EEL5840: Elements of Machine Intelligence · PDF fileEEL5840: Elements of Machine Intelligence...

Nov-9-15—10:05 AM

1 University of Florida EEL 5840 – Class #30 – Fall 2015

© Dr. A. Antonio Arroyo

EEL5840: Elements of Machine Intelligence

Urban NaviGator Announcements

•  Reading Assignment: >  Nilsson chapters 15-16

•  Announcements: >  LISP and Extra Credit Project

Assigned >  2nd Exam Thu 12/03

•  Today s Handouts in WWW >  Outline for Class 30 >  www.mil.ufl.edu/eel5840 >  Software and Notes




Urban NaviGator Today s Menu

•  See Class 29 Slides 3-6 •  Resolution-Refutation Graphs •  Resolution-Refutation Examples •  English Word Problems

Nov-9-15—10:05 AM




Urban NaviGator Predicate Calculus •  Resolution-Refutation Strategies

>  Breadth-First: All of the first-level resolvents are computed first, then the second-level resolvents, and so on. (A first-level resolvent is one between two clauses in the base set; an ith level resolvent is one whose deepest parent is an (i-1)th level resolvent)

>  Set-of-Support: At least one parent of each resolvent is selected from among the clauses resulting from the negation of the goal wff or from their descendants (the set of support)

>  Unit-Preference:Select a single literal clause to be a parent in each resolution

>  Linear-Input: Each resolvent has at least 1 parent belonging to the base set.

>  Ancestry-Filtered: Each resolvent has a parent that is either in the base set or that is an ancestor of the other parent





>  The strategies are better visualized by drawing a so-called Refutation Graph—place each resolvent as the root of an inverted tree with each parent being one of the leafs.

>  Example:

(4)’ ~I(z) ∨ R(z) (3b)’ I(A)

(5) R(A) (1)’ ~R(x) ∨ L(x)

(6) L(A) (2)’ ~D(y) ∨ ~L(y)

(7) ~D(A) (3a)’ D(A)

(8) Nil

Nov-9-15—10:05 AM





>  The strategies are better visualized by drawing a so-called Refutation Graph—place each resolvent as the root of an inverted tree with each parent being one of the leafs.

>  Example:

(4)’ ~I(z) ∨ R(z) Ans(z) (3b)’ I(A) {A/z}

(5) R(A) Ans(A) (1)’ ~R(x) ∨ L(x) {A/x}

(6) L(A) Ans(A) (2)’ ~D(y) ∨ ~L(y) {A/y}

(7) ~D(A) Ans(A) (3a)’ D(A)

(8) Ans(A) {Provided the substitutions are consistent}

A set of substitutions {si} where each substitution si consists of termi/vari is consistent iff we form a set U1 of all of the numerators {termi} and a set U2 of all the denominators {vari} and set U1 unifies with U2. A set J1 unifies with set J2 if we can find a substitution σ such that J1=J2σ, σ≠nil. σ is called the unifier of J1 & J2.





>  Example: (4)’ ~I(z) ∨ R(z) (3b)’ I(A) {A/z}

(5) R(A) (1)’ ~R(x) ∨ L(x) {A/x}

(6) L(A) (2)’ ~D(y) ∨ ~L(y) {A/y}

(7) ~D(A) (3a)’ D(A)

(8) Nil Clearly this is a set-of-support resolution because at least one parent of each of the

resolvents is selected from among the clauses resulting from the negation of the goal wff or from their descendants (the set of support)

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>  Example: (3b)’ I(A) (4)’ ~I(z) ∨ R(z)

{A/z}

(5) R(A) (1)’ ~R(x) ∨ L(x) {A/x}

(6) L(A) (2)’ ~D(y) ∨ ~L(y) {A/y}

(7) ~D(A) (3a)’ D(A)

(8) Nil Clearly this is a unit-preference resolution because at least one parent of each of the

resolvents is a unit clause.





> Example:

(4)’ ~I(z) ∨ R(z) (3b)’ I(A) {A/z}

(5) R(A) (1)’ ~R(x) ∨ L(x) {A/x}

(6) L(A) (2)’ ~D(y) ∨ ~L(y) {A/y}

(7) ~D(A) (3a)’ D(A)

(8) Nil Clearly this is a linear input resolution because at least one parent of each

of the resolvents is selected from the base set

Nov-9-15—10:05 AM





>  Example: (4) ~Q(z) ∨ ~P(z) (2) Q(x) ∨ ~P(x)

(5) * ~P(z) (3) ~Q(y) ∨ P(y)

(6) ~Q(z) (1) Q(w) ∨ P(w)

(7) P(z)

(8) Nil {z/x, z/y, z/w} Clearly this is an ancestry-filtered resolution because at least one parent of each of the resolvents is selected from the base set or is an ancestor of the other parent. The resolvent marked with an * is called a merge. Completeness is preserved if the ancestors that are used are limited to merges.




Urban NaviGator Predicate Calculus Mike, Eric, and Tony belong to the Alpine Club. Every member of the Alpine Club is either a skier or a mountain climber or both. No mountain climber likes rain, and all skiers like snow. Eric dislikes whatever Mike likes and likes whatever Mike dislikes. Mike likes rain and snow. Is there a member of the Alpine club that is a mountain climber and not a skier?

1.  Member(Mike, Alpineclub) 2.  Member(Eric, Alpineclub) 3.  Member(Tony, Alpineclub) 4.  (∀x)[Member(x, Alpineclub) → (Skier(x) ∨ MC(x))] 5.  ~(∃ x)[MC(x) ∧ Likes(x, Rain)] 6.  (∀x)[Skier(x) → Likes(x, Snow)] 7.  (∀x)[Likes(Mike, x) ≡ ~Likes(Eric, x)] 8.  Likes(Mike, Rain) ∧ Likes(Mike, Snow)

9.  (∃ x)[Member(x, Alpineclub) ∧ MC(x) ∧ ~Skier(x)]

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Urban NaviGator Predicate Calculus •  The clause forms of these sentences are: •  1. Member(Mike, Alpineclub) •  2. Member(Eric, Alpineclub) •  3. Member(Tony, Alpineclub) •  4. ~Member(x1, Alpineclub) ∨ Skier(x1) ∨ MC(x1) •  5. ~MC(x2) ∨ ~Likes(x2, Rain) •  6. ~Skier(x3) ∨ Likes(x3, Snow) •  7a. ~Likes(Mike, x4) ∨ ~Likes(Eric, x4) •  7b. Likes(Mike, x5) ∨ Likes(Eric, x5) •  8a. Likes(Mike, Rain) •  8b. Likes(Mike, Snow) •  The clause form of the negation of the theorem to be proved (with Ans literal) is: •  9. ~Member(z, Alpineclub) ∨ ~MC(z) ∨ Skier(z) ∨ Ans(z) •  The refutation is: •  {Substitute z for x1 in 4 to yield ~Member(z, Alpineclub) ∨ Skier(z) ∨ MC(z)} •  10. ~Member(z, Alpineclub) ∨ Skier(z) ∨ Ans(z) resolving 9 with 4 {z/x1} [2 merges] •  11. Skier(Eric) ∨ Ans(Eric) resolving 10 with 2 {Eric/z} •  12. Likes(Eric, Snow) ∨ Ans(Eric) resolving 11 with 6 {Eric/x3} •  13. ~Likes(Mike, Snow) ∨ Ans(Eric) resolving 12 with 7a {Snow/x4} •  14. Ans(Eric) resolving 13 with 8b {z/x1, Eric/z, Eric/x3, Snow/x4}




Urban NaviGator Predicate Calculus •  Resolution-Refutation Examples

Example RF-1 Resolution-Refutation

–  Given the following database axioms about dogs: We know that horses are faster than dogs and that there is a Collie that is faster than any rabbit. We know that Camarero is a horse, Energizer is a rabbit, and that Pooch is a Collie. Is Camarero faster than Energizer?

–  Solve by drawing a Refutation Graph resulting from a complete strategy. (Make sure you mark clearly the required substitutions).

[Required: (1) Represent the axioms/goal in the Predicate Calculus. (2) Represent any additional commonsense knowledge needed to solve the

problem using Predicate Calculus, (3) Define your strategy, i.e., My strategy is _________________ (4) A Refutation Graph, (5) Describe how your graph meets the strategy requirements]

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Urban NaviGator Predicate Calculus >  We know that horses are faster than dogs and that there is a

Collie that is faster than any rabbit. We know that Camarero is a horse, Energizer is a rabbit, and that Pooch is a Collie. Is Camarero faster than Energizer?

Step 1: Define the domain D of interest. Let D contain horses, dogs and rabbits.

Step 2: Write the given information (axioms) in the predicate calculus including the unknown wff w.

1. (∀x) (∀y) [{Horse(x) ∧ Dog(y)} → Faster(x,y)] 2. (∃y)[Collie(y) ∧ {(∀z) Rabbit(z) → Faster(y,z)}] 3. Horse(Camarero) 4. Rabbit(Energizer) 5. Collie(Pooch) 8. Faster(Camarero, Energizer) ≡ w, the goal wff




Urban NaviGator Predicate Calculus Step 3: Additional commonsense (world) knowledge.

6. (∀x) (∀y) (∀z) [{Faster(x,y) ∧ Faster(y,z)} → Faster(x,z)] 7. (∀y) [Collie(y) → Dog(y)]

Step 4: Select a complete strategy {choose among breadth-first, set-of-support, and ancestry-filtered} Let us use set-of-support

Step5: Put your axioms and negation of the goal in clause form The axioms set Γ includes: 1’. ~Horse(x) ∨ ~Dog(y) ∨ Faster(x,y) 2’. Collie(Pooch) 2’’. ~Rabbit(z) ∨ Faster(Pooch,z) 3. Horse(Camarero) 4. Rabbit(Energizer) 5. Collie(Pooch) 6. ~Faster(x,y) ∨ ~Faster(y,z) ∨ Faster(x,z) 7. ~Collie(y) ∨ Dog(y) and the negation of the goal is 8. ~Faster(Camarero, Energizer)

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Urban NaviGator Predicate Calculus 8. ~Faster(Camarero, Energizer) 6. ~Faster(x,y) ∨ ~Faster(y,z) ∨ Faster(x,z)

9. ~Faster(Camarero,y)∨~Faster(y,Energizer) 1 . ~Horse(x)∨~Dog(y)∨Faster(x,y)

10. ~Horse(Camarero)∨~Dog(y)∨~Faster(y,Energizer) 3. Horse(Camarero)

11. ~Dog(y)∨~Faster(y,Energizer) 7. ~Collie(y) ∨ Dog(y)

12. ~Collie(y)∨~Faster(y,Energizer) 5. Collie(Pooch)

13. ~Faster(Pooch,Energizer) 2.’’ ~Rabbit(z)∨Faster(Pooch,z)

14. ~Rabbit(Energizer) 4. Rabbit(Energizer)

15. nil






Try an alternative expression for 2 1. (∀x) (∀y) [{Horse(x) ∧ Dog(y)} → Faster(x,y)] 2. (∃y) (∀z)[{Collie(y) ∧ Rabbit(z)} → Faster(y,z)] 3. Horse(Camarero) 4. Rabbit(Energizer) 5. Collie(Pooch) 6. (∀x) (∀y) (∀z) [{Faster(x,y) ∧ Faster(y,z)} → Faster(x,z)] 7. (∀y) [Collie(y) → Dog(y)] 8. Faster(Camarero, Energizer) ≡ w, the goal wff

Nov-9-15—10:05 AM




Urban NaviGator Predicate Calculus Step 5: Put your axioms and negation of the goal in clause form

The axioms set Γ includes: 1’. ~Horse(x) ∨ ~Dog(y) ∨ Faster(x,y) 2’. ~Collie(Pooch) ∨ ~Rabbit(z) ∨ Faster(Pooch,z) 3. Horse(Camarero) 4. Rabbit(Energizer) 5. Collie(Pooch) 6. ~Faster(x,y) ∨ ~Faster(y,z)∨Faster(x,z) 7. ~Collie(y) ∨ Dog(y) and the negation of the goal is 8. ~Faster(Camarero, Energizer)









13. ~Faster(Pooch,Energizer) 2.’’ ~Collie(Pooch) ∨~Rabbit(z)∨Faster(Pooch,z)

14. ~Collie(Pooch)∨~Rabbit(Energizer) 4. Rabbit(Energizer)

15. ~Collie(Pooch) 5. Collie(Pooch)

16. nil

Nov-9-15—10:05 AM






Try another alternative expression for 2 1. (∀x) (∀y) [{Horse(x) ∧ Dog(y)} → Faster(x,y)] (∃y)[Collie(y) ∧ {(∀z) Rabbit(z) → Faster(y,z)}] {original} 2. (∃y)[Collie(y) ∧ (∀z) {Rabbit(z) ∧ Faster(y,z)}] 3. Horse(Camarero) 4. Rabbit(Energizer) 5. Collie(Pooch) 6. (∀x) (∀y) (∀z) [{Faster(x,y) ∧ Faster(y,z)} → Faster(x,z)] 7. (∀y) [Collie(y) → Dog(y)] 8. Faster(Camarero, Energizer) ≡ w, the goal wff




Urban NaviGator Predicate Calculus Step5: Put your axioms and negation of the goal in clause form

The axioms set Δ includes: 1’. ~Horse(x) ∨ ~Dog(y) ∨ Faster(x,y) 2’. Collie(Pooch) 2”. Rabbit(z) 2’’’. Faster(Pooch,z) 3. Horse(Camarero) 4. Rabbit(Energizer) 5. Collie(Pooch) 6. ~Faster(x,y) ∨ ~Faster(y,z)∨Faster(x,z) 7. ~Collie(y) ∨ Dog(y) and the negation of the goal is 8. ~Faster(Camarero, Energizer)

Everybody is a rabbit! This is inconsistent in our domain of horses, dogs, and rabbits. With z=Pooch and since z∈D we would need Rabbit(Pooch) which is false!

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13. ~Faster(Pooch,Energizer) 2.’’’ Faster(Pooch,z)

14. nil

You get nil but only because everybody is a rabbit, that is (∀z) Rabbit(z) is true, but this is false in our domain! Energizer/z works but the set Γ is inconsistent…Luck!





Example RF-2: If Peter goes wherever John goes and if John is in Miami, where is Peter?

1. (∀x) [At(John,x) → At(Peter,x)] 2. At(John,Miami)

Goal wff 3. (∃z)At(Peter,z)

Negate Goal 3.’ (∀z)~At(Peter,z)

Convert to clauses 1.’ ~At(John,x) ∨ At(Peter,x) 2.’ At(John,Miami)

Perform Resolution Obtain ℜ1 from 3.’ and 1.’ ℜ1:~At(John,x) s1={z/x} Obtain ℜ2 from ℜ1 and 2.’ ℜ2: Nil s2={Miami/z} Answer: Yes there is a place where Peter is. {But note that we want more than a yes answer—we want to know where!}

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Example RF-2: If Peter goes wherever John goes and if John is in Miami, where is Peter?

The wff we derived is: ~At(Peter,z) but we also know that s2={Miami/z}, that is we derived (∃z)At(Peter,z) with z=Miami

Suppose we add a tag-along literal, either Ans(z) or At(Peter,z), to the goal wff and suppose that whatever substitutions we do to the clause that contains the tag-along literal, we also do to the tag-along literal, but the tag-along literal never participates in Resolution (just tags along)

Perform Resolution 3.’ ~At(Peter,z) ∨ At(Peter,z) Obtain ℜ1 from 3.’ and 1.’ ℜ1: [~At(John,x) ∨ At(Peter,z)] s1={z/x} Obtain ℜ2 from ℜ1 and 2.’ ℜ2: Nil ∨ At(Peter,z) s2={Miami/z} New Answer: Yes, At(Peter,Miami)

Do you like this better? –  This process is called Answer Extraction (Green, 1969) ‒  *Note: the form of the answer literal is identical to the goal wff.




Urban NaviGator

The End!

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