1 Simulation of Uniform Distribution on Surfaces Giuseppe Melfi Université de Neuchâtel Espace de l’Europe, 4 2002 Neuchâtel.
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1
Simulation of Uniform Distribution on Surfaces
Giuseppe Melfi
Université de Neuchâtel
Espace de l’Europe, 4
2002 Neuchâtel
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Introduction
Random distributions are quite usual in nature. In particular:
• Environmental sciences
• Geology
• Botanics
• Meteorology
are concerned
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Distribution A
Distribution of trees in a typical cultivated field.
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Distribution B
Distribution of trees in a typical intensive production. For the same surface and the same minimal distance, there are 15% more trees.
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Distribution C
Distribution of trees in a plane forest. Uniform random distribution on a plane.
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Problem: How to simulate a distribution of points
• In a nonplanar surface
• Such that points are distributed according to a random uniform distribution, namely the quantity of points for distinct unities of surface area (independently of gradient) follows a Poisson distribution X
ekX k
k
!)Pr(
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Input and tools
• The input of such a problem is a function
D compact, f supposed to be differentiable. This function describes the surface
• The basic tool is a (pseudo-) random number generator.
2,: RDRDf
DyxRyxfyxS ),(:),(,, 3
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Algorithm 1Step 1:
Generation of N points in D
• D is bounded, so
• Random points in the box
can be partly inbedded in D.
• This procedure allows us to simulate an arbitrary number of uniformily distributed points in D, say N, denoted
).,(),( dcbaD
),( 212 nn uu )1,0()1,0(
).,),...(,( 11 NN yxyx
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Step 2: Random assignment
• We assign to each point in D a random number w in (0,1).
• We have that w1, w2, …,wN are drawn according to a uniform distribution.
• This will be employed to select points on the basis of a suitable probability of selection.
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Step 3: Uniformizer coefficient
• The region corresponds into the surface S to a region whose area can be approximated by
• We compute
yxy
yxf
x
yxf
2
00
2
00 ),(),(1
22
1max
y
f
x
fM
D
)()( 00 yyxx
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Step 4: Points selection
• The probability of (xi, yi, f(xi, yi)) to be selected must be proportional to the quantity
• The point (xi, yi, f(xi, yi)) is selected if
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1
y
f
x
f
M
y
yxf
x
yxf
w
iiii
i
22),(),(
1
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Remarks
• If S does not come from a bivariate function, but is simply a compact surface (e.g., a sphere), this approach is possible by Dini’s theorem.
• If D is bounded but not necessarily compact, it suffices that
is bounded.
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1
y
f
x
f
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Some examples
• Let
f(x,y)=6exp{-(x2+y2)}
• Let
D=(-3,3)x(-3,3)
• We apply the preceding algorithm. We have 1000 points in D. A selection of these points will appear in simulation.
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A uniform distribution on the surface S={(x,y,6exp{-x2-y2})}
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Another example
• Let
f(x,y)=x2-y2
• Let
D=(-1,1)x(-1,1)
Again, 1000 points have been used.
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Uniform distribution on the hyperboloid S = {(x,y, x2-y2)}
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Uniform distribution on the surface S={(x,y,6arctan x)}
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Under another perspective S={(x,y,6arctan x)}
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Uniform distribution on the surface S={(x,y,(x2+y2)/2)}
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How to simulate non uniform distributions on surfaces
Density can depend on
• slope
• orientation
• punctual function
These factors correspond to a positive function z(x,y) describing their punctual influence.
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Algorithm 2
• Step 1: Generation of random points in D
• Step 2: Random assignment
• Step 3: Compute
• Step 4: (xi,yi,f(xi,yi)) is selected if
22
1),(maxy
f
x
fyxzM
D
22),(),(
1),(
y
yxf
x
yxf
M
yxzw iiiiii
i
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Non uniform distribution: an example
• Let f(x,y)=6 exp{-(x2+y2)} It is the surface considered in first example• Let z1(x,y)=3-|3-f(x,y)| This corresponds to give more probability
to points for which f(x,y)=3• Let z2(x,y)=exp{-f(x,y)2}
In this case we give a probability of Gaussian type, depending on value of f(x,y)
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A non uniform distribution on S={(x,y,6 exp{-x2-y2})} using z1
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A non uniform distribution on S={(x,y,6 exp{-x2-y2})} using z2
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… and with less points
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Non uniform distribution on S = {(x,y, x2-y2)}
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With a normal vertical distribution
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Non uniform distribution on S={(x,y,6arctan x)}
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Another non uniform distribution on
S={(x,y,6arctan x)}
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Non uniform distribution on S={(x,y,(x2+y2)/2)}
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Further ideas
• A quantity of interest Q can depend on reciprocal distance of points
• on disposition of points in a neighbourood of each point
• A suitable model for an estimation of Q by Monte Carlo methods could be imagined.
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