This splitting is
recursively repeated for the emerging subspaces, leading to the below shown curve. An important fact is
the so called full inclusion: whenever we further split a subinterval, the mapping to the corresponding finer
sub squares stays in the former interval.
Repeating the partitioning symmetrically (for all subspaces) again and again leads to 22n subintervals / sub
squares in case of a 2-dimensional target space. (n: # of partitioning steps)
1/
2/
4
3/
4
4/
4
4
0
1
1/
16
2/
16
2
1
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0,0
3/
16
4/
1,1
16
1,1
16/
16
3
4
3
1
2
4
0,0
16
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2.c Peano’s Space-Filling Curve
The Italian mathematician Giuseppe Peano
(1858-1932) was the first one who
constructed (only algebraically) a space
filling curve in 1890. According to him we
split the starting domain  into 3m sub areas.
0
9
3/
9
4/
9
3
Recursive partitioning leads to finer curves
(compare with Hilbert’s curve). Symmetrical
partitioning leads to 3mn sub areas.
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2/
5/
9
6/
9
7/
9
8/
9
1,1
The Peano curve goes always from 0 to 1 in
all dimensions. The way we run through the
target space is up to us.Two different ways
are marked with blue / red numbers. Only
the first and the last subspaces are mapped in
a fixed way.
m: # dim()
n: # of partitioning steps
1
1/
4
9
7
8
9
2
5
8
6
5
4
1
6
7
1
2
3
0,0
7
2.c Peano curve in 3-D
We begin with the unit cube. The Peano
curve begins in (0,0,0) and ends in (1,1,1).
Now we begin to split this cube into finer
sub cubes. We can select the dimension we
want to begin with.
1/
1,1,1
0,0,0
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0,0,0
2/
3,0,0
3,1,1
1,1,1
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2.c Peano curve in 3-D (continued)
Now the emerging sub spaces are split in
one on the two remaining dimensions.
1/
3,1,1
After splitting the sub spaces in the last
dimension we end up with 27 equally sized
cubes (symmetrical case, here: 7 cuboids.)
1/
1,1,1
1,1/3,1
3,1,1
1,1,1
1,1/3,1
2/ ,0,2/
3
3
1,1/3,1/3
0,0,0
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2/
3,0,0
0,0,0
2/
3,0,0
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2.d Usage in numerical simulations
•
So far we’ve only spoken about the generation
of space filling curves. But how can this be of
advantage for the data transfer in numerical
simulations?
•
To answer this question let’s have a short look
at a simple grid.
Within a square we need the values at the
corner points of that square. When we go
through this square following e.g. Peano’s
space filling curve we can build up inner lines
of data points, which are processed in a
forward or backward order.
•
This scheme corresponds to a stack, on which
we can only access the very most upper
element (forward: writing to stack, backward:
reading from stack). In our example we would
need 2 stacks (blue and red).
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1,1
0,0
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2.d Example for stack usage
Let’s examine what’s happening in the squares.
In the first square we need the data points 1,2,5,6.
Stacks after processing the first cell:
stack 1: 5
6
2
stack 2: 1
In the second square the points 2,3,6,7 are needed.
Stacks after processing the second cell:
stack 1: 5
6
7
13
14
15
16
9
10
11
12
5
6
7
8
2
3
4
2
3
stack 2: 1
Now we jump to the 5th square, points 6,7,10,11
1
have to be accessed. Stacks after processing the fifth cell:
6
stack 1: 5
stack 2:
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1
2
3
4
8
12
11
10
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2.d Usage with adaptable grids
To guarantee the usage of this idea for complex computations it has to hold for adaptable
grids, too.
We start with the first square on the coarsest grid and recursively refine where needed
(remark: full inclusion). This leads to a top-down approach where we first process the cells
on the deeper level before finishing the cells on the coarser level.  top-down depth-first
(When entering cell 6 first we first process cell 6, then cells 7 to 15 are finished and
afterwards we return to the coarse level and proceed with cell 16.)
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1
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19
16
6
5
2
3
4
15 14 13
10 11 12
9 8 7
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2.d Stacks for adaptable grids
Points 1 and 2 are used to calculate cells on both coarse and fine level. When working with hierarchical
grid we only have the points 1a and 2a. If we would only use two stacks we would have to store the points
3 and 4 to the same stack as points 1a and 2a. When finishing fine cell 9 (fine level) point 4 would lie on
top of the stack containing point 2a, cell 3 (coarse level) couldn’t access point 2a. We handle this problem
by introducing two new stacks. We use 2 point stacks (0D) and 2 line stacks (1D).
When working with a generating system we have different point values on all levels (points 1a,1b and 2a,
2b).
7
8
6
5
1a
1
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4
2a
2
1b
3
30
29
28
27
26
25
31
32
33
22
23
24
36
35
34
21
20
19
2b
1
2
3
16
17
18
3
6
5
4
15
14
13
4
7
8
9
10
11
12
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2.d Stacks in adaptable grids & generating
systems
As already mentioned, with generating systems we have point values on different levels. To
deal with these problems the so far used number of stacks is insufficient. One can show that
the number of stacks can be calculated the way shown below.
0,1
21
1,1
0,0,1
201
1,0,1
02
22
12
002
202
102
0,0
20
1,0
0,0,0
200
1,0,0
We numerate the points according to their value in the Cartesian coordinate system. The connecting lines
e.g. 02 are numerated the following way (element-wise):
- if the corresponding point coordinates are equal: take their value
- else write a 2
This leads to the following formula for the # of stacks: 3dim
dim = 2  9 stacks (4 point stacks, 4 line stacks, 1 plane stacks = in/output)
dim = 3  27 stacks (8 point stacks, 12 line stacks, 6 plane stacks, 1 volume stack = in/output)
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2.d Conclusion
•
•
•
Peano-curves allow efficient stack usage for n-dimensional problems. (Hilbertcurves show this advantage only for the 2-dimensional case, as the higher
dimensional curves don’t show exploitable behaviour.)
The number of needed stacks is independent of the problem size.
 computation speed-up due to efficient cache usage
(CPU pre-fetching unit can efficiently load the needed data when organised in
an easy way.)
The only drawback is the recursive splitting into 3n subintervals when using
Peano curves which lead to more points, but we gain more than we loose here.
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3.a Standard nodal basis for FEM-analysis
FEM – analysis uses a set of independent piece-wise linear functions to describe
the solution. We can write the function u(x) as follows:
n
ui: function value at position I
u ( x)   ui i ( x)
i: hut-function which has the value one at position i, zero
i 1
at the neighbouring positions (i-1, i+1) and is linear
between
ith hut-function
0
xi-1
xi
xi+1
1
A set of linear independent hutfunctions forms a basis of the
solution space. (boundaries not
treated)
0
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3.a Resulting LSE (Linear System of Equations)
We now have a short look on the one-dimensional Poisson equation using the
standard discretization approach.
u   f  x 
A  u  x   b x 
 1
1
0
0
0 
2


1
1
1

0
0 
 2
2


A 0
1
1
1
0 
2
2
 0
0
1
1
1 

2
2
 0
0
0
1
1 
2


The problem condition depends on the ration max/ min. max goes towards the row
sum (= 2) and min tends towards zero, therefore the resulting condition is rather
poor. This causes slow convergence.
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3.a Drawbacks of standard approach
•
Slow convergence rate as already shown due to bad eigenvalues. Moreover the
condition depends on the problem size!
•
When we have to examine a certain area in more detail, i.e. when we need
locally a higher resolution we cannot reuse the already performed work. We
have to calculate new hut-functions.
 Use hierarchical basis to compensate these two effects.
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3.b Hierarchical Basis
•
•
•
A possibility to deal with the mentioned
drawbacks is to use a hierarchical basis.
Unlike with the standard approach we
now use all the hut-functions on the
coarser levels when refining. On finer
levels hut-functions are only created at
points which don’t define functions on
higher levels.
These hut-functions still build up a
basis. One can uniquely describe the
function u(x) by summing up the
different hut-functions on the different
levels.
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standard basis
hierarchical basis
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3.b Interpretation of hierarchical basis
•
•
•
The picture below shows a simple interpolation of a parabola using standard basis (left
side) and hierarchical basis (right side).
The coefficients needed for the normal approach are the function values at the
corresponding positions. (This is also the reason why one cannot locally refine and reuse
older values.)
For the hierarchical basis only the first coefficient equals the corresponding function
value (= 1 in below example). The coefficients for the finer levels just describe the
difference between the real function value and the one described by all previous hutfunctions. This supports reuse when locally refining.
coefficients: ¾ , 1 , ¾
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coefficients: ¼ , 1 , ¼
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3.b Representation in hierarchical basis
Let’s remember the representation of the function u(x) using the nodal basis:
n
u ( x)   ui i ( x)
i 1
The same function u(x) can be represented using a hierarchical basis:
n
u ( x)   vi i ( x)
i 1
Both representations describe the same function. We can define a mapping
transforming the normal representation into the hierarchical one:
v  H u
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3.b Transformation into hierarchical basis
When transforming the coefficients of the standard description into the
hierarchical basis we use the following formula:
v2 k 1  u2 k 1 
u2 k  2  u2 k
2
hierarchical difference (coarse – fine level)
 coefficient for hierarchical basis
So we can find the entries (-1/2, 1, -1/2) in
each row of the transformation matrix H.
(We have to use this formula recursively to
calculate the coefficients on all levels.)
v2k+1
u2k+2
u2k+1
u2k
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3.b Transformation into hierarchical basis (cont.)
Let’s examine as example the parabola defined by the points (0,0), (1,1) and (2,0).
We are using 7 equally distributed inner points.
coefficients for
hierarchical
basis
(v1,…,v7)’
 0   1
 0  

 
 1   1 2

 
 1 4   1 2
1 4   

 
1 16  1 2
1 16 

 
1 16 
1 16 

 
transformation matrix H
1
1 2
1 2
1
1
1 2
1 2
1
1 2
1 2
1
1
1 2
1 2
v4 v2 v5 v1 v6 v3 v7
0
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1
u1 u2
1
0
  0 
1   7 16 
 1 2  3 4 
 

 15 16
 1 2   1 
 

 15 16
  34 
 

  7 16 
 1 2  0 
coefficients for
standard basis
(u1,…,u7)’
u7
1
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3.b Usage in numerical schemes
We want to solve more effectively A  u ( x)  b( x) .
To do so we replace the standard representation of u by its hierarchical one.
u  T v
A  u  b  A  T  v   b
When we now multiply the equation from the left by T’ we’ll end up with a new
equation for v(x) (this will be a very nice one). The transformation into u(x) is then
easily performed.
A  u  b  A  T  v   b  T ' A  T   v  T 'b
 A2  v  b2
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3.b Transforming v(x) into u(x)
Again we consider the transformation for our parabola example.
transformation matrix T
coefficients for
standard basis
(u1,…,u7)’
u1 u2
0
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 0   1
 7 16  

 
 34  

 
15 16 
 1 

 
15 16 
 34  

 
 7 16  
 0  

 
14 12
12 1
34 12
1
34
12
12
1
14
12
1
1
1
1
u7
  0 
  0 
 

  1 
 

 1 4 
  1 4 
 

 1 16
 1 16
 

1  1 16
 1 16
 

coefficients for
hierarchical
basis
(v1,…,v7)’
v4 v2 v5 v1 v6 v3 v7
1
0
1
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3.b Calculating T’AT
Again we consider the
transformation at our parabola
example. The resulting matrix
is diagonal. After multiplying
the old right-hand-side of the
equation with T’ we can
directly read of the result with
respect to the hierarchical
system v(x). Finally we only
have to transform this back to
the standard basis u(x).
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1 4



1
2




12


T ' A  T  
1



1


1 


1

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3.b Hierarchical basis in 2 D
Higher dimensional hut-functions are
the product of one dimensional hutfunctions.
•
•
•
define hut-function on coarsest level
refine one direction
refine the other direction
Possibilities of refinement-order:
•
•
first refine completely one dimension
before working on the other
always refine both dimensions before
going to a finer level
We can transfer this idea to higher
dimensional problems.
One can split the dimensions differently.
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3.b Matrix T’AT for n-dimensional case
•
•
For the one dimensional case the matrix T’AT is diagonal.
Higher dimensional problems
– T’AT is not diagonal any longer
– T’AT is still “nicer” than A when using nodal basis
better condition  faster convergence
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3.b Advantages of hierarchical basis
•
•
Possibility to locally refine the system and reuse older values.
Faster convergence than the standard approach (optimal for 1-dimensional
problems).
(We need more sophisticated algorithms to perform the necessary matrix
multiplications in an effective way.)
•
Remark:
Although hierarchical basis are always much more effective than the standard
representation they can be beaten when dealing with higher dimensional
problems.

This leads us to generating systems.
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3.c Generating system
•
•
•
So far we restricted ourselves to either nodal or hierarchical hut-functions
forming a basis to describe the function u(x).
We can also build up u(x) using the necessary linearly independent functions
and additionally as much linearly dependent functions as we like to use.
This representation is not any longer unique, but we don’t need uniqueness.
We are only interested in describing the function u(x).
• Example:
We get the same vector a by either using a unique representation or one of
various possible representations in a generating system.
 3
1
 0
 0
 
 
 
 
a   2  3   0  2  1  4   0
 4
 0
 0
1
 
 
 
 
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 3
1
 0
 0
1
 
 
 
 
 
a   2  1  0  2   1   2   0   2   0
 4
 0
 0
1
1
 
 
 
 
 
 3
1
 0
 0
1
 
 
 
 
 
a   2   0   0   2   1   1  0   3   0 
 4
 0
 0
1
1
 
 
 
 
 
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3.c Generating system
•
The normal hierarchical basis only
allows refinement on nodes which
aren’t already defined by hutfunctions on coarser levels.
hierarchical basis
•
A generating system refines
additionally on already described
nodes (red graphs). It can also be
thought of as the sum of all nodal
hut-functions on all levels of
observation.
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generating system
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3.c Generating system
As already mentioned the condition of a matrix depends on the ratio of max/ min.
To be precise: min means the smallest non-zero eigenvalue.
• When using a generating system we introduce lots of dependent lines in the
matrix and therefore get a corresponding number of eigenvalues which equal
zero. The gain lies in the smallest non-zero eigenvalue. It’s not any longer
dependent on the problem size but a constant (independent of dimension).
• The solution with respect to the generating system is not unique. Depending on
the start value for the iterative solver we get a different solution. But when
transforming these solutions back into a unique representation we end up with
the same value, i.e. the calculated solutions from the generating ansatz
depending on different initial values can only vary within the composition of
the linear dependent functions.
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