ku naru shoh-mei no michi
DESCRIPTION
The way of the empty proof. Ku naru shoh-mei no michi. L’art de la preuve vide. If you have the right definition you have a simpler proof. If you have the right data structures, you have a simpler algorithm. If you have the right…. Power of Elimination. Tarski’s Meta-theorem - PowerPoint PPT PresentationTRANSCRIPT
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Ku naru shoh-mei no michi
The way of the empty proof
L’art de la preuve vide
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If you have the right definition youhave a simpler proof.
If you have the right data structures,you have a simpler algorithm.
If you have the right…..
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Power of Elimination
Tarski’s Meta-theoremTo any formula Ф(X1, X2,……,Xm) in the
vocabulary {0,1,+,.,=,<} one can effectivelyassociate two objects:
(i) a quantifier free formula θ(X1,……,Xm) in the same vocabulary and
(ii) a proof of the equivalence Фθ that uses the axioms for real closed fields.
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x ax2 + b + c = 0
If and only ifb2 – 4ac > 0
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Resolution: Tarski’s Meta-theorem for Logic
Elimination not restricted to algebra andgeometry
2x - 3y ≤ 5-3x + y ≤ -1
3(2x - 3y ≤ 5)2(-3x + y ≤ -1)
6x - 9y ≤ 15-6x + 2y ≤ -2 6x - 9y ≤ 15-6x + 2y ≤ -2___________________________________________________________________________________________
0x -7y ≤ 13
2x - 3y ≤ 5-3x + y ≤ -1
6x - 9y ≤ 15-6x + 2y ≤ -2___________________________________________________________________________________________
-7y ≤ 13
P(x) V Q(x)-P(a) V R(y) P(x) V Q(x)-P(a) V R(y) P(a) V Q(x)-P(a) V R(y)
_____________________________________________________________________________________________________________________________
x = a Q(a) V R(y)
P(a) V Q(x)-P(a) V R(y)
_______________________________________________________________________________
Q(a) V R(y)
P(x) V Q(x)-P(a) V R(y)
________________________________________________________________________________
x = a
P(a) V Q(x)-P(a) V R(y)
_______________________________________________________________________________
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Used in automated theorem proving (algebra,geometry & logic)But…
However…Can be used in special cases to prove theoremsby hand
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F0 F1 F2 Fn ….. P P P
F3 F4 P P P P?
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The elimination of the proof is an idealseldom reached
Elimination gives us the heart of the proof
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Markov’s Ergodic Theorem (1906)
Any irreducible, finite, aperiodic Markov Chain has all states Ergodic (reachable at any time in the future) and has a unique stationary distribution, which is a probability vector.
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x1
.45
1x2
x3 x4
.25
.6
.75.55.4
Probabilities after 15 stepsX1 = .33X2 = .26X3 = .26X4 = .13
30 stepsX1 = .33 X2 = .26X3 = .23X4 = .16
100 stepsX1 = .37 X2 = .29X3 = .21X4 = .13
500 stepsX1 = .38 X2 = .28X3 = .21X4 = .13
1000 stepsX1 = .38 X2 = .28X3 = .21X4 = .13
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x1
.45
1x2
x3 x4
.25
.6
.75.55.4
Probabilities after 15 stepsX1 = .46X2 = .20X3 = .26X4 = .06
30 stepsX1 = .36 X2 = .26X3 = .23X4 = .13
100 stepsX1 = .38 X2 = .28X3 = .21X4 = .13
500 stepsX1 = .38 X2 = .28X3 = .21X4 = .13
1000 stepsX1 = .38 X2 = .28X3 = .21X4 = .13
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You can view the problem in three different ways:
• Principal eigenvector problem• Classical Linear Programming problem• Elimination problem
p11x1 + p21x2 + p31x3 = x1 p12x1 + p22x2 + p32x3 = x2
p13x1 + p23x2 + p33x3 = x3
∑xi = 1 xi ≥ 0
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Difficulties as an Eigenvector Problem
• Notion of convergence• Deal with complex numbers• Uniqueness of solution
• Need to use theorems: Perron-Frobenius, Chapman, Kolmogoroff, Cauchy… and/or restrictive hypotheses….
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Symbolic Gaussian Elimination
System of two variables:
p11x1 + p21x2 = x1
p12x1 + p22x2 = x2
∑xi = 1
With Maple we find:x1 = p21/(p21 + p12)
x2 = p12/(p21 + p12)
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x1 = p21/(p21 + p12)
x2 = p12/(p21 + p12)
x1 x2
p12
p21
p11
p22
x1 = p21/(p21 + p12)
x2 = p12/(p21 + p12)
x1 = p21/(p21 + p12)
x2 = p12/(p21 + p12)
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Symbolic Gaussian EliminationThree variables:
p11x1 + p21x2 + p31x3 = x1
p12x1 + p22x2 + p32x3 = x2
p13x1 + p23x2 + p33x3 = x3
∑xi = 1
With Maple we find:x1 = (p31p21 + p31p23 + p32p21) / Σ
x2 = (p13p32 + p12p31 + p12p32) / Σ
x3 = (p13p21 + p12p23 + p13p23) / Σ
Σ = (p31p21 + p31p23 + p32p21 + p13p32 + p12p31 + p12p32 +
p13p21 + p12p23 + p13p23)
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x1
p12
p21p11
x2p22
x3
p33
p13p23
p31 p32
x1 = (p31p21 + p23p31 + p32p21) / Σx2 = (p13p32 + p31p12 + p12p32) / Σx3 = (p21p13 + p12p23 + p13p23) / Σ
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x1
p12
p21p11
x2p22
x3
p33
p13p23
p31 p32
x1 = (p31p21 + p23p31 + p32p21) / Σx2 = (p13p32 + p31p12 + p12p32) / Σx3 = (p21p13 + p12p23 + p13p23) / Σ
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x1
p12
p21p11
x2p22
x3
p33
p13p23
p31 p32
x1 = (p31p21 + p23p31 + p32p21) / Σx2 = (p13p32 + p31p12 + p12p32) / Σx3 = (p21p13 + p12p23 + p13p23) / Σ
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x1
p12
p21p11
x2p22
x3
p33
p13p23
p31 p32
x1 = (p31p21 + p23p31 + p32p21) / Σx2 = (p13p32 + p31p12 + p12p32) / Σx3 = (p21p13 + p12p23 + p13p23) / Σ
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x1
p12
p21p11
x2p22
x3
p33
p13p23
p31 p32
x1 = (p31p21 + p23p31 + p32p21) / Σx2 = (p13p32 + p31p12 + p12p32) / Σx3 = (p21p13 + p12p23 + p13p23) / Σ
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Do you see the LIGHT?
James Brown (The Blues Brothers)
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Symbolic Gaussian Elimination
System of four variables:
p21x2 + p31x3 + p41x4 = x1
p12x1 + p42x4 = x2
p13x1 = x3
p34x3= x4
∑xi = 1
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x1
p12
p21x2
x3
p31
x4
p41
p34
p42p13
x1 = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 / Σx2 = p31p41p12 + p31p42p12 + p34p41p12 + p34p42p12 + p13p34p42 / Σx3 = p41p21p13 + p42p21p13 / Σx4 = p21p13p34 / Σ
Σ = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 + p31p41p12 + p31p42p12 + p34p41p12
+ p34p42p12 + p13p34p42 + p41p21p13 + p42p21p13 + p21p13p34
With Maple we find:
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x1
p12
p21x2
x3
p31
x4
p41
p34
p42p13
x1 = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 / Σx2 = p31p41p12 + p31p42p12 + p34p41p12 + p34p42p12 + p13p34p42 / Σx3 = p41p21p13 + p42p21p13 / Σx4 = p21p13p34 / Σ
Σ = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 + p31p41p12 + p31p42p12 + p34p41p12
+ p34p42p12 + p13p34p42 + p41p21p13 + p42p21p13 + p21p13p34
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x1
p12
p21x2
x3
p31
x4
p41
p34
p42p13
x1 = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 / Σx2 = p31p41p12 + p31p42p12 + p34p41p12 + p34p42p12 + p13p34p42 / Σx3 = p41p21p13 + p42p21p13 / Σx4 = p21p13p34 / Σ
Σ = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 + p31p41p12 + p31p42p12 + p34p41p12
+ p34p42p12 + p13p34p42 + p41p21p13 + p42p21p13 + p21p13p34
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x1
p12
p21x2
x3
p31
x4
p41
p34
p42p13
x1 = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 / Σx2 = p31p41p12 + p31p42p12 + p34p41p12 + p34p42p12 + p13p34p42 / Σx3 = p41p21p13 + p42p21p13 / Σx4 = p21p13p34 / Σ
Σ = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 + p31p41p12 + p31p42p12 + p34p41p12
+ p34p42p12 + p13p34p42 + p41p21p13 + p42p21p13 + p21p13p34
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x1
p12
p21x2
x3
p31
x4
p41
p34
p42p13
x1 = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 / Σx2 = p31p41p12 + p31p42p12 + p34p41p12 + p34p42p12 + p13p34p42 / Σx3 = p41p21p13 + p42p21p13 / Σx4 = p21p13p34 / Σ
Σ = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 + p31p41p12 + p31p42p12 + p34p41p12
+ p34p42p12 + p13p34p42 + p41p21p13 + p42p21p13 + p21p13p34
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x1
p12
p21x2
x3
p31
x4
p41
p34
p42p13
x1 = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 / Σx2 = p31p41p12 + p31p42p12 + p34p41p12 + p34p42p12 + p13p34p42 / Σx3 = p41p21p13 + p42p21p13 / Σx4 = p21p13p34 / Σ
Σ = p21p34p41 + p34p42p21 + p21p31p41 + p31p42p21 + p31p41p12 + p31p42p12 + p34p41p12
+ p34p42p12 + p13p34p42 + p41p21p13 + p42p21p13 + p21p13p34
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Ergodic Theorem Revisited
If there exists a reverse spanning tree in a graph of theMarkov chain associated to a stochastic system, then:
(a)the stochastic system admits the followingprobability vector as a solution:
(b) the solution is unique.(c) the conditions {xi ≥ 0}i=1,n are redundant and thesolution can be computed by Gaussian elimination.
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Internet Sites• Kleinberg• Google• SALSA
• In degree heuristic
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Markov Chain as a Conservation System
x1
p12
p21p11x2
p22
x3
p33
p13 p 23
p31p32
p11x1 + p21x2 + p31x3 = x1(p11 + p12 + p13) p12x1 + p22x2 + p32x3 = x2(p21 + p22 + p23)p13x1 + p23x2 + p33x3 = x3(p31 + p32 + p33)
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Kirchoff’s Current Law• The sum of currents
flowing towards a node is equal to the sum of currents flowing away from the node.
i31 + i21 = i12 + i13
i32 + i12 = i21
i13 = i31 + i32
1 2
3
i13
i31 i32
i12
i21
i3 + i2 = i1 + i4
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Kirchoff’s Matrix Tree Theorem (1847)
For an n-vertex digraph, define an n x n matrix Asuch that A[i,j] = 1 if there is an edge from i to j,for all i ≠ j, and the diagonal entries are suchthat the row sums are 0.
Let A(k) be the matrix obtained from A bydeleting row k and column k. Then the absolutevalue of the determinant of A(k) is the numberof spanning trees rooted at k (edges directedtowards vertex k)
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Two theorems for the price of one!!
Differences• Kirchoff theorem perform n gaussian
eliminations• Revised version – only two gaussian
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Minimax Theorem
• Fundamental Theorem in Game Theory• Von Neumann & Kuhn
• Minimax Theorem brings certainty into the world of probabilistic game theory.
• Applications in Computer Science, Economics & Business, Biology, etc.
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Minimax Theorem
max(x + 2y)
-3x + y ≤ 2 2x + 3y ≤ 4 4x + 2y ≤ 1 -x ≤ 0 -y ≤ 0
0 ≤ z ≤ 1
max value z can take is min value of the right hand side
z = x + 2y
-3z + 7y ≤ 2 2z - y ≤ 4 4z - 6y ≤ 1 -z + 2y ≤ 0 -y ≤ 0
2z ≤ 2 3z ≤ 8 -z ≤ 010z ≤ 1911z ≤ 30 -3z ≤ 2 min value z can take is max value of
the left hand side
x = z – 2y
z ≤ 2/2 z ≤ 8/3 -z ≤ 0 z ≤ 19/10 z ≤ 30/11-3z ≤ 2
z ≤ 2/2 z ≤ 8/3 -z ≤ 0 z ≤ 19/10 z ≤ 30/11-3z ≤ 2
z ≤ 2/2 z ≤ 8/3 0 ≤ z z ≤ 19/10 z ≤ 30/11-2/3 ≤ z
z ≤ 2/2 z ≤ 8/3 0 ≤ z z ≤ 19/10 z ≤ 30/11-2/3 ≤ z
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Duality Theorem
If the primal problem has an optimal solution,
x* = (x1*, x2
*,……,xn*)
then the dual also has an optimal solution,
y* = (y1*, y2
*,……,ym*)
and
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Géométrie
Geometry
Γεωμετρία
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Κανένας δεν εισάγει εκτός αν ξέρει τη γεωμετρία
Πλάτων
Nobody enters unless he knows GeometryPlato
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Personne ne sort s’ il ne connait laGéométrie
κανένας δεν παίρνει από εδώ εκτός αν ξέρει τη γεωμετρία Jean-Louis L.
Nobody gets out unless he knows Geometry
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A polyhedron is definedas the intersection of afinite number of linearhalfspaces.
2x + y ≤ 4 -x +3y ≤ 2-3x - 4y ≤ 5 -x - 6y ≤ 4
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A polytope Q is defined as a convex hull of a finite collection of points.
x = ∑λixi
y = ∑λiyi
∑λi = 1 λi ≥ 0
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Minkowski(1896)-Steinitz(1916)-Farkas(1906)-Weyl(1935) Theorem
Q is a polytope if and only ifit is a bounded polyhedron.
• Extension by Charnes & Cooper (1958)
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“This classical result is an outstanding example ofa fact which is completely obvious to geometricintuition, but wields important algebraic contentand is not trivial to prove.”
R.T. Rockafeller
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A polytope Q is defined as a convex hull of a finite collection of points.
λ1 + 2λ2 + λ3 = xλ1 + 2λ2 + 3λ3 = yλ1 + λ2 + λ3 = 1λ1 ≥ 0 λ2 ≥ 0 λ3 ≥ 0
y
x
p3(1,3)
p2(2,2)
p1(1,1)
A polyhedron is defined as the intersection of a finite number of linear halfspaces.
λ1 = 1 - λ2 - λ3
x + 2λ3 = y 2 - x - λ3 ≥ 0 x - 1 ≥ 0 λ3 ≥ 0
λ1 + 2λ2 + λ3 = xλ1 + 2λ2 + 3λ3 = yλ1 ≥ 0 λ2 ≥ 0 λ3 ≥ 0
1 + λ2 = x 1 + λ2 + 2λ3 = y 1 - λ2 - λ3 ≥ 0 λ2 ≥ 0 λ3 ≥ 0
λ2 = x - 1
2 - x - λ3 ≥ 0 x - 1 ≥ 0 λ3 ≥ 0
1 + λ2 + 2λ3 = y 1 - λ2 - λ3 ≥ 0 λ2 ≥ 0 λ3 ≥ 0
λ3 = (y - x) / 2
x ≤ 4 - yx ≥ 1y ≥ x
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References• Kirchhoff, G. "Über die Auflösung der Gleichungen, auf welche man bei
der untersuchung der linearen verteilung galvanischer Ströme geführt wird." Ann. Phys. Chem. 72, 497-508, 1847.
• А. А. Марков. "Распространение закона больших чисел на величины, зависящие друг от друга". "Известия Физико-математического общества при Казанском университете", 2-я серия, том 15, ст. 135-156, 1906.
• H. Minkowski, Geometrie der Zahlen (Leipzig, 1896).• J. Farkas, "Theorie der einfachen Ungleichungen," J. F. Reine u. Ang. Mat.,
124, 1-27, 1902.• H. Weyl, "Elementare Theorie der konvexen Polyeder," Comm. Helvet., 7,
290-306, 1935.• A. Charnes, W. W. Cooper, “The Strong Minkowski Farkas-Weyl Theorem
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References• E. Steinitz. Bedingt Konvergente Reihen und Konvexe Systeme. J. reine
angew. Math., 146:1-52, 1916.• K. Jeev, J-L. Lassez: Symbolic Stochastic Systems. MSV/AMCS 2004: 321-
328• V. Chandru, J-L. Lassez: Qualitative Theorem Proving in Linear Constraints.
Verification: Theory and Practice 2003: 395-406• Jean-Louis Lassez: From LP to LP: Programming with Constraints. DBPL
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