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

= + ℛ = λ ˆ 𝒠 + (1 ⊗ λ) ˇ 𝒠 0 ⊘ λ ⊘ 1

Our goal: Answering ‘Does this gamble lie in that cone?’

= + ℛ = λ ˆ 𝒠 + (1 ⊗ λ) ˇ 𝒠 f

?

= λ√︂

g∈ ˆ Dˆ

νgg + (1 ⊗ λ)√︂

g∈ ˇ Dˇ

νgg 0 ⊘ λ ⊘ 1 and ν > 0

Our goal: Answering ‘Does this gamble lie in that cone?’

= + ℛ = λ ˆ 𝒠 + (1 ⊗ λ) ˇ 𝒠 f

?

= λ√︂

g∈ ˆ Dˆ

νgg + (1 ⊗ λ)√︂

g∈ ˇ Dˇ

νgg 0 ⊘ λ ⊘ 1 and ν > 0

Our goal: Answering ‘Does this gamble lie in that cone?’

= + ℛ = λ ˆ 𝒠 + (1 ⊗ λ) ˇ 𝒠 f

?

= λ√︂

g∈ ˆ Dˆ

νgg + (1 ⊗ λ)√︂

g∈ ˇ Dˇ

νgg 0 ⊘ λ ⊘ 1 and ν > 0

Our goal: Answering ‘Does this gamble lie in that cone?’

= + ℛ = λ ˆ 𝒠 + (1 ⊗ λ) ˇ 𝒠 f

?

= λ√︂

g∈ ˆ Dˆ

νgg + (1 ⊗ λ)√︂

g∈ ˇ Dˇ

νgg 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... 2n 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... 2n 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... 2n 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

find λD ∈ [0, 1] and νD ∈ (R>0)D for all 𝒠 in ℛ0 s.t.

√︂

D∈R0 λD = 1

and

√︂

D∈R0 λD

√︂

g∈DνD,gg ⊘

⊙ f

The general, nonlinear problem with strict inequalities

find λD ∈ [0, 1] and νD ∈ (R>0)D for all 𝒠 in ℛ0 s.t.

√︂

D∈R0 λD = 1

and

√︂

D∈R0 λD

√︂

g∈DνD,gg ⊘

⊙ f h ⊘ ⊙ f means h(ω) ⊘ f (ω) for ω in ΩΓ h(ω) ⊙ f (ω) for ω in Ω∆ with ΩΓ ∪ Ω∆ = Ω

The general, nonlinear problem with strict inequalities

find λD ∈ [0, 1] and νD ∈ (R>0)D for all 𝒠 in ℛ0 s.t.

√︂

D∈R0 λD ⊙ 1

and

√︂

D∈R0 λD

√︂

g∈DνD,gg ⊘

⊙ f h ⊘ ⊙ f means h(ω) ⊘ f (ω) for ω in ΩΓ h(ω) ⊙ f (ω) for ω in Ω∆ with ΩΓ ∪ Ω∆ = Ω

The general, nonlinear problem

find λD ∈ [0, 1] and τD ∈ (R≥1)D for all 𝒠 in ℛ0 and σ ∈ R≥1 s.t.

√︂

D∈R0 λD ⊙ 1

and

√︂

D∈R0 λD

√︂

g∈DτD,gg ⊘

⊙ σf h ⊘ ⊙ f means h(ω) ⊘ f (ω) for ω in ΩΓ h(ω) ⊙ f (ω) for ω in Ω∆ with ΩΓ ∪ Ω∆ = Ω

The general problem, with isolated nonlinearities

find λD ∈ [0, 1] and µD ∈ (R≥0)D for all 𝒠 in ℛ0 and σ ∈ R≥1 s.t.

√︂

D∈R0 λD ⊙ 1

and

√︂

D∈R0

√︂

g∈DµD,gg ⊘

⊙ σf λD ⊘ µD ⊘ λDµD for all 𝒠 in ℛ0 h ⊘ ⊙ f means h(ω) ⊘ f (ω) for ω in ΩΓ h(ω) ⊙ f (ω) for ω in Ω∆ with ΩΓ ∪ Ω∆ = Ω

The general problem, with isolated nonlinearities

find λD ∈ [0, 1] and µD ∈ (R≥0)D for all 𝒠 in ℛ0 and σ ∈ R≥1 s.t.

√︂

D∈R0 λD ⊙ 1

and

√︂

D∈R0

√︂

g∈DµD,gg ⊘

⊙ σf λD ⊘ µD ⊘ λDµD for all 𝒠 in ℛ0 (forces λD ∈ ¶0, 1♢) h ⊘ ⊙ f means h(ω) ⊘ f (ω) for ω in ΩΓ h(ω) ⊙ f (ω) for ω in Ω∆ with ΩΓ ∪ Ω∆ = 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∈Ri λD

s.t. λD ∈ [0, 1] and µD ∈ (R≥0)D for all 𝒠 in ℛi and σ ∈ R≥1

√︂

D∈Ri λD ⊙ 1

and

√︂

D∈Ri

√︂

g∈DµD,gg ⊘

⊙ σf λD ⊘ µD for all 𝒠 in ℛi.

The plain CONEstrip algorithm

• Init. Set i := 0.
• Iter. Does the linear programming problem below have a solution (¯

λ, ¯ µ, ¯ σ)? max.

√︂

D∈Ri λD

s.t. λD ∈ [0, 1] and µD ∈ (R≥0)D for all 𝒠 in ℛi and σ ∈ R≥1

√︂

D∈Ri λD ⊙ 1

and

√︂

D∈Ri

√︂

g∈DµD,gg ⊘

⊙ σf λD ⊘ µD for all 𝒠 in ℛi.

• No. f /

∈ ℛ0. Stop.

The plain CONEstrip algorithm

• Init. Set i := 0.
• Iter. Does the linear programming problem below have a solution (¯

λ, ¯ µ, ¯ σ)? max.

√︂

D∈Ri λD

s.t. λD ∈ [0, 1] and µD ∈ (R≥0)D for all 𝒠 in ℛi and σ ∈ R≥1

√︂

D∈Ri λD ⊙ 1

and

√︂

D∈Ri

√︂

g∈DµD,gg ⊘

⊙ σf λD ⊘ µD for all 𝒠 in ℛi.

• No. f /

∈ ℛ0. Stop.

• Yes. Let 𝒭 := ¶𝒠 ∈ ℛi : ¯

λD = 0♢ and set ℛi+1 := ℛi \ 𝒭.

The plain CONEstrip algorithm

• Init. Set i := 0.
• Iter. Does the linear programming problem below have a solution (¯

λ, ¯ µ, ¯ σ)? max.

√︂

D∈Ri λD

s.t. λD ∈ [0, 1] and µD ∈ (R≥0)D for all 𝒠 in ℛi and σ ∈ R≥1

√︂

D∈Ri λD ⊙ 1

and

√︂

D∈Ri

√︂

g∈DµD,gg ⊘

⊙ σf λD ⊘ µD for all 𝒠 in ℛi.

• No. f /

∈ ℛ0. Stop.

• Yes. Let 𝒭 := ¶𝒠 ∈ ℛi : ¯

λD = 0♢ and set ℛi+1 := ℛi \ 𝒭. Is ¶𝒠 ∈ 𝒭: ¯ µD = 0♢ = 𝒭?

The plain CONEstrip algorithm

• Init. Set i := 0.
• Iter. Does the linear programming problem below have a solution (¯

λ, ¯ µ, ¯ σ)? max.

√︂

D∈Ri λD

s.t. λD ∈ [0, 1] and µD ∈ (R≥0)D for all 𝒠 in ℛi and σ ∈ R≥1

√︂

D∈Ri λD ⊙ 1

and

√︂

D∈Ri

√︂

g∈DµD,gg ⊘

⊙ σf λD ⊘ µD for all 𝒠 in ℛi.

• No. f /

∈ ℛ0. Stop.

• Yes. Let 𝒭 := ¶𝒠 ∈ ℛi : ¯

λD = 0♢ and set ℛi+1 := ℛi \ 𝒭. Is ¶𝒠 ∈ 𝒭: ¯ µD = 0♢ = 𝒭?

• Yes. Set t := i + 1; f ∈ Rt ⊆ R0. Stop.

The plain CONEstrip algorithm

• Init. Set i := 0.
• Iter. Does the linear programming problem below have a solution (¯

λ, ¯ µ, ¯ σ)? max.

√︂

D∈Ri λD

s.t. λD ∈ [0, 1] and µD ∈ (R≥0)D for all 𝒠 in ℛi and σ ∈ R≥1

√︂

D∈Ri λD ⊙ 1

and

√︂

D∈Ri

√︂

g∈DµD,gg ⊘

⊙ σf λD ⊘ µD for all 𝒠 in ℛi.

• No. f /

∈ ℛ0. Stop.

• Yes. Let 𝒭 := ¶𝒠 ∈ ℛi : ¯

λD = 0♢ and set ℛi+1 := ℛi \ 𝒭. Is ¶𝒠 ∈ 𝒭: ¯ µD = 0♢ = 𝒭?

• Yes. Set t := i + 1; f ∈ Rt ⊆ R0. Stop.
• No. Increase i’s value by 1. Reiterate.

Structured gambles

Assume g =

∑︂

φ∈Φ

gφφ for all g in ¶f ♢ ∪

⋃︂

ℛ0, with Φ a finite set of basic functions.

Structured gambles generate structured problems

Assume g =

∑︂

φ∈Φ

gφφ for all g in ¶f ♢ ∪

⋃︂

ℛ0, with Φ a finite set of basic functions. Then

∑︂

D∈Ri

∑︂

g∈D

µD,gg ⊘ ⊙ σf iff

∑︂

φ∈Φ

κφφ ⊘ ⊙ 0 when κφ :=

⎞ ∑︂

D∈Ri

∑︂

g∈D

µD,ggφ

⎡

⊗ σfφ for all φ in Φ.

Structured gambles generate structured problems

Assume g =

∑︂

φ∈Φ

gφφ for all g in ¶f ♢ ∪

⋃︂

ℛ0, with Φ a finite set of basic functions. Then

∑︂

D∈Ri

∑︂

g∈D

µD,gg ⊘ ⊙ σf iff

∑︂

φ∈Φ

κφφ ⊘ ⊙ 0 when κφ :=

⎞ ∑︂

D∈Ri

∑︂

g∈D

µD,ggφ

⎡

⊗ σfφ for all φ in Φ. 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

∑︂

φ∈Φ

κφφ ⊘ ⊙ 0 (this relaxes the problem)

Row generation: adding constraints back

Remove

∑︂

φ∈Φ

κφφ ⊘ ⊙ 0 (this relaxes the problem) Iteratively, add back

∑︂

φ∈Φ

κφφ(ω) ⊘ ⊙ 0 for some wisely chosen ω in Ω:

Row generation: adding constraints back intelligently

Remove

∑︂

φ∈Φ

κφφ ⊘ ⊙ 0 (this relaxes the problem) Iteratively, add back

∑︂

φ∈Φ

κφφ(ω) ⊘ ⊙ 0 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.

Row generation: adding constraints back intelligently

Remove

∑︂

φ∈Φ

κφφ ⊘ ⊙ 0 (this relaxes the problem) Iteratively, add back

∑︂

φ∈Φ

κφφ(ω) ⊘ ⊙ 0 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.

◮ A ‘deep cut’ is generated by solving the problem

argmaxω∈Ω

⧹︄ ⧹︄ ⧹︄ ∑︂

φ∈Φ

¯ κφφ(ω)

⧹︄ ⧹︄ ⧹︄

Row generation: adding constraints back intelligently

Remove

∑︂

φ∈Φ

κφφ ⊘ ⊙ 0 (this relaxes the problem) Iteratively, add back

∑︂

φ∈Φ

κφφ(ω) ⊘ ⊙ 0 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.

◮ A ‘deep cut’ is generated by solving the problem

argmaxω∈Ω

⧹︄ ⧹︄ ⧹︄ ∑︂

φ∈Φ

¯ κφφ(ω)

⧹︄ ⧹︄ ⧹︄

◮ Iteration stops when no feasible ¯

κ or no violated constraint ω can be found.

Propositional sentences

Combinations of ¶0, 1♢-valued ‘literals’ βℓ with ℓ in a finite index set ℒ Operations: disjunction ∨, conjunction ∧, and the negation ¬

Propositional sentences

Combinations of ¶0, 1♢-valued ‘literals’ βℓ with ℓ in a finite index set ℒ Operations: disjunction ∨, conjunction ∧, and the negation ¬ Example: 𝜚(β) := (β♠ ∨ ¬β♣) ∧ β♥ with ℒ := ¶♠, ♣, ♡♢

Propositional sentences represent events

Combinations of ¶0, 1♢-valued ‘literals’ βℓ with ℓ in a finite index set ℒ Operations: disjunction ∨, conjunction ∧, and the negation ¬ Example: 𝜚(β) := (β♠ ∨ ¬β♣) ∧ β♥ with ℒ := ¶♠, ♣, ♡♢ Possibility space: ΩΓ := ¶β ∈ ¶0, 1♢L : χΓ(β) = 1♢ Ω∆ := ¶β ∈ ¶0, 1♢L : χ∆(β) = 1♢

Propositional sentences as basic functions

Combinations of ¶0, 1♢-valued ‘literals’ βℓ with ℓ in a finite index set ℒ Operations: disjunction ∨, conjunction ∧, and the negation ¬ Example: 𝜚(β) := (β♠ ∨ ¬β♣) ∧ β♥ with ℒ := ¶♠, ♣, ♡♢ Possibility space: ΩΓ := ¶β ∈ ¶0, 1♢L : χΓ(β) = 1♢ Ω∆ := ¶β ∈ ¶0, 1♢L : χ∆(β) = 1♢ Basic functions are indicators of events: Φ ⊆ ¶0, 1♢L ⊃ ¶0, 1♢ Assume, w.l.o.g., that φ(β) = βφ for all φ in Φ.

Row generation in a propositional context

Form of constraints to generate:

∑︂

φ∈Φ

κφ ¯ βφ ⊘ ⊙ 0 So truth assignments ¯ β in ¶0, 1♢L must be generated.

Row generation in a propositional context: any cut

Form of constraints to generate:

∑︂

φ∈Φ

κφ ¯ βφ ⊘ ⊙ 0 So truth assignments ¯ β in ¶0, 1♢L must be generated. By solving a SAT instance: find β ∈ ¶0, 1♢L such that χΓ(β) = 1

• r

find β ∈ ¶0, 1♢L such that χ∆(β) = 1

Row generation in a propositional context: deep cuts

Form of constraints to generate:

∑︂

φ∈Φ

κφ ¯ βφ ⊘ ⊙ 0 So truth assignments ¯ β in ¶0, 1♢L must be generated. By solving a SAT instance: find β ∈ ¶0, 1♢L such that χΓ(β) = 1

• r

find β ∈ ¶0, 1♢L such that χ∆(β) = 1 By solving a WPMaxSAT instance: maximize

∑︂

φ∈Φ

¯ κφβφ subject to β ∈ ¶0, 1♢L χΓ(β) = 1

• r

minimize

∑︂

φ∈Φ

¯ κφβφ subject to β ∈ ¶0, 1♢L χ∆(β) = 1

The propositional CONEstrip algorithm

• Init. Set i := 0. Is χΓ or χ∆ satisfiable?

The propositional CONEstrip algorithm

• Init. Set i := 0. Is χΓ or χ∆ satisfiable?
• No. The original problem is not well-posed. Stop.

The propositional CONEstrip algorithm

• Init. Set i := 0. Is χΓ or χ∆ satisfiable?
• No. The original problem is not well-posed. Stop.

Yes.

• If ¯

γ satisfies χΓ, set Γ0 := {¯ γ}; otherwise set Γ0 := ∅ and ¯ γ := 0.

• If ¯

δ satisfies χ∆, set ∆0 := {¯ δ}; otherwise set ∆0 := ∅ and ¯ δ := 0.

The propositional CONEstrip algorithm

• Iter. Master problem:

Does the linear program below have a solution (¯ λ, ¯ µ, ¯ σ, ¯ κ)? max.

√︂

D∈Ri λD

s.t. λD ∈ [0, 1] and µ ∈ (R≥0)D for all 𝒠 in ℛi and σ ∈ R≥1

√︂

D∈Ri λD ⊙ 1

and

∮︂√︂

φ∈Φ κφ ¯

βφ ⊘ 0 for all ¯ β in Γi

√︂

φ∈Φ κφ ¯

βφ ⊙ 0 for all ¯ β in ∆i λD ⊘ µD for all 𝒠 in ℛi wh. κφ := (√︂

D∈R0

√︂

g∈D µD,ggφ) ⊗ σfφ ∈ R for all φ in Φ.

The propositional CONEstrip algorithm

• Iter. Master problem:

Does the linear program below have a solution (¯ λ, ¯ µ, ¯ σ, ¯ κ)? max.

√︂

D∈Ri λD

s.t. λD ∈ [0, 1] and µ ∈ (R≥0)D for all 𝒠 in ℛi and σ ∈ R≥1

√︂

D∈Ri λD ⊙ 1

and

∮︂√︂

φ∈Φ κφ ¯

βφ ⊘ 0 for all ¯ β in Γi

√︂

φ∈Φ κφ ¯

βφ ⊙ 0 for all ¯ β in ∆i λD ⊘ µD for all 𝒠 in ℛi wh. κφ := (√︂

D∈R0

√︂

g∈D µD,ggφ) ⊗ σfφ ∈ R for all φ in Φ.

• No. f /

∈ ℛ0. Stop.

The propositional CONEstrip algorithm

• Iter. Master problem:

Does the linear program below have a solution (¯ λ, ¯ µ, ¯ σ, ¯ κ)? max.

√︂

D∈Ri λD

s.t. λD ∈ [0, 1] and µ ∈ (R≥0)D for all 𝒠 in ℛi and σ ∈ R≥1

√︂

D∈Ri λD ⊙ 1

and

∮︂√︂

φ∈Φ κφ ¯

βφ ⊘ 0 for all ¯ β in Γi

√︂

φ∈Φ κφ ¯

βφ ⊙ 0 for all ¯ β in ∆i λD ⊘ µD for all 𝒠 in ℛi wh. κφ := (√︂

D∈R0

√︂

g∈D µD,ggφ) ⊗ σfφ ∈ R for all φ in Φ.

• No. f /

∈ ℛ0. Stop.

• Yes. Let 𝒭 := ¶𝒠 ∈ ℛi : ¯

λD = 0♢ and set ℛi+1 := ℛi \ 𝒭.

The propositional CONEstrip algorithm

• Iter. Row generation:
• If Γi ̸= ∅, let ¯

γ be a maximizer of √︂

φ∈Φ ¯

κφβφ under χΓ; set Γi+1 := Γi ∪ ¶¯ γ♢.

• If ∆i ̸= ∅, let ¯

δ be a minimizer of √︂

φ∈Φ ¯

κφβφ under χ∆; set ∆i+1 := ∆i ∪ ¶¯ δ♢.

The propositional CONEstrip algorithm

• Iter. Row generation:
• If Γi ̸= ∅, let ¯

γ be a maximizer of √︂

φ∈Φ ¯

κφβφ under χΓ; set Γi+1 := Γi ∪ ¶¯ γ♢.

• If ∆i ̸= ∅, let ¯

δ be a minimizer of √︂

φ∈Φ ¯

κφβφ under χ∆; set ∆i+1 := ∆i ∪ ¶¯ δ♢.

Is √︂

φ∈Φ ¯

κφ¯ γφ ⊘ 0 ⊘ √︂

φ∈Φ ¯

κφ¯ δφ and ¶𝒠 ∈ 𝒭: ¯ µD = 0♢ = 𝒭?

The propositional CONEstrip algorithm

• Iter. Row generation:
• If Γi ̸= ∅, let ¯

γ be a maximizer of √︂

φ∈Φ ¯

κφβφ under χΓ; set Γi+1 := Γi ∪ ¶¯ γ♢.

• If ∆i ̸= ∅, let ¯

δ be a minimizer of √︂

φ∈Φ ¯

κφβφ under χ∆; set ∆i+1 := ∆i ∪ ¶¯ δ♢.

Is √︂

φ∈Φ ¯

κφ¯ γφ ⊘ 0 ⊘ √︂

φ∈Φ ¯

κφ¯ δφ and ¶𝒠 ∈ 𝒭: ¯ µD = 0♢ = 𝒭?

• Yes. Set t := i + 1; f ∈ ℛt ⊖ ℛ0. Stop.

The propositional CONEstrip algorithm

• Iter. Row generation:
• If Γi ̸= ∅, let ¯

γ be a maximizer of √︂

φ∈Φ ¯

κφβφ under χΓ; set Γi+1 := Γi ∪ ¶¯ γ♢.

• If ∆i ̸= ∅, let ¯

δ be a minimizer of √︂

φ∈Φ ¯

κφβφ under χ∆; set ∆i+1 := ∆i ∪ ¶¯ δ♢.

Is √︂

φ∈Φ ¯

κφ¯ γφ ⊘ 0 ⊘ √︂

φ∈Φ ¯

κφ¯ δφ and ¶𝒠 ∈ 𝒭: ¯ µD = 0♢ = 𝒭?

• Yes. Set t := i + 1; f ∈ ℛt ⊖ ℛ0. Stop.
• No. Increase i’s value by 1. Reiterate.

Thank you!

Questions?