A propositional CONEstrip algorithm Erik Quaeghebeur Centrum - PowerPoint PPT Presentation

A propositional CONEstrip algorithm Erik Quaeghebeur Centrum Wiskunde & Informatica Amsterdam, the Netherlands

Which cones? General cones of gambles. ℛ

Decomposing a general cone into open cones = + ˆ ˇ ℛ = λ 𝒠 + (1 ⊗ λ ) 𝒠 0 ⊘ λ ⊘ 1

Our goal: Answering ‘ Does this gamble lie in that cone? ’ = + ˆ ˇ ℛ = λ 𝒠 + (1 ⊗ λ ) 𝒠 ? f = λ √︂ D ˆ ν g g + (1 ⊗ λ ) √︂ D ˇ ν g g g ∈ ˆ g ∈ ˇ 0 ⊘ λ ⊘ 1 and ν > 0

Our goal: Answering ‘ Does this gamble lie in that cone? ’ = + ˆ ˇ ℛ = λ 𝒠 + (1 ⊗ λ ) 𝒠 ? f = λ √︂ D ˆ ν g g + (1 ⊗ λ ) √︂ D ˇ ν g g g ∈ ˆ g ∈ ˇ 0 ⊘ λ ⊘ 1 and ν > 0

Our goal: Answering ‘ Does this gamble lie in that cone? ’ = + ˆ ˇ ℛ = λ 𝒠 + (1 ⊗ λ ) 𝒠 ? f = λ √︂ D ˆ ν g g + (1 ⊗ λ ) √︂ D ˇ ν g g g ∈ ˆ g ∈ ˇ 0 ⊘ λ ⊘ 1 and ν > 0

Our goal: Answering ‘ Does this gamble lie in that cone? ’ = + ˆ ˇ ℛ = λ 𝒠 + (1 ⊗ λ ) 𝒠 ? f = λ √︂ D ˆ ν g g + (1 ⊗ λ ) √︂ D ˇ ν g g g ∈ ˆ g ∈ ˇ 0 ⊘ λ ⊘ 1 and ν > 0 Why? Finitary (imprecise) probabilistic deductive inference reduces to such feasibility problems or optimization variants thereof.

Problem: The cones are really big How big? With dimension exponential in the number of events involved in the assessment represented by the cone.

Problem: The cones are really big How big? With dimension exponential in the number of events involved in the assessment represented by the cone. So for n events... 2 n dimensions

Solution: Only deal with a small subset of dimensions How big? With dimension exponential in the number of events involved in the assessment represented by the cone. So for n events... 2 n dimensions How many? Polynomial in n

Solution: Only deal with a small subset of dimensions How big? With dimension exponential in the number of events involved in the assessment represented by the cone. So for n events... 2 n dimensions How many? Polynomial in n Any downsides or trade-offs? Computationally ‘hard’ to find the right dimensions.

The general, nonlinear problem with strict inequalities ν D ∈ ( R > 0 ) D find λ D ∈ [0 , 1] and for all 𝒠 in ℛ 0 s.t. √︂ D∈R 0 λ D = 1 and √︂ D∈R 0 λ D √︂ g ∈D ν D , g g ⊘ ⊙ f

The general, nonlinear problem with strict inequalities ν D ∈ ( R > 0 ) D find λ D ∈ [0 , 1] and for all 𝒠 in ℛ 0 s.t. √︂ D∈R 0 λ D = 1 and √︂ D∈R 0 λ D √︂ g ∈D ν D , g g ⊘ ⊙ f h ( ω ) ⊘ f ( ω ) for ω in Ω Γ h ⊘ ⊙ f means with Ω Γ ∪ Ω ∆ = Ω h ( ω ) ⊙ f ( ω ) for ω in Ω ∆

The general, nonlinear problem with strict inequalities ν D ∈ ( R > 0 ) D find λ D ∈ [0 , 1] and for all 𝒠 in ℛ 0 s.t. √︂ D∈R 0 λ D ⊙ 1 and √︂ D∈R 0 λ D √︂ g ∈D ν D , g g ⊘ ⊙ f h ( ω ) ⊘ f ( ω ) for ω in Ω Γ h ⊘ ⊙ f means with Ω Γ ∪ Ω ∆ = Ω h ( ω ) ⊙ f ( ω ) for ω in Ω ∆

The general, nonlinear problem τ D ∈ ( R ≥ 1 ) D find λ D ∈ [0 , 1] and for all 𝒠 in ℛ 0 and σ ∈ R ≥ 1 s.t. √︂ D∈R 0 λ D ⊙ 1 and √︂ D∈R 0 λ D √︂ g ∈D τ D , g g ⊘ ⊙ σ f h ( ω ) ⊘ f ( ω ) for ω in Ω Γ h ⊘ ⊙ f means with Ω Γ ∪ Ω ∆ = Ω h ( ω ) ⊙ f ( ω ) for ω in Ω ∆

The general problem, with isolated nonlinearities µ D ∈ ( R ≥ 0 ) D find λ D ∈ [0 , 1] and for all 𝒠 in ℛ 0 and σ ∈ R ≥ 1 s.t. √︂ D∈R 0 λ D ⊙ 1 and √︂ √︂ g ∈D µ D , g g ⊘ ⊙ σ f D∈R 0 λ D ⊘ µ D ⊘ λ D µ D for all 𝒠 in ℛ 0 h ( ω ) ⊘ f ( ω ) for ω in Ω Γ h ⊘ ⊙ f means with Ω Γ ∪ Ω ∆ = Ω h ( ω ) ⊙ f ( ω ) for ω in Ω ∆

The general problem, with isolated nonlinearities µ D ∈ ( R ≥ 0 ) D find λ D ∈ [0 , 1] and for all 𝒠 in ℛ 0 and σ ∈ R ≥ 1 s.t. √︂ D∈R 0 λ D ⊙ 1 and √︂ √︂ g ∈D µ D , g g ⊘ ⊙ σ f D∈R 0 λ D ⊘ µ D ⊘ λ D µ D for all 𝒠 in ℛ 0 (forces λ D ∈ ¶ 0 , 1 ♢ ) h ( ω ) ⊘ f ( ω ) for ω in Ω Γ h ⊘ ⊙ f means with Ω Γ ∪ Ω ∆ = Ω h ( ω ) ⊙ f ( ω ) for ω in Ω ∆

The plain CONEstrip algorithm Init . Set i := 0 .

The plain CONEstrip algorithm Init . Set i := 0 . Iter . Does the linear programming problem below have a solution ( ¯ λ, ¯ µ, ¯ σ ) ? max. √︂ D∈R i λ D µ D ∈ ( R ≥ 0 ) D s.t. λ D ∈ [0 , 1] and for all 𝒠 in ℛ i and σ ∈ R ≥ 1 √︂ D∈R i λ D ⊙ 1 and √︂ √︂ g ∈D µ D , g g ⊘ ⊙ σ f D∈R i λ D ⊘ µ D for all 𝒠 in ℛ i .

The plain CONEstrip algorithm Init . Set i := 0 . Iter . Does the linear programming problem below have a solution ( ¯ λ, ¯ µ, ¯ σ ) ? max. √︂ D∈R i λ D µ D ∈ ( R ≥ 0 ) D s.t. λ D ∈ [0 , 1] and for all 𝒠 in ℛ i and σ ∈ R ≥ 1 √︂ D∈R i λ D ⊙ 1 and √︂ √︂ g ∈D µ D , g g ⊘ ⊙ σ f D∈R i λ D ⊘ µ D for all 𝒠 in ℛ i . No . f / ∈ ℛ 0 . Stop .

The plain CONEstrip algorithm Init . Set i := 0 . Iter . Does the linear programming problem below have a solution ( ¯ λ, ¯ µ, ¯ σ ) ? max. √︂ D∈R i λ D µ D ∈ ( R ≥ 0 ) D s.t. λ D ∈ [0 , 1] and for all 𝒠 in ℛ i and σ ∈ R ≥ 1 √︂ D∈R i λ D ⊙ 1 and √︂ √︂ g ∈D µ D , g g ⊘ ⊙ σ f D∈R i λ D ⊘ µ D for all 𝒠 in ℛ i . No . f / ∈ ℛ 0 . Stop . Yes . Let 𝒭 := ¶𝒠 ∈ ℛ i : ¯ λ D = 0 ♢ and set ℛ i +1 := ℛ i \ 𝒭 .

The plain CONEstrip algorithm Init . Set i := 0 . Iter . Does the linear programming problem below have a solution ( ¯ λ, ¯ µ, ¯ σ ) ? max. √︂ D∈R i λ D µ D ∈ ( R ≥ 0 ) D s.t. λ D ∈ [0 , 1] and for all 𝒠 in ℛ i and σ ∈ R ≥ 1 √︂ D∈R i λ D ⊙ 1 and √︂ √︂ g ∈D µ D , g g ⊘ ⊙ σ f D∈R i λ D ⊘ µ D for all 𝒠 in ℛ i . No . f / ∈ ℛ 0 . Stop . Yes . Let 𝒭 := ¶𝒠 ∈ ℛ i : ¯ λ D = 0 ♢ and set ℛ i +1 := ℛ i \ 𝒭 . Is ¶𝒠 ∈ 𝒭 : ¯ µ D = 0 ♢ = 𝒭 ?

The plain CONEstrip algorithm Init . Set i := 0 . Iter . Does the linear programming problem below have a solution ( ¯ λ, ¯ µ, ¯ σ ) ? max. √︂ D∈R i λ D µ D ∈ ( R ≥ 0 ) D s.t. λ D ∈ [0 , 1] and for all 𝒠 in ℛ i and σ ∈ R ≥ 1 √︂ D∈R i λ D ⊙ 1 and √︂ √︂ g ∈D µ D , g g ⊘ ⊙ σ f D∈R i λ D ⊘ µ D for all 𝒠 in ℛ i . No . f / ∈ ℛ 0 . Stop . Yes . Let 𝒭 := ¶𝒠 ∈ ℛ i : ¯ λ D = 0 ♢ and set ℛ i +1 := ℛ i \ 𝒭 . Is ¶𝒠 ∈ 𝒭 : ¯ µ D = 0 ♢ = 𝒭 ? Yes . Set t := i + 1 ; f ∈ R t ⊆ R 0 . Stop .

The plain CONEstrip algorithm Init . Set i := 0 . Iter . Does the linear programming problem below have a solution ( ¯ λ, ¯ µ, ¯ σ ) ? max. √︂ D∈R i λ D µ D ∈ ( R ≥ 0 ) D s.t. λ D ∈ [0 , 1] and for all 𝒠 in ℛ i and σ ∈ R ≥ 1 √︂ D∈R i λ D ⊙ 1 and √︂ √︂ g ∈D µ D , g g ⊘ ⊙ σ f D∈R i λ D ⊘ µ D for all 𝒠 in ℛ i . No . f / ∈ ℛ 0 . Stop . Yes . Let 𝒭 := ¶𝒠 ∈ ℛ i : ¯ λ D = 0 ♢ and set ℛ i +1 := ℛ i \ 𝒭 . Is ¶𝒠 ∈ 𝒭 : ¯ µ D = 0 ♢ = 𝒭 ? Yes . Set t := i + 1 ; f ∈ R t ⊆ R 0 . Stop . No . Increase i ’s value by 1 . Reiterate .

Structured gambles Assume ∑︂ ⋃︂ g = g φ φ for all g in ¶ f ♢ ∪ ℛ 0 , φ ∈ Φ with Φ a finite set of basic functions.

Structured gambles generate structured problems Assume ∑︂ ⋃︂ g = g φ φ for all g in ¶ f ♢ ∪ ℛ 0 , φ ∈ Φ with Φ a finite set of basic functions. Then ∑︂ ∑︂ ∑︂ µ D , g g ⊘ ⊙ σ f iff κ φ φ ⊘ ⊙ 0 D∈R i g ∈D φ ∈ Φ when ⎞ ∑︂ ⎡ ∑︂ κ φ := µ D , g g φ ⊗ σ f φ for all φ in Φ . D∈R i g ∈D

Structured gambles generate structured problems Assume ∑︂ ⋃︂ g = g φ φ for all g in ¶ f ♢ ∪ ℛ 0 , φ ∈ Φ with Φ a finite set of basic functions. Then ∑︂ ∑︂ ∑︂ µ D , g g ⊘ ⊙ σ f iff κ φ φ ⊘ ⊙ 0 D∈R i g ∈D φ ∈ Φ when ⎞ ∑︂ ⎡ ∑︂ κ φ := µ D , g g φ ⊗ σ f φ for all φ in Φ . D∈R i g ∈D Why? The now explicit structure can be exploited to solve the problem more efficiently even though we have added constraints.

Row generation: removing constraints Remove ∑︂ κ φ φ ⊘ ⊙ 0 φ ∈ Φ (this relaxes the problem)

Row generation: adding constraints back Remove ∑︂ κ φ φ ⊘ ⊙ 0 φ ∈ Φ (this relaxes the problem) Iteratively, add back ∑︂ κ φ φ ( ω ) ⊘ ⊙ 0 φ ∈ Φ for some wisely chosen ω in Ω :

Row generation: adding constraints back intelligently Remove ∑︂ κ φ φ ⊘ ⊙ 0 φ ∈ Φ (this relaxes the problem) Iteratively, add back ∑︂ κ φ φ ( ω ) ⊘ ⊙ 0 φ ∈ Φ for some wisely chosen ω in Ω : ◮ Each iteration, a solution ¯ κ is obtained; choose the next ω such that ¯ κ violates the corresponding constraint.