speech mea cpnl tuto2014 - ENSTA Bretagne · 2014. 4. 3. · speech_mea_cpnl_tuto2014.dvi Author:...
Transcript of speech mea cpnl tuto2014 - ENSTA Bretagne · 2014. 4. 3. · speech_mea_cpnl_tuto2014.dvi Author:...
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Interval methods: a tutorialLuc Jaulin, Labsticc, IHSEV, ENSTA-Bretagne
http://www.ensta-bretagne.fr/jaulin/
April 3, 2014, CNAM Paris.
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1 Interval analysis
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Problem. Given f : Rn → R and a box [x] ⊂ Rn, provethat
∀x ∈ [x] , f (x) ≥ 0.
Interval arithmetic can solve efficiently this problem.
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Example. Is the function
f (x) = x1x2 − (x1 + x2) cosx2 + sinx1 · sinx2 + 2
always positive for x1, x2 ∈ [−1, 1] ?
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Interval arithmetic
[−1, 3] + [2, 5] = [1, 8],[−1, 3] · [2, 5] = [−5, 15],abs ([−7, 1]) = [0, 7]
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The interval extension of
f (x1, x2) = x1 · x2 − (x1 + x2) · cosx2+sinx1 · sinx2 + 2
is
[f ] ([x1] , [x2]) = [x1] · [x2]− ([x1] + [x2]) · cos [x2]+ sin [x1] · sin [x2] + 2.
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Theorem (Moore, 1970)
[f ] ([x]) ⊂ R+⇒ ∀x ∈ [x] , f (x) ≥ 0
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2 Computing with sets
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Subsets X ⊂ Rn can be bracketted by subpavings :
X− ⊂ X ⊂ X+.
which can be obtained using interval calculus
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Example.
X = {(x1, x2)��� x21 + x
22 + sin (x1 + x2) ∈ [4, 9]}.
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3 Contractors
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The operator C : IRn → IRn is a contractor for the
equation f (x) = 0, if�C([x]) ⊂ [x] (contractance)x ∈ [x] and f (x) = 0⇒ x ∈ C([x]) (consistence)
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Example. Consider the primitive equation:
x2 = sinx1.
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C is monotonic if [x] ⊂ [y]⇒ C([x]) ⊂ C([y])C is idempotent if C (C([x])) = C([x])
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Contractor algebra
intersection (C1 ∩ C2) ([x]) def= C1 ([x]) ∩ C2 ([x])union (C1 ∪ C2) ([x]) def= [C1 ([x]) ∪ C2 ([x])]composition (C1 ◦ C2) ([x]) def= C1 (C2 ([x]))reiteration C∞ def
= C ◦ C ◦ C ◦ . . .
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Contractor associated with a database
The robot with coordinates (x1, x2) is in the water.
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Building contractors for equations
Consider the primitive equation
x1 + x2 = x3
with x1 ∈ [x1], x2 ∈ [x2], x3 ∈ [x3] .
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We have
x3 = x1 + x2⇒ x3 ∈ [x3] ∩ ([x1] + [x2]) // forwardx1 = x3 − x2⇒ x1 ∈ [x1] ∩ ([x3]− [x2]) // backwardx2 = x3 − x1⇒ x2 ∈ [x2] ∩ ([x3]− [x1]) // backward
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The contractor associated with x1 + x2 = x3 is thus
C
[x1][x2][x3]
=
[x1] ∩ ([x3]− [x2])[x2] ∩ ([x3]− [x1])[x3] ∩ ([x1] + [x2])
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4 Solver
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Example. Solve the system
y = x2
y =√x.
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We build two contractors
C1 :�[y] = [y] ∩ [x]2[x] = [x] ∩
[y]
associated with y = x2
C2 :�[y] = [y] ∩
[x]
[x] = [x] ∩ [y]2associated with y =
√x
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Contractor graph
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