Control Algorithms 2Chapter 6

Production Systems

A Model of Computation

Emil Post (40’s): production systems as a formal theory of computation. Equivalent to a Turing machine. Set of rewrite rules for strings.

Newell and Simon (60’s, 70’s, 80’s): General Problem Solver

John Anderson, Newell and Simon (80’s): learning models, ACT*, SOAR

Everyone (80’s): Expert systems

Components

1. Set of rewrite rulesS NP VPLHS: Condition PartRHS: Action Part

Components

2. Working Memory--Contains the current state of the world--Contains pattern that is matched against the condition of the production--When a match occurs, an action is performed

Components

3. Recognize-Act Cycle--Isolate a subset of productions whose conditions match patterns in working memory: conflict set--Choose one of them

---Fire---Change contents of working memory

--Stop when there are no matches

Example: Production system to generate the set of palindromes over the alphabet {0,1}

Productions1. N 0N02. N 1N13. N 04. N 15. N λ

Iteration Working Memory Conflict Set Fired0 N 1,2,3,4,511 0N0 1,2,3,4,512 00N00 1,2,3,4,523 001N100 1,2,3,4,534 0010100

Knight’s Tour As a Production System

Given a 3X3 matrixWhat squares can a knight land on

What values of X, Y satisfy mv(X,Y) X,Y are elements of {1,2,…,9)

1 2 3

4 5 6

7 8 9

1. mv(1,8) 7. mv(4,9) 13. mv(8,3)

2. mv(1,6) 8. mv(4,3) 14. mv(8,1)

3. mv(2,9) 9. mv(6,1) 15. mv(9,2)

4. mv(2,7) 10. mv(6,7) 16. mv(9,4)

5. mv(3,4) 11. mv(7,2)

6. mv(3,8) 12. mv(7,6)

The General Case

),(),(),(( yxpathyzpathzxmvzyx

)),(( xxpathx

Changes

1. Every expression of the form mv(x,y) becomes on(x) on(y)

2. Use no path expression3. Working memory is the current state and

goal state4. Conflict set is the set of rules that match

the current state5. Apply all rules until the current state

equals the goal state

Productions

1. mv(1,8) 7. mv(4,9) 13. mv(8,3)

2. mv(1,6) 8. mv(4,3) 14. mv(8,1)

3. mv(2,9) 9. mv(6,1) 15. mv(9,2)

4. mv(2,7) 10. mv(6,7) 16. mv(9,4)

5. mv(3,4) 11. mv(7,2)

6. mv(3,8) 12. mv(7,6)

1. on(1) -> on(8) 7. on(4) -> on(9) 13. on(8) -> on(3)

2. on(1) -> on(6) 8. on(4) -> on(3) 14. on(8) -> on(1)

3. on(2) -> on(9) 9. on(6) -> on(1) 15. on(9) -> on(2)

4. on(2) -> on(7) 10. on(6) -> on(7) 16. on(9) -> on(4)

5. on(3) -> on(4) 11. on(7) -> on(2)

6. on(3) -> on(8) 12. on(7) -> on(6)

Can We Get from 1 to 2?

Iteration --Working Memory-- Conflict Set FiredCurrent Goal

0 1 2 1,2 11 8 2 13,14 132 3 2 5,6 5

3 4 2 7,8 74 9 2 15,16 155 2 2 Halt

Pattern Search

path(1,2) {1/x,2/y}mv(1,z)^path(z,2) {8/z}mv(1,8)^path(8,2) mv(8,z)^path(z,2) {3/z} mv(8,3)^path(3,2) mv(3,z)^path(z,2) {4/z} mv(3,4)^path(4,2) mv(4,z)^path(z,2) {9/z} mv(4,9)^path(9,2) mv(9,z)^path(z,2) {2/z} mv(9,2)^path(2,2) t t t tt

Now look at working memory in the production system

Equivalences

Production System Pattern SearchProductions mvworking memory path(X,Y)Fire lowest numbered production Choose first rule that unifies

Conclusion:Production Systems and pattern search are

almost equivalent

Almost?

Production systems have no loop detection mechanism

Solution:Invent two new productions1. assert(X) causes X to be stored in WM2. been(X) is T if X has been visited3. assert(been(X)) records in wm that

we’ve already visited X

Can be expressed in PC notation like this

),(),(^

))(()^(),((

yxpathyzpath

zbeenassertzbeenzxmvzyx

)),(( xxpathx

Can We Get from 1 to 7?

Iteration --Working Memory-- Conflict Set FiredCurrent Goal been

0 1 7 1 1,2 11 8 7 8 13,14 132 3 7 3 5,6 5

3 4 7 4 7,8 74 9 7 9 15,16 155 2 7 2 3,4 3

(firing 3 causes been(9) to fail)2 7 2 4 47 7 7

Notice that this search is data driven

Can Also Be Goal Driven

Instead of starting with current state=1 and goal = 7

Start with current state = 7 and goal = 1

Works great for a 3x3 matrix

What about 8x8?Either enumerate all moves or encode them8 possible situations1. d(2),r(1) 5. u(2),r(1)2. d(2),l(1) 6. u(2),l(1)3. d(1),r(2) 7. u(1),r(2)4. d(1),l(2) 8. u(1),l(2)

Not applicable everywhere

Situation have preconditions:Pre: row <=6, col <=7Situation 1: d(2),r(1)

Requires 4 new functionssq(r,c) returns cell number, left to right, top to

bottom where r is row number, c is column number

plus(r,2) returns r + 2eq(X,Y) T if X = Ylte(X,Y) T if X<=Y

Encoding of situation 1: d(2),r(1)

mv(sq(R,C),sq(Nr,Nc))

lte(R,6)^eq(Nr,plus(R,2)) ^lte(C,7)

^eq(Nc,plus(c,1) There are 7 more just like this

Control Loop for Knight’s Tour

)),(),,((

))),((()),(()),(),,(((

)),(),,(((

),(),,(((

ncnrsqzczrsqpath

zczrsqbeenassertzczrsqbeenzczrsqcrsqmvzczr

ncnrsqcrsqpathncnrcr

crsqcrsqpathcr

Strength of Production Systems

1. Said to model human cognition2. Separation of knowledge from control3. Natural mapping onto state space

search4. Modularity of production rules5. Simple Tracing and explanation—

compare a rule with a line of c++ code6. Language independence

But

Production systems are easily rendered in prolog

We’ll consider several versions of the knight’s tour

Global Record of Squares Visited

Put Squares Visited on a List

Use a Stack to Display Path to Goal

Use of Queue to Keep Track of Visited Squares

Display Backtracking

Try: path(1,W)◦ 1 is on the stack◦ Base case succeeds: w = 1◦ Now path1 is invoked, Start = 1◦ Move(1,6) succeeds, stack: [6,1]◦ Base case succeeds: w = 6, path is 1,6◦ Now path1 invoked:, Start = 6◦ Move (6,1) fails because 1 has been visited◦ Move(6,7) succeeds: stack:[7,6,1]◦ Base case succeeds: w = 7, path is 1,6,7◦ This process continues until all possible states have been visited◦ Occurs when path is 1,6,7,2,9,4,3,8◦ Now we backtrack from 8◦ But everything has been tried until we return to second move predicate starting with 1

(move(1,8))◦ Now we go through the entire process again

Cut

!◦Always succeeds the first time it is encountered◦When backtracked to, it causes the entire goal

in which it was contained to failWithout ! (4 2 path moves)With ! (2 2 path moves, because the first move cannot be backtracked to)

Farmer Problem

A farmer (f) has a dog (d), a goat (g),and a cabbage (c)

A river runs North and SouthThe farmer has a boat that can hold only

the farmer and one other itemWithout the farmer

◦The goat will eat the cabbage◦The dog will eat the goat

How does the farmer (and his cohort) cross the river

Define a predicate:state(F,D,G,C)Where F,D,G,C can be set to e or w

indicating the side of the river each is on.

As State-Space

st(w,w,w,w)

st(e,e,w,w) st(e,w,e,w) s(e,w,w,e)

st(w,w,e,w)

s(e,e,e,w) st(e,w,e,e)

etc.

Constructing a Move Predicate

st(e,e,-,-) st(w,w,-,-)Means Farmer and dog went from east to

westCan be rewritten:mv(st(X,X,G,C),st(Y,Y,G,C))

Facts

opp(e,w)opp(w,e)

Givingmv(st(X,X,G,C),st(Y,Y,G,C)) :- opp(X,Y).opp(e,w).opp(w,e).

In Motion

Suppose: st(e,e,G,C)mv(st(X,X,G,C), st(Y,Y,G,C)) :- opp(X,Y).opp(e,w).opp(w,e). w,wmv(st(e,e,G,C),st(Y,Y,G,C)) opp(X,Y)

{e/X, w/Y)

opp(e,w)

This is one of 4 move predicates (3 items to move + return trip alone)

Unsafe

Goat and cabbage are together◦unsafe(st(X,D,Y,Y)) if x != y◦unsafe(st(X,D,Y,Y)) :- opp(X,Y)

Dog and goat are together◦Unsafe(st(X,Y,Y,C) if x != y◦Unsafe(st(X,Y,Y,C) :- opp(X,Y)

Never move to an unsafe state

mv(st(X,X,G,C),st(Y,Y,G,C)) :- opp(X,Y), not(unsafe(st(Y,Y,G,C))).

Traversal Mechanism

Use the stack mechanism from Knight3

Farmer Problem